Differential-Equations-by-Zill-3rd-Edition-Solutions-Manual(ciitbooks.blogspot.com)_text.pdf

1

Introduction to Differential Equations Exercises

1.1

1.

Second-order; linear.

2.

Third-order; nonlinear because of (dy/dx) 4

3.

First-order; nonlinear because of yy'.

4.

First-order; linear.

5.

Fourth-order; linear.

.

6. Second-order; nonlinear because of sin y.

(Sy/dx2^

7.

Second-order; nonbnear because of

8.

2 Second-order; nonlinear because of 1/r

9.

Third-order; linear.

.

.

2

10. First-order, nonlinear because of y x /2

11.

From y = e~

12.

From y = 8 we obtain y

13.

From y =

we

obtain y' 1

e

3*

+

lOe

2*

=

0,

= -\e~ x l 2

6

6

5

5

.

+ y= ~e~ x l 2 + e~^ 2 = + 4(8) = 32.

Then

2y'

we obtain dyjdx

From y = - — -e -20i we obtain

$+

(3e

=

20j,

31

=

<&//
0.

+ iy = = 3e 3r + 20e 2a\ Then

so that y'

^ - 2» = 14.

.

+ 20^) - 2 20t

24e

24e-

,

20t

(e

31

+

lOe

21 )

=

e

3r

so that

2

0(g-^- 20£ )=24.

at

15.

From y

=

5tan5x we obtain

y'

—

25sec 2 5i. Then

2 y = 25sec 5z = 25 1

16.

From y =

(y/x

+ ci) 2

we obtain

y'

=

2

(l .+ tan

2 (./c

+

5ar)

ci)

=

25

+

(5tan5x) 2

=

25

+

2 j/

.

/2y^, so that

(^ + Cl) \

17.

From

j/

=

- sin x — ^cosx 2

+

lOe

x

we obtain

?/

= -cosi +

2 y'

+ y — ^cosx +

2

^

sin

-1 x — lOe ^

+

^ sinz

—

lOe *.

Then

2

^ sins — ^ cosi + 10e

-1 ^

=

sinx.

—

1

Exercises

1.

18. First write the differential equation in the form 2xy 2

x y 19.

20.

+ V2 =

1

we obtain 2xy + {x +

ci

First write the differential equation in the

so that

~2y/x

From y

=

+

x

=

2x~

%

=

21. Implicitly differentiating

i/

=

=

y'

=

2 j/

+

(x

ci

{x 2 —x

^

(y')

+ xy' =

3

ifx>0 „ Smce if*
= xlnx

we obtain

P=

24. Differentiating

y'

we obtain y

x > ~~

(l

+ +

=

+

x

—

c\/2y.

Then

4y

(2x

,

=

see that

ifx>0 if ,
}

{-2., Then

\y\

at )

?/

a

2

ae at -

1+bae*' 2

——

25. Implicitly differentiating In

(2

-

26. Differentiating y

W =

e

-

x

=

x

2 ,

-co < x <

itiBaPP-ent that v

— —y =

cc,

and

f-~

,

=VM-

1.

ab Cl e at

acje at

1

ln(2 — x)

+

„.

h

[

bcK**

—r

1

— ln(l — x) = f we

^

(3lj£.

obtain

ct^C

—

1-

i).

edt + cê

x

we obtain

X

2

- e-^V - Ixer 3? [ Jo

y'

=

we obtain

fccie"')

bCl e

we

y

f»(l+fci^)-^j

^=

,

y.

ifx<0

,

= 1 + lnx.

ac\e at / (l

dp

Then

2x~ 3

C V + y(y'f= -^ + ^ = -=y.

if

*

l 1

c ij

'

From y

Implicitly differentiating

0.

= —l/x 2 we obtain y' =

—2y/x. From y

V

23.

—

1

y

0.

form

so that

1,

2xy'

r— (x, vlfl't-x,

+ 2y}

y'.

we obtain

1

—

2yj y'

(x 2

+

e

t2

dt

-

2c-l XfT

x*

-

1

- 2xe~^

[' e^dt

-

2cixe~ x2

-

Jo

Substituting into the differential equation, we have 1

y

+

2xy

=

1

- 2xe~ x *

f e^dt - 2c xe"

x'

l

+

2xe~ x *

Jo

-x - y — — 2

27. First write the differential equation in the form y

—+

—

x

^ z

(x

m y) 2

and

+ 2c\xe~ x% =

1.

2 .

xy

Tfj/1

-r

dt

j

JO

implicit differentiation gives 2ci (x

+

y)(l

+

y')

Then cAx +

=

v/x xv' ,

xe

2

y)

—V „

—

xe y ' x implies

+ ev

I

'

x .

Solving

Exercises

we obtain

for y'

y

+ y)-

2c\ (x

2s

e
FVom y

+ y' -

y" 29.

=

FVom

y

that y"

c,??1

=

\2y

=

Sx

e

-6y'

+

c2 e~

ix

we obtain y =

we obtain

+

=

13y

From y - e^+se 21 we obtain

31.

From

j/

=

coshx

32.

From y

=

C\

33.

From y =

35.

+

From y =

ci

we obtain

c\

+

I

3e ix cos 2x

=

^ = 3e y'

2t

—

2e

3x

+2ic 3ir and

=

y'

= — 5ci

sinhi

ix

=

and y"

2x and y"

sin

cosx

+

-1

x

+ tan 2).

we obtain

x

9cie

—

From

3/

=

that x 2 y"

37. From y 38.

39.

From y =

— — 9ci

y'»

_

y"

-

+2X 2

From u* =

From y

=

+

3y"

cos 3s

c2

9c 2 cos

9y

=

In

3y'

—

=

-2bc\ cos 5s, so that y"

{x

+ c-iY

,

= — 1 + sins ln( sec x + + y = tanx. and y"

=

2c2S

-3 ,

1

f0, <

,

= - sin(lnx) + cos(lnx) and

—

0.

=

0.

iy')

2

tan x) and

x^-^

so that

y"

+ 25y = +

so that y"

y.

j!

= + 2ÔX

0.

—- cos(hix) — — sin(lnx), so

3a:

=

+ 2s In £

3s

and y"

= 5 + 21ns

so that x 2 y"

r

+ 4e we obtain y' = 3ci cos 3s — 3c 2 sin 3s + 4e + 4e s and y'" = — 27ci cos 3x + 27c2 sin 3s + 4e^,

— 3xy' + 4y =

0.

x ,

so that

,

0.

dx

j/

+V

x<0 '

„ 2

sin 2x, so

=

x 2 ex

+ 2xe x

y"

=

x2ex

,

+

4xe x 2e x and ,

y'"

=

xV + Sse 1 + 6e x

,

so

-y = 0.

X

(-x 2

Vie1*

4"|-4— +4y - 0.

+ sinhs =

coshx

obtain y'

Then y"

y'

s + 4x 2 we obtain

dx 2

(

y'

obtain

—

sin 3a;

+ 9j/ _

From y = c\x dx*

42.

+

3s

ci sin

—

0.

From y = x 2 e x we obtain

1

41.

+ 2y =

xy'

— x 2 + x 2 In x we

y"

that y'"

40.

scos(lns) we obtain

-

lGcae- 4 *, so that

+

so that

(fcr

36.

2x

5e 31 cos 2x

Se^+tee 2 *

and y"

+ cj

= — cis -2

y'

+ coshx

—-— and y" =

1

+ tans) we

^=

sin5x and y"

we obtain y =

c2

ln(sec

cox

y'

we obtain

cos x ln(secs

= tans +

y"

+ sinh x

cos 5s

From y — —

4c-2 e~

0.

30.

34.

3ci
'

x 2 - xy

^

0.

cos 2x

In Is

-

ix

1

_

+y

x 28.

1.

x >

s<0

=

y'

+ c\

c.\

in

x + 8s, y"

=

cis

-1

+ 8,

and

y'"

= — C\X~ 2

2

we obtain y

,

=

f-2s,

x >

1 2s,

we obtain y

, '

=

so that

a;<0

f0, <

x<0

i

so that

..

3

,

(y

xy

,,9

V—

—2y = f°/

0.

x<0

,

so that

Exercises 1.1

43.

=

From y y'

=

+

cx

c2

we obtain

2kx so that xy'

+

(y

y'

=

2

1

)

—

c so that xy'

+

(y')

2

=

cx

+ 4k) —

2

y implies that x k(l

—

+

Then

0.

= — 1/4

A:

=

From y

V-

fcx

2

we obtain

produces a singular

solution.

44.

From y = cx y'

=

4- \/l 4- e2

-xjy. Then

1

we obtain y

#

for y

xy'

0,

= r

+

c so that xy'

— = -x

+

(y')

+ ^/l + 2

45.

+ y2 —

implied by x 2

By

= —1

inspection, y

is

^

together with y

1

a singular

=

2

From x 2 + y 2

y.

=

1

we obtain

2

w — =!/.

1

+ -7= =

2

y is

(y')

vy

2

The

condition

-1 < x <

1

"

0.

Note that

solution.

this is the "solution"

obtained by computing

the limit as c approaches infinity of the one-parameter family of solutions.

46.

The

=

function y

f

47.

48.

right

=

From y

hand

-V4 -

limit

is

we obtain

x

—2,) and hence

= me"

11

y'

—

0.

does not exist at x

=m

and y"

2

mx Then y" e .

m = 2 and m =

—

for all x,

From y =

e™

we obtain

for all

not continuous at

j/

=

—

(the left

hand

limit is 2

0
,

(m _ 2 )(m -

>

ml >

y'

x2

m 2 e mx _ 5me mz + fe m* = Since e mx

Since e

-2 < x < is

I

and the

s/T^,

i

Thus y

3.

mi and

rae

=

y"

m 2 e m;E

e .

2x

and y

Then y"

5y'

3)e

=

e

+

ml

3x

=

implies

0.

are solutions.

4- lOj/ 4-

m 2 e mx + IQme™ + 25e mi = (m + 5) 2 e mi = x, m = Thus, y = e 51 a solution. 5.

=

6y

25y

=

implies

0.

is

m~ l

= mx and y" = m(m— l)x m_2 and substituting into the differential equation we obtain m m m(m - l)x - x = (m 2 - m - l) x m = 0. Solving ro 2 - m - 1 = we obtain m = (l ± i/o") /2.

49. Using y'

Thus, two solutions of the differential equation on the interval

50. Using

j/

+ 4y

and y"

.

m(m Thus,

y 51.

—

m=

x~* and y

It is easily y?,

—4,

-

arO+v^)/^

= m(m— l)x m ~ 2 and substituting into the differential equation we obtain = [m(m — 1) + 6m + 4]x m The right side will be zero provided m satisfies

= mx"1-1

x 2 y" + Sxy'

0
1)

—1 and two

=

x _i

shown that

2c!X4-3c2 x 2 and

+ 6m + 4 =

m 2 + 5m + 4 =

(m + 4)(m

solutions of the differential equation

4- 1)

=

0.

on the interval

< x<

00 are

.

y\ 3/3'

=

x 2 and

j/2

=

x 3 arc

= 2a + 6c 2 x so that

solutions. If

x 2 t/3' - 4xy'3

are solutions.

4

4-

y3

6y3

= cjyi + c2 y2 = cix 2 4- c2 x 3 then = 0. Hence cxj/i, c2 y2, and yi 4-3/2

Exercises 1.2

52.

shown that

easily

It is

=

y\

x 2 and

=

5/2

x 3 are

solutions. If

=

y

=

ciyi

2cix

+ 2ci

—

1

then y

2c\ so

2

(uM + 2U- =

that zy'

+ 2c\^y

2c x x

2

= ~x + 2

y'

53.

(a) y

~~

+

so that

= ~^-x 2

ji

?

and

for c 2

/

and

+ —^— =

so that xy'

+ 2^y.

=

The sum

on the body

of the forces acting

,

,

mo —

,

second law we have

s

,

kv

mg —

is

from g

„ Using

,

= k/R 2 =

,

,

(b)

a

„

„

,

,

we

m—

=

find k

= gR 2

gR 2

dv dr —

,

dr dt 4.

(a)

=

3/

;/i

cm

,

t

The sum

of the forces acting

ma —

—

kv

m -t-t at 1

d^r

k dr

dt 2

m

(b) Letting

1

6.

By

=

is

on the

kv where k k

or

—kv 2

—

mv

1

dt

&= _

a

is

—

+

=

1

then

2

are solutions.

1/ 2

— —1

or y

a constant of proportionality,

From

a.

3,

a

=

=

k

^=

gR 2

dr

rl

=

ma —

SR 2

gR ___ = 2

d2 r

or

.

0.

kv, so

by Newton's second law,

^5-. Thus, using v l

= ~.

r

and

r

—

=

kv

t1

dt

v,

the equation

is

part (b) becomes

Problem

=

m ^Xr2

or

dt

mv

=

a,

which

is

the

100,

and

1.

=

+ dt*

equation

1

dt

E(t).

at

we obtain R^- + ^-q =

is

+ Ri =

we obtain

E(t)

dt

~=— dt

g

j/

1/1

+

r2

and ^ dt

differential

and

2x

2

Kirchoff's second law

The

= -x 2 /2 +

= -c 2 x

+ mg.

dv

satellite is

C

at

7.

i/2

(c)

dv

„ = 0oru-

From Problem

.

_ gR

dt

R =

equation

i

+

(-C2/2)a: a then y'

c2 y2,

dt

5. Since

-

.

<^r J t > u -^j and part (a) we obtain

Part (b) becomes

(c)

c 2 y2

acts in a direction opposite to the motion.

dv

m— =

from Newton's second law we obtain (a)

If

1.

=

j,

°

1

dt 3.

^

c2

dt 2.

If

1.

Thus, none of

and the minus sign indicates that the resistance

„ Newton

c,

(b) no real solution

Exercises 1.

y

for C]

Aw

E(t).

J2gh. Using

.4o

=

jt

('-—')

\

'

=

127

32 this becomes

dh dt

-

0.6tt/36 '

100

8nVh rvr-— 64/t = - --„„,- - - —- Vh tt

6000

750

.

— 36

,

^=

10

2

=

'

Exercises 1.2

8.

The

differential equation

and 5

=

is

^=_

0.6j4o

^f—

0.6tt/576

dt= The


— = — —A—w

—

To

32.

find

Aw

we

x

solve

2

0.6(8)

=

- \l2gh

:

is

"4(576}

We

.

v

at

and g

tt

=

=

7r(2)

2

=

47r,

+

(5

—

h)

1

—

have An

=

2

A^ ^ "480 ^-

25 where

a:

-

VlO/i

ft

2

we obtain dft

0.6^/144

_

A

A w = ir(10/i -

ir(10A

-

/i

/l

/g^r

2

^=

1

=

-

30A(H>

)

10.

The


is

A'(t)

=

kA(t) where k

11.

The


is

x'{t)

=

r

12.

Equating Newton's law with the net forces

fcx(i)

144

represents the is

From

h.

Thus

).

2

-

—

(——] \12/

iv

radius of the circular area of the surface of the water whose depth

x

=

f^r)

32 this becomes
9.

A§ =

/^ft. Using

>

31>A (10 -

'

h)

0.

>

where

in

1

A)

0.

—

the x- and ^-directions gives

and

2

d y m— j = — 13.

respectively-

Prom Newton's second law

in the x-direction

we have

2

d x m— dt2

md

2

y

A 15.

To

=

respectively,

and k >

0,

see from the figure that

tan20

=

,

,

Id 11

r-

.

rfi

is

Ady

k(a~x)(j3—x) where a and

rfi

=

20, tan

1

—^ =

2tan0

- tan-^

x

x

L down

-, so

- =

j/

y

—-

2ldx/dy)

1

f 7T

I

L-~~ and i

(dx dy) 2

6

^

~dt

are the given

— — tan — — 9 \2 ax

,

dy lcl

to the x-axis.

x dy — — and y

=

1

.

better understand the problem extend the line

tan?i

dx —dx = — — v dt

,

x'(t)


and B,

,

= ~ mg ksm0a = ~ m9 ~ k = ~ mg ~ ~ vTt

-k¥

The

,

we have

In the y-direction

14.

„

—kcos6 — —k

-

2

)

=

amounts of chemicals

Then we

cot

G.

Now

'V y

—

(dx\ „ ( dx\ = +2y \dy ) \dyj

—

L

i.

Exercises 1.2

16.

We

have from Archimedes' principle

upward

It

force of water

on barrel

then follows from Newton's second law that

=

weight of water displaced

=

(62.4)

=

(62.4)7r(s/2) y

x (volume of water displaced) 2

—^-^ =

—15. 6tts

g dt 1 g 17.

=

w

32 and

~ &i

'

m gives

Dividing by

2

15.67rs y.

2

or

i(

dt 2

w iere '

^-f 4 dt

M

the mass of the earth and k\

is

= --4£

of motion with his law of gravitation,

k\M = gR j/B is

,

3

=

w

q wnere

where k

,

=

k\M. The constant k

= gR

2

or k

.

we obtain

a constant of proportionality.

is

is

R is the

gR 2 where ,

y

the earth. This follows from the fact that on the surface of the earth y 2

15.67T8

+

the weight of the barrel in pounds.

is

By combining Newton's second law

m ^f" =

=

=

If t

is

=R

radius of

^" = m9<

so that ^l -

the time at which burnout occurs, then y(0)

the distance from the earth's surface to the rocket at the time of burnout,

= R + yg, where and y'(0) — Vg is

the corresponding velocity at that time. 18. Substituting into the differential equation

we obtain —(mo

—

at)g

=

du — at)— + M—a)

(mo

or

dt

i

(mo

dv

— at)— = i

,

ab

— m^g + agt.

at

19.

By

the Pythagorean

Theorem the

slope of the tangent line

is y'

^

—

.

2

20. (a)

We

have

M

r

=

\l»

M •So 3

-dr and

= ^SR

3

Then

.

r

,

M

r

where

The

3

w2 =

differential

=

is

2

part (a)

we have

d r — m —-^ at 4

dA — = k(M — dA — = ki(M - A) A).

dt

22.

The

differential equation is

dt

-^ and

3

k-^j-

equation

r

Mm/R _

2

d r . From rr = ma = m-pr and at*

21.

T

it

Mm_ (b)

M

3

- y2

kÂ.

—k

mM mM _ r R Aa

— ^ dt*

—

kM = 2 r —w r R6

,

Chapter

Review Exercises

1

Chapter

Review Exercises

1

2

1.

First-order; ordinary; nonlinear because of y

2.

Third-order; ordinary; nonlinear because of tAnxy.

3.

Second-order; partial.

4.

Second-order; ordinary; linear.

5.

From y = x + tunx we obtain y'

+

2xy

From y

=

c\

have 6.

= 2+x +y 2

2 so that x y"

7.

+

From y

—

cie

y'"

=

cje

and

x

xy'

+y=

+ -

c2 eT

— sm2x +

1 sec 2 x tan

obtain

= —

[c2Cos(lnx)

%c i e

2x

we obtain

so that y'"

=

cosh2x we obtain

=

y'

-

cye x

—

-y'

2y"

+

C2B~ X

y

=

x2

10. y

=

e

11. y

=

\x 2

12. y

=

2

13. y

=

e* p

14.

= V*

15. y

=

sinx, y

17. For

all

x2 - x 18. If is

19.

\x\

—

-

<

1

> -1. That

2 and

\y\

also true for

The

differential

>

\x\

2,

>

2

Aw

ff

16. y

>

2y

is,

x

—1. Avoiding

<

or x

>

2

<

then (dyfdx)

and

equation

r corresponding to

is

\y\

<

2

x we

ci sin(lnx)j

and

c\ sinflna;)]

left-

y"

—

,

c\e x

+ c 2 &~ x + Acê 2*,

—

16 cosh 2x so that

—

e

\Qy

=

0.

5x

x

and right-hand derivatives, we then must have

1.

and the

^~ — —^-\/2gh. dt Aw v

satisfies

+ 2
differential

equation has no real solutions. This

2.

r = dh dt

20.

16 sin 2x

=

values of y, y 2

sec

+ 2y = 6.

9.

=

+ tan 2 x =

1

0.

+

cosx, y

—

+ C2 sinflnx) + ci cos(ln x) —

From

=

Using

x

cos(lnx)

[ci

y'

8.

y

x.

.

+ Qe* x + 3

x

=

and y"

2

— —j

C2e~ x

x

= 1 + sec 2 x,

+ C2sin(lni) we

cosfrnz)

y"

y'

.

rr-

_ "

From Newton's second law we obtain

Thus

We r

have An

=

-. 4

To

y— _ _25y^S 9

1/4

= \mg — 2

Aw

we note

that the radius

Then

"

loir/* 3 / 2

u— mi 2

8

find

^ and A w = —^r--

4^/25 V at

—

=

or at

16

f1 >

— VSu). '

2

First-Order Differential Equations Exercises

1.

For f(x, y) in

y

2

^

we have

df — = —y~^. Thus ay 2

j/)

For f(x, y)

=

y/xy we have

= — we ^

region where x 4.

For f(x, y)

—

5.

For /fx, u)

=

have

~=-

For f(x,y)

=

.

we have

s

4-y 2

—x— —y

differential

differential

,

we have

^ 7

<

equation will have a unique solution

equation will have a unique solution in any

equation will have a unique solution in

differential

2

2x y

*

Thus the

.

- y2 ) —2 < y <

equation will have a unique

differential

(4

—2,

—

-3xV ——

9/ — =

j

or y

2,

>

2.

Thus the

.

differential equation will

have a unique

-

—1. 2

2

.

r 5 2

^

Thus the

1.

3/ = dy

9/ we have 7— jx^ + y dy

—

Thus the

-r-

solution in any region where y

For f(x,y)

0.

Thus the

=

have

2

7.

^

0.

+ y we

x

™=

solution in any region where y 6.

the differential equation will have a unique solution

i

any rectangular region of the plane where y

2. For /(i,

3.

=

2.1

=

—-2x y (x 2

^ ,

Thus the

-

differential


+ y2 f

solution in any region not containing (0,0). 8. For fix, y)

—+—Xy

=

we have

y-x

solution in any region 9.

For f(x,y)

—

df — dy

=

where y < x

3

a;? cosy we have x ha

~2x (y-x) 2

.

where y >

or

df — = —x dy

2

siny.

Thus the

differential


differential


x.

Thus the

solution in the entire plane.

10. For f(x,y)

=

(x

-

l)e

v/l' x ~ l)

we have

unique solution in any region where x 11.

Two

solutions are y

= =

12.

Two

solutions are y

13.

The

solution

is

y

=

0,

and y

=

and y

=

which

is

2

3

x

2

= e^ ^ dy

^

\

1-1 *.

Thus the

differential

equation will have a

.

.

.

(Also,

any constant multiple of x 2

unique by Theorem

2.1.

is

a solution.)

1

Exercises

14.

A

function satisfying the differential equation and the initial condition

}{x,y)

=

—

\y

—

1| is

is

is

All of these solutions satisfy the initial condition y(0)

=

0.

Since

(a)

1

-\-y 2

=

y'

it is

and

c,

from which we see that y

not different iable at x

— (tanx) = ax

(b) Since

equation and the

sec

2

Since

(d)

Since

tanx

x =

and

tan x

tanO

discontinuous for

\y\

<

3

=

undefined for x

is

and continuous on — 1 <

For Problems 17-20 we identify f(x,y)

Theorem

1

a solution of xy

The

2.1

—

does not apply.

y

for all values of

piecewise defined function

is

not

0.

2

+

1

value problem on the interval — 1 < x <

is

Theorem

=

0,

y

=

tana; satisfies the differential

tt/2,

y

=

tana:

y

—

tana:

initial condition.

different iable

is

so

Although

1.

has a unique solution through every point in the plane.

—2 < tt/2 < 2 and tanx interval -2 < x < 2.

(c)

=

1,

=

y

is

derivative with respect to y are continuous everywhere in the plane,

its partial

the differential equation

f(x,y)

=

not continuous at y

cx

we have

cx

a solution since 16.

df/dy

continuous,

—

15. For y c.

2.

=

2

yjy

a;

<

1,

is

not a solution on the

a solution of the

is

initial

1.

-

9 and

and that df/dy

is

df/dy = y 2 /\Jy 2 - 9

discontinuous for

\y\

<

3.

We

•

We

further note that

then apply

2.1.

17.

The


has a unique solution at (1,4).

18.

The


is

not guaranteed to have a unique solution at (5,3).

19.

The


is

not guaranteed to have a unique solution at

20.

The


is

not guaranteed to have a unique solution at (-1,

(2,

—3). 1).

Exercises 2.2 In

many

of the following problems

solve for g(y)

implies g(y) 1.

2.

3.

we

we exponentiate both

— ±e e^ c

T '.

Letting c\

From dy =

sin 5a: da:

From dy =

(x

+

l)

2

encounter an expression of the form In

will

sides of the equation.

— ±e c we

we obtain y

obtain g(y)

= —-cos 5a; + c.

dx we obtain y

o

=

\{x

+

l)

3

+ c.

o

3x From dy - -e -31 dx we obtain y = \e~ +

c.

10

This yields

=

x

c\e^ \

\g{y)\

=

\g(y)\

e^

+e

= f(x) + c. To = e c e^ which

Exercises 2.2

4.

5.

\dx

From dy =

=

From dy

=

/ 1

5

+

\

2xe~ x dx we obtain y

From dy

7.

From - dy

4 = — dx

8.

From — dy

— —2x dx

we obtain

II

In

we obtain

=x+

dx we obtain y

i + 1/

\

1

6.

I

c.

x

-dx =

+

x

—- +

we obtain y —

xl x +6

+

x

5 In

+

II

c.

= — 2xe~ x + 2e~ x + c.

\y\

=

In

\y\

4 In

=

|x|

+c

or y

=

cjx 4

2

+c

or y

=

Cie

—a:

.

_:c .

2/

9.

2

From

—^— dy = - dx we obtain y+i

x

In \y

c.

\-

x

y*

10.

— x

From —= dy — —rz dx we obtain y~ 2 —

+

=

11

In Ixl

+c

—

11.

From y 2 dy = ( —x£ + —~\ dx we obtain -y3 — - + \x 3 x x)

12.

From ^- +

13.

From

e

-2

"^ = e^dx we

14.

From ye"dy

15'.

From

^

2

16.

17.

From (~K2 Vy

From

19.

From

y

31

e~

*

—

2

— — dx we

=

X

~ dS =

we obtain

ln|2

+c

Ixl

= —3 + 3xln Ixl + cix.

xy 3

or

c.

+ e~ x + ê~ 3x =

ey

+ y2 =

In |4

|

—+

+ x2 + c

y

—

tan""

— In |x +

1|

+ c.

we obtain

C\x.

|

1

x

c.

or 2

+ y2 =

ci (4

+ x 2 ).

+ c.

y

obtain y 2

=

x

~h 1

= \dx we l\

- dy yj » )

-r* dy y

+ 3)

+ 2e 3x =

dx we obtain ye -

—+—x^ 1

iy

In

+1=

+ y 2 = — cosx + c.

\y\

y

)

11

From

From

+

obtain In

obtain 3e~

= j—pdx

l\ dy

+2+

(2y 21.

dj/

2

dy

\* 20.

(e~ x

)

y

{

=

+

From 2y dy

18.

= smxdx we

dy

2y^j

or y

2

k dr

=

obtain

2

\-y

- y + In U

|y

(4x

+

az

2y a

2y

+

=

fcf

+ ln|y| IWI =

3

— 4x +

+c

or

2

dx we obtain

we obtain 5 =

= —- + c

or

2 ^y -

3

+

2 -—-^ 2 5)

1|

2

2 x lnxdx we obtain ^-

— —

+

=

-

1

y

+ In |y +

X — ln|x| - -x 3 9 1

1

1

3

1|

= — — + c\.

+ c.

+ c. 5

kT'.

J 1

22.

From

23.

From or

dQ = kdt we

p ^ p2

——= =

dP =

cie'.

obtain In |Q

+ j^Tp)

Solving for

3 i

^=

we have

-

70|

weobtain

P= 11

In

Q - 70 =

cie

|P|-ln |1-.P|

=

kt .

(+c so that

In

j^—p = *+c

—— —

-

Exercises 2.2

24.

From

25.

From

^dN =

(te

t+2

-

—^— dy =

5— dx

or sint/di/

x

sec-1

esc j/

we obtain

l) df

From

= —

2i/dy

—

r^r~ dx

cos" ix e 2y

27.

Fromey

— y, -~-dy =

+ ye~ y + e~ y =

2x

,

dx

cosx 2 cos x

31.

From

-

From

y

— -dx

From 2y dy —

33. From

—

(2x

+

3

=

-

\dy

+3

y

1

From y^tl dy y -

=

1

y

35.

From

—

^

(2cos-! cot

36.

y

From

—-

^-r

j/

= — cos x +

1)

—

I-

sin

dx

dy 2 sin y cos y

—

„\

= -2sinxdx we obtain

ye y )dy '

In sec y |

=

|

xsinx

(e y

+

cosx

4-

I)"

=

1

*

+ c.

(e l

1

+

l)"

(l+y 2 )^ =

dx or

f

—j— - —

=

a:

2

?/-3y

+

1/2

2 a;

)

+ c.

+

[

1 + 4/

V

= x-51n|x + 4|+c

f 1 H

=

(l

]

dx we obtain

+ x + c.

^—

)

dy

=

^

5— dy

=

dj/

^— \dy=(l

[l

^

dx or

)

+ c.

+ ^-y = ^ln|l+x|-iln|l-xl + c.

dx we obtain y 2 dx or

?

?

2

'

(l V

sin

x dx or

cos-'y

dx we obtain

=

or

dx we + —^—] 3/

2;

=

}

^ — cos^y-l

dy

x'y .

obtain

g ~

or

cie

cie'~».

= - esc 2 ydy =

sin

xdx we

obtain

c.

dy

sec y—

obtain

= — - sec 2 3x + c.

2 j/

-

e"

+ 21n|y-l|=x + 51nlx-3| + c -

c.

2

2

x-3

/

,

dx we obtain

j,-51n|y-3|

34.

obtain

2

1

+4

t

^

we obtain -

= -

1)

- +

'

l)

ln|y+l| 32.

t+2

+ c.

+

(e*

=

dy

e

+ sin2x + c\.

dx or

cosx

From

30.

2x

2 sin x cosx

=

29.

= xdy 2

-

6

sin

lj

=

= — tan 3x sec 2 3xda: we

From tan ydy = xcosx dx we obtain

(e^

3

= -cos 2 xdx = --(1 + cos2x) dx we

4 cosy

or

28.

——+

ie'+

2

- cosy = -ix - - sin 2x + c 26.

=

In \N\

x cos y — cos x sin y

esc

2j/

dy

=

cos x dx.

=

sin

x cos « 4- cos x sin y we

Then -

In

|

esc 2y

2

12

—

cot 2y|

find sec ydy

=

sin

x

+ c.

=

2 sin y cos x

t

Exercises 2.2

37.

From xdx =

38.

FVom

dy we obtain -x

.

lie ^

=

dy

.

,—

sin

dx we obtain J^

.

y/4-x*

+

39.

^

=

2,

=

From —zdy

+

ex

jr

dx

e~ x

+

40. To integrate dx/ {x

-Jx)

dx

r

———

Thus, from

rfw

{e x

make r

=

/

—

—

/

r-

+c

y

—

or y

+ y2 —

sin

-1

sin

- +

1

dx we obtain

Y+

ci

1-

f

)

c.

2

——

1

ttt

1

tan

-1 e

x

+

c.

i/

the substitution

2udu = -j—

—

=

1

r

/

u2 =

—2du— =

dx we obtain

,

Then 2udu = dx and

x.

—

\

/ u+1 +c = 21n(^+l) + c.

,

21n

y

,

+

2 In (,75

=

1)

2 In (\/x~

+

l)

+c

or

+

e y )'

y/y+l=ci (v^+l). x

sin

41. From

+ cos x

1

(1 is

e~ y

+ e")(l + cosx) = (l

+

e M )(l

+ cosi)

1

42. FVom l

+

.„

=

(2{/)

we

€^

=

we

* 2

+

FVom -dy

=

(1

2y

=

= — —1 tan _1

7r/4,

= 4idi we

—

=

3e

The

c

obtain

i)df

we obtain

is

.

The

*

45. From

dx

2

obtain tan

*

y2

-

dy 1

=

-

x2

—

1

-l

is

+

_1

9

tan

=

a;

!2

=

2x 2

+

c.

+1=

2x 2

+

\/2'

c

or y

yV + 1 2

=t—

-t

2

+

-1

=

x

tan

-1

dx or ^ (—^— 2\y-l -

+

4y

x

— y

=

~ +

c.

4y

is

Using y(0)

=

y

=

3e

—

or

=

is

'-' 2

find c

=

V2- The

Using y(l)

=

3

1

/2-i/2

=

1

.

=

3e -(t-i)

we

x = tan ^4y

—

we

3

2 cie'""' ^

Using x(7r/4)

= J

1)

ci.

= ^.

1

solution of the initial-value problem

= 4dy we

2y


In


46. From

tan

or

2

-1 ' 2


_!

*j

x +

J/

c

+ c or

dx we obtain


44.

The

4.

Irtfl

(z8) I

find Ci

dy

,

=

find ci

tan'^ + tan" 43. FVom

— lnfl + cosx) =

obtain

e"

4.

1

=

1

Using y(0)

— tan Using y{l)

1

—x l

—+ — dy we

dy — -

+

c\.

—

dy

,„ 2

1

=

dx

^— ^ f 2Vx-l x

+

find c

find

2

/2.

= — 3tt/4. The

— dx we obtain

lJ

we

—

.

Exercises 2.2

\y-l\~

In

The 47.

In \y

From -

From 1

From

\

dy

—

=

find ci

- In \x +

1|

find ci

=

=

e

_1

The

3

(b)

If 1/(0)

=

3.

3 then »

1 -I-

,

—

dy

]

=

(a)

If 3/(0)

-

(b)

If 2,(0)

=

(c)

Ify(l/2)

In

-

|1

2

_

2

+

3

e

1

theny

=

1,

then y

=

0.

= l/2theny =

|a;|

=

+

%

^

-

-

c or 1

-

is 1

2y

is

xy x

=

2y

c or

c\e~'*

= -Ae~ 2x

.

xy

=

=

-1-1 ^.

e

c\e~ l ^ x

Using

.

Using y(0)

=

5/2 we

2x

+

-

or y

=

2e~

.

6x

ce

.

3

Iy

—

1

1

—

=

lnlyl

In

+c

la:]

=

or y

— 1-cix

Another

.

^.

52.

By

inspection a singular solution

is

y

=

0.

53.

The

singular solution

54. Separating variables

t,

=

1

satisfies

we obtain

1.01

1

-

-rs

=

dx.

we obtain

C

^


.

Setting z

we obtain

^

=

y -

solution

=

1)^+ and y

^—

Then

1)

we obtain c — —100. The


— tan X +

the initial-value problem.

-.

(y

+

In

T

1.

1

= —— —

\y\

x.

0.

—

=

=

2y|

obtain In

y

y

=

i

is

=

or y 1

efa „ a

inspection a singular solution

and y

In

1

+

61 " 2

By

—

x

= 2 we find c = 0.

Using y{2)

I

r

51.

x

1

—+

= i <£r we

y /

1

y

=

= ltheny =

Ify(l/3)

is

+

2

=

+ c.

\

+

-^.

= 0theny =

solution

y

obtain y

y

3

\y-

1

x


-~ + -^—) dy = dx we y-3)

|

—

I

-4.

\y +

1

=

-

y y is

—+

or -


~

dx we obtain

iy

Ifi,(0)

From

The

.

+c

1|

we obtain

~)

~~

f ~~2

(a)

(c)

50.

-

In |i

= ~—y~ ^ =

(it,

1

49.

=

l\


— 1) = —1 we 48.

+

=

—

01 1

is

14

dx.

=

c

=

Then

x

+

c

—— ——

X

^

^

we obtain

=

y

— 1

el1

0.

—+

=

x

.

Setting

c

-

100

l^tan

The lOy

5 In

and y

-1

10(y

—

1)

=

1

+

c

= 1 + — tan — \

solution

-

11

=

is

x

y

and

+ c.

Setting x

X

—

Exercises 2.3

and y

=

=

57. Let u 1

tan

58. Let u or u 2

~u 2y

=

x

=

x

+ «y

=

=

i

(1

+y

x

-

=

2x

-

lOy

The

solution

=

+

1

2x

+ ci

=

1

+

61. Let u

2^/u

=

62. Let u

-e~

u

= x

=

=

— -

y

+c y

x

2x

+

4x

+ ci,

+c

—

or tan(x

1

1

+

=

1

u2

=

or

c)

x.

or

—+—u^ £

\

—-

=

-

dx

tan (a:

+ c) — x —

udu =

or

=

ifu

^—

1

+ 1/] +

dx.

Thus

1.

Thus \u2

dx.

=

+c

x

2

ti

and 2(x

+ dy/dx. Then

+ y) —

sec(x

+3=

=

udu —

2

tan « or cos 2

+ y) =

sin2(x

+ ci

Ax

Thus

dx.

or

-e^"^ = 5

x

+

1

=

sinu or

= dx

or (sec

=

j/)

x

+

rfu

=

dx. Multiplying

u - tan it sec u) du

2

v

+

+ sin u

1

=

Thus

dx.

'

c.

2.

Then

^+2 =2+

1.

Then

— + dx

du =

dx.

Thus

dx.

Thus

+ c.

x

=

du/dx

™ — dx

du

u

— dy/dx du/dx dy

3 so that

5 so that

and

—

+ c\

and 2\/w - 2x

- x+

-

11

~ dx

tan(x

Then

dy/dx.

sinu) we have

— secu = x + c

=

-

10y-9

Then

dy/tt Then + dy/dx.

cos-'

tanii

4- 1

5 In

is

cfa/x.

and £ + y

+ sin 2u =

so that du/dx

sin«)/(l

1

=

y so that du/dx

+ sin 2(x + y) = =

+ c),

2

+ y)

2u

0.

du/dx

du/dx

so that

=

1

tan(x

a, and (x

+

5 In

so that

1

+ coru =

+

2x

+

=

c

+ isin2it-x + cor

60. Let u

by

x + y

u

u

59. Let

we obtain

1

dy/dx

-

1

=

1

+

e" or e~ v du

=

c.

Exercises ° 1.

Since f(tx,ty)

=

3

(tx)

+

2(tx)(ty}

2

=

-

3 t

f{x,y), the function

is

homogeneous of degree

3.

fx 2.

Since f(tx, ty)

—

3.

Since f(tx, ty)

=

sjtx

+ ty (4tx + 3ty) =

3 2 t ^ f(x, y),

=

(txV jt;;) 2

t2

f& V)<

the function the function

is

homogeneous of degree homogeneous of degree

is

3/2.

2.

i

4. Since f(tx,ty)

—

Since f(tx.ty)

=

5.

(x

(tx)

6.

Since f{tx> ty)

=

3

cos

+y

1

-f(x,y), the function

—=

("cos

—X — -

x+y

for

any

-

x

-\-

y

homogeneous

n, the function is not

X sin

is

of degree

2

4-

x

=

f(x, y), the function

is

homogeneous

homogeneous.

of degree 0.

—1.

Exercises 2.3

2 7. Since f(tx,ty) \nt

8.

Since f(tx, ty)

—

+ inx 2 —

^nix

J (i^^

3 In ty

Since f{tx, ty)

=

10. Since f(tx, ty)

—

9.

11. Letting y

=

^j(tx

In

ty

+

2

l)

for

an y n

f(x,y), the function ^ ne functi on

^

i

s

is

-j$f{ x < f).

#

n

t

(x

tne function

+y+

l)

2

for

is

any

homogeneous of degree —2.

n,

the function

— ux)dx + x{udx + xdu) — dx

4-

x du

=

dw

=

f-

a:

a:

=

In |x|

+m=

c

In \x\

+y =

ex.

ux we have (a:

4-

ux) dx

+ x(udx +

(1

+

2u)(ix

—x + In

13. Letting

0.

not homogeneous.

ux we have {x

12. Letting y

homogeneous of degree

y

=

+

+

+ In y) =

2(ln(

|:r|

+

=

+ xdu = +

l

2u 2u\

=

c

x 2 + 2xy

=

ci.

In

^

xdu)

+

[1

x = vy we have

+ y (fv) + vydv +

1

{y

(j/

—

(w

In |w

-

1|

-

2v

-

2vj/)
=

+

=

l)

2 1)

i— + In

v —

16

1

dy

y |j/|

=

c

is

not homogeneous.

Exercises 2.3

^-1

In

(x

14. Letting x

=

+ \ny —

x/y-l

V

- y) In

|

a:

-

y\

-y=

c(x

- y).

vy we have

y{vdy

+ ydv) y dv

2{vy

+ y) dy =

-

+ 2) % =

(v

tit)

+

v lnjv

= ux we

-

2

—=
2

y

+ 2| -

- +

In

15. Letting y

c

In \y

\

=

c

-

In

\y\

=

c

a;

+

2j/

=

ci?/

2

have 2

[u x

2

+ ux 2 )
2

+ xdu) =

x {udx

(is

— 3
(is

2=0

—=

In |z| H

y In

=

c

+-=c

lnld

16. Letting y

.

[a:j

+x=

cy.

ux we have 2

(u x

2

+ ux 2 )

dx

+ x 2 (udx + xdu) -

(u

2

+ 2u)dx + xdu = dx

du

x In \x\

+

^

In |u|

-

u(u

^

In ju

+ 2)

=

+ 2| =

c

Exercises 2.3

u+2

ci

\x

x

=

2

x y

17. Letting y

= iiiwe

cj(y

+

2x).

have

(ux

- x)dx (u

2

+ x) (ti dx + x du) =

(ux

+

l)

+

dx

+

x(u

du

1)

+ — + -~u — + dx

1

,

du

-

ln\x\

+

Inz

^ln (u 2

^+

2

ln(z 2

18. Letting y

=

+

l)

=

_

1

U-'

a:

=

1

+ tan"

lj+2tan

!/

_1

=

^ =

+ y 2 ) +2tan~ - =

c

ci

1

ci.

ux we have

(i

+ 3ux) dx (u

(3x 2

-

+ ux){u dx +

+ X In

|rc|

dx

l)

+ x(u + ti

7

lj

=

3)

du

=

r

+ w

3

du

=

—

In |ti+l|

=

c

-

Cl

(u-l)(u

+ 21n |u —

x du)

+

l)

z(u-l) 2 !1+

( !/

18

1

-x) 2 =

c 1 (y

+ x).

Exercises 2.3

19. Letting y

=

ux we have

—uxdx +

(x

+

{x

+x

+ xdu) =

~Jux)(udx )

+ u 3/2 dx =

du

u!

\

x

-2u- ^ + lnH+lii|x| = l

In \y/x\

+

y(\n\y\

20. Letting y

= ux we

In \x\

=

2y[x]y

-

=

4x.

2

c)'

+

c

have

x\j\

+ u 2 dx-x 2 du = du

dx

In

III

—

i+

In

u

+ u2

y/l

+

+

1

21. Letting

c

= - c

\/l

+ u2 =

c\x

2

+ x2 =

c\x

y?/

x — vy we have

2v

2

ydv -

(v

2u

-In

3 It;

3

+

l]

2

+

,

dv lj

1

-

In

dy

=

dy — =

\y\

,\2/3

=

c

2 .

Exercises 2.3

25

+

£T

Letting y

— ux we

=c 2 y 3

+ y°) =c2 y s

x*

12.

-

l]

.

have

x

4

+ uV)

dx

-

3

+ x du) =

2x ux(u dx 2

u2 -

l)

=

dx - 2xudu

—x — —1u du dx

(u 2

In

2

(K

:3.

Letting y

— ux we

-

I

I2

set

+

I

U

a-

- \f

u2 -

1

)lnM+x 3 = C (» -x a 2

have (x

2

+ u 2 x 2 ^ dx -

+ xdu) =

2 ux {udx

=

dx — ux du dx

u du —

x In

|a:|

— û 2 =

2la\x\-(y/x)

4.

Letting

j?

=

2

=

c

c.

ux we have 3

(u x

3

+

3 a:

+ u2x3 }

+ xdu) =

dx - u 2 x 3 {udx (1

4-

w 2)

da:

-

u 2 x du

—+

dx ~

x In

In

txl

|a;|

20

—u+

- - +

1

tan

tan

an =

-

u2

=

-1

-1

u=

c

- =

c.

Exercises 2.3

25. Letting x

=

vy we have

y(vdy

+ ydv) -

+ 4j/e" 2 ")

(vy

dj/

=

ydv - 4e" 2 "rf{/ =

|e

2

"-4In |y| =

e^'v - 8]n\y\ 26. Letting y

— ux we

2

e~ u

+ u 2 x 2 ) da: -

ux

2

(utte

+ x du) = — ux du —

e~" dx

—i -«c x In

j/

= ux we

= a.

have (;r

27. Letting

c

In

|x|

\x\

-

n

- we" + (y

-

du

=

e"

=

c

x

=

ex.

x)e y t

have (ux

+ x cot u) rfx — x(u dx + x du) — cot u dx

— x du =

—x — tan u du — In

28. Letting

t/

= tii*e

|x|

+

cosu|

=

c

y x cos — x

=

c.

In |

have

ux In u dx — x(u dx + x du) = (it

In

u — u) dx

dx x

— x du —

du u In u — u

=

Exercises 2.3

-

ln|x|

In

|

=

c

=

c\

x

=

Cl

lnu

=

cjx

s

=

xe

-

lnu

1|

x

—

lnu

29. Letting y

= ux we

1

I

i

n

V x

_

l

+ In 2 +

1

1+C2X .

have 2 (a;

+ ux 2 -

=

u 2 x 2 ) dx + ux 2 {udx + xdu) (1

+ u) lix + xudu =

—x + uu +dur = dx

i

In

|x|

+u—

+

In |u

u

+

1|

1

=

c

=

cie

=

ciie"/*

a;

^ +

1

a;

30. Letting y

= ux we (x

3

have

+ ux 2 + 3u 2 x 2 )

dx

-

(x

(l

In

Ixl

2

+

2ux

+ u2 )

-

2

dx

)

-

(udx i(l

+ x du) =

+

2u) rfu

1

+

x

1

+ v?T

tan" 1 u

-

In f 1

2u

,

du

=

=

+ u2 ) = c 1 1

+u

2

=c ie tH1

"

,u

xîf + x )^™ 2

22

1

*!*.

Exercises 2.3

31. Letting y

=

ux we have

+ u 2 x du =

dx

dx

Vu 2 du =

Q

x 1

+ û 3 =

In |x|

3r3

=

Using 32. Letting y

2

we

= ux we

find c\

=

8.

The

2

+

2

2u x

2 )

+u

2

In

udu

|x|

1

- ^ln

In

|x|

+ y3 =

Sx 3

+ u2

+u *

we

find ci

=

1/2.

The

=

+u 2 ) =

(l

l

=

3

- ux du =

dx

)

x

33. Letting y

is 3a:

dx-ux 2 (udx + xdu) =

eta:

1

c\J?.

solution of the initial- value problem

(l

=

y

have (x

Using y(-l)

=

3

+

ln|x|

c

4

—

ci

=

ci

2

{y

solution of the initial-value

+ * 2 ). problem

we have {Zux

2

+ u 2 x 2 ) dx (u

2

2

2x (u dx

+ u)

cfz

£

-

2 In

|w|

- 2xdu 2du

dx

ln|x|

+ xdu) =

u(u

+ 2 In \u + x(u

x(y

+ x) 2 =

+

ciy

2 .

+

_

q

l)

1\

l)

=

c

=

ci

2

is

2x 4

=

y

2

+

a;

2 .

.

Exercises 2.3

Using

j/(l)

34. Letting

x

= —2 we

=

=

find c\

1/4.

The


+ ydv) —

2

(v y

2

djf

=

1 dj/

=

In Ivl

=

c

=

Ciy

+ vy\Jv 2 y 2 + y 2 ^ ydv —

+

\jv 2

—

In

\7 + x2 3/(0)

35. Letting y

4x(y

+

\Jx

+ x)

vy we have vy 2 (vdy

Using

is

=

1

we

= ux we

find ci

=

The

1.

1

+ y l =ciy 2


is

.

x

2

+ y2

have (x

+ uxe") dx -

xe u (u dx

+ x du) =

=

dx — xe"

—

e

"du =

x

-

In |x|

ln\x\

Using y(l) 36. Letting

x

=

=

we

find c

=

~

e"

=

c

x

=

c.

e»'

—1. The solution of the initial-value problem

vy we have 3/(1;

dy

+ y dv) +

(y cos

3/

— vy) dy =

v

+ cos dv

sec v

t>

(fy

= =

-\

V In sec v I

+ tsnv\ + In \y = \

/ sec

y \

24

x — y

I-

tan

z\ — — ]

v)

c

c\

is In \x\

= e^1

— Exercises 2.3

(zr x\ sec - + tan -1=2. vj

y 37. Letting y

=

ux we have 2

(u x

2

+ 3uz 2 ) dx -

(ix

1

+ ux r) (udx + xdu) = — u dx — x(4 + u) du =

—

dx 4 + u — + du = X u In

\x\

+

4In

|uj

,

+u= xu 4 y

Using y(l) 38. Letting y

=

1

we

~ ux we

find cj

=

e.

The

3

) da:

-

2a? (udx

=

ua:

= rA x 3 e~yîs

=

u4

~ xê i v l x

.

+ xdu) =

we

find ci

=

e'^

2 .

The

du

2a:

dx

2du

a:

Ji''

In

39. Letting y

-ti

have

u 3 da: -

=

c\e


(uV + 2ua;

Using y(l)

4

=

c

=

r=

\x\

H

t

—

c


is a;

we have (x

— ux —

v?t 2 xj dx

+

(i

+

vâ:) (udx

dx

+ xdu) =

+ x (l + V") du =

^+ In

x

(l

+

du

=

+ u + û3 ^ 2 =

c

O 3 2 3a: / lna:

+

2

'ix^ y

+

2y

3'2

=

3 2 c,a: / .

=

e

_x2 /!/ 2 +i/2

Exercises 2.3

Using y(l)

=

1

we

find c\

=

5.

The


(Note: Since the solution involves yfx

40. Letting

x

=

vy

,

x >

and we do not need an absolute

+ ydv) + vy(\nvy —

Iny —

y du

=

1)

+ ulnudy =
w In v

dj,

+

-

yln

41. Letting x

=

e

we

find ci

= — e. The solution

3

ci.

of the initial-value problem

+ y du) +

2

(u y

2

+

m/

ydw +

2

+

(jj

+

2 t,

)

l)

2

dv (v

+

v

+

x

+y

\)

is

1

we

find c

{x

=

-

dy

=

dy

=

y .

in

.

tf

=

c

1

y

=

dy

2

1

j/(0)

c

vy we have

y (n dy

Using

=Q

y

+ In |y| =

In |ln |f ||

=

ve

we have

y(vdy

Using y(l)

is

+ In \y\ =

c.

-1. The solution of the initial-value problem

+ y) In \y\ = y -

(x

+ y) 26

or

(x

+

y) In

\y\

=

is

-x.

yki

Exercises 2.3

L2.

— ux we

Letting y

have

+

{^/x

dx — x{ u dx

y/ux J

+ x du) =

du

(tc

1

a:

+ 2,/u

=

nil in \x\ =

=

f /

—-=

/rfa

Using y(l)

3.

Letting x

=

=

we

find c\

—

1.

x

The

+

2^/y

=

I

u=

+

2,1-

+c

2

c,e


is

vy we have

2

vy-

\Jy

-vy 2

-

y(«d?/

J

vT— v

In

In

Using &(l/2)

=

1

we

find c

=

\/2

.

The

|y|

+

ydu)

=

— y dv =

dy

dv

y

Vl - v -1-

2\/l --f

= =

+ 2^1 -x/y =

c

c.


\n\y\

+ 2yJl-x/y=V2.

2tdt

= '-* 1,l|1+a|+c I

Vx

r 3/2

—d« =

«

du

-

is

Exercises 2.3

— ux we

44. Letting y

have

x(u dx

+

x du) — (ux a:

x cosh u) dx

du

— cosh u dx =

sech u _1

tan

1

tan

=

Using 45.

we

From i = vy we

Using M(vy,y)

find c

=

n y M(v,

1)

(sinhu)

^sinh

and the

+ y dv) +

and N(vy,y)

=

[vM(v,

1)

+

2 y N(v,

2/

=

r cos

(?

and

?/

=

r sin

In

|x|

=

c

—

In

|x|

=

c.

N(v,

1)]

dy

vAf(u, 1)

we obtain dx =

1)

+

is

tan

-1

^sinh

-

^

=

In \x\.

equation becomes

N(vy, y) dy

=

0.

and simplifying we have

1)

+ ydv) + y

M(u,

From i

j

—

differential

M(vy, y)(v dy

n y M(v, l)(vdy

46.

—

—=

—

ciu


= vdy + ydv

obtain

=

The

0.

—

+

n

N(v, l)dy

+ yM(v, du

1)

=

dv

=

Q

=

0.

JV(t>, 1)

cos
—

r sin

and dy

—

sin fldr

+ r cos

fldfl.

Using Af(i, y)

= M(r cos 0,r sin 0) =

T

Wfx,

= N(r cos 9, r sin 9 =

r

n

M(cos9, sinf)

and y)

)

n

N(cos 9, sin #)

the differential equation becomes

r

n M(cos9, sin9)(cos9dr

Simplifying

[M(cos 9,

-r sin fldfl) + r n A^cosfl, sin0)(sin 0dr + r cos0df>) =

0.

we have

sin 9) cos 9

+

AT (cos 6, sin 0) sin 5] dr

dr r

Af(cosfl, sin

ff)

- [rM(cos 9,

sinfl

sin 8) sin

- rN(cos 9, sin 9) cos 0] d9 =

- r N (cos 0, sin 9) cos 6 ^ _ + N(cob 8, sin9) sin 9

M(cos9, sin.9) cos9

28

Q

Exercises 2.4

= yn M

47. Using M(x, y)

^

,

1

j

and N(x, y) - y n

48.

let

1

j

^

u = y/x, then by homogeneity f(x,y) = x"f

(l,

Mx

dy dx

we

If

,

+ yn N

n y

or

M

M^

iy'

l)

N(x/y,

1)

(

partial derivatives,

-c

(

,ljdx

,

we obtain

=

lj dy

x

\y |j

=i"/(l,u). Using the chain

rule for

we obtain

and

du

9j/

du

dy

\x/

du

Then

=

nx n /(l, u) = nx n f

(l.

|) =

tf).

Exercises 2.4 1.

Let Af /i'(jf)

2.

3.

=

and JV

=

3y+7sothat

Afj,

=

=

3y

+

7,

and h(y)

= -y 2 +

7y.

The

From/^ - 2i-l we obtain / = x 2 -x+h(y),

War.

3

solution

is

x2 - x

M = 2x + y and N — — x — Then M = and N Let M = 5x + 4y and N = 4x - 8y so that M = 4 Let

6y.

1

y

x

3

v

/ 4.

= 2x-l

= ^x + 4xy+h(y), h'{y) = 2

Let A/

fx is

=

=

siny

siny

—

ysina: and

— ysini we

obtain

1 2 = xsmy + ycosx — ~y

c.

-Sy

=

3 ,

and

zosx

=

4

-2i/

+ xcosy —

.

y

/ = isiny + ycosx + frf!/),

3

+ -y 2 +

7y

=

c.

=

—1, so the equation

=

JVj..

The

solution

so that /i'(y)

=

=

Prom /a

5i

+

4y we obtain

^x +4xy - 2yA = 2

is

My =

not exact.

is

cosy

—y, and

—

=

sinx

=

2

^y

.

Nx

The

.

c.

From

solution

Exercises 2.4

5.

Let

=

/ 6.

=

Let

2y2 x - 3 and

x y — 3x 2

2

=

Let A/ JV^

7.

M-

+ h(y),

4X3 —

—

3j/sin3.c

M=x

2

—

y

2

N

and

2yx2

=

h'(y)

— 3 sin 3x. The

2

1/x

=

JV

4,

+

N

y/x 2 and

=

=

and h{y)

equation

M

4 so that

is

=

1y.

= Nx

4xy

The

solution

1/x

+

—

2y

=

y

2

x y

is

=

From

.

—

2

-

we

3

+ 4j/ = c. = — 3sin3x —

3x

M

cos3x so that

2

2y x

y

obtain

1/x 2 and

not exact.

= -2y

2

x - 2xy so that Afy

Nx =

and

2x ~

The equation

2y.

is

not

exact. 8.

/ 9.

M—

+ lnx + y/x and JV = -1 + lnx so that My = l/x = Nx Prom /B = -1 + lnx we obtain = -y + y\nx + k(y), h'(x) = 1 + ini, and k(y) = x\nx. The solution is — + y lnx + ilni = c.

Let

=

y

=y 3 -

fx

solution

10. Let

M

3

-

- x we

3

2

+y

xy

—

-

sinx

sinx

2 Jf

is

J/

2

x3

+

y

3

obtain

=

and JV h'(y)

M

=

+

0,

=

and Afy) l/y

+

Afj,

=

The

0.

2

3y

=

=

is

~x 4

= 0,

V(jf)

-

y

6x 2

and h(y)

2ie I and

-

M = I-3/1+

if

and

JV

so that

—

0.

JL/

y

The

N — x so that

=

- 31n|x[ + xy + h(y),

/

=

x

+ y + xy -

16. Let

M = xy

From fy = solution

is

+

1

lay

= -2x/y 2 =

solution

Af„

=

I

=

x2

is

Nx

+

2

3\n \xy\ sinh x

e"

=

+y

1

-3/j/ + x so that

h'(y)

=

0.

The

ye~ xv and

JV,

=

\ay.

The

,

and h(y)

x 3 ^3

=

2x/y we obtain

cy.

= JV,;. From /x = y - 6x 2 - 2xex we obtain - 0. The solution is xy - 2x 3 - 2xe + 2e* = c.

M„ =

3 1

cosh x and

=

1

=

JVX

.

From fx

y - 3 In

N = e y + 2Tj/cosha; so that

2xj/coshx we obtain /

2 3

is

From

.

3:

From fx = 3x 2 y + e" we 3 2 is x y + xe v — y — c.

The

= 1 -3/x + y we solution

obtain

is

c.

2

+ + xj/ 2 cosh 1 =

ey

=

ey

+

xj/

2

coshx

+

A/s

=

2a:?/

/((j/), ^'(j/)

sinh ar

=

0,

+ 2y cosh x =

and h(y)

=

JVX

.

0.

The

0.

The

c.

2

(l

1

solution

=

and h(y)

0,

From

.

M = r ^ - 1/ + 9x and JV = x ^ so that M = 3xV = we obtain / = \x y - \ arctan(32:) + h(y), /e = iV- 1/f + $x

17. Let

Nx

=

+ xy 3 = c.

.

x

2|fsinx

From fx = x 3 + y3 we obtain

.

= xy-2x 3 -2xe x + 2e* + h(y), h'{y) = 0, and h{y) 2 3 2 = A^. Let Jtf = Zx y + e" and JV = x + le" - 2y so that Afj, = "ix + 3 2 v — — — — = y The solution obtain / x y + xe + h{y), h'{y) 2y, and

15. Let

-

=

h{y), h'(y)

JVX

solution

x In y so that JWU

/

14.

+

3y

2

c.

3xy 2 so that

=

=

2j/cosx so that JUV

2 2 / = xy 3 + y cosx - ^x

= -X 2 /y 2

2x/y and JV

h

13. Let JW

3xy

2

not exact.

is

=

=

=

JV

M = y\ny - e'^ and N =

equation 12. Let

x and

- ^x 2 =

cos 2

= ^x 4 + xy 3 + h{y),

11. Let

/

.

j/

Let A/

/

l

3

)

y

3

2

3

)

— arctan(3x} =

c.

30

JV,.

From

h'(y)

=

0,

and h(y)

-

Exercises 2.4

M = -2y and TV = 5y-2x so that M

18. Let

—

h'(y)

19.

= -2 =

y

and

5y,

= ^y 2 The

/i(y)

.

solution

— 2xy +

is

From

TV^.

2

jjy

=

/j

= -2y we obtain / =

c.

M = tanx - sin x sin y and TV = cos x cosy so that M

Let

= — sin x cosy = + cos x sin y + h(y), k'(y) = y

/x = tanx solution

x sin y we obtain / =

sin

In sec x\

is

|

+

=

cos x sin y

secx|

In j

TVX

.

From

and h(y)

0,

=

0.

The

c.

= 3xcos3x + sin3x- 3 and TV = 2y + 5 so that My = = TVX From = 3xcos3x + sin3x - 3 we obtain / = xsin3x - 3x + h{y), h'{y) = 2y + 5, and h(y) = y 2 + 5y. The solution is xsin3x — 3x + y 2 + 5y = c. 2 = 4x 3 + 4xy we obtain Let M = 4x 3 + 4xy and TV = 2x + 2y - 1 so that My = 4x = Nx From 2 i 2 4 2 2 f = x + 2x y + /i(y), h'{y) = 2y - 1, and ft(y) = y -y. The solution is x + 2x y + y - y = c.

20. Let A/

21.

—2xy+h(y),

.

.

M = 2ysinxcosx - y + 2y

22. Let

2 xy2 e

and

= -x +

TV

My - 2sinxcosx -

1

sin

+ Axy z e

2

x

xyl

+ Axye^

so that

+ Aye 1^ =

Nx

.

From fs = 2y sin x cos x - y + 2y 2 e xy2 we obtain / = ysin 2 x — xy + 2e xy2

=

h{y)

The

0.

solution

M = 4x y - 15x 3

23. Let

is

2

y sin x

- y and

2

TV

— xy + 2e

=

xy2

=

x 4 + 3y 2 - x

4

x y

is

24. Let

M

v

/=

—

5x

3

—

M — l/x -

4

2

In \x\

+ y = c. l/x 2 - y/ (x 2 + y 2 )

- x 2)

2

(y

---

/ {x

2

2

+ y2) =

arctan

x

[

-

]

+

Nx

so that TW„

3

and TV

=

From /a

.

My),

h'[y)

=

ye"

=

2

— — —

+ 2xy + y 2 and

TV

=

-x 3

-I-

+ xy 2 — y =

x 2 y + xy2

l/x

4-

,

arctan

2xy + a: 2 -

we obtain / = -x 3 + x 2 y + xy 2 +

-x 3 + x 2 y

= 4x 3 -

h'{y)

2

+ x/

ye y and

x

M=x

—

3y

1

2

=

Nx

.

and h{y)

,

—

y

3 .

The

solution

fx

l/x 2

+ y2 ) -

My) =

so that

2 y/ (x

+ y2 )

-

ey

—

c.

2(x + y)

=

ye y

.

we obtain

The

solution

is

-

y

If

c.

=^

y(l)

=

1

[

—

|

+ ye y —

e

8

\yj 1

so that

hly), h'(y)

then c

=

—

M

y

=

—1, and h(y)

W^.

From

= — y.. The

=

x 2 + 2xy+y 2

general solution

is

4/3 and the solution of the initial-value problem

is

-

M = e x + y and TV = 2 + x + ye x

and

From

M„ = 1 = TVj.. From fx = e x + y we obtain / = e + xy + h(y), h'{y) = 2 + ye", and h(y) — 2y + ye y — y. The general solution is x e + xy + 2y + ye" — e y = c. If y(0) = 1 then c = 3 and the solution of the initial-value problem is e x + xy + 2y + ye" — e" = 3.

26. Let

0,

\y/ In Ixl

25. Let

=

3

xy

4-

h(y), h'{y)

c.

fx = 4x y — 15x — y we obtain / = x y — 5x — xy + h{y), 3

+

y

so that

Exercises 2.4

M

27. Let

=

Ay

=

obtain /

4xy

+

28. Let

- 5x +

x 2

—y=

c.

If

y

=

8.

+ 3y + 3y 2 —

5i

M = x/2y

4

N=

and

3-2

then c 29.

+ Ax h'(y) =

y{— 1)

- x2 )

M

=

y

4

= Nx Prom = 4y + 2x - 5 we 2 = 3y - y. The general solution is .

-

1,

and h{y)

—

2 then c

=

8 and the solution of the initial-value problem

5

so that

/y

6y

3

=

and

so that

1

M„ = — 2x/j/ 5 = The

and the solution

x

general solution

problem

of the initial-value

— 4y

is

2

x/2y4 we obtain

^ ^3

-

If

c.

y(l)

=

1

3 5 — s = — 4 2y

—

7

is

=

A^. From /x

a-2

= —5/4

M=

2

= ^,

h'{y)

6y

h(y),

(3?/

3

= ^4 + Hv),

/

N=

and

5 2

x 2 — 5x

+

4xy

is

+

4xy

—

2

2x -

+

-

q

l

= 2ysinx — x 3 + ln so that Ms = 2j/cos£ — 3x 2 = A^. From 2 2 3 2 2 = j/lny — /e = y cosx — 3:r y — 2x we obtain / = y sin x — x y — x + h(y), h'{y) — lny, and 2 3 2 The general solution is y sin x — x y — x + y In y — y — c. If y(0) = e then c = and the solution 2 3 2 of the initial- value problem is y sin x — x y — x + y In y — y = 0. Let

cosx — 3x 2 y — 2x and

2

4/

A'

3/

j/.

30. Let

M=y

=

fx

2

+

J/

2

-1 xy 2 — ycosx — tan y

is

the initial-value problem A/j,

M Equating M

32. Equating 33.

y y

34. Equating 35. Since /x

=

/j,

/,

=

37. Let

/

=

38. Let

/

=

-

—

2 J,

= 2

3i y 3

M

4xy

+

e

x

Nx =

and A^

-

xe x "

=

^+

+

5

=

+ h(y),

= -y/x 2

{x

2

''

2

A'

and

h'{y)

=

A'x

=

M

i-

1

/2

(a:

If

-1

y(0)

y

and A^

+

ke

x

3

+

4j/

Ay

N=

3 ,

=

y

+ y)

'

and

+

=

y

4 .

lny.

The

=

e

fc

xy

2 +» "

Let N(x, y)

1/x so that

=

1

My

+ sinx =

.

From

= -tan ~' V- The

^(t/)

'

Nx

and the solution

it/4

=

.

=

10.

A;

=

9.

+ xy 2 + - + h(x)

so that

=

2y

1/2

18x?/

2

=

x 1/2 + ^ ^

A^.

(x

2

In |x

2

+ y|+/i{x)

From

j,)"

.

=

6xy 3 we obtain

solution of the differential equation

M

= -l/x a = Ax From /x = -y/x 2 we

is

.


32

so that

1

+

The

y

= — 5.

1.

= y-^V/ 3 + =

of

4

we obtain /

9x 2 y 2 so that

2y

= — 20i/ 3 ~ xycosxy — s'mxy we obtain

we obtain

eXV

=

J^~2 then c — —1 —

we obtain k

we obtain /

and

l/y

B

=

= —1 — ^

1

+ y)' + h'(x).

=

M

2kxy 2 — siny we obtain

(z< »)

2

+x

'

c.

so that

ft'(y)

+ iOxy 3

Sy 2

Axy

+ 2xy + 1/x

ti( x )- Let

= h'(y) =

6xy 3 and

= — + h(y),

and

18xy 2 — s'my and

M{x,y) = y 1

y- 1/2 x'/ 2

M

xy — ycosx

+ 4fcxi/ 3

2 3j/

N(x,y)

+

is

—

— tan

2

= — xy cos xy — siuxy + 4% 3

My =

ye*"

36. Since /x

=

+ y2 )

1/ (l

we obtain / = ly 2 - ycosx +

y sinx

general solution

31. Equating

N = 2xy - cosx -

ysinx and

4-

is

— +ln

obtain \y\

=

c.

Exercises 2.5

M = —x

39. Let

From /y

=

2x y cosx we obtain /

M

/y =

=

xe

x

equation

xye x

+

2yex

+

is

M = 2xy

x 2 y 2 cos x

2 x y e

+

=x = c.

2

N

=

ye x and

we obtain /


N ~ 2x 2 ycosx so that My =

2xy 2 cosx and

2

differential

40. Let

smx +

2 y'i

=

2

—

h'{y)

+

2ye x so that jWs

x

+

h(x), h'{y)

2

e

=

0,

—

and h(y)

0,

xe 1

+y

xy
+ h{y),

y coax

— 2x 2 ys'mx + 4xycos:c =

xe x

=

The

2ye x

+

=

and h(y)

0.

0.

+

e

.

solution of the

=

x

The

+ y 2 e x = c. JV = 2x 2 y so

Nx

A^.

FVom

solution of the

xye x

= 2xy 2 3x 2 we obtain + 3a: 2 and that My = 4xy = A^. FVom 2 2 3 = 0. The solution of the differential equation is / = x y + x + k(y), h'(y) = 0, and 3 x + x = c2 2 2 = (j/ 2 + 2xy - x 2 ) / (y 2 + 2xy + x 2 ) so that Let M = (x 2 + 2xj/ - y ) / (x + 2xj/ + ) and N

41. Let

2

-+-

V

42.

j/

My =

-Axy/{x 2y

/

=

x2

+ y2 =

x

H

+ y}

3

=

=

A^. From /E

[x 2

+ 2xy + y 2 -

2y

2 )

/(x

+ y) 2

we obtain

2

h /i(y), /i'(y)

x+y

+ y) M — —g(x) and

=

=

-1, and k(y)

The

-y.


c(x

43. Identifying

N = h(y)

we

from

see that exactness follows

M

v

=

=

Nx

.

_

Exercises 2.5 1.

For

y'

-

+

— oo < 3.

For for

4.

For for

5.

For

x <

7.

an integrating factor

=

+-y=

< y'

=

e~

5x

— dx

so that

=

5x y]

fe 1

and y

=

ce

5x

for

J

factor

is

ef 2dx

=

e

2x

4~

so that

dx

2x \«

=

y]

<

_2x

and y

=

ce

and w

=

- +

for

J

oo.

+ 4y = - an 3 —oo < x < oo.

For y'

x <

oo.

+y =

e

integrating factor

Z/x an integrating

3x

is

=

e/ 4
e

41

le^y]

so that

ox

factor


is

e/*

2

^

=

is

*e

<

=

x 2 so that

e* so that

=

-e

~

[e*j/]

2

e

4x

ax

ce

_4t

3

=

[x y]

=

41

3

J

1

3x and y

and y

=

=

31

ê 4

-

+ cx~ 2

+ ce~ x

for

oo.

—y =

e

x an integrating factor

-oo < x <

CO.

2

—

For ]f+3x y for

an integrating

y'

y'

/ 5dl

e

is

oo.

2y

-oo < x < 6.

=

by

-co < x < 2. For y'

is

—oo < x <

is

x 2 an integrating factor oo.

e'f^ =

is e /

3l2liE

e~ x so that

=

—

[e

-:I j(]

3

e^ so that

e

dx

y

= =

1

and y 3

=

xe x

x 2 eI and y

+ ce x

for

1

— —+ce~ x .

o

3

Exercises 2.5

8.

+ 2xy —

For y'

x 3 an integrating factor

ef 2xdl

is

—

— \e^y dx

so that

=

x

3 x2 e

and

I

1

y

1

= -x 2 —

9. For y'

-

+ ce -x

+ —~ y =

10. For

y - 2y 1

and y 11. For

,

—00 < x <

for

-i an integrating X1

X

for

1

00.

factor

1 r tic e/' / '

is

=

x so that

for

=

x

2

+5

1 1 2 = ~-x -7,%


—t + 11

— + — x = -2y an mt dy 2y — —x = an dy

ce

2xr

for

is

e'^ 2dx

_0° < x <

integrating factor

y

=

+— y — ~ x

sinx

\J\

15. For y'

16. Fory'

and y 17. For

y

=

y'

— dx

-x\j\

for

is

=

x3

+

-l

for

+ ccosx (cotx)y

[(sinx) y]

=

~

e~ 2x so that

_2ir [e

=

y]

x 2 e~ 21

+ 5e _2x

00

ef^ 2

^=

y

l i2

— ay

so that

fy

3 2 xl = -2y ' — f

1/2

L

and

00.

is

e

S dy

—

e

y so that

dy

is

e/t 1 /*)***

''xl

fe

=

ye * and x

= — y — 1+ce"

J

L

= iso

that

[xy]

=

xsinx and

ax

00.


+ x2

= -|

(l

is

is

e

j>/0+z 2 )l
go tnat 1/2

+ x 2 ) + c (l + x 2 ) eJ^^

1

^^ =

1

for

+

-co < x <

e* so that

[1

ax

1

for

< x<

is

+ e x y\ =

and

eJP*VC**-i)]
=

x* -

1

so that

[(x

3

-

l) yl

=

00.


— tt/2 < x <

2 cos

00.

00.


sec

=

and y


-co < x <

+ (tanx)y =

sinx

18. For j/

= -x

y

+ -|^-y = =

<

< x<

ior

+e

+ e1

x

x

—-—1 y = 1

1

+ x,

„

.

+—

+ x2 y =

+

y

sinx an integrating factor

x ^ 1

<

for

c

+

— — In x +

00.

cos x

x

14. For y'

1 ^2

y an integrating factor

—00 < y <

13. For if

y

x

00.

— —-y 2 + cy~

12. For

= ~

[^f]

dx

2?i

x

4~

is

_

gec

^ go na ^ (.

uX

7r/2.


2 sinx cos x and y

e/tanstte

=

sinx

-+-

is

ccscx

eJ for

34

coix
=

sini so that it.

[(sec x) y]

=

sec

2

x and

Exercises 2.5

+ -y =

1

19. For y

x

2

-


1

pJW x

is

)

dx

=

— dx

x 4 so that

a:

y

111 — -x + cx

=

-a;* 7

for

i

=

x 6 - x* and

J

oo.

5

= x an integrating factor 7~— + xj V

-

20. For t/

4

4

\x y\

e -J>/(l

is

r

= x+

i) e

(

-* go that

(1

d — dx

\,

22

.

l)e

- 2T

1

y'+

21. For

3/

+

\(x

=

For

1 ;r

2

x2

w'

+

+

-~\

+

x

=

24. For y'

+

=

so that

2smx COS 3^

J-

^-

+ ^ dy \ |

and x

—

[(1

27. For

y

=

28. For

ox 29. For

= -x-

.

and

i/

—+— x+

2x

3

-

-

+

ce

1

e /ll+(2/^)l^

is

„

x+

-1 < x < oo

for 1

= x2 ei

^.a^j =

so that

e 2*

^

00.

*sin2x an integrating factor

= — -cos2x + c

for

is

<

x <

is

eS

e f^ + (

lS

x )\ dx

=

xe* so that

oo.

3

y

tanx(l

=

|x

2

=

y]

-cosx) an

e«

tan x

integrating factor

— sinx

— cosx) 2 =

and y(l


— + - -s H e>

1

4 y2

ce _B

y

s2

<

for

w


is

<

e

is

In

j

sec x\

e^ l+(2/y ^ dy =

is

=

sec

2

x

=

_

(i

C0S x}

2

+ cosx 4-c for so that

— dy

<

x <

2

fv e"xl L

tt/2.

=

J

oo.

e"^ 3

^

dx

~ x~ 3

_3 — tJ = —-— and x+l (2

so that

[x

i/l

(ix

—1 <

1)

+ cx 3

jj

=

~- an integrating factor

ce~ ix for

H

[(sinx)«l

/[aein

1

+

^

~ dx

sinx so that

n/2.

x*

—

^H- ^3 + 3x

=

2 y

x 4 ~ i 3 In

e~

=

y

< x<

for

CMtxdx

J

x

]

xx +

1

x esc x an integrating factor

y)

1 1 y - — -e

y

2

cosx)

+-

1

2

=

sec

+ ccscx

secx

(

y

e

x

sin2x and xex y

=

-r 1

2

and y

26. For

= -

y

xj

23. For y'+(cotx)y

25. For

l)e

< X<

T=

(l

+

^ an integrating factor

*- tor

H

[xe y]

=

j,

ce _I

\

4~ ax

/

x(x

1

^

+

(l

=

1

yJ

for

< x<

x <

oo.

is

e/P+W*)]-**

is

e

=

xe 31 so that

^

[xe

3x 3/]

=

1

and

oo.

x y'

+

\(x

+

y x l)e y]

^ — -x

=

^1—

an integrating

=

2x and (x

=

4t/

5

+

lje

1 ?/

=

x2

factor

+c


for

is

J\(x+2)/(x+V]dx

-1 < x <

e~

=

(

x + i)^ so t^t

oo.

fWvWv = «~ 4

so that

4 fi/'"

xl

=

Ay and

— Exercises 2.5

x

=

2y

6

+ cy4

=

=

ze

1

—

+ ce 1

1

1

1

y

y 2

=

ey

+

-5

-x

1-

for

=

x

]

r rf

—00 <

a:

<

is

e

x

—d

so that

is

=

J"***

e

<

for

y <

=

35. For

2

-

ey

-e"

!T

r seed

=

[r(sec0

36. For

c

3/

+ ^ do

—+

x \e

=

y]

e

x

+

-5

y

is

1

e

ef^y+

is

^

e^ 2

y

<

- IIP =

+ sin0

1

-2

At

e~

x

—

and

+ e'

x

so that

[e

=

j/]

(l

^

—

e

^ and

^^ v

dy

=

ye y * so that

ye y *x

=

dyl

2ye»

dy

=

a

2

— ayL> = y

so that

\y x]

and

ey

and

00.

cosS an integrating factor

+ tan0)] = (2i

<

for

2

-

„ ex

00.


2

+ x In x

00.

2 an integrating factor

a

xe x

J

L

.

y

x

\x y)

dx

oo.

=

eJ

=

2

so that

y)

- + -ce _ y

dx 34. For

e^ 2 ^^ = x 2

is

for

sinhx an integrating factor

\

=

+c

—00 < x <

for

)

— + \2y + dy

and x

2

- an integrating factor

+ e'

= ^xe x +

33. For

7a;

factor

-21

e

+ e~ x + ce~~ x

=

y

-

ex

e~ x lnfe1

an integrating

- ex + ^- lnx -

1

+y=

y'

32. For 1/

3/

oo.

xx

and i 2 !/

y

y <

+ - y = - (e x + In x)

30. For y'

31. For

<

for

r(secf?

is

sec

eJ

^^ =

+ tan#} -


is

9

gee $

+ tan#

-cos0 +

e-^

2 '" 1 '

so that

c for -tt/2

<

<

tt/2

.

(I_1

&=

so that

e

dt d_

=

Pe

{At

2

-

2)e'

"'

and

P = 2 + ce'"'*

-00 <

for

i

<

00.

Jt

+

37. For

— — x+

2

438. For

+

\{x

i/

39. For

^

y'

[e"

4

2) y\

t — 1 - l)y =

+

+ 2)-

{x

=

5(i

+ 2) 2

—y = x —

+

and (x

=

y

x(x

(coshi)y

inha;

y]

=

+

=

l)


and y

=

|(»

+

2)" 1


+

is

JW^ x+2

dx

+ 2)- 4

for

)\

c(z

eSW^-

is

1) for

-1<

lOcoshx an integrating

lOtcoshiJe""

1131

and y

=

10

{

x

x<

+ 2) 4

-2 < x <

^=-+ 3

1

+ c(x+

1

=

so that

00.

x\ so that 1

1.

factor

is

+ ce _sinhi: 36

e/

coshllil

for

=

-co <

e

a ' nbx

x<

00.

so that

dx

.i

1

=1 + lV

Exercises 2.5

+ 2x = Ze ^ ay

40. For for

—oo <

<

y


ef 2dy

is

=

e

2y

so that

2y

-f- \e i

20 an integrating factor

-oo < i <

—

2y

and y

=

3 xe *

^ dt

43. For

x

-

e

J,

=

eJ™*

is

hx

e

3x

31

—

/L

2 then c

= —2

and y

=

4

)

an integrating

factor

is

e~f 2dx

2a

- ix 2 e 2x + ce 21

for

-oo < x <

integrating factor

~ — — x = 2y an dy

is e

—

2e~ 5x

<

45. For 1/

?/

=

<

y

<

t

=

oo. If i{0)

integrating factor

=

e

2 then c

+ ccosx

1 cos x 4

-00 < x < jm

=

then c

io

iQ

=

/(Vs^J/

e

is

5 then c

=

— jt/2 < x <

for

=

-49/5 and x is e

ir/2. If


e

is

2y

2

47. For

—-

jfcT

-7-

y

dy

50

ce

ftt

1

=

e

y(0)

and

if]

=

xex — x

=

and

—+

F,

- —\e~ mlL

(i \

H

and 2 - 2« 2

2

.

til

+

49

=

= —1

=

_2at

J

\

S ecx

then

- lS

so that

= —1

c

so that

e

-7e l5

=

[(sees)

—

and y [e

ax

-50fc an integrating factor

foT

+-

x

[e

_ *6

x cos x

sin

—

cosx.

= andQ =

Q|

L

cosx and

ce x

*

J

.

1

=

)

x/

and y

+ ce" 51

and

3

-x —

is e J"

t_fc)lfc

=

e~

fc(

—

so that

+

-00 <

f

<

00. If

= - e~x

y

T(0)

=

150 then


c

is

= e

150 and

/(l+2/*)
T=

£

x

c

1

—ê

for

00. If y{\)

—

then c

= -1 and

50

V =

y

=

[7\r fct l

at

+

4

- —y.

at

T=

=

- E/R and i =

I so that

/ tflnirfx

-I^ Adx =

= -7 then c= -7

IfQ(O)

00.

^

£ [ie^l

so that

ax

for

=

+ ce~ 2y

J

_2x so that

=

/W^< = e «A

cos 2 x an integrating factor

— -5x g =

46. For

=

00. If y(l)

+ (tanx)?/ =

sin

20e5x and «

ey

.

oo. If j/(0)

y

for

=

n.t L

-oo <

for

R

44. For

e

=

=

4~ U^y] dx

so that

J,

— + Ce- Rt

=

(e

= | ^ ^ L

+

/it

i

oo. If y(Q)

—

y'

42. For

and x

3y

3e

J

l

for

=

x\

dy

oo.

+ 5y =

41. For y'

y

l

+

150e fct

_fc(

and

.

— [xVyl dx

so that

e

-50fce

J

L

1

x

=

2x

J

x .

1

x-

49. For 1/

+

—+— x

y

= x

1

+

an integrating factor X

3;

c

= hix "x+1 x+lx+1

and v V

=

50. For

I

hix

—

+

x

+

u'

+ — v — — ex

x

1

x

x

+ 1

is

e/[V(*+i)H*

=

^

+

1

so that

[(x

dx

1

for

00. If yll)

=

10 then c

=

21

+

1) w]

=

hix

and

21

x

+

1

an inteeratine factor

is

e^

l^

x da ^

—

x so that

~ dx

\xy\ 1

1

—

e

x

and y

xx

= —e + — 11

Exercises 2.5

< x<

for

51. For

—

(a:

52. For

y

j/

=

53. For

+

2

y'

y

=

cx for 2

<

.

z(x -

=

2)y

+

2 then c

=

-

2

and y

e


-e*

e

H

e^ 2/xi x

is

—

2

1

=

= -

2 )\ dx

-

—

for

<

a:

=

— tt <

— + -x — ay

oo. If j/(3)

=

6 then c

an integrating factor x

<

0.

If

y(—ir/2)

-


1

e^

is

1

54. For y

— dx

00. If t/(5)

+

(sec

tana:

fe

2

x'jy

-

yl

sec

y=l- 4e- tani 55. For y'

+ 2y =

2

= — ie

2 then c

sec

2

t8JI1

=

8

=

and z

dy

and y

=

1

y(0)

=

8

2

y

+ ce~ tani

/(a;)


is e

then ci

= — 1/2

and

21

=

x-2 s inx

y so that

(

is

z)dx

sec

-7- [yx]

y'

+y =

=

1

/(a:)

then

For

y'

+ 2xy —

e

tani


e

is

37

tt/2. If y(0)

ci

=

and

=

C2

<

a;

je^

—

3;

+ ci,

\ -e*

for continuity

=

.i

+

e

is

x >

x2

1,

C2

< >

a;

1.

a:

= <

2e so that 1;

1.

so that

+c u 0
I

I

C2,

0
we must have

1-1

f(x) an integrating factor

x>3.

,

so that

fe*

C2

=

and

X >

,

38

= -y + -

for

y

g0 that

= -3

3.

we must have

Ue 6 -lle- 21

f

57.

y and x

?

=

^=

y(0)

[(sin x) y]

so that

X >

pC

If

=

dy

=

for -tt/2

< f

56. For

so that

.

C2,

for continuity

and

T

.

.

If

1

-y+ -


=

= -1 and y = - esc x.

c

^

x

i-2

2x

=

2 and y

/™ tKitc =

e

is

then

1

=

y

< y<

^dx

- so that

2)

(cotx)y

ccscx

=

go. If y(l)

1.

^ so that

then c

= -4

and

.

Exercises 2.6

=

If 3/(0)

2 then c\

=

3/2 and

3 = -e + 1

we must have

for continuity

C2

\{ + \e- x \

58.

,

For

+

y'

{


+x

is 1

=

then

ci

=

and

* >

.

'

1

so that

fix

If y(Q)

7=^

1;

°S r<1

fi£?-

2x = —^y +x 1

2

0
.

so that

Z

z

for continuity

a

0
+ ci,

we must have

C2

=

1

so that

2(1+^)

Exercises 2.6 1.

From

y'

xx

+ — y = iy

x3w = x3 + 2.

From e

z

4.

—

—

y

j/

e

31

y'

y

y'

—

^1

+

5.

From

im =

y'

In

+

+c

=

if

y

3

and 3x

y

-1

=

y

+

=

=

u>

-3

_,

we obtain

-iu

= —

x

.

An

integrating factor

is

x 3 so that

An

integrating factor

is

e

x

c or 2

and 1

,

u;

= — In x H x

1 .

x

so that

^ — 3w = —

3x.

An

integrating factor

is

e~ 3x so that

ax

=x+

y~ 3

and

y

+ w = -e ^ dx

- *.

we obtain

y

u>

=

i+

— y^ we 1

= ~xe x + e1 + c

or y

+ ^ ax

we obtain

= -re 1 + ce

— — y = — -^y 2 a;

it)

or

\e~

xe 1 so that xe*w

and

y

= xy 4

xe~ 3x

=

From

+

and

= 1 + cx -3

x 2

w = — -e 2* + c

Z) From e"

y'

3

c or

2

or y _1

y

—x c

_1

ce

3x .

obtain

= —H

we obtain

—+

4-

—^ w =

—1.

An

integrating factor

is

c

1

h

x

^1

-e _a \ x

^ + —w =

.

An

integrating factor

is

x so

that

.

Exercises 2.6

integrating factor

+ xl

1

7.

n trom u

3

2

/

x

= -^w

y

or

From £

9.

3 X /2

y'

+ y = y~ l,/2

W=

e

3 x /2

Prom

=

yx

10.

= 2-

From

3/

so that

2 3/

e

3x

w—

y

3

^

w =

and

2x

12. Identify P(x) x2 ~ x

13. Identify P{x)

j f j,

{0 j

=

û;

=

2

+

/2

2ef

and

uj

= -3x 1/2 + cor

=

-2, Q(x)

3l

An

.

=

—

= —y

h

yx

x so

is

5

7 and

2

An

/2

.

x 2 (? l so that

is

integrating factor

=

If

.

6

= j + 7fi -to/2

j,3/2

.

that

—5 x~

= --x _1 +

3

integrating factor

4 then c

2

=

=

y

y

3

3

we obtain

=

-

1

x,

Q(x)

=

2

x x so that e ~ w

= -4/x 2 is

,

=-

J"

=

xe

ce~ sx

=

12

""dx

If

.

r-

e" '*

is

= -1

then c

and

~+

2x 2

(?(x)

=

1/

1/a;,

=

4 and y3

+ 4)uj =

Then

x.

+c

or

^+

-1.

u=

= -3x 2 + 4x 3 / 2

.

integrating factor

=

1.

5

fxe x2

~x

dx

.

+c

Thus, y

Thus, y

.

-

tu

-1.

An

x

\

~l

integrating

= 1 + u.

^ + f-- + xj

Then 1

-2. Then

An

-x.

—& X ~ X

4

=

An

+ 2x)w =

(-1

— —\x^ + cor u — --x + cx -3J and R(x)

x~ s / 2

+ u.

2

dx

=

is

1/3

-1/x, and R{x)

x 3 so that x 3 w

,

then c

1

(-1

Thus,

.

integrating factor

2

=

y{\)

4

=

= ~x. An

-^-w 2x

Then

1.

^—

u=

or

=

-1, and

Q{x)

— ^ ax

= ~Zx 2 + cx 3 ' 2

-1, and R(x)

w = -ê 31 + c

integrating factor

14. Identify P(x)

— and y"

=

c

^

K

r

integrating factor

.

2,

+corr' = 2-j + ce~f

/2

+ c fl + x 2 )/ v

1

5

we obtain —

x

2

so that e

e

- then

ay 2

=

x

x

^+

we obtain

= x + ce -3x/2

3/2

= — ^l/~ 2

—

x" 3 ' 2 w

is

—w =

=

y(l)

If

.

=

—9,. An

6

I-

-3

2

3

factor

— ax

,

c or y

- e V/2.

11. Identify P(x) is

°

y

+

=

x^

lilt we obtain — .

w—

+

1

5

y x

e^l^w = -ye"

so that

x-

y

+ xl

= --x -1 + cx~ 8

3

and

+ cot

ay

1

.

and

jT

5 8.

i

1

x^

= --x 5 + c

x% w

—^—^ =

r so that

is

~ + ^-4xjw =

=

— 1 2

I-

u.

An

integrating

=

+ u.

2.

-1 factor

is

xe

2:13

15. Identify Pfx)

=

e

integrating factor

16. Identify

P (x) =

2x

so that 2x

is

,

e

Q(x) x

sec 3 x,

=

1

so that e

Q(x)

2l

w = — -e + x

2e x

,

or

1/

c

or u

and R(x)

=

.

2x

=

and

w — —e x +

= — tanx,

+c

ce x 1.

+ ^ dx

Then

=

=

40

1.

x

—

Then

.

(1

+

Thus, y

Thus, y 2e*

-

x

x

2e )w

=

-1.

An

= — e x + u.

l

-j—

+

(— tan x

+

2 tanx)to

= — 1. An

e

1

Exercises 2.6

integrating factor

w secx = — In Thus, y

—

is

e

=

is 1

=

=

solution

20. Let y

i/

=

=

solution

23. Let y is

—

—

R(x)

—

y\

37

5,

+ c or

+c

or

u=

—a:

where

~

t~

xy'

+

xy'

filf)

—

+

3t

2

f(y')

u

=

[— cos x In sec a; |

where

and y

=

/(()

=

where

2t

3

f(t)

and

1,

=

2

(~ 2 .

2

xy

4-

ccosx]

-l

family of solutions

2

=

j/

27x 2

2

-t-

—

Ax

+ t2 A

and

= —t

y

=

cx

—

is

cx

+1—

In c.

The

singular

ma:. is

y

+

c~

2 .

The

singular solution

The

singular solution

.

family of solutions

4f

is

y

—

cx

—

3

.

j/

=

family of solutions

3

~ + (6-6)w = -1 ax

= —3 + u.

A

.

— —4 — 2t

integrating factor for

Ini or

= — t3 A

=

A

t.

An

-3.

^— + (5 — 4)u; = —

= -2 + u.

2

.

family of solutions or y

= — ^(2 +

+ f(y') where /(() = — e'. A family of solutions x — e and y = — e' + Je or y = In x — x.

is

y

—

+ c2 The

+

Ac

c .

The

singular solution

In c.

The

singular solution

cx

.

singular

2

4)

1

xy

+ tana:]

integrating factor for

—

or 4y

or 27y

=

yi

Thus, y

.

3t~

An

—2.

Thus, y

.

—x + c = 1 — In

=

=

1

—

u

and y

where /(f)

given by x

is

l

and

1,

=

f(t)

x = 2t~ 3 and y

is

.

y

=

cx

—

e

!

t

given by

24. Let y is

—

or

6,

+ f{y')

given by x

22. Let y

+ tana;} + c

~

+ f(y')

xy'

sec a;

Q{x)

given by x

is

given by

21. Let is

=

xy1

Q(x)

— —

9,

w=

so that

19. Let y

is

6,

1 so that e x w

18. Identify P(x)

|

+ u.

tan x

17. Identify P(x)

x so that

sec

is

a:

1

+

given by x

f{y') where f(t)

= —y

and y

—

Int.

= \nt—

=

1

A

family of solutions

ory = ln('— —

[P(x)

—

is

y

=

cx

+

1.

+ R(x)yf\ + [Q(a> + 2 Vl R(x)u + R(x)u 2

+

}

dy\

du

dy

—

u

Exercises 2.6

26.

Assume

du that

ax

1

If

y

28. (a)

+ y 2 - Q(x)y Assume that

—

[Ft{y

F{» -

—

2

P(x)

=

by


27.

— Ru o

(Q + 2yiR)u

-;

F=

c )l

x?/, y')

and

29.

If

y

=

is

WW — then ^ = w ax

=

and y

— cx, c) —

and F(y

(f ™ c — 0- Assuming = F(y - cy, c) = 0. 1

)

—

1

FYom i =

=

1

/{()

— tf'(t) — ~ */'{*) +

of y

=

V+

=

j>

xy

and substituting

c

=

-/'(t) and y

Substituting into y

'

'

and w" - Q(x)w'

W

Then, by the chain

0.

we have

^

Ft

~WW

"

-

y'

3

xy')

= +

c

- P(x)w =

0.

rule,

and

(y')

2

+

5

=

0.

j/

=

By

(a) a family of

2

=

+ /(c) =

Multiply the given

.

+ c + 5 = 0.

3

cx)

cx + f(c) then y

the identity cx

30.

(y

ndu dw — = —u — ax

— + (Q + 2yiR)w = —R. ax

(b) Write the differential equation in the form (y solutions

so that

dx

obtaining

F(t, s)

w = u _i

let

+ f(y') =

xy'

/(f)

+ f{i/)

into the differential equation

ic

tf(t)

we

find f(t)-tf'(t)

an

is

+ f(y') we

obtain

+ /(c).

-

which

xy'

we obtain

identity.

£ =

^

=

xf

=

=

+ /(().

=

Since x

f

for /"(()

-/'(*), this

^

0.

becomes

Thus, the parametric equations form a solution

/(?/)

Exercises 2.7, 1.

Let u

=

„

e iV

.

„ Then

du „dy — = 2e — and dx ay 2s

linear with integrating factor

,

—

x 2 Thus .

x 2 u = 2xlnx- 2x + 2. Let

u

=

lny.

Then

— = ax

with integrating factor e

1 .

Let u

=

ye*.

Then y

=

ue~ x dx

~ ax

=

ie 1

ue~ x and dy

+

= 21nx

2

c

ii]

or

(1

[e

x

u]

=

e

+ ce" 1

21

x

u =

x 2 e 2y = 2xlnx - 2x

21nx 5— —

.

This equation

or

In

y

e~ x du)

42

—

is

x*

+u= ^ dx

+ c. e*.

This equation

and

= ê1 + ce~ K

= — ue~ x dx + e~ x du,

+ u){—ue~ x dx +

2

I

and

and the equation becomes

u

3.

[x

y dx

Thus

— dx

rfii -

,

,

the equation becomes

.

and the equation becomes or

(1

+ u) du = u z dx.

is

linear

Exercises

2.

Separating variables and integrating we find

— — + In |u| =

x

—~ +

+c

4.

Let u

=—

ln|y|

+x=

x

+c

yln

\y\

=

e~

x

+ cy.

i/6

ZL

=

so that x

=

uy and dx

u dy

4-

The equation becomes

y du.

V

(2

2

+ e~ u )(udy + ydu) + 2(l -u)dy =

-+*-

2,

2e u 2eu --

Writing this in form

+ «| +ln

In |2e"

5.

Let u

—

=

c

we

6.

Let

w= x

+u

+ x4

-e _u = x 2 +

find

so that -r-

=

1

dx

+

This

is

Let

u

—

a

=

=>

+ u) =

cl

y

2

is

so that

4ui 3 di

+ x 4 du. The

3

c or

-r-

.

-e~ y t

xi

=

5

e"

x2

e

,

Let

c

2t,e

1 /*'

+ x - c.

equation becomes

=

2x dx.

+ c.


dx

+w+

1

=

uV T

— + u = uV*. dx

or

j

w = u _1

to obtain

dx

w—

e

-3*.

An.

_I so e ,

^ = 2™'.


~ +u= — ax

2

—

1.

An

integrating factor

so

^[e*u]

8.

=

+

e""rfu

or

rfx

x

0.

y

find

p ^2e*/»

a Bernoulli equation and we use the substitution

integrating factor

7.

and integrating we

- 4x w = 2x ^ dx

1

dx

=0

dv — du + — =

xig

y

and dy

4ux 3 Integrating

dy

y(2e

=»

= ux 4

so that

-^7

=

|s/|

+1 du H +u

e~ u

+

or

u

=_

—y+

1

(j?

+

so that

x

~)

e*

c

du dy — =— dx dx

.

*

u= _

(xV - xe 1 + e x ) + c =>

2

y


du — =u— dx

1

+ xu+l

du or

dx

2 u = xu

.

- -x a + x -

1

+ ce".

is

—

7

Exercises

This

2.

a Bernoulli equation and we use the substitution

is

integrating factor x

e

is

x

w = -x+l+ce~ x

=s- u

=

(1

ax 9.

Let

—

1

(ill

dx

—

u

I

dx

integrating factor

2csc2y

so

is x,

x d — ax

10. Let

=

r

An

2.

-— =

so that

2

y dy — tany dx

sec

rfu

u — mitany}

u = x2 w

=>

,

.

2x

xuj

— = x -— dx dx 2

so that

1

u

—+w =

to obtain

dx

—x.

An

so

,

= -xe x

-^-\e w\

w —

I-

dy — dx

i

xx

—

H

.

.

-x + ce -1 )^ - 1. 1

(1

du -r-

—

2x

—u

or

dx

— c

=

ln(tau y)

x

H

— = u dx


2xy.

=

y

The equation becomes x

.

c

=

li

- x + ce -1 ) -1

2

+

=

x

.

1.

Separating variables

and integrating we have

du

+

u2 11. Let

u = x3 y3

=

=>

,

dx

1

u

tan

= 3x y ^ + 3x y ^ dx dx 3

so that

tan

=

du

2

2

3 .

The equation becomes ^x 3

6x

--

2

=s>

it

=

2x

3

-

+

9In|x|

==* x 3 y 2

c

x Let u factor

= is

e.

y

x

so that u'

-2

u = x

+y

integrating

The

=

e y y'.

=

2

fx" ul 1


= - => u =

x2

= 1 + y'.

solution

Let w

~

=

v

u

— sinxdx

du

e

u

= — cosx +

— sinysinhxdx + sin ^ cosh x = e.

cosh x so that du is

u=

c or

wi

dx y — = — —h2ylnx dy x dy cfti

so that

— = xe" —dy— =

y dy

=

— 2u = x 2

ey

=*•

iz'

=

=f-

e


2

15.

= 2x ^ dx

3

—

3.

Separating

2x

3

-

91n

or u'

x

+ c.

|x|

u—

x.

An

integrating

e

=

x 2 ln

III

+ cx 2

.

_u sin x. Separating variables and

we have

sin y

„

xu'

+ cx 2

In Ixj

x

J

so that v!

e

14. Let u

+ c.

so

,

— dx 13. Let

x y

and integrating we have

variables

12.

2

1

=1+c

l

or

ey

.

or

c

cos y cosh

I+y

x du

dy

y dy

dx

we have

=

+ c =^

,

l-2xlnx.

—

ye y

ti

ye"

—

ev

44

2

y In x

=

ye"

—

e"

=

0.


y

du

c.

x efy. The equation becomes du

—— = y-—


= — cosx +

+ c.

—

7

u

Exercises

16. Let

An

du dy — = smy — dx dx

= —cosy so that

u

integrating factor

=

u

—1 u X

-^L- = -dx => 18. Let

=

ii

1

y so that

u

tan

=

u'

du ax

_

1

01 -

x

u

= — xe

.

— —xe + cx

=5-

=

cost/

xe 1

—

ex.

.

~ y". The equation

y' so that u'

9 u — ~x e

,

1/x, so

is

A. dx 17. Let

du Ihe equation becomes xdx

_, .

2.

-1

becomes

u=-2 + ci =*

y'

u'

= -u -

=

tan(ci

-

=>

i)

Thus

separable.

is

j/

=

In cos(ci

-

x)\

|


y".

which

1

— — u = u2

u'

which

,

Bernoulli.

is

+ c%

Using the

'X'

substitution

w = u _1

ax

-M = -x=> » = 11 d

=

19. Let u

jd

y' so

=

that u'

—

we obtain — dx

=u

d

f

jx

i 2

iix

=

it

j/ so

|

x

w—

du

dx

,,2

™-2

—

=f>

tt

=

A

1

—

1

_

U

y' so that

with /(()

+

t

it'

—

y".

A

.

u

=

t/ so

—î— = 1 + u1

that u'

dx

=^

=

i/

=

2 ±\/c i - X

_

c\x

+

11

+ c2

/

I

u

„

1

3

u = —

is

x

ci

- x2

,

,

which

= -in

n\

i

C1

- x 2 +C3

Bernoulli.

is

.

|

Using

so

2

x

+ u2 =

x\

and u

=

3 /2

)

'^^

= x + ci

Separating variables we obtain

0.

if U^f—+ ci ^x + 1/ci

=

xv!

1

+

x

+

(u')

3

+

+

(l

+ cfj,

3

—

t

t

(

\cix

ci

a 1

/

1.

y

This

—

— 3* 2 ) =

is

^Cix 1

+

a Clairaut equation 2

4f

+ 3 .

(l

+ cfj x +

~~5\~^)

=

v!

=

tan(x

1

+ u2

+ ci)

.


j/

= — In

I

cj.

Eliminating the

x\ 5 /2

24 /

==> u

xa

+ C2.

u — C\X

is

_

^/C j

_


_1

-j

.

1

——

= — 3t 2

= 1 + 4 --

tan

^

2x

J

family of solutions

y".

u

z

The equation becomes u

,

parameter we obtain u

1

-1 + -|i2

^

given by x

is

integrating factor

c

=

w>

i In Icii +

=

x, so

x

1

X

y

3

singular solution

22. Let

An

.

v!

The equation becomes x 2 w'

j/'.

I'

tl"

21. Let

=

_

=

«

1

J

that u'

=>

is

|

=i* 20. Let

- x2

c\

1

-=

integrating factor


y".

„ = -2x

1

w]

L

An

—1.

x

=>

.

,

+ —w =

we obtain

cos(x

+ Cj)| + C2-

7 Exercises

=

23. Let u

2.

=

so that v!

y'


y".

,,,,,, In u = In +c

—U = -
1

=

xu'

,

u=

a;


u.

>

=

y

cio;

C2£ +C3.

a;

24. Let

=

1/

—=

=

In |u|

=

du — dy

ti

u'

+ ut&ax =

+ c =>

In cosxj j

so that y

if


y".

—(tana;) dx

it

=

25. Let u

=

y' so that u'

=

u


0.

—

du

The equation becomes w—

.

=*

cicosi

I-

=

2yu

y

=

c\

ainx

we

obtain

+ C2-


0.

dy

we

obtain

du u*

26. Let u

= du

11

=u

so that y

11

=

dy

T

27.

We /

[I

=

0,

need to solve

+ ir J

= u

u

— dy

y

2

=

1-

(1

+

/

or (1

11

=>

C]

,

y

c\y

=

—

1

2

—— )dy = T

#

+ ci

ci

_


ciy-l

0)

I

—

/

j/

y

=^

dx (forci

+

du = «. u— dy

V

+

=s-

dy

=

dx

—y + 4

1

and another solution

(y')

+u

—-

1

— y*

The equation becomes y

.

1

= — dx |l

=>

c

V

then ydy

->\3/2

h

t/it

=

u

— If ci

1

J = + „2y ay

2

=

^ ]

J

=

we have

— = dx

=>

Let u

-y = — x + c%

is

=

y'

so that u'

=

y".


/u v"

du ^-

.

Separating variables and using the substitution u

tjz d$

co$6dQ

Vl -

=

x

=>

x-1

46

=

x =f-

sin#

=

a;

/

—

s-r

.

d$

=

x

3;

=

tan0

7

Exercises 2.8

Exercises 2.8 1.

=

Identify x for

=

n

0,

=

y

1, 2, 3, ...

and

1,

2.

j/

=

Identify

—

n (x)

eT

*

=

0, j/o

1,

1

n

As n 3.

=

—

»

1, 2, 3,

- X + ^X 2

ra

=

.

and

/((,i/n-i(i))

= 1 + x + x 2 + jU 3

=

0.

— —1 — x + »

Vo

1, 2, 3, ...

=

4.

=

7i

=

i«(x)

=

=

l

+

1

a;

+ ^x 2 - gz 3

- X + ^x 2 - jU 3 +

^x 4

.

2e

Picard's formula

2

+ ^x 3 +

^x 4

=

1

+

=

1

+ x + x 2 + ^x 3 + -^x 4 +

x

+

x

^x

z .

/(*.£n-i{'))

=

2ty„_i((). Picard's formula

j/„(x)

=

1

we

is

is

+ 2 £tyn ^{t)dt

find

+ x2

!fl(x)

=

nix) =

1

+ x 2 + ^x 4

y4 (x)

= 1 + x 2 + ^x 4 + ix 6 + ^x 8

0,

yo

e

*

=

1, 2, 3, ...

.

+

x

1 .

0,

and /(t,yn -i(0)

Iterating

»(«)

1

2

+ ^x 4 + \x e

1

—

j/„_!(t)(ii

+ yn -i(t))dt

j*(t

ysix)

yn (x) for

=1-

=

oo, j/n(x)

Identify xo

1,

Iterating

.

= 1-

find

y 2 (x)

oo, yn (x)

y3 (x)

= * + yn-i(*)-

+ x + ^x 2

yi(x)

As n —

we

=

1

y„{x)

.

Iterating

.

is

x

t/i(x)

Identify xn

for

.

.

Picard's formula

find

yn (x) for

= -yn -i[t).

= l-x

»(x) = oo,

we

Iterating

.

!fl(x)

As n —*

/(t,j/„_i(())

we

=t-

2fj/n _i(f).

= -x 2 -2

Picard's formula

is

*Hn-l(0*

find

= ^x 2 - Ix 4

»(*)

= ix 2 - ix 4 +

^x 6 - ±x*.

.

5 .

Exercises 2.8

As n -> 5.

Identify to for

6.

oo,

n =

=

-> -

- -e

=

and

yo

0,

1, 2, 3, ...

=

Identify xq

= we

Iterating

.

=

0, j/o

0,

.

and

1,

yn (x)

forn

=

3,...

l, 2,

W

(a:)

y2 (x)

As n —> 7.

(a)

Identify xo 3/

n (x)

—

x

=

2e

=

1

FVom dy

(c)

The Maclaurin

=

(l

2e*

= 2e'-l-£ y n -

-

=

2e*

+ x + \x 2

y 4 (x)

=

1

1

=

t/a(x}

If

yo

(b)

If

yo(x)

=

oo, yn (x)

—*

0.

is

{*) dt

1

-

- x - \x 2 - ^x 3

1

+ x + \x 2 + \x 3 + ^-x4

.

24

o

I

.

=

and

0,

/(i,

=

=

=

x

and y(0)

=

y2(z)

series for tana;

is

=1+

n _j(())

j/

1, 2, 3,

.

.

Iterating

.

.

Picard's formula

1

+ -x3

yn{x)

,

=

x

we

is

find

12

1

+ -x 3 + ~x 5 + —x 7

we use separation of variables to obtain y 1

2

17

O

i.0

did

i+ -x 3 + —x 5 + -tt^x

1

for \x\

H

k then the iterants are k times the iterants given in Problem

(a)

= - jf

As n —*

0.

formula

j/„_i(t). Picard's

y3 (x)

-

=

>/3(z)

x

+ ya )

(b)

-

—

y„(x)

is

find

y\-\{t)dt for n

-\-

=

x

=

0*0

3/2

I

yo

0,

y\{x)

8.

=

x yn (x) —> e

oo,

we

Iterating

.

=

find yi{x)

/((, j/„_i(())

-J/n-i(0- Picard's formula

<

.

—

tans.

ir/2.

3.

x then

m(x)

=

x

+

2

kW = x + ,

-i

-x* 4

2

=

,

i

.

4

8

7

2.4c8

7

2 3

x

3

+

x

5

+

Ts

mx

16q

Chapter 2 Review Exercises 1.

For f(x,y)

—

(25

— x2 —

y

2 )

we obtain fy (x,y) = 2y

solution for any point (xq, yo) in the region x 2.

y

2

+y < 2

=

48

(25

— x2 —

25 or x

2

y

2

so there will be a unique

^

+y > 2

25.

Chapter 2 Review Exercises

=

y

3.

False; since

4.

True; since f(x,y)

5-

(a) linear in

a solution.

is

=

(y

—

l)

3

—

and fy {x,y)

—

3(y

x

are continuous everywhere in the plane.

homogeneous, exact

y,

(d) Bernoulli in x

(e) separable

(f) separable, Ricatti

(h) homogeneous

x

(g) linear in

Bernoulli

homogeneous, exact, Bernoulli

(j)

x and

(k) linear in (1)

2

(b) linear in

(c) Clair aut

(i)

l)

y, exact, separable,

exact, linear in

homogeneous

(m) homogeneous

y

(n) separable

(o) Clairaut

(p) Ricatti 6. Separating variables

cos

7.

2

y V +

xdx —

1

we obtain 1 1 —x + —

dy

,


we

„

.

=

2i

sin

-1,In

4

2

1

/j

- ^y 2 = 4'

ij/ 2 hi a

2

8.

y(l)

=

c=

1,

-1/4. The solution

2y

is

2

ln

|y|

Write the differential equation in the form y equation, so

let

x

=

Then dx — udy +

uy.

2x

+ sin2j; =

-

e

21n(y 2

+

l)

+c.

obtain

ylnydy = xe x dx If

=>

l)+c

(j/

- y2 -

In

xe*

4xe x - 4e x

— dx = \x\n

V ydu and the

ylnu(udy + ydu) — (uylnu —

y

1

+ c.

I.

j

This

dy.

is

a homogeneous

V

\

J differential equation becomes

y) dy

ylnudu — —dy.

or

Separating variables we obtain In

ji

du

=—

uln

|u|

— u = — In \y + c

y x(ln x

9.

The equation

is

homogeneous, so

becomes ux (udx + xdu) = 2

- - = -ln|y| +

*ln

\

y

let 2

(3u x

2

—

In y)

y

=

+x

ux. 2 J

— x = — y In

|y|

+

cy.

Then dy = udx + xdu and the

dx ot

uxdu =

c

y

^2u

2

+


Separating variables we

l) dx.

obtain

u

dx

j\n(2u 2 +

l)

=lnx +

+ 1 = Cl x 4 2^ x i

If

y{— 1)

=

2 then

c\

=

c

2y

2

=>

2

2u

+ x2 =

9 and the solution of the initial-value problem

+l=

Cl x

is

cix

4

6

2y

.

2

+ x2 =

9x 6

.

1


10.

The

^ + 6x +


^

„

.

tii

dw — dx

6 H

+

The

or e"

— dy

=

lM =

12y

2 .

= —1

=

then c

u = xy so

12. Let

5

is

integrating factor

dx y-r-

„, .

The

...

...

l)y

3

= -3x 3 + c.

/dw — \dy

„

equation becomes e"

differential

2

=

I2y dy

i/

is

+

dx

(u

x

I

12w + xe" = --2

\

.,

I

J

+y-

=

2

fx

3 4j/

+

c

=

of the initial-value

2 x - 2x) dx

+ 2x) e 1 =>

4

- -320

e

I

e*"

=

4y

problem

3

is

+ c.

e

xy

=

4y

+ 5.

3

equation becomes

=

=

u

dy 8x — + —-y =

±[(x*+4) y\=2x(x* + 4f => then c

=

e"

or

^+w=x

2

=>

+

+ 2x.

e*, so

13. Write the equation in the form

= -1

+

==* (6x

dy

and the solution

-T-leû]

If 3,(0)

so

1,

u 3 we obtain

also exact.)

= xdy + y dx. The differential

that du

du -

An

1

w =

Using


e"du If 3/(0}

+

Bernoulli.

-^r + r^-T

=> ™ =

-9* a

du — =x+ dy

that

« 6x

is

is

1


u = xv so

11. Let

+

6a:

* . f . integrating factor

a An

.

+

1

-£-[(fa {Note:

9l2

w=

+

6x

2

= —~~—ry

y

1

and y

=

2x „

.

+

c

An

=

+

[x>

2

i e*

+

^+

+4

=

An

integrating factor

2x

y.

£

-e _I

.

n

/

(aT

dx

=

integrating factor

- - 320


y

y

c

is

is

(ar

\i

+ 41

,

so

= \ + c (x 2 + 4)

e~ 2y so ,

dy

dy 15.

The

[ *

e

-2

"x1

=

ye"


dw

4

dx

x

w = — x 1 An .

.

— dx If

[x

-4

wl '

<-

y(l)

=

1

2

"

=> e-*i = ~»c-^2

1

=

is

w=

Bernoulli. Using r

.

,.

mtegrating factor

=t- x~ 4 io

is

x

= - lnx + e

-4

_1

y

=

1

and y

=

(x 4

-

=t-

x 4 lnx)

2

"

+ c =>

4

x

= -\y 2

we obtain -xy 2

\ 4

^+

4y

=

=t- y

—

(cx

+ ce 2".

x 4 y 2 or

so

,

u-

x then c

V

1 .

50

- -x 4 lnx +

cx

4

v

4

— x4 lnx) '

.


16. Writing the differential equation in the form y

=

with /(f)

(t

+

l)

A

2 .

family of solutions


is

y

=

is

=

xy'

y

=

i ^+ ]

— ^ (cos u) 2

cos u du

=

2x dx

cx

+

+

(c

I)

2

see that If

.

it is

=

y(0)

a Clairaut equation

then c

= -1

and the

—x.

17. Write the equation in the form -i [cos £ !r \ y } dx

the differential equation becomes

+ {y' + l) 2 we

=

a:

and

•

Then — -

u=x

sin

—^2

=

x

A

we obtain


0.

— = \y — and dx

2 dx

y

+x — ^ dx

sin

u — -Lz

let

+- c.

y 18. Let is

e

— dx

u— y

1

x

=

ti'


?/'.

1

^]

=

19. Identify xq

xe

z

—

e

i-

0, t/o

=

x

u — xe x -

1,

and

e

x

+ c\

yn _i

/((,

(())

yn(^) for

20.

n=

n

=

...

1, 2, 3,

From dy =

for

=

x

—

x -

\

+ c\e

u'

u ov

v!

+u=

x.

An

integrating factor

so

,

[e

so that

(4

—

.

Iterating

we

=

i

l

+

1

i

2

=

y'

+ y 2 -i(t).

^

3

x

=

l

+ x + -x s

y 2 (x)

=

1

+ x + x2 +

—

y=

Picard's formula

-x 2 -x-cie

is

+ _(%n-l(()^

3 we obtain

+ ^x 4 + ?/

=

2

~x 5 + ^x 7

+ e _2a\

.

Picard's formula

is

so that

yi{x)

= 3~2x

y2(x)

= 3-2x + 2x 2

y3(x)

=3-

2x

+ 2x 2 -

yi (x)

=3-

2x

+ 2x 2 -

4

3

4

3

2

4 2!

and j/n (z) -* 2

-

find

yi(x)

2y) dx and y(0)

1, 2, 3, ...

=

=t-

+ e -21

as

n

—

>

oo.

3!

4!

X

+ C2-

3

Applications of First-Order

Equations

Differential Exercises 1.

= ax we obtain y' — —

From y

Prom 3i y

3.

— —x dx

'

=

—

=

—

.

so that the differential equation of the orthogonal family

is


is

c\e~

e ClX

.

ylny

=

2,

-—

— .

=

= -dx

+ x2 =

C2-

2^/y so that the differential equation of the orthogonal family

and

\--y\dy = xdx

Then

U

3 2 4j/ /

+ 3i =

c2

is

.

1


ia

we obtain y '— 1

cf

x

11—dx =

2x

1

dx and y

we obtain

y'

Then y\ny dy 3

cix

we

= x2 +

2 In \y[

2 j/

+ c2

.

obtain

2

=

=

2x

—x -

-

= —xdx y'

2x so that the differential equation of the orthogonal family

and y 2

we obtain y = — y

=

and

/

y

=

y'

and 2y 2

2 y 1 we obtain y =

1

Then ydy

.

we obtain

Then -dy

.

—

y

2 }

—x

=

y'

xy

+ y2 =

V — 2x

=

From y «'

c2

3 = --

y'

= -xdx

Then 2^/ydy

-y

Prom y = 1

=

C2-

we obtain

— c\

(x

—^j2

=-

2

Then 2ydy

.

From

v

9.

cji

=

is y'

8.

=

From 2i 2

y

=

y

y

+ x2 =

we obtain

cj

Then Zy - 4x

=_

1

7.

=

From ax 2 +y 2 y

6.

4- 4j/

and y 2

1

ia

o

FVom y y'

5.

4 -.

Prom y y'

4.

=


x

Then ydy 2.

3.1

—

=

o^x.


is

+ ciso that the differential equation of the orthogonal family 2

and 2y \ay

— y 2 = —2x 2 + ci.


Then 3ydy = -2xdx and Zy 2

is

+ 2x 2 =

3y

52

ca.

is

Exercises

10.

From y a = C\X b we obtain y'

^

=

y'

3.


— — 3£ Then bydy = —axdx and

by

.

+ ax =

2

2

is

C2-

by 11.

=

Prom y y'

—

x2

=

r

£ 12.

=

From y

1

is y'

13.

we obtain

-

=

x

family

y'

— y2

(1

dy

}

c?.


— 2xdx =

4cix we obtain y'

—s—-—*

y

— 3x 2 —

and 3y

y

3

=

C2-

- 2x 2

2

so that the differential equation of the orthogonal

2xy

2xy_

=

2x z

This

z

y

is

a homogeneous differential equation

Let y

.

=

ux so that

.

^2

du

g

=

^£ —

z2

From x 2 +y 2 — 2c\x we obtain y'

2xy

=

„1

This

a

- y1

x-

is

+ y 2 «i|y| =

y'

2 2 = y— -x

—

^.2

— u -2 — In |u| =

=>

is

=

= u + xv! Then 2

14.

—

=

y'

u3

is

2x

Then

-

+ y2 =

y'


c\x

From 2x 2 is

=

y'

Then y 2 dy = —x 2 dx and x 3 +

.

2

we obtain

+ CjX

1

In |x|

cry

+ c =^

^

—

n

If

— n x \) = '

I

mar

\

+c

2 .


2xy

a homogeneous

differential

equation

.

x

Let

=

uy so that

dx dv — = u+w— dv dy

Then

y

15.

From

y

family

3

is

+ 3x 2 y = y'

=

x2

ci

we obtain

=

+

—+

.

This

is

~2

]

=

i2

c2

+ y 2 = c 2 y.

2xy

r

x*

+ y2

--- - --

j/

1

[

so that the differential equation of the orthogonal 5 £

y

a homogeneous

differential

equation

Let y

.

2xy

=

ux so that y1

= u+xu

1 .

Then -

16.

2u

J..

7j

rfu wu,

_ = —

^ —

_

r

-In — ill

I-, j.

1

From y 2 — x 2 = c\x 3 we obtain

_ - ..21 — = u —

3/'

=

,_ i_i

In |x| 111 j,

3y2

_

+r c ,

>-

—f .

-

2

_ a? J-

/\1 f

i

y~ \ W

I

=

ci

=> .

_a — ..a x y

=

c\x

x2 so that the differential equation of the orthogonal

2xy

53

1

Exercises

3.

is y'

=

family 1

y

——

2xy t-j Zy

x2

—u + xu'.

Then

2

dx — I

l-3u —sdu= 3 ti + 3u ,

y (x

—

From y

+

1 >

y

^—x

xz

we obtain

2

+ 3y 2 ) =

y'

—x dx

2

(x

ci

= l

differential

,

==>

+

„ ,i d

+

|

In \u

equation

3y

x y

=a

+ 3, 3 = {x

2

=

+ 3u 2 = )

c 1 (x''

+ 3y 2 )

ux

so that

In \x \+c

/

^

2

=*

==> y

)

,

,

21n(l

I

2

2

-

3u

I

Let y

.

1

1

+ 6xV+V)

.

= so that the differential equation of the orthogonal family is 1

+ x-

2 l + x = l+SL Then .

2xy

From

+ x2

-

=

2 dx ==> y

x

—+—x we

=

i/

l

=

2ydy 18.

a homogeneous

^=^(1 + 6^ + 9^)

1 +

17.

is

2 + 9u 2 — -12u ; du= u + 3u3

l

=^

This

.

— —y 2

obtain y'

1

In \x\

+ -x 2 + c

=

2

2y

2 hi

2

Ixl

+ x 2 + ci.


is

c\

1

y 19.

=

Then y 2 dy — dx and y s = 3x +

.

From 4y

+ x 2 + 1 + c\e 2v = 4

orthogonal family

H

is

—x

— — [xy\=x rf

I"

20.

~ —x —

FVom y

= — dx

+

1

dy

family

is

21.

1

=

.

V 22.

From y family

23.

2

From y'

=

is

y

sinh y

„

tanhy

=

^'

— +x = — dy

= -ye v

2 y dy

=

+

y ,

An

y.

evx

==s-

y'

=

is

x4

so

,

~

y

x so that the

integrating factor

-ye y +

ev

+

c

l_i.2.„-4 = ---x +cx

differential

.

.

+ tan x)

y so that the x

x dx and 2y 3

we obtain

cii we obtain .

integrating factor

^4^1 x y=-x 4_1 --x 6 +c +c

we obtain

= —e v cos 2 x. Then

=

An

x.

so that the differential equation of the 1

x

cix

ln(ci 1

y x\

or

=

+ xi —

ey

is

=s> x

,

equation of the orthogonal

so

= -y + 1 + ce

3

Then

=

[e

1

=

dx

+x

=

?/

Ay

-x~5

we obtain

^

In

y

y

1

=

From y

x

1

d — dy

f

cie

3

j/

we obtain

c.

y'

Then tanhudw

y'

equation of the orthogonal family

is

+ c.

=

3x 3

=

e~ y sec x so that the differential equation of the orthogonal

e~ v dy

=

differential

2

= —cos 2 xdx

and 4e~ y

=

2x

+ sin2x + C2-

taimy so that the differential equation of the orthogonal family

x

= — xdx

and

2 In

j

coshyl

54

+ x2 =

ci.

is

1

Exercises

24.

v'

25.

=

From y

—-

From i 1 ^ 3

26.

y

.

+

y

= —

is y'

family

l

~a

28.

From

=

=

r

family

=

y'

is

is

r

2ci

y'

and y

=

—^

4~[ex y]

~

2

=

x

|

x

+ c2

is

.


'

2- "

j/

^=

x5 / 3

+ c2

.

—

x.

An

x e y

=

2e

1

+ 3e~ x

x2

=

'a

+ c2

.


+ y-1

-xe 1 =^

3cix we obtain y'

2 In cosx\

orthogonal family


1

= 3andy = 2-

+

3x

—

Then y + y — e

differential equation of the

5

i 1 "" dx and y'

^

2

x 2/ 3 dx and y

x y.

2

=

dy

-

=

y'

we obtain

—x—

5 then c

3xt/ a

From family

29.

=

u(0)

dy

y

we obtain

ci

Cie H

1

2'3

Then

dx

If

y cot x so that the

we obtain

cj

Then y

.

=

x

=

t/

=

From x

—

1 ^3

From x" + ya — is y'

27.

is

+

y'

= - tanxdx

Then ydy

.

family

x we obtain

ci sin

3.

integrating factor

x

->xe*

+ c =»

y

is

e

x ,

so

= 2-x + ce^ 1

.

.

= —_


Zx*y

y.

Then dy

=

cos0 we obtain

dd — = tan dr

6.

3x 2 ydx and r

^ dr

=

—cot

j/

=

ce

1 .

If

=

y(0)

10 then c

=

10 and y

=

lOe1

.


Then

cot 0d9

=

dr

—

=i-

In lsin#|

=

ln|r|

+c

=4-

=

r

cisin#.

r

=

30. From r

ci(l+cos#) we obtain r

family

= — dr

is

r

l+cos0

,„

sin

31.

From family

2

is

= r

sin 6


Ifien

-z

l

+

cos0

dr 1

-

r— dr

20.

.

.

.

,

=

tan20 so that the

differential


Then

-tan20(f0=— r

.

r

cos

cisin20 we obtain

^ = - cot

dr

sintf

r

r

i

dr sinfl

.

(

—=

-ln|cos20| 2

55

=

Inr

+c

r

2

=

C! cos20.

Exercises

32.

ci

—

From r

1

....

family

is

3.

de — =— dr

r

+ coa 9

1

,

From r

r^-

=

=

ci sec

sinfl

dS l+cos0 — — — — — so that the sm8 dr -

differential equation of the orthogonal °

:

— dr

sinfl => — s
we obtain

6

r

Then

.

+ cos0

l

dr — r

=

d(? sinf?

33.

we obtain

+ cosS

= cot ^ dr

r

-

In

|1

- cosfl] =

+c

lnr

1

=

==> r

ci:

r

1

-

cos#"


is

-tanfl. Then

dr

dr —

-cot8~

=> — In

=

sin SI

I

In

Irl

+c

—

r

c\c&c8.

r

34.

From

r

=

C[e

fl

we obtain

=

r ^j-

1

dr r

= _l. ^ dr


is

Then

—

=

-dB

-9

=

+c

ln\r\

=

r

c\e~

e .

r

35. See the figures for this problem in the answer section in the text. Let

measured from the positive

x—

tangent line to a

axis, of the

member

be the angle of

inclination,

of the given family,

and

ip

the

angle of inclination of the tangent to a trajectory. At the point where the curves intersect, the angle

between the tangents

a.

is

Now, the slope of the tangent

Depending on how the angle a

is

we

chosen,

line to

any event, using the ;r— periodicity of the tangent function and the dy — — tan^J = dx

tan{/3

tan/J±tana ± a) = -— 1

T tan p tan a

36. Since the differential equation of the original family

is

/(x, y)

is

y'

—=

= 1

T y/x

.

i

TV

This

is

(j>

+

is

fact that f{x,y)

,

= —

,

x

homogeneous so

dy/dx

—

=

Troi0 — a

f(x, y) ± tan a IT f[x, y) tan a

r

isogonal family

a trajectory

+a=

have

will either

=

—

tantp.

— tt.

In

tan/3,

r

.

the differential equation of the let

y

=

ux.

Then

t/

= u + xv!

and 2

,

xu —

±1 ± u — =>

lû

±

Itu + v?

J,

" dti

=

~ dx

=*

±tan

u- ^

1

1

±2tan

_1

|-ln^l + |^ =

21n|x|+ci

37. Since the differential equation of the original family

the isogonal family

is

y'

=

yjx

±=.— \/3

— =

lTV3si/i

y

i

In (l

is

f{x y) y

\/3 x

y=-

x^VSy 56

.

This

is

+ u2 ) =

=

In

\x\

+c

±2 tan ~ - In

y = -

,

(x

2

+ y 2) =

a.

the differential equation of

X

homogeneous so

let

y

=

ux.

Then

1

Exercises

y'

= u + xu'

3.

and

=

xu

~

1

,

1

\/3

T + u2

±-~ tan ±

2

dx

j

x

1

1

u- ^

+ u2 ) =

In (l

!/

tan-^-lnfl +

=

In

+c

\x\

21n

+ci

|x|

"V3

=*

±-j= tan

-1

38. Since the differential equation of the original family 1

the isogonal family y'

= u + xu'

y

is

V^ X

^ }^^- = V ^ 3 v^j ^ l^y/VSx x^fy/ V3

=

±l/
,

+U

±y/3 tan

=

This

.

2

+ y 2) = = -

f(x,y)

is

-lT«/v^, r ~— aii = => ±/ ±vJ— 1

from y 2

is

In (x

,

Cl

.

the differential equation of

homogeneous so

let

y

=

ux.

Then

and

xu —

39.

-

1

ci(2x

+ ci)

we obtain

c\

-1

==

±2\/3 tan

=*•

±2^3

= -x±

u- I

-1

tan" 1

yjx 2

2

In

—

da:

+ v?) =

(1

^ -

In

^ -

In (x

^1

+ 2

In

=

+ y2 ) =

Ixl

+c

21n|x|

+

ci

ci-

+ y 2 and x

-\ + y

\yj

\

y

i.

Vt

Self-orthogonality follows from the fact that the product of these derivatives

40. Ftom

+

ci

+ l

y = —

FYom

jt

—

we obtain

ci

= -

—1.

yy*

i

ci

results in exactly the

41.

1

is

r

c\G cos

orthogonal family

t

same equation. This shows that the family

and y

is

-r-

ax

=

=

*

cie sint

— w

+x

.

we obtain

This

dy — dx

is satisfied

x+y x by x

—y

=

.

is

Then

C2e -i cos

self-orthogonal.

the differential equation of the

t

and y

=

coe _t sin i.

42.

Exercises

3.

We

-

have

tb<

ifo

=?

so that

=

tani/ji

+

tan (^2

2

coti/^

—

—

tan ^2

2/

\

Exercises 3.2 1.

Let

P=

P{i) be the population at time

P=

obtain

Poe

kt .

= 2PQ we

Using P(5)

find

we have 3

=

- 4P we

Setting P(t)

n

e

'

P

Setting

=10,000 and

=

e

=

f

=

3.

Let

P= 4.

P=

fct

Problem

=

taa

=

obtain

JV =

Ne

JV(10)

=

JV

=

will

N = e

fci .

= 4P

kt .

=

fl,B<

(ln2)t

W

1

Setting P(i)

.

=

3Po

3 — t=^£« 7.9 years. *

5

=j.

t

=

10 years.

we obtain

=»

=

Po

61n2

10,000e-°'

Prom dP/dt

t.

k=

575 we find

=

=

kt

«

6597.5.

t

and Wo the

=

initial

2000 we find

= P =

and P{0)

^lnl. 15. Then P(30)

400 and iV(10)

=

we

1/2

=

0.1

=

500 we obtain

500e 3taL15 r* 760<9^p>

number. From dN( dt

400 =

N

3k e

or e*

=

JV{0)

=

= kN we

(400//V

)V3

.

= 01

find

97 we find

=

50

in

fc

=

=

fe

^

From dN/dt —

t.

When 90%

In 1/2.

kN

and

we obtain

of the lead has decayed, 0.1

^

*

t.

£ In 0.97.

ho.!

From dN/dt Then

JV(24)

(

=

=

kt

=

»

grams

and N(Q)

=

iooetV«)("»ojr)M

10.96 hours.

100

=

we obtain

100 (

.9 7 )4

Problem 6 we obtain

go,!,,**

1

we have

be the amount at time

JV(t)

7. Setting JV(i)

e

2000 we then have

Using JV(3.3)

=

P=P

(ln2)t/5

5 In

^

=i^

n4

N(t) be the amount of lead at time

Using JV(6)

and

= kP we

From dP/dt

population.

«26,390.

Using JV(3)

remain. Setting JV(f)

Let JV

1 In 2

^

N(t) be the number of bacteria at time

et (V3.3)ln(i/2)

6.

=

Using P(10)

.

Let JV

Let

i

P(f) be the population at time

500e

From

5.

Poe 21n2

(ln2)t/5

3 in

10,000

Then P(10)

= —

111 J ln3

=

fc

initial

have

4 2.

——

2) t / 5

and Pq the

f,

H = lnl

—

t

58

=

«

136.5 hours.

N = 100e w „ ggg mg

.

Exercises 3.2

8.

The

= kA

solution of dA/dt

=A

A(t)

is

2

=A

=

Then Ai

.

*(t,-i 2 )=ln£ ^2

= £ A Solving .40/2

kt

e

e*' for

we obtain

t,

i

=

=

A(ti)

Aoe*' 1

A2 =

,

= Aoe^

A(ti)

and

=> ife^-J—to^. ~ (tl

-(ln2)/fc. It follows that

At

(2)

~

( t2

=

t

-

In(Ai/A2} 9.

=

be the

I(t)

£ln

FVom dS/dt

=

rS we obtain S

,

.25,

If

Sa =

(b)

If

S(t) =$10,000 then

(c)

5

$5000 and r

A=

Assume

that

12.

Assume

that dT/dt

and

c

14.

Aoe*' and

fc

16.

=

=

Assume Ldi/dt

=

+

Ri

=

10) so that

=

=

io

L =

i(t)

=

g

=

and 19. For

j t

=

fttfg/df

+

+ ce"

i(.005)

=

(l/c)g

c

=

=

(l/c)g

2001

and

.1472 amps.

= i

io

+

=

=

and £{(),

=

As

and

i

=

60

=

A(t)

10 +

T = =

If

then

.145-4

T(\)

i

ce**. If

=

t

wl5,600 years.

=

55° and T(5)

=

+

100

50,

=

T{0)

15° then

=

t

ce*'.

90° implies

sinwt-

Z'w'

=

30° then k

= -|ln2

(

=

and £(t)

70° and T(l/2)

=

50° then-c

=

60

3.06 minutes.

=

If J"(0)

=

82.1 seconds.

50 so that

= =

20° and T(\)

=

i

§

T(i)

If

+ce~ sm

.

22° then 98° then

If t{0)

=

-» co

=

i(0)

iQ

so that

-r^^^rcoswf + ce +

&

Rt/L .

.

+

it 2

C=

200,

fa-

/t =

-200cet

If

i^sinwt, and

IO

-4 ,

=

and £(t)

100 so that q

=

1/100

+ ce -60

2001 .

C=

-6

5 x 10

[f *(0)

=

-4

,

and £(t)

then c

=

=

200 so that

-1/500, g(.0O5)

=

.003 coulombs,

have g -> 1/1000.

we

is

+ cje"*/ 10

*.

51*.

1000,

20 the differential equation

= 6e" 10

T=

.1,

R=

E(t),

- -1/100

then c

1/1000

[e*/

So-

3/5.

E(t), E(t)

we obtain

Assume Rdg/dt +

Assume

then

.25/

$6665.45.

36.67°. If T(t)

100) so that

£(f),

-3/5 and limt—oo

Since i(0)

18.

=

T = 5 + ce*'.

ln(39/40) so that T(t)

Assume Ldi/dt + Ri =

If ?(0)

=

64.4611°.

L 2 uj 2 + R 2

17.

=

where 5(0)

-.00012378.

jfc{T-5) so that

Assume that dT/dt = k(T -

then c

rt

and 1(3)

fc/

12 years.

=

fc

21n(2/3) so that T{1)

= c = -80 and t = 145.7 seconds. 15.

=

t

dT/dt = k(T -

that

=

e

=

If (*//<**

-

.

5.75% then 5(5)

59.4611 so that T(0)

Assume and k

-

=S

.00098/

/

$6651.82

=s

=

=

and /(15)

fa)

=

the thickness, and 1(0)

t

=

I e

kt

11.

13.

intensity,

k

/ 10.

=

Let I

.

20 di/dt If *{0)

-

+ 1i = then

120.

a=

An

integrating factor

-60 and

i

=

60

-

10

is e*/

60e-'/ 10

,

.

so

Exercises 3.2

For

At

>

t

*

=

20 the differentia! equation 20 we want c2 e

-2

=

Setting g{0)

fci

=

-

dt

-

+

=

2i

=

so that c 2

- 60e-'/ 10 j 60

\60( e

2

-l)

e

and

-

60 (e 2

=

i

-'/l0,

m

.

Thus

l).

<

,

c2 e~^

<

t

20;

t>20.

-CIn

fc2 t

we

ijo

=

+

we obtain


dq

20di/dt

- 60e~ 3

60

i( * )

Ea - q/C

is

find c2

—

=

~ fjii

—

1

80

/

21.

Prom dA/dt =

-

4

4 = 200-170e~ 22.

From dA/dt =

23.

From dA/dt

4= 24.

1000-

4(0)

25.

=

26.

=

10

5Q0

——

200

A=

we obtain

100

0.18

= -12/400 and empty

after

+ ce"'/ 50

ce"-'/

A =

-10 -

_

5)t

The tank

.

+

-

empty

24

4)r

(o

=

.

if

A(0)

=

30 then c

i/ck2

= -170

and

=

oil

50 .

If

1000

4(0)

+

^

=

ce"*''

30 then c 100

If

.

we obtain

=

4(0)

A =

30 and

=

1000

SOe'^ 50

A=

then c

-

1ft

=

.

-1000 and

+

c(100

If

4(0)

-

t)

2 .

If

in 100 minutes.

we obtain

4=

50

+ ( + c(50 + £)" 2

.

=

10 then

=

12 then

f

64.38 pounds.

44/(400

4(60)

is

—— _ = 3-~+—

44

= 3-

4

is

obtain

^

-

-100,000 and 4(30)

tank

=

.

From dA/dt = c

,4

\

.

10 - A/100 lOOOe '/ 100

dt c

we obtain

=

then c

dA

From

-4/50

/ 50

- 4/50 we

^=

From

t

ki

-

t)

we obtain

14.1 gallons.

4 =

.06(400

The percentage

400 minutes.

60

-

t)

+

c(400

-

f)

4 .

If

4(0)

of alcohol after 60 minutes

is

4.1%.

The

Exercises 3.2

27.

(a)

From mdv/dt

= mg — kv we obtain v = gm/k + ce

the solution of the initial-value problem

—

(b)

As

(c)

From ds/dt

(

oo the limiting velocity

*

=

=

v and s(0)

28.

From

d£/
= -E/RC

m

gm\

(

V dC/df = kA{Ce -

30. From

31. (a) From dP/df (b)

If

>

fci

as t

—

*

=

(*i

-

P—

&2 then

=

and £(ti)

»

_kt/

Eq we obtain

- Co we

C) and C(0)

k2 )P we obtain oo as

t

m

m

9

I

—

*

oo.

P=

vq

e( (l

- ( V fic

so that x ~>

A/B

33.

P{0)

ffcet*'-*")'

Po

From

r

2

In [P|

i

Po then

dO

for every

t.

If &i

<

=

fcsinl

+

c

—

— P=

-

,

cie

frsin K

ci

-P

- (L/m) dt we v

'

and

P=

^

— — — —— — —

obtain

A=

*

\ I

2

r d$

=

* 1 e'* 1+ 2 '

^ 2

2/9i

is

+

— / mJa +

(ki

4

=

_^_+ ce k2 ki

^2

>

0,

d(

=

^ 2

foJA

=

l

i

10

—(6 - a).)

m

k\M. x,n

and

,

J

If

l

~l

s

Pee*™'-

'

integrating factor

.4(0)

>

2 .

+

Using

P—

then

/>

34. Write the differential equation in the form dA/dt

Then an

.

= p( 0)>

where Pq

P=

then

as t -» oo.

oo.

k cos

=

and

.

i

If

— gm/A;

m\

C = Cs + (C - Ct )e- kAi * v

obtain

=

If

£= S

32. Separating variables we obtain

=

=

gm/k.

is

= A-Bx&nA X(0) - Owe obtain x - A/B - {AfB)e~ Bt = A/2B then T = (In 2)1 B.

Jf(T)

vq then c

is

FVom dXjdt If

29.

=

v(Q)

we obtain

so

am

*"/*". If

&2

t

-(*i+t>)t.

+

= we find c = --™^-' and A =

,

(l_

the materia! will never be completely memorized.

61

e

-(*i+fc2)i).

As

(

—

*

oo,

A

kiM

Exercises 3.3

Exercises 3.3 1.

~=

Prom

=

C(l - .0005C) and C(0)

1

we obtain

\G

d£

C = ^°^Q5 1 e 2.

3.

N=

1

(ltT

1,000,000 as

^ = P{a ^ + — P= — + From ^ =

-

c

atfe*"

1

bEe\ a

-

5.

=

_7

f ^

'

=

P(t)

=

obtain

E

where

~ bc > t

=P

P(0)

From Problem

5

then c

=

we have

^= 2 Setting d?P/dt

X{5) as

8.

=

Let A"

t

—

From

^

A;

=

= '

X

(t)

6ce

| -

=

P = eê - ™

-

" Q^- w - ce W e

and iV

Since

-

=

5 °°

P=

-

^'°°IL

_

so that

W

.

52.9months.

—^ln|a- bin P\ = + Q t

^

c

—

^ ^-^^ =

^=

of

b

2

obtain

t

C

^=

&(120

at time

of

A—

,

•

as

t

and

where k

- X) 2 X(0) =

B—

=

= a/b- lnP we

fee"" v

(1/6) ln(a/f>

-

-

l)

In Po)

- 2X)(150 -

.

'

and

P= e^ 6-1

If

X(0)

29.3 grams.

Now

A").

=

.

and

180A:t

amount

Then X(20)

P= eê - ™"".

so that

" so that

and

150 - 150e X=— Q^

fc(150

dt .000095238.

Va*0,

.00014,

dt*

and using

co, so that the

and the amount 9.

^^-^

In Pq.

be the amount of

10 then

—=

=

oo.

f

VfezM ) d P = dt a-bP + P - c /

f \

df

7.

=

t

-» 2000 as

an arbitrary constant.

is

P(a - bin P) we obtain

If

.7033, b

and

.0005C /

MOfc+

5000 we obtain

500,000 then

- cP"*) we 1

1

=

1000 we have a

P) and P(0)

If

N=

500 we obtain

C

dC =

—

1

-6

(a)

(b) 6.

rrr-7

10

fcP)

dt

=

JV(1)

-t oo.

t

From

1834 supermarkets, and

bN) and N(0)

^ =50,000 and

^=P

From

P 4.

^ = N(a -

From

Um

-

t

=

Then °( 1Q )

'

+

= t

•

.0001259, and X(20)

>

and the amount of

0,

and X(5)

33.3 grams

—

=

oo. If

and

X(()

fc(a-X)(/3-X), and X(Q)

=

X

=

-

10

then

62

»

30 as

we obtain

-» 150 as

75 then

dt

B—

i

=

(

t

—

X =

X -» 60

oo.

*

~+— w

150

150At

-* co so that the

amount

here

1

of

A—

•

70minutes.

Xp^ Xj^

(iMz^L + \ct —

=

~

—

)

dX =

kdt so that

— Exercises 3.3

X= 10.

„

—= dX

From

k(a

a=

If

.

,

- ae^-ftto

-^dX = kdt

-

then

- X)(0 - X)(j - X) we

X = a-

and

- X)*

(a

kt+c

obtain

at

,(/?-a)( 7

1

a-X +

-Q)

0-X +

(a-0)( 1 -0)

l-X

(a- 7 )(/3-7)

dX =

kdt

so that

-1

-

{{0

11. (a)

As

.

a)( 7

?/

—

(b) Using

-

,

In

a)

=

32 ft/s and

#=

-

/?)(7

+

-/?)'"

Then

.

4000(5280)

we

ft

7760

ss

-1

„. - X\

- 7 )0? - 7)

(a

'

= 2gR

Uq

= v^2)(4000X5280}

/2(0.165)(32)(1080)

,„

.

In \0

oo we assume that v ~>

*

ff

=

v

(a

'

«o

(c)

-1

X„> +

a-

and «o

=

\a\f~X\ =

fci

+ c.

V^gli.

find

» 25067

36765.2 ft/s

«w

mi/hr.

5291 mi/hr

ft/s as

v

12.

p

13.

^=^

From

=

sinh ^-x.

From y(l)

dx l

=

+

1

FVom

u(0)

=

and

vi

=

i>2

,

\dx }

then y

=

^

and

,

p =

= ^x 2 —

^ dx

ln|x|

^

we obtain p

— cosh ^-x +

=

and

,

=

y'(0)

follows that y

1 it

= —\l+(^r) V2 \

p

,

=

y'(l)

—

If ^j.

14.

k (t 4

From

- T^) we

^(1)

=

From

ôt

and

w|

'

1/(47%)

1/(27%)

T-Tm

T + Tm

T 2 + 7"2

ft(0)

e

wx

^

so that

p=-2 (x n/v

*

- x~ v ^).

If

'

V2 then

VlV2

+

l/M In

25

l-vi/v2\

l

+ p2 =

.

^

and

,/l

-

'

v\

obtain

so that

15.

+

—

—

we obtain

1

2[vi/v 2

1

+

-

T + T„

-2tan- 1

20 we obtain

dT = kdt

~m = 4T^H +

c-

-i

ft

=

["^20 \

-

.

50/

If

h(t)

-

then

*

= 50^20 seconds.

—

}

Exercises 3.3

16.

If

h

=

,1/2

=

then r 1 ! 2

we

table of integrals

(2/xr

1=

-,

dx

'

+ 2fcr 2 )

-9

-

From

If

>

/i

W

then

find

^ 17.

= Jtyt + c.

-r 3/3

yfZjidb so that

and A(0)

=

/2

10

-

In

(fiXJTr +

we obtain 10

/(cscfl

^) =

- smO)

dB

=

+ 2/ir

=

dr

dt.

From a

=

lOsintf.

+ c.

t

= - dx

where y

^100

Then 10 J100 - y _ +

From

m dv ^

=

-

fa>

2

and w(0)

=

-\/l00-y 2 =

wo

we obtain

1

— yjk/mgv

v

+ ^gjk

so that

1/2?

+

=

vo

+ ^gjk

vq

—
e2

Divide this equation by e 2 \Zs k /mt and multiply by v

di

(g)

From x

—

(dx\ \dyj

I

,

i and

dy

The

equation.

20.

_ + 2y— =

solution

2

x

is

dx

+ 2y-

x

2

=

=

dy

dx

2 u-i'so

cy

+

.v

^

that

dw — — = dy

-c 2 which

2x

»t

21. Using

2

+ y2 = /

dt

—7 In y + 5y = a In x — 22.

From y

[l

+

dy

(y*)

=

2 ]

=

as

t

—»

oo.

2

1 /'di/A

,

(

j

"

1

,

a Clairaut

dx

+y

y

H

dy

,

2

=

±dy.

+ y2

yjx 2

+c I

_|_

/

1

»

dx dw — — and w = y —4 \dy dy dy

x \/k 2

dy dx dy — =— — we obtain dx dt

— ^mg/k

to see that v

and the quadratic formula we obtain

_-y± ijx 2 + y 2

±J/

t

4

dy

^

î

— yjmg/k

or

Then

dt

a family of parabolas.

is

,

=

dv

+ tjk/mgv

1

v — yjmgjk

19. Let *

:r..

f

1/

18.

2

v

10 In

k

/3x

$y\

dy

=

——

/& [

-

fix

|

dx. Using

x

>

kd

- ^sin20 + c.

and

t/

>

\

+a

we obtain dz =

2fcsin0cos0 d$,

dx

=

-^L=dy.

2k

If

y

- ^cos20^ 64

=

fcsin

d0,

2

then

and

x

=

we have


If

23.

= when

x

then c

2^-^dt =

From

(a)

=

9

0.

^

-^f-smOdS and

(It

(it

= t

^ dt J

=

]

(b) Solving (

^

^ (cos

=

J

\dt J

,„

-r-

v cos

=

.

^

- cos Po

=

0(0)

we

O

obtain

~ cosô).

I

r

=

rffl

0,

^p(cos0

and separating variables we obtain

cos #o) for ~jt dt

i

V 2 ff

•'Co

—

l

H~

r° /

I

etc

T/i

/2T ™ T = *2,/—

,

dt

/

or

Jo

/

fl

°

/

V

.

.

,

- cos (?o

vcos

9 JO

.


+ Cij =

From y(x 3 family

— —x 2 y 2

y'


1

—

is y'

we obtain

3

,,

The orthogonal

.

,,

trajectories are

arjr 2.

From y family

Y'

=

-

16x C2

-

-

,„

.

Then

Ay

- 16x = ^ dy

3

c%.

-j

differential

=

-Ay and x

+

64

4


+

qje

16*

If

.

z(0)

=

then

-- 1

=

64 3.

—x =

3

we obtain y — Ay — I6x so that the

= 4x+ 1 + cie 41 is

y

From y — family

2

= a(x —

1

is

y

=

—, 2(!/

4.

From

5.

From

=

^f-

2

we obtain

y'

=

^ — -— x

The —^-r 2)

and P(0)

jfcP

^=

I)

orthogonal trajectories are (y

-

2)

2

=

x

- -x2 + c 2

.

2

= P we

0.018P and P{0)


1

=

P=

obtain

4 billion

Poe*'. If

we obtain

P=

P(T) 4e

= 2P

0i8t

then

T=

so that P(45)

]-

=

ln2.

8.99 billion.

dt

6.

Let

A —

A(t) be the volume of

CO2

at time

From

t.

dA — =

1.2

-

A = ,4

7.

4.8

+

-» 4.8 ft

From

—

3

11.2e"*/

4 .

Since A(10)

=

5.7ft

3 ,

A —

and

=

-4(0)

16ft 3

we

obtain

4

CtZ

the concentration

As

0.017%.

is

t

-> 00 we have

or 0.06%.

=

—

kjx(a

x)

we obtain

|

—

\ x

dt

1

\dx

a - xJ

=

k\ dt

so that x

=

r-;

l

dy — = K2xy we obtain ,

.

at

lnM^lnll+d^'l+c

or

y

=

c 2 (l

+c

1

e

ak > t h/k ' .

)

+ cie"*

1'

.

From

.


,

8.

dv dv dv — = v — so that m~— = —mg — kv dy

.

,

(a) Let

,

at

and v(0) - v height

v%

The

(a)

is

m — 2k

~~

=

h we see that the velocity equation

^-=]n\(l+B)T-{BT T 1 + D

1

Since T{0)

=

7i

<

0,

'

-

'

1+B ^

rs = T2 + B(7i -

solve f 1 \

10

2i

=

then the

-

——

^

=

+T2

=

)\

C3

7i

=

maximum find that

^Jmgjk

k[(l

kt

—=-r =k r +

[BTi

+c

+ B)T -

(BT\

dt.

Then

2)

and

=

T(t)

^"J

^ + 0.2i =

4.

3

+ C3 e

fc

<

1+B >'.

-f-

— and so

+B

+

+B

+B

1

w=

and lim Tit)

^+ BJ +

2 .

1

^T2 + BT^B

UmT,

+ 7b)].

- r2

t-00 T),

=

.

+ B)T -

=

^f^/'


20

10/ dt

.

10

Then

t

10

-iln|40-2i| = — In Since i(0)

<

t

<

=

10.

—

^-t=

1

we must have

lim e ^ t-00

—

is V{

dT -.

1

40

w

If

-

,

B(Ti - T)}

MtJ

first

^

Using y{0)

•>

impact

at

1

We

-

.

is

(1

10.

~ 2ky/m

7

the square root of the result obtained in Problem 27 of Exercises 3.2.


fc

e

dv — = — mg — kv dy

*

2

Since

trra

—

.

^ = k[T-T -

(c)

becomes

mg + kvn dv = mg — kv «(0) = 0, and y(0) = we From mv dy mg 2 t*y/ m — oo we see that the terminal velocity is v = ^ Letting y

In

e

differentia!

(b) Since v

i

mg "|

-

v2

follows that

it

_

f\

(b) Setting y

9.

=

is ft

„

This

2

dt

we must have For

r

>

ci

=

|10

2/"/l0

.

—

t\

+c

V40 -

or

Solving for

i

we

4 *"*'

2

°£ < (

'

10 .

I 20,

f

66

>

10

=

=

get

10 the equation for the current becomes 0.2i

it,*)-/

2i

=

ci(10 At

4 or

— i

-

gt 2

=

t).

,

20.

Thus

20

4

Linear Differential Equations of Higher Order

__ 1.

From y =

=

that ci 2.

We

have

The 3.

4.

5.

6.

c\e

x

i/(0)

is

4 eye *

=

so that ci

=

=

ci

=

=

find y'

cie

x

—

czeT x

e (e

+ cê -1

0,

-

e~ x

we

find

x

=

3/5 and c 2

=

rffO) ^

j (e 1

y

=

The

2/5.

is

4cie

4E

y

_1

y(0)

= ê 1 — ê

=

+ C2 =

c\

e/ (e 2

- cae -1 Then

y(0)

=

1

.

solution

is

y

3 41

= -e

+

=

cj

—

=

so

1

-

a=

=

0, y'(0)

-1

so that

+ c2 e - l).

c1e 2

Then

'.

—1/2. The solution

From y

= ax + Qxhix

that c\

=

3 and C2

From

=

ci

jf

=

=

we

find y'

=

Ci

—4. The solution

+ c2 x 2 we

find

=

x

is

y{0)

=

a 2 (x}

we have

In this case

ci

-

l)

and

+ cj =

1,

C2

= -e/(e 2 -

1/(0}

=

l).

- C2 =

4ci

2

2

-e~*. D

x and y = 2x 2

1

y

ci

=

2c2X.

at

x=

=

0,

+ C2(l + lnx). Then is

y =

Then

j/(0)

Theorem

0, t/{0)

=

=

=

c\

=

2c2

3, y'(l)

= a + C2 = -1

so

3x — 4x In x.

2cj

•

=

ci

4.1 is

=

=

0, y'(0)

=

and

y'{0)

=

1 is

not

not violated. so c\

—

and c2

is

arbitrary.

Two

solutions

2 .

we have y(0) = c\ — 1, y'{\) — 2c2 = 6 so that cj = 1 and c 2 = 3. The solution is 2 3:r Theorem 4.1 does not apply because y and y' are evaluated at different points. + y From y = cie^cosz +- C2e x sinx we find y' = cie x (— sinx + cosx) + 026^(003 x + sinx). (a) We have y(0) = c\ = 1, y'(0) = c\ + C2 = so that ci = 1 and c2 = -1. The solution is In this case

=

9.

y

4.1

+ C2 cos x + C3 sin x we find y' = —C2 sin x + C3 cos x and y" = — 2 cos x — C3 sin x. Then j/(tt) = ci - C2 = 0, j/'fjr) = -C3 = 2, j/"(tt) = c 2 = -1 so that ci = -1, c 2 = -1, and cz = -2. The solution is y = —1 — cosx — 2 sinx. From y

are y 8.

=

= ci+c2 =

possible. Since 7.

+ C2e~ x we

1/2 and C2

solution

From y

Exercises

1

.

— e x sin x. We have y(0) = ci = 1, j/(jt) = — cie* = —1, which is not possible. We have y(Q) = ci = 1, j/(tt/2) = c2 e"'' 2 = 1 so that ci = 1 and c2 = e^ 2 The solution is x v 2 x y = e cosx + e~ / e sinx. We have y(0) = ci = 0, y{n) = -c\e" = so that ci = and c 2 is arbitrary. Solutions are y

(b) (c)

(d)

=

e

x

cos x

.

y

= c^sinx,

for

any

real

numbers

c2

.

Exercises

4.

10. (a)

We

have y(-l)

(b)

We

have y(0)

(d)

We

y

is

—x—

12. Since oi(x)

=

from

1/

=

=

and

From y and

—

C2 sin

sin5A

+3 =

+ c2 + 3 =

c\

which

1,

4,

which

not possible.

is

=

16. Since (1)0

any

Ax we have

The problem

From

0, y(7r)

=

+

ci cos Att

=

0,

the family of solutions

Ax + 01 sin Ax we have

=

ci

=

0,

y(5)

or

The problem

0.

(3)x

+

(0)x

+

(l)(4x

1

(OJe

=

-

3x

2 )

+

2

+

(1)1

(3)(x

—

(1) sin 2

+

2

(-2)cos x

1)

the graphs of /i(x)

c\

cos 5A

integer. (If

=

A

=

0,

+

{l)(x

=

2

will

=

—

=

be

A

similar

and /2 (x)

=2+

functions &re linearly independent since

23.

W(x

l

'\x 2 ) =

x1/2

x2

\x-W

2x

(l)sinhx

= -x 3/2

=

require that ci

j^t, - 2+x —

W

(l

+

x, x,

x2 ) =

+x 1

x

x2

1

2x 2

+

so that c\

y =

is

0.)

|x|

1

_I

0,

=

the functions are linearly dependent.

v

(l/2)e

c\


3*

+

so that

=

0.)

argument shows that

-H-+-

22. Since (-l/2)e*

0,

linearly dependent.

see that the functions are linearly independent since

The

=

Thus we

they can not be multiples of each other.

21.

tt/2.


+ 3) =

+x

0.

The

Thus we require

+ c 2 sin 5A =

/

1.

2.

0.

y

is

=


— x

^

the family of solutions


+

(l)cos x

=

have nontrivial solutions when c2

n a nonzero

rnr/b for 2

will

y{0)

x <

C2 sin A7r

have nontrivial solutions when 02

A

A =

-f

will

=

c\

(If

=

(— 4)x

—

y(0)

arbitrary and

and c2

— tt/2 < x <

set of functions containing f(x)

19. Since

a = -1

or A be a nonzero integer.

5A

18. Since (l)cos2x

15 so that

the problem has a unique solution for

0.

17. Since (—1/5)5

+3 =

is

—

=

+

16c 2

so that ci

and xq

c 2 sin A7T

+

+

=

3

— oo <

C2 sin

c\ cos

4ci

not possible.

the problem has a unique solution for

+

=

=

y(2)

3,

is

=

Ax

15. Since (-4)i

we

—

y(l)

2 and xq

tana:

ci cos

that shiAtt

20.

+ c%

•

= c\ + c2 + 3 = = —x 2 + x 4 + 3.

11. Since a2{x)

14.

ci

+ c 2 + 3 = 0,

have y(l)

solution

ci

=

ci

We have y(0) = ci-0 + c2-0 + 3 = 3, y(l) — c\ + c2 + 2 4 C2 = — 3 — c\. Solutions axe y = c\x — {c\ + 3)x + 3.

(c)

13.

=

=

4-


for

< x<

£1

68

00.

t

—— t

= 2^0.

Exercises

24.

1

W(l+x,x 3 ) =

+x

x3

sin

=

25. tV(sina:, cscx)

-

sec

2

cot

—

x

e 27.

W^.e-V

4

-

*)

e

x

esc

2

e

1

-e"

28. IV

(a:, a;

In

a;,

i 2 lnx) =

4e

- -30e 4x #

4ar

x

—

need c\f[x) 30.

(a)

+ C2f{x) —

The graphs

for a// values of

-co < x <

for

is

on — oo < x <

oo.

x(2

+ lnx) ^

=

/2

0,

for

linearly

oo.

x

=

0.

3

*

We

the interval.

in

as shown.

dependent on any interval containing x

Obviously, 5-

Hence, /i and f2 are linearly

4-H-

independent on (—00,00).

>

oo.

.

a constant multiple of the other

neither function

(b) For x

7r/2.

+ 21nx

and f2 are

and f2 are

of fi

r*

-

+ 2x In £ =

3

i 29. No, this does not imply that fi

x 2 lna;

+ In x

1

for

jr.

4*

xlnx

1

oo.

for

= -2secxcscx ^

.

x

-oo < x <

x e

1

for

= -2cotx ^

cot x

esc

tanx

=

x

esc

a;

cosx

26. W(tanx,cotx)

= x 2 (3 + 2x) ^

3x 2

1

4.

x 2 and

x1

-x<

2x

-2x

3^

W (fi.fi) =

so

= -2x 3 + 2x 3 =

x2

x2

=

„

2x

We

0.

2x 3

-

=

2x 3

0.

For x

<

0,

2x conclude that

W(fi,f2 )

=

2 f2 = -x and

for all real values

of x.

31.

(a)

(b)

If If

y

=

y

-

1/x then y

= -1/x 2

c/x then y"

-2^ =

1

=

32. (a) Clearly y\ (b)

If

J/"

y

=

+

yi

+ 2

(j/)

1

1/2

and

=

1

;/2

=

+lnx

= -J +

In

and y"

implies that c a;

satisfy

jj"

+

-c= 2

(j/)

=

0.

so that c

=

for c 2

^

or

1.

69

2y

3

0,

+1, or -1.

0.

1 + {y') 2 = zl —y1 + ^ =

x

-

=

2/x 3 so that y" 3

then y"

± % x'

~

0.

If

ji

=

cijfi

+ c2 jft =

ct

+ C2lnx

then

1

Exercises

33.

The

.

4.

W(e—oo < x <

for

oo.

The

general solution

The

3x

,e

4x

linearly independent since

)= 7e*/0

is

=

y 34.

and are

functions satisfy the differential equation

cie~

3l

+ cae

and are


4:t .

linearly independent since

Whoosh 2x,sinh2x)

— oo <

for

x <

oo.

The

general solution

is

=

ci

y 35.

The

36.

— oo < x <

The

for

oo.

The general

1

(e

solution

oo.

The

general solution

=

cie


oo.

The

general solution

The

37 cie cos 2x

3

x/2

The

< x <

oo.

The

general solution

/

pje

1

sin 2x.

and are

,x

4

=

)

x

.

independent since

linearly 6

^0

=

ci#

3

+ C2X

4 .

=

1/x

?

is

=

ci cos(ln x)

+ C2 sin(ln x)

functions satisfy the differential equation and are linearly independent since

W for

2x


y 39.

+

+ c2 xe x ' 2

W(cos(Lnx),sin(lnx)) for

2e

is

y 38.

=

sin 2x)

is

w(x for

=

y

is

x


-co < x <

The

+ C2 sinh 2x.

cosh 2x

cos 2x, e

y 37.

2


W for

=

<

x <

oo.

The

(x,

general solution

y

x~

2 ,

x"

2

In

x)

=

9x"

6

^

is

=

c\x

+ C2X

-2

70

+ C3X -2 lnx.

Exercises

40.

The

and are


linearly

W(X, x,cosx, sinx) for

— oo <

x <

The

oo.

general solution

=

y 41.

42.

is

+ C2X + C3 cos x + C4 sin x.

The

=

functions yi

and yp

The

functions

The

(a)

=

functions y\

and yp

=

coax and yi

=

e

i

2x

and

e

is

a particular solution of the nonhomogeneous equation.

ze 2* form a fundamental

=

3/2

+x—

2 2x

2 is

the homogeneous

sinx form a fundamental set of solutions of the homogeneous

xsinz+fcosa;) bi(cosa;)

=

3/1

and yp

equation,

45.

1

equation, and yp

functions y\

equation,

44.

independent since

= e 2* and y% = e 51 form a fundamental set of solutions of — Ge x is a particular solution of the nonhomogeneous equation.

The

equation,

43.

c\

=

4.

set of solutions of

the homogeneous

a particular solution of the nonhomogeneous equation.

= x~ f 2 and 3/2 — z -1 form a fundamental set of solutions of the homogeneous — j$x 2 — g£ is a particular solution of the nonhomogeneous equation. l

From the graphs

— x 3 and

of yi

—

ys

|x|

3

we

see 3

that the functions are linearly independent since they

cannot be multiples of each other.

H-H

(b)

If

x

j/i

>

then

3/2

= x3

and W(y\,y2)

x"

=

3x

(c)

(d)

x3

-x*

3x 2

-3x 2

=

x"

2

3x 2

The

functions Yi

(e)

The

(f )

Neither

function y

a 2(a0

46. Assume y\

^

is

=

=

a:

3

W [x x3

= x are solutions = 4 ^ for —00 < 2

and Y2 3

,x

2 ^

a:

satisfies y(0)

3/2

0.

If

x <

then

3/2

= —x 3

and

=

and

y'(0)

—

=s

x2

of

x y" — 4xj/

x <

is

zero at x

2

= 0.

+ 63/ =

0.

They

are linearly

00.

0.

the general solution since we form a general solution on an interval for which for every

satisfies

x

y(xo)

in the interval.

=

1

and

y'(xo)

4.2 they are linearly independent since

and

=

0.

Part (b) does not violate Theorem 4.4 since 02(2 )

independent since

3/1

y-l*l a

3*

= x 3 solves x 2 y" — Axy' + &y = 0. To show 3 3 that ya = \x\ is a solution let 3/2 — x for x > and 3 for x < 0. let j/2 = -x that

It

y-i

shown

easily

It is

form a fundamental

=

W

and

3/2

(1/1,1/2)

=

on

/.

set of solutions

satisfies y(a;o)

yi(x)i/ 2 (x)

-

=

and y'(xo)

j/i(a:)yj(x)

=

1

= at

1. a:

By Theorem

=

xq.

Thus,

1

Exercises

4.

+ a-ilf + avy =

47. (a) Assume y\ and y 2 are solutions of a2y"

0.

W

If

(3/1,1/2)

=

yu/2 — y[y2 then

dW to

=

yi (021/2

+ a u/2 + ao^) - fa

=

-ffl(0)-B2(0)

The equation

(b)

is

x

(c)

Let

(d)

Prom

We

+ a iv\ + a oyi)

°-

The

first-order linear.

is

a 3j//

solution

is

W = ce^-H" '"^" ^'''

=

W = ce" O

ai(l)/as(()idI

x

in

We

we

part (c) for

see that

if

W(xq) =

every x in / since

=1-

identify 02(3)

identify 02(1)

47(c)

is

an exponential

=

x 2 and oj(i)

W

then

=

have a 2 y" 02(2/1

and

where

c

=

for every

z

in J.

a;

and ai(x)

=

1.

—2x. Then from Abel's formula

Then from

If

W(x

)

^

then

in

Problem 47 we have

-

xa

the alternative form of Abel's formula in Problem

we have

=

We

',

function.

ki

k3

k2

ki

lni|

e

50.

1

= ^(^5.

obtainitlg c

,

1

49.

2

1

a constant.

W^ 48.

in part (a)

(

1/1

+ i/2

(ft!**

-

+ an/j + oq =

+ y2)" + fll(yi is

k2 k3

*o

=

(fok4

-k2 k 3 )e- ]nx + iDX

<>

)e^M = (*1*4~W*0 x

E\ and 03^'

+V2)'

+ =

+a =

+ (<*2I/i

E%.

Then

+ aij/i + ao) +

a response of the system to the input E\

+

(U2J/2

+ a if2 + «o) = E\ + &2

Ei-

Exercises 4.2 In Problems 1-10 (4)

1.

from

we use reduction

Define y

=

u(x)

-

Now

we

use formula

1 so y'

If v>

of order to find a secoond solution. In Problems 11-30

the text.

=

u',

y"

=

u",

and

y"

+ By' =

— u' we obtain the first- order equation ti/+5u; = [e

5l

w]

—

gives

72

u"

+ 5m' =

0.

which has the integrating factor e

5l

tu

=

c.

e

5

/

=

e

5x .

Exercises 4.2

w=

Therefore 2. Define

=

t/

=

u'

u(x)

ce 5x

and u

=

=

second solution

and

u",

—

Define y

=

u{x)e 2tie

xe

2*

2x

ce

1

it

=

ce x

y"

=

e

and

A

.

0.

gives

l

e

second solution

=

ui

=

is j/2

=

c

-1 .

c.

e

E .

so

+ u'e 2*,

=

Therefore u"

=

=

u'

5*.

e

which has the integrating factor e~/ dx

ax

w=

—

is t/2

= u"-u' =

i/'-i/

=

=

-r1- [e

y'

4.

A

.

we obtain the first-order equation u/— to

ti'

Therefore

3/2

=

y"

«',

Now

3.

c\e 5x

so

1

j,'

If u)

—

=

and u

V+ V+

a

+

cix

4e

2

Taking

C2.

4e

=

ci

2l

1

- 4y' + Ay =

and

u,

and

—

C2

we

see that

4e

2l

u"

=

0.

a second solution

is

2*.

=

Deiing y

u(x)xe 1 so l/

=

(1

-

ije-'u

+ ie -I u',

x

=

j/"

xe~ u"

+ 2(1 - x)e~ x u' -

- x)e _ *u,

(2

and

+ If if

=

we obtain

v!

2j/'

+

=

j/

e

_a:

(a:«"

the first-order equation

+

w

2u')

=

u"

or

+ -u' =

0.

x

—w= 2

which has the integrating factor

H

a:

e*ff/'

=

x1 Now .

—

[x

2

w]

=

A

second solution

gives

x2 w

—

c.

£IX

w=

Therefore 5.

Define y

=

v!

=

c/x 2 and u

=

ci/x.

is j/2

= — xe _I =

e

-2 .

u(x) cos 4x so y'

= — 4w sin 4x + u' cos Ax,

y"

=

u" cos 4x

—

8u' sin

4x

—

16u cos Ax

and y" If

e

w=

-8

/

u'

+

=

16y

we obtain the

tan4tiit

=

2

cos 4x.

(cos4x)u" -8(sin4x)u'

6.

Define

w=

y =

u'

=

—

u"

or

8(tan 4x)tu

-

=

8(tan4x)u'

=

0.


Now

^ Therefore

w'

first-order equation

=

2

[(cos

csec 4x and u

2

Ax)w]

=

c\

=

tan4x.

gives

A

(cos

2

4x)w

second solution

—

c.

is 3/2

= tan4xcos4x =

u(x) sin3x so y'

=

3u cos 3x

+ u' sin 3x,

y"

—

u" sin 3x

+ 61/' cos 3x — 9u sin 3x,

sin4x.

1

Exercises 4.2

10. Define y

—

u(x)e x / 3 so

\e x !\

1

y

+ e*/V,

t/"

-

e*/

o

V

4-

^WV + Jeû o y

and 6j/"

If ui

=

u'

e (S/6)M

we obtain the

=

+ y'-?/ =

e

l/3

w=

=

w'

11. Identifying P(x)

ce

-51 ^ 6

second solution


and u

= — 7/x we V2

A

+

w'

first-order equation

=

|iu

is j/2

=

second solution

=

=

gives

C\e~^ x ^.

A

e

In

—

second solution

3/2

second solution

W= = =

c.

is 3/2

=

e

-W 6 e s/3 — e - T / 2

In

\x\.

=

x

=

x

j^—dx =

2

.

x

2

x"'

/

j

x

_6

dx

=

-7^ 3

-3 .

=

dx

=

=

lna

7

a:)

lnx

x(lnx)

1.

we have

=

is

=

w=

1/x we have

A second solution is

15. Identifying P{x)

0.

|x|.

(In

A

=

6

2/x we have

is j/2

=

5x/6

fe -f-(7Mdx fl — dx — x dx — x § j J

%

^W7h^r 14. Identifying P(x)

§u'


-/(2/,)

+

have

—

=

y2

A

u"

or

Now

e 5*/6.

~-[e**/* VJ]

Therefore

+ 5u')=0

(6u"

2(1

(*

x 1 ?2

+ x)/ +

!)

+

(l

- 2x - x 2 ) we have

e

-/2(l+x)dx/(l^- I 2 )

/

^+ (l

.

1)

-

2x

(x

+

+

l)

- x2 l)

[-7TT-

2

a

dx

2

dx

=

(x

= -2 -

+ 1

x"

1)

-

= (!+!)/

/ x.

(x

+

dx -

iy

dx

.

Exercises 4.2

A

second solution


is

y%

=

= -2x/

+ x 4-

x2

2.

- x 2 ) we

(l

have

B = / c -/-^Mi--*}^ = | e -Ki-A

second solution


is j/2

=

j

= — 1/x

— 1

.

=

A

second solution


e

=

= — 3/x

/ j i"1

^

A

second solution


=

4x/(l

y2

A

second solution


V2

=

=

is

^

+

+

A

second solution


is

J yi

/

J

- / ~3dx/x

xô^x)

J

x a x^snrfmx) .

—

x)

dX

=

x3

^îiWH f

X

x)

=

e

-i*

—

j

e~ 2x [i e 2r

+ xe 2x -

±e

.

e

-*+ln(H-z)

-

X]

2

,

dx^xj

(

du——e

x

l

+

a: )

e

-»

—

X.

,

dx

^

= x]

^ / e -«

x

u—

,

dx

e

x ,

dx,

dv

=

-^dx, v x*

-

e

dx

we have

— dxj + x j — dx = -e

= -1/x

1

x 2 sin(lnx).

dx

~

=

d:

we have

2x)

,

x

xsin(lnx)

= — 2cos(lnx).

= e-^ J(l+ 2x)e 2*dx = 1/2 — x.

x

^—-e

e

X)

-g-JxAj/d+a) zj—g—dx-

x

a!

i 2 sin(lnx).

fe~ = x fe~ -~y dx-\-x j j

=

=

snv^ln x)

= e -2xy

=

|

.

-dx

j-

x 2 cos(ln x) tan(ln

is t/2

+ 1-x 1

fa-li 11

we have

X C03

~

(

xcosflnx).

.

=

| ] -^

=

fa

-j.-dx/x

-

[xsinflna:)] [— cot(Inx)]

is j/2

(

we have

= xsaiâx)

y2

)

— x)\.

+

In (1

a

x .

we have

fe -S-d*/*

m = xj

-

^

dx

76

=

—

fdx =

xj

xln|xj.

=— x

-

+

e -

Exercises 4.2

A

second solution


=

is j/2

=

xln|x].

we have B -J0
A

second solution


=

^2

is

x5

= — 5/x

.

we have

-5 — /(lnx) —

fx— 5

(ir/l

x

A

second solution


?/2

=

=

cos(ln x)

6

= x3

y%

is

dx

r^-

st;

2

=

In

second solution


/ -stt 6

2

J x (lnx)

dx

I

—

r

s-t;

=

is ;/2

=

dx

cos(ln z)

/

J

=

is 3/2

x3

=

is j/a

x

A

e

3x

— —

,

3/2

=

e"

J

3x

+

I

28. Identifying P(x) e

e31

dx

^

is 3/2

=

-a;

3 .

dx

r

—(3;

= +

3x

cos(lnx) tanflnx)

=

sin(lnx).

e

~ J ~ ldx/x

xi

.

*

,:,)-

in

.

, I

_7 X

in

/ /

1

_n\\

dx

= J*f

1

i

1

we have

1)

^

=

e 3*

|

(3x

=

+

g3x

^_ xe

-3,

_ |

efa

e _3^

_

=

|_

+ 2.

l)/x

- f -{x+l)dx/x 5^

=

cos-'flnx)

-d» = e*/

second solution

=

)

Lux/

.

= -(9x + 6)/(3x +

(

\

2

m-^j =

1

have

„-/ J -7ds/x


—

-

.

= —7/x we

second solution

—

f

have

2

A

x

\

-^•^(.^-(^^ 26. Identifying P(x)

In

sin(lnx).

= — 4/x we

second solution

I

x

1/x we have

r

A

=

.

J cos-'flnx)

A

i

we have -

dx

—

e

x

j

gi+Ini

—

dx

=

77

e

1

f

j xe

x

dx

=

e?{~xe

1

-

e

x )

= —x -

1.

-

e

.

Exercises 4.2

A

second solution


=

is 1/2

x

second solution

=

is j/2

=

-(2

have

= j e' i =


1.

= — 3tana: we y2

A

+

—

sec

3tanxdx

x tan x

+ ij/z

+

x tan a;

sec

+ In

-

In sec x j

sec x

|

3lnstKX

^J

dx

dx

= Jsec 3 xdx

+ tanx|.

+ tan x\

we have

W = | e" J "P^W 1 ^ = | e 21nl+ *di = 1 1 V dx = A

second solution

31. Identifying /"(i)

A

=

second solution

general solution

is 1/2

=

—

+

2a;

yp

=

y%

is

=

x.


=

e^

1 .

1

We

we have

The

see

by observation that a particular solution

=

y2

= —3

QAe

31

-

x .

is

yp

—

—1/2. The

3 (SAe *}

=

e

+ 246

= -4

e

_a!

-.

We

.

see

by observation that a particular

is

x

= a + cieT x + x.

31

=

5e

3 *.

^

j

dx

=

e

= Ae 3x Then .

Thus y


+ c2 e'!I -

we have

a particular solution we try yp 3

!I

"

cie

= j e'f^dx =

general solution

y2 find

2) e

we have

y

To

-2x +

is

32. Identifying P(x) is

2

2) e*.

y

solution

(x

=

A= c1 e

I

/ -4 dx

y'

J

=

5/2 and yp

+ c 2 e 2x +

we have -

x

r e

V2

78

Ax

e

x

dx

=

e

2x

3 3/le *, y"

=

\
fe

x -

3*.

.

— 9Ae 3x and

The

,

general solution

is

Exercises 4.3

A

second solution

is e

and 0-4a + 3(ax + b)

A

particular solution

3x .

To

find

a particular solution we try yp

If j/2

=

J/i

/

ax

+ b. Then

= 3ax-4o + 36 = x. Then 3a = 1 and -4a + 36 = yp = ga: + | and the general solution is

=

so a

= j/J,

o, j/£

1/3 and 6

=

=

0,

4/9.

is

!/

35.

=

=

cl e

I

+ C2e JI +

-a;

+

-.

3— ^ then

and

=

Z£ e - /

^_4

e

-/

+

v\

y\

4 y{

e

-/

+ y« / J

so that y'i

+

Py'2

+ Qv2 =

(y'{

+

Py[

+

—~

£^

re -fPdx Qyi)

I

dx

vt

dx

=

°-

Exercises 4.3

m=

11.

m = -1/4 so that y = c\ + cit~ x 4 From 2m 2 — 5m — we obtain m = and m = 5/2 so that y = Ci + oie" x 2 Prom m 2 — 36 = we obtain m = 6 and m = — 6 so that y = cie 61 + C2e~ 6x From ro - 8 = we obtain m - 2\/2 and m - -2^ so that y - cie 2 ^ 1 + c2 e _2v 21 From m 2 + 9 = we obtain m = 3i and m = — 3i so that y = ci cos 3x + ci sin From 3m 2 + 1 = we obtain m = and m = — i/y/Z so that y = ci cosx/y/3 + 02 sini/v'S. ~ m — 6 = we obtain m = 3 and m = —2 so that y = cie 31 + C2e~ 2x From From m 2 — 3m + 2 = we obtain m = 1 and ro = 2 so that y = cie 1 + cae a *. From m + 8m + 16 = we obtain m = — 4 and m = — 4 so that — c\eT Ax + C2xe~ ix From m 2 — 10m + 25 = we obtain m = 5 and m = 5 so that y — cie 5x + C2ie 5x From m 2 + 3m-5 = we obtain m = -3/2± %/3§/2 so that y = Cie (-3+^>/2 + C2e (-3-VS)*/2_

12.

From

1.

2. 3. 4. 5. 6. 7.

8. 9.

10.

13. 14.

From 4m 2 +

we obtain

m=

and

^

.

^

.

.

'

2

.

3:r.

.

2

1/

.

.

m = -2 ± %/5 so that y = qe'" 24 ^) 1 + C2^~ 2 ~^ x From 12m 2 - 5m - 2 = we obtain m = -1/4 and m = 2/3 so that y = cie -1 4 + c2 e 2* /3 From 8m 2 + 2m — 1 = we obtain m — 1/4 and m - —1/2 so that y = c\e x i A + cae~ x l 2 m 2 + 4m - 1 =

we obtain

.

/

.

.

.

Exercises 4.3

— 4m + 5 =

15.

From

in 2

16.

From

2m 2 - 3m + 4 =

=

Prom

3m 2 + 2m +

=

1

From 2m 2 + 2m

3l/4

+1=

=

e"

19.

From

1 /3

=

m3 — 4m 2 — 5m =

e

( Cl

From 4m3 + 4m 2

21.

From

m3 — 1 =

we obtain ?/

22.

From

m 3 + 5m

2

=

=

m= cê*

we obtain

1

=

=

From

m 3 — 5m 2 + 3m + 9 =

From

+ e~ x/2

=

we obtain

25.

From

m3 + m2 — 2 =

we obtain y

26.

From

m3 — m 2 — 4 =

we obtain j/

27.

Prom

=

m 3 + 3m + 3m + 1 = 2

lx

c\e

c2 sin

5j:

+ c3 e

±

.

m=

cje~

2x

m—

—5

—2,

m = —1 ±

=

x

+

x

m=2

and

+ e~ x ^

we obtain

=

1

2,

cos x

i

.

.

and

m = —3 so that

.

so that

+ C3 sin x).

m = —1/2 ± \/7 i/2

(c 2 cos\/73;/2

m—

cie"

(c2

.

so that

+ c2 e 2x + c3 e- 3x

and e~

so that

-5 *.

m=

1

c\e

— 1/2

m = 3, and m = 3 so that

—1,

m—

so that

_3;

= QeT* + c 2 e 3x + c 3 ie 31

=

—1

+ c3 sini/3a;/2)

+ C23: + cje

c\

.

^/Si/2 so that

m=

?/

^2 x/3)

m=

—1/2, and

and

0,

.

+ C2sin:r/2). and

5,

—1/2

m—

we obtain y

c2 e

+

(c2Cos vâ:/2

m — 0, m =

m 3 + 3m 2 — 4m — 12 =

+ C2sina:).

V23 x/4)

c2 sin

+ ci&~ x ' 2 + csxe~ x / 2

m=

and

j,

24.

+

ci

ci

y 23.

e^fci cosx

so that

cos:r/2

(ci

m = 0, m =

we obtain j/

± i/2

-1/2

m = 0, m =

we obtain

+m =

=

\/2i/3 so that

cos y/2 z/3

_I ^ 2

y 20.

±

-1/3

m=

we obtain y

?/

V^3 x/4 +

(ci cos

m=

we obtain y

18.

e

so that

i

m = 3/4 ± a/23 i/ 4 so that

we obtain y

17.

m= 2±

we obtain

—1,

m—

+ c 2 xe _:c + 80

so that

+ C3sin\/7a;/2)

m=

—1, and c3 x

2

e~ x

.

—1

so that

Exerci99S 4.3

28.

From

m 3 - 6m 2 + 12m - 8 =

=

y 29.

m4 + m 3 + m 2 =

From

=

30.

FVom

m 4 — 2m 2 + 1 =

From 16m4

y 32.

m 4 — 7m

From

=

—

2

=

m

From

=

18

=

we obtain

=

m 5 — 2m 4 + l7ro 3 =

From

=

38.

m 2 + 16 = c;

40.

2, C2

c\ 4- C2

From

=

m=

—1

.

so that

.

V3x/2 + CiXs'mV^x/2.

m = ±V5i so that

—3, and

_2a:

m = ±2i so that

-2, and

+ c 4 cos2a: + Cssin2x.

m = 0, m =

0,

m = 0,

and

m=

1

±

so that

4i

+ C2X + C3X 2 + e x (ci cos 4x + C5 sin 4x). we obtain

=,

we obtain c\

-

C2

—

m + 6m + 5 =

m

=

cie"

1

m=

m=

-1,

-1,

x + C2ze~ I + c3 e + axe* + c5 e~ 5x

we obtain

3 then c\

1,

c2

m= so ci

=

m=

1,

and

m=

1,

and

(c4 cos

1

and

=

0,

-d

m=

1, ca

m= -

=

—1 0,

bo that y

and y

-1 and

5c 2

4:c

2x

.

-1/2, and

m = 2 ± 2i 80 that

+ c5 sin 2z).

+ c 2 sin 4i.

If

y(0)

=

2 and j/(0)

= —2

^sin4x.

=

=

m=

3, so ci

— 8m + 17 = we obtain m = 4 ± i so = —1 then c\ = 4, 4q + c 2 = —1, so ci = 4, 2

2iC

m — ±4i so that y = c\ cos

we obtain

+

m = 0, m = 0, m =

= c\+&ix + c3 e~ x ^ + e

we obtain

2

3/(0)

From j;'(0)

=

m —1=

From

and

c3 e

= -1/2, and y = 2cos4i -

2

then 39.

+C2e 2x +

+ 12m3 + 8m 2 = y

then

-JZi/2 so that

so that

From 2m 5 - 7m 4

From

±

+ c4 sin y/lxfty

—1, and

c$x cos

m = 2, m =

0,

2

y

37.

c\

-1/2

+ C2e~ 3x + C3 cos \f2 x + asm V2x.

m=

ci

m=

3,

m 5 + 5m 4 - 2m 3 - 10m + m + 5 =

From

m = —5 36.

3x

we obtain

y 35.

c\e

x/2

.

m = ±y/3i/2 and m = ±\Z%if2 so that

m=

we obtain

m = 2 so that

+ C2xe x + c$e~ x + Cixe~ x

+

y 34.

x

m=

and

m — 1, m =

1,

c\ cos

- 16m =

6

c\e

we obtain

y 33.

m—

and

2,

+ ca^re 2* + c$x 2 e 2x

(c3 cos \/3

a;

we obtain

+ 24m 2 + 9 =

21

+ c 2 + e" 1 ' 2

ci

y 31.

cie

m = 0, m = 0,

we obtain

y

m = 2, m =

we obtain

=

e

=

c\e

x

+ C2e~ x

.

If

y(0)

=

1

y'{0)

=

If j/(0)

—

and

1

1

-5

.

so that y

3/4, c 2

that y c2

=

e

4a

-

— ae~ x + C2e -5x

-3/4, and y

(ci

=

\e~ x

cosx

+ C2sinx).

=

(4cosi

—17, and y

e

4l

—

If

.

-

fe

y{0)

-5 *.

—

17sinx).

4 and

Exercises 4.3

41.

FVom2m 2 -2m+l = Owe obtain to = l/2±i/2 so that y = e^ 2 (ci cosa:/2+C2 sinx/2). Ifv(O) = -1 and j/(0) = then c\ — — 1, \c\ + ^c2 = 0, so cj = —1, c 2 = 1, and y — (ain^x — cos jx).

42.

FVom i/(0)

43.

m 2 — 2m + =

10 then

=

1

=

c\

m

we obtain

5, ci

+ 01 =

=

=

10 so c\

m

and

1

= 1 so that = 5, and y =

5, 01

m

From i/fl)

From

=

m

2

1

then Cie

+ c2 e =

2

+

1

then -c\ 47.

=

— 3m +

2

+

1

=

0,

—z-C2

——c\ + -c

0,

1

=

2

=

so ci

48. From

If

=

-

5/36, c2

-5/36, c3

=

0, y'(O)

=

0,

ci

so ci

+ c2 =

=

m 3 + 2m 2 — 5m — 6 =

y(0)

=

-1/6, c2

=

=

m3 - 8 =

=

we obtain

y{0)

=

and

y'(0) C!

so ci

=

-1/6, c 2

=

1

then

=

Cl e

-ci

1/10,

_;c

+ 2c2 -

=

cie

2iC

xl 2

c\e~

=

cie -2

+

cae

+

— e~ 10

+

ci

6

e

lx

15

21

+

e

_I

=

1/6, c3

- -1/2V3, and

c2

cos \/3 2

+

\/3c3

=

-1,

-7,

—3

so that

+ 9c3 =

1,

*

+ C3 sin V3x^

2c 2

.

- 2V3c3 =

= -ê 21 +e~ x ^cosV3a; - ^=sin\/3x^ 82

y(l)

=

and

21-2

2

c\+C2e~ 6x +C3xe~ 6x

3x

-

If

.

a;

=

4c2

1

31 ' 2

= — V3~ cos +sinx.

so that

4ci

=

Ify(0)

.

then

0,

-

i

.

= -e +e If y(ir/3) = and 2/(jt/3) = and y

,

.

0,

5 and

+ c 2 shi\/7x/2).

37-1

3a;

=

=

-fe^+ê

2*.

6 *.

m=

and

2,

y(0)

+c 2 e 3x/2

andy =

12C3

Jaw"

+ c3 e"

3c3

—

= --e- x +

+ c2 =

2ci

=

and

-1, and y"(0)

y

+ c2 e

cos\/7x/2

1

e

If

.

(ci

and y

1,

-

36C2

1,

m = 2 and m = - 1 ± \/3 y

If

=

=

C2xe x

andm = —6 so that y =

—6,

=

C3

=

c2

,

+

11/4,

4- c 2 sin 2.

c2

m = —1, m =

y

+ c 2 + C3 = 0,

+

1

= & - |e" 61 +

and y

we obtain

v V

From

3;

— — \/3,

so c\

2,

-6C2

0,

1/6,

and y"{0)

1/15, C3

= -e~

so cj

Fromm 3 +12m 2 +36m = we obtain m = 0, m = If y(0) - 0, ^(0) = 1, and y"(0) = -7 then ci

49.

2c2 e

=

2 so that y

1/

^/3

=

+

C]e

m=

and

1 2

m = ±i so that = ci cos

we obtain

^^3

m=

we obtain 2

3/2 so that y

-7/4, c 2

x

+ 5xe*.

5e*

1/

4m2 -4m- 3 = Owe obtain m = -1/2 andm = andi/{0) = 5thenci+C2 = 1, -£ci + §c 2 = 5, soc! =

46.

c\e

Fromm 2 +m+2 = Owe obtain m = -l/2±V7t/2 so that = e - ^ 2 If y(0) = and y'(0) = then ci = and c 2 = so that y = 0.

44. From

45.

=

3/

.

0,

.

Exercises 4.3

50.

m4 = Owe obtain y = Cl + c2 x + c3 x + ax 2

From then

=

ci

=

2, C2

=

2c3

3,

6q =

4,

m4 - 3m 3 + 3m 2 - m =

From y

—

x

x

+ c 2 e + C2xe + C4X

c\

+ C2 =

0,

=

2,

Cl

=

so ci

2,

=

C2

—2, C3

2

=

e

If

.

y(0)

+ C3 =

C2

a—

Prom m* — 1 If

y(0)

=

=

=

-

0, y'(0)

0, £/"(0)

+ C2 + C3 =

Cl

so ci

m=

we obtain

1/4, C2

=

=

=

-1/4, eg

1,

0,

=

y(l)

54.

m 2 — 10m + 25 =

=

m

From then

55.

ci

+

=

ci

—

4

and y

m +1=

From

ci

=

—2, c 2

1,

cie

5

-

0, C4

+ C2e

—

y"{Q)

=

4,

andy"'(0)

=

5

m = 0, m =

3 .

o

m=

1,

y"(0)

=

+ 2C3 + 2C4 =

1,

y'(0) C2

0,

=

§z:

0,

and

1, 1,

and C2

m=

so that

1

?/"(0)

=

1

then

+ 3C3 + 6C4 =

1,

Cl

=

=

1

then

+ C4 =

0,

Cl

y"'(0)

5

=

+ C2 -

C3

=

0,

Cl

— C2 —

=

C4

1,

-1/2, and _

4

m=

0,

1

1

= -e*--e 5

so ci

=

1,

m=

±2i so that

m=

±i so that y

2

m=5

and

.

--sini.

4

ao that y

C2

=

—1, and y

-

ci

cos2i

1/

+

= cie 5x + C2xe 5x — e 5^ — xe hx

If

.

y(0)

=

1

and

.

C2sin2x.

If j/{0)

=

y(it]

=

y'{ir/2)

=

and

c% sin 1x.

we obtain

=

and y

=

cosi

ci

+

C2sinx.

If

y'(0)

=

and

1

= — 2cosx.

-l = 0we obtain m = 1 and m = — 1 so that = cie 1 + c 2 e~ z or y = C3 cosh x + cy sinh x. = 1 and j/'(l) = then ci = 1, ci sinh 1 + C2 cosh 1 = 0, so ci = 1, C2 = — sinh 1/ cosh 1 and 2

From

1/

If y{0)

- coshx ,

y

(m-4){m+5) 2 =

58. Since

fm +

Q (m

sinhl

2

,

smb. a;

cosh

57. Since

=

—

cosh x cosh

1

From the

— 6m +

solution yi

=

dividing the polynomial

1

—

loj

= m 3 — ^-m 2 + 7m + 5

e

x

2

coshfx

cosh

+

is

y'"


5 !/

=

1)

.

1

15?/'

— IOO3/ =

0.

is

0.

= 1 is a root of the auxiliary equation. Now, by m — 1 gives m 2 — 8m + 17. Therefore m = 4±i are

we conclude that mi

m — 9m + 25m— 17 3

73/'

———

sinh x sinh 1

-r~cosh 1

m 3 +6m 2 -15m-100 the differential equation y'"-y!/" +

59.

3,

m = — 1, and m = ±i so that y = ciex +C2e~ x +c$ cosx+tysinx.

we obtain

we obtain

2

then 56.

then 2

=

+ 3x + 2x 2 +

0,

1

Prom

=

3:

y 53.

y'(0)

2,

= 2-2e I + 2xe -^rV.

and

Cl

0,

=

Hy{0)

—1/2, and

y 52.

2

we obtain

x

.

and

5,

j/

51.

3

Exercises 4.3

the remaining roots of the auxiliary equation, and the general solution of the differential equation is

y 60.

From

the solution y\

=

c\e

- e'^cosx we

x

4-

e

ta

(c2Cosx

m 3 + 6m 2 + m-34 by [m- (-4 + i)][m- (-4 =

y

62.

Prom

64. Since

FVom

_4x

m 2 + 16 =

a differential equation

(m —

3

m +1=

(m 2

4

Now

sinx.

-4 - J

are roots of the

dividing the polynomial

m 2 + 8m + 17 gives m- 2.

=

i)]

e

m2 =

and

i

Therefore

m3 = 2


+ c2 e _4l sinx + cê1*.

cosx

m 2 — 3m — 18, a differential equation is y" — 3y' — ISy = 0.

3)

a m 7) = m — 7m (m — 3)(m + 3) = m 2 —

63. Since

65.

=

cie

(m — 6)(m +

2

=

-4l

and the general solution of the

the third root of the auxiliary equation,

61. Since

mi = -4 +

conclude that

Hence another solution must be y%

auxiliary equation.

is

4- C3 sinx).

is

+

y"

16y

=

a differential equation

,

9,

a

v^m + l) (m + \f2rn. + /

/

l/v 2±i/v 2

l)

and

y"'

is

equation

differential 2

m=

0.

is

—

y"

7y"

—

9y

=

0.

=

0.

we obtain ro

=

-l/\/2

± i/v^

so that ^

_

€

i/v^ f Cl coa

_L 1 + ca sin

x^

+

m.2

—

e"

x

^ ^3

cos

+ C4 sin ^= x

3;

Exercises 4.4 1.

From yp

=

m2

-+-

3m + 2 =

we

find

mi — —1 and

A. Substituting into the differential equation

y 2.

From yp

=

4m 2 + 9 =

we

find

mj = — § i and

From

cie

we obtain

+ c2 e~

m2 =

1

1.

21

=

m

2

— 10m +

25

=

= Ax + B. Substituting p Then A = § B yp = j/

,

we

find

3

Then

yc

mi =

m2

into the differential

fx

+

f

,

y

3

ci cos -a:

—

2A —

+ c2 sin -x +

=

cie

51

84

and we assume yp

=

3 and

c2 sin

Then

j4

§x and we assume

=

jj

,

yp

=

| and

5

-

5.

+ c 2 xe 0;t + gx +

|x +

15.

.

cie

1

Sl

25-4

and

=

+ c 2 e 2x Then A = 3,

6.

c\ cos

9A =

Then yc = equation we obtain

=

c\& x

+ 3.

A. Substituting into the differential equation we obtain y

3.

=

-1

Then yc

—2.

J-

=

+

c 2 xe

51

30 and

and we assume — 10-4 + 255 = 3.

,

Exercises 4.4

4.

Prom

m2 + m — 6 =

we

mi — —3 and mi =

find

= Ax + B. Substituting into the differential A = -3 B = yv = £x - ^ and yp

,

,

=

+ cje 2 * +

3 cie" *

^

-

.

\m 2 + m + 1 = we find mi = ma = 0. Then yc — cie _2a: + C2xe~ 2x and we assume 2 ?/p = Ax +JBx + C. Substituting into the differential equation we obtain 4 = 1, 24 + 5 = -2, and \A + B + C = 0. Then A = 1, B = -4, C = \ yp = x 2 - 4x + | and From

,

,

j/

6.

+ ci& 2x and we assume = 2 and A — 65 = 0. Then

,

y 5.

3x

Then yc — c\e equation we obtain —64 2.

From

m

2

- 8m +

=

cie

-21

+ c2 xe~ 2x + x 2 -

4x

+

^

.

= we find mi = 2 + 4i and ni2 = 2 - 4i. Then yc — e 2l (ci cos 4i + ci sin 4x) = Ax 2 + Bx + C + (Dx + E)^ Substituting into the differential equation we

20

and we assume yp

.

obtain

24 - 85 + 2QC =

-6D + 13E = -164 + 205 =

Then

4=

5,

B = 4, C = ^ D = ,

y

7.

From

m2

+

3

=

=

e

we

2l

(cicos4x

find

£ = - ]§

-2,

+

C2sin4x)

3x

— (Ax + Bx + <7)e + 12C = 0, 124 + 125 - 0, = (-4x 2 + 4x - §) e 3x and

24 + 65 yp

i/

8.

From 4m 2 - 4m -3

=

= we

= -26

20A

=

yp

=

100.

5x 2

ci

cos \/3x

find

mi =

+ |

.

+ 5x 2 + 4x +

^ + f-2x - ^] yc

=

e

and

1 .

cicos\/3x

+

Substituting into the differential equation

and 124

=

Then A = -4,

-48.

+ 4x -

2 C2Sin-\/3x+ (|-4x

and

1 {§) e

+ 4x + ft + (-2x -

mi — V3i and mi = — v^3i- Then

2

and we aaaume yp

,

13D

m,2

= -5 Then •

yc

=

cie

e

B =

4,

C2sim/33:

we obtain

C = -f

3x

3l '' a

.

+ C2e~

1: '' 2

and we assume

yp = 4coe2x + 5sin2x. Substituting into the differential equation we obtain —19 — 8B = 84-195 = 0. Then 4B= - Jg, yp = - cos2i - Jgsm2x, and

^

y

=

cie

3l/2

+ c 2 e" l/2 -

^

cos 2x

-

^

sin 2x.

1

and

Exercises 4.4

9.

Prom

m2 — m

=

we

—

mi

find

and m2 -

1

we obtain

Substituting into the differential equation y 10.

=

C]e

x

m

+

C<2

= —A =

c\e x

Then yc

0.

+ C2

Then A

—3.

=

Ax.

= 3x

and

and we assume yp

=

3,

yp

+ 3x.

+ 2m = we find mj = —2 and m2 = 0. Then yc = cie _2x + c2 and we assume 2x 2 Substituting into the differential equation we obtain 2A + IB = 5, yp = Ax + Bx + Cxe~ 4A = 2, and ~2C = -1. Then 4 = § B = 2, C = \ yp = \x 2 + 2x + |ie _2x and Prom

2

.

,

y 11.

=

cie~

m2

.

yp

m2

-

— Axe ix

.

16

=

m.2

]-x

\

+ 2x + ^-xe~ 2x

2

-

differential

.

Then yc — c\e x ^ 2 + C2xe x / 2 and we assume equation we obtain \A = 3 and 2B = 1. Then

= ciW 2 + c 2 xe^ 2 +

y

From

+ C2 +

,

,

12.

2x

— m + \ = we find mi = 2 x/2 = A + Bx e Substituting into the yp = 12 + j^V 2 and A = 12, B = £ Prom

,

,

=

we

find

m\ =

Substituti:ig into the differential

+

±xV

2 .

Then yc = cie 4x + C2e~ ix and we assume equation we obtain 8.4 = 2. Then A = | yp — jxe^

mi =

4 and

12

-4.

,

and

=

y 13.

From yp

=

-4,4

m2 + 4 = Accos2:r

=

3.

we

find

mi =

+ Bx sin 2a:.

Then A = ~\

,

+ C2e

- 4l

+

^

= — 2i. Then

and mi

e

yc

41 .

=

ci

cos2x

-f

C2sin2:e

and we assume

m\ =

yp

=

=

c\

4B =

and

-|a:cos2:z, and

cos 2x

+ C2 sin 2x — —x cos 2x.

m2 = — 2i. Then yc = c\ cos 2x + C2 sin 2x and we assume = (4i + Bi + Cx) cos 2x + (Dx + Ex 2 + Fx) sin 2x. Substituting into the differential equation

From yp

m2 + 4 =

4l

Substituting into the differential equation we obtain

B = 0, y

14.

2i

Cl e

3

we

2

find

2i

and 3

we obtain

25 + AF =

+ 8£ = 12D =

-4C + 2.E = -3

-8B + 6Z> =

-12A86

1.

Exercises 4.4

Then

A=

-fa,

B=

0,

C=§

=

ci

cos

=

,

0,

£=

F=

fa,

0,

and y 15.

From

m2 + 1

=

we

„ + + C2Sin2x

2a;

mi —

find

and

i

f

1

\

12

= (-^x 3 + fx) cos 2x + ^x 2 sin2x,

25 \ In — x cos2x + — i' sm2x. 62 lo

n

xJ +

)

J

— —

r«2

p

j/

Then yc

i.

=

c\

+

cosx

c^sinx and we assume

= (Ax + Bx) cosx + (Cx + Dx)s'mx. Substituting into the differential equation we obtain 4C = 0, 2A + 2D = 0, -4A = 2, and -2B + 2C^0. Then ,4 = -\ B = 0, C = 0, £> = £ yp = — ja co8X + gxsinx, and 2

yp

2

—

,2

—

y

16.

From

m2

— 5m =

we

ci

,

1 2 1 + C2Sinx — -x cos x + -ism x. .

cosx

mi =

find

,

5 and

m2 =

Then yc =

0.

c\e 5x

+

and we assume

ca

= Ax + Bx + Cx + Dx. Substituting into the differential equation we obtain — 20A = 2, 12A-15B = -4, 65— IOC = -1, and 2C-5D = 6. Then A = -fa S = $ C = D = -|g 4

yp

2

3

,

*

= "A* 4 + fr 3 +

=

v

17.

From

m 2 — 2m + 5 =

we

^

,

,

,

SB*. -»d cie

find

kt

1

+ C2

mi =

x

+ 2i

1

4

53

14

, + —x

and m2

697

,

x

H

x.

= 1 — 2i. Then yc =

we assume yp = Axe? cos 2x + Sate* sin 2a:. Substituting into the 4S = 1 and -4A = 0. Then A = 0, B = \ yp = ^xe 1 sin 2r, and

x e (c\ cos

differential

2x + ca sin 2x) and

equation we obtain

,

=

y

18.

From

m2

— 2m +

2

=

A + 2B =

1

?/

19.

From

m2

+ 2m +

1

=

=

(ci

cos2x

+ C2sin2x) + ^xe I sin2x.

find

= Ae and -2A + B =

and we assume yp

2

= 1 — i. Then y c = e*(ci cosx + C2sinx) mi = 1 + i and 2x cosx + Be sin x. Substituting into the differential equation we obtain

we 2x

e

e

we

-3. Then x

cosx

(ci

find

,4=|,B=-£,yp = ^cosx- ^sinx

+ C2 sin x) +

mi =

TO2

=

7

1

5

5

-e 2l cosx — -e^sinx.

Then y c = c\er x

—1.

— A cosx + Bsinx + Ccos2x + Dsin2x. Substituting into the 2.8 = 0, -2A= 1, -3C + 4£> = 3, and -AC - 3D = 0. Then A = y p = —j cosx — ^cos2x + ^ sin 2a;, and yp

y

=

c\e~

+ C2xe -T —

x

19—

- cosx

20.

From

m 2 + 2m -

assume yp

24

—

we

find

= A + (Bx 2 + Cx)e 4x

.

—

cos2x

Jo

2.

mi — -6 and m2 =

4.

+

+

C2xe~ x and we assume

differential

equation we obtain

D=

B = Q,C =

12 — sin2x. 2o

Then y c

—

cie"


87

and

61

+

x and we

C2t*

we obtain —24,4

=

16,

Exercises 4.4

2S+10C =

=

20S

-2, and

A--\,B = -^,C=-^,yv = -§-(^x 2 + fax) e 4*,

-1. Then

and e

21.

From

m 3 — Qm 2 =

we

mi =

find

2 yp = Ax + B cos x + C sin x. 6j?-C = -1, andB + 6C = 0.

=

7712

and

—

013

Then yc —

6.

+ cix + cge 8 *

c\

= 3, = -±x 2 - £ cosx + ^ sinx, we obtain - 12A


A = -\ B =

Then

,

C=^

,

yp

,

and we assume

and

=

?/

—6 cos x + — 1

1 + C2X + C3e Br - ~x 2 -

Ci

4

22.

From

m3 -2m 2 -4m + 8 =

we

find

2

=

y 23.

From

-B=

0,

= Ax

and

= we B + Cx 3 e x

4-

6C =

Then

-4.

= 24.

From

m 3 — m 3 — 4m+4 =

-3B =

-1, and

we

we obtain

C4X sin x and

A=

1,

B= y

26.

m4 — m 2 =

=

we

1

-+-

2

sin x.

/

=

Then yc

2 2x cie * +C2xe +C3e~ 2x


.

+ caxe 21

4-

mi =

find

(±x

cae"

21

2

) e

Then yc

1.

.4

cie

2

find

=

B=

-1,

-3,

C = -§

+ c 2 xe + c3 xV - x i:

mi

=

1,

ma =

2,

3

,

24^4

=

c\e

.

=

cie

x

c%xe x

+ C3X 2 e* and we obtain —A = 1,

+

*/„

- -x -

- -x 3 e x

- fxV, and

3

.

and m3 — —2. Then yc

=

cie

I

+ c2e 2a: +C3e -2x

+ c2 e 2x + C3e _2lc +

7 4

+

^xe* 3

+

7 xe

5,

21 .

4

= i and TO2 = ran = —i- Then yc — ci cosx + C2sinx + + Bx + C. Substituting into the differential equation 4^4 + C = 1. Then A = 1, B = -2, C = -3, y = x 2 - 2x - 3, and p

find

mi =

013

we assume yp = Ax 2

-2, and

=

x

6

2x


.

we obtain

and

^x 3 - ^x

+

m3 =

—

rri2

- ^x 2 )

3

.

m 4 + 2m 2 + =

C3X cos x

m = 2 and m3 = -2.

= -^,yp =

J

y

From

j

/

= 4+Bxe I +Cxe 2x Substituting into the differential equation we obtain AA = AC = 1. Then -4 = |,fl = J, £7=^,^ = 5 + ^xe 1 + Jaw 2 *, and

and we assume yp

25.

21

m 3 — 3m 2 + 3m — 1

we assume yp 3-4

cje

mi =

2x

and we assume yp = (^4x + Sx )e and 6A + 8B = 0. Then A^\,B 3

o

ci cos

x

4-

C2 sin

x

4-

C31 cos 1

+

C4X sin x

+ x2 —

2x —

3.

mi — m,2 = 0, ro3 = 1, and 014 = —1. Then yc = c\ + C2X + c$e x + c\e~ x and we assume yp = Ax + Bx 2 + (Cx 2 + Dx)e~ x Substituting into the differential equation we obtain -6A - 4, -IB = 0, 10C - 2D = 0, and -4C = 2. Then ,4 = -| B = 0, C = -\ From

we

find 3

.

,

,

D--\,yp=

3

-\x - [\x 2 + y

=

ci

§x) e~

x ,

and

+ c2 x + cge 1 + c 4 e _I - ^x 3 68

^x 2 + ^x^ e~ x

.

.

.

Exercises 4.4

27.

= 4 — 4cos2x. From rr? + 1 = We write 8 sin we find m,\ — i and rug — — i. Then yc = c\ cos x + C2 sin x and we assume yp — A + B cos 1x + C sin 2x. Substituting into the differential equation we obtain A = 4, -3B = -4, and -3C = 0- Then A = 4, S = § C = 0, and = 4+ | cos2x. j/p a;

,

28.

We write sinxcos2x = ^sin3x — ^sinx. From m 2 + 1 = we find mi = i and m 2 = — i. Then + Dx sin x. Substituting c = C] cos u + C2 sin i and we assume yp = A cos 3x + B sin 3x + Cx cos = = = \,2D 0, and -2C = Then A = 0, into the differential equation we obtain -8A 0, -SB B= C = \ £> = 0, and p = -^sin3i + \xoasx. We have yc = c\ cos2x + C2Sin2x and we assume yp = A. Substituting into the differential equation we find A = —\. Thus y = c\ cos2x + C2sin2x — \ Prom the initial conditions we obtain c\ = and C2 — V2 so y = V2 sin 2x - \ a:

j/

.

j/

,

,

29.

.

,

30.

We

=

have yc

cie

-2*

A=

equation we find

From the

initial

+ cie x l 2

and we assume yp = Ax 2 + Bx + C. Substituting into the differential = -19, and C = -37. Thus y = ae~ 2x + cze*/ 2 - 7x 2 - 19x - 37.

B

-7,

conditions

we obtain y

31.

=

A

-2z

5

=

+

and

oi

186 gl/2

_

= Hp 7x 2

,

so

_ 19x _

3?

5

We have yc = c\e~ x ^ +C2 and we assume yp — Ax 2 +Bx. Substituting into the differential equation we find A — -3 and B = 30. Thus y = c\e~ x ^ + C2 - 3x 2 + 30x. Prom the initial conditions we obtain c\ = 200 and ci - -200, so y

32.

e

ci

We

=

have yc

differential

From the

c\e

-2x

+

equation we

initial

200' I/5 - 200 - 3x 2

=

C2xe~ 2x and we assume yp

find

conditions

A =

^ and

we obtain y

=

c\

2e~ 2x

B = =

2

§

and

— (Ax 3 + Bx 2 )e~ 2x

Thus y

.

—

ci

+ 9xe~ 2x +

+ 30x.

9,

=

cie'

2*

+

.

c2 xe~

Substituting into the + ( B x 3 + §i 2 ) e -2*.

2x

so

Qx 3 + \x 2 )

33.

We

have yc

=

e~ 2x (c\ cosx

equation wj find obtain ci

+ c2 smx)

A = ^ Thus y =

= -10 and

C2

=

9,

We

have yc

=

Cie

equation we find

1

+ C2e _I

A= a

|

=

e~

(c\ cos

x

+

c
.

.

so

y 34.

= Ae~ 4x Substituting into the differential sinx) + 7e~ 4x From the initial conditions we

and we assume yp 2x

=

e

~ 2x

(- 10 cos x

and we assume yp

+ 9 sin x + 7e~ 4x

= Axe x +

Bxe~~ x

.

)

Substituting into the differential

and B = — \ Thus

cie

.

x

+ c2 e _I + \xe x - \xe~ x =

cje

1

+ C2e _3: + ^ sinhx.

Exercises 4.4

Prom

35.

We

the

conditions

initial

have x c

=

the

ci

—

y

—

7e

We

= Q cos uit + we

A=

have yc

=

~ sinhx.

=

c\cosx

+

ci

C2sina:

- 72

+ C2 sinwf +

coswf

=

ci

fb/(w

itt

15

7

=

7—= (w 2

-

7

)

cos 7*. ^ 2

the

initial

38.

We

have

j/c

=

cie

-1


+

C2e

we

(

= —\ t

J

= Ax coax + Bxsinx + C7cos2i + £>sin2:r. A = 0, B — ^,C = 0, and D = 5 Thus .

^-x sin

+

x

~

x

—

.

j/

cie~

x

+ C2e 3x -

,

the

conditions

initial

y

39.

We

have yc

=

ci

+

c-ie?


we

+

we obtain -*

1

=

e

F0

+

y

=

3

55 3.

G

52

csxe x and

find

c\

and

C2

— sm

„

2x.

H

3

—^

C-

=^

2,

=

+ C2C 1 + c 3 xe x +

—12, and

Thus sin 2x.

00

n

4

.

„

90

-

= Ax + Bx 2 e x + Ce 5x

C= 2a:

j

.

Substituting into the

so

,

7

1

.

^3-65 COs2l -65 S1Il2x

A= ci

x

sin

~ cos 2x 00

i -

we assume yp

B=

3:

and

j

From

so

,

= A + Bcos2x + C sin 21.

p

,

2a:.

I

1

H

2

A = — 5 B = —^ =

= —\

C2

4

sin

O

and

sin

a:

and we assume

find

if

c\

6

3x

so

0,

+ 7—; — sr- COS 7*. w 7

COSwt

J

+ C2 sin x +

x

ci cos

1 = — - cos

y

=

and we assume yp

we obtain

conditions

)

C2

it

From


Thus

- 7 2 and

2

Substituting into the differential equation we find

y

= A cos 7* + B sin 7*.

B = 0-

and

)

/2w)fcoswi.

(_F

and we assume xp

Fo/(w z

~~

We

+

-

2

x— 37.

x

5e~

(fb/2
we obtain

conditions

initial

-

C2 sin uit

find

x the

—

,


Prom

x

+ C2 smart and we assume x p = Atcoswt + Btsinuit. Substituting into the = — fo/2w and B = 0. Thusx = c\ Coswt+C2sinuJi— (Fo/2u>)t coswt. conditions we obtain c\ = and C2 — Fq/2w 2 so

initial

have i e

—5, so

we find A

x 36.

=

7 and C2

cosw(

c\


Prom

we obtain

Thus

- l2a;V + ê 5 *.

.


Exercises 4.4

FVom

the initial conditions

we obtain y

40.

We

=

have yc

cie

-2x

=

11

Substituting into the differential

=

11, C2

=

—11, and C3

- Ue x + 9xe x + 2x-

x e (c2 coa

+

=

ci

+ cssim/Zx) equation we find A =

I2x

2 x e

+

ê

\

B — —|

,

the

initial

»

41.

We

have yc

—

conditions

2 = ~

i

c\

and

t/(l)

' 2x

+

eX

("S

c\

— —

cos

C2

,

— —H

^

^x

= Ax + B + Cxe -21 C = | Thus

5

t

and

+

2

-xe~ 7a

=

C3

+

sin

-

.

.

.

so

,

r - \ + h'*-

= A 2 + Bx + C. Substituting into the differential Thus = cj cos x + C2 sin x + x 2 — 1. Prom 3/(0) = 5

cosz + C2sin:E and we assume yp

A=

equation we find

=

e

we obtain

.

and

,

1

FVom

5x

and we assume yp

\/Zx

= qe" 21 + e I (c 2 cos \/3a; + c 3 sin\/3x) + -x -

y

so

9,

1,

B = 0,

and

C=—

1.

j/

we obtain ci

(cos l)ci

Solving this system we find ci

=

-

1

=

5

+ sin(l)c2 =

6 and C2

= —6 cot 1. The

=

—

0.

solution of the boundary-value problem

is

y 42.

We

have yc

=

e*{c\

cosx

equation we find A =

1

+ C2sinx)

and

B=

0.

6 cos a;

6(cot l)s'mx

+ x2 ~

1.

and we assume yv = Ax + B. Substituting into the differential Thus y = ex {a cosa:+C2sin:r)+:E. From y(0) = and y{ir) ?r ---

we obtain

—

7T

Solving this system we find c\

problem

=

and

C2 is

We

have yc

=

C\

cos

2a;

+

e^ci

=

any

7T.

real

number. The solution of the boundary-value

On

(7r/2,

oo)

—

c%e

x

sinx

+ x.

C2sin2i and we assume yp

into the differential equation [0, jt/2].

=

is

y 43.

ci

we

find .4

we have y =

=

c3 cos 2x

and

B—

= j

.

oleosa:

Thus y

+ c 4 sin 2x. FVom

+ Bsuix =

y(0)

c\ cos

=

1

on

and

Substituting

[0,ir/2].

+

2x

ca sin 2x

y'(0)

=

2

+

we

3 sin

x on

obtain

Exercises 4.4

=

Solving this system we find c\

Now

x=

continuity of y at

1

and

— 5

H

7T

— 1

,

sin

7T

H

6

—1 +

j

= — C3.

Hence

|

—

Thus y

.

C3

=

|

.

t

— =

.

sin

3

Continuity of 5

—|

— 2C4. Then

04

=

| sin

2x

+

| sin

x on

C3 cos

tt

+

C4 sin

7r

2 j/

1

at x

=

implies

7T

— 2smx + - cos7r + - cos — = — ZC3 shitt + or

+

cos 2x

7r/2 implies

cos

or

=

c-i

cost

2c4

| and the solution of the boundary-value problem

!cos2x |cos2x

g sin 2a;

+

+

^ sin

x >

|sin2x,

is

tt/2

tt/2

Exercises 4.5 =

1.

{D +

3.

(3D 2 -

5.

(D 3 - 4D 2

7.

9D 2 -

9.

D

2

-

5)y

+

5£>

4

l)y

=

+ 5D) =

=

4£>

9sinx e

x

4x

(3D - 2)(3D

-

12

+ 2)

= (D - 6){D + 2)

+ 8)y =

+3

2.

(4D

4.

(D 3 - 2D 2

+ 7D-G)y = l- sinx

6.

{D 4 - 2D 2

+ D) =

8.

D

2

-

5

a:

e" 3 *

+ e 2*

= (d - V5) (D + y/E)

10.

2D 2 - 3D -

2

= {2D +

1){D

11.

D 3 + 10D 2 + 25D = D{D + 5) 2

12.

D 3 + 4Z> = D{D 2 + 4)

13.

D 3 +2D 2 -13D+10 =

14.

D 3 + 4D 2 + ZD = D(D +

16.

D4

(D-l)(D-2)(D+5)

17.

D 4 + 8D = D{D + 2){D 2 -2D + 4) D 4 y = D 4 (10:e 3 - 2x) = D 3 (30x - 2) =

18.

(2D -

=

{2D -

l)4e x i 2

19.

(D - 2)(D +

5)4e 2 *

— (D -

20.

(D + 64)(2cos8i - 5sin8x)

15.

2

\)y

a

£> a (603:)

= %De x l 2 - 4frc / 2 =

=

- 8D 2 +

D(60)

Ae x l 2

-

D4

because of x

3

1)(D

=

+ 20e 2x = (D - 2)28e 2 * = 56e 2 * - S&e 2* = = D(-16sin8i - 40cos8x) + 64(2cos8z - 5sin8x)

2)(8e 21

)

22.

92

D5

2)

+ 3)

= (D - 2) 2 {D + 4) 2

=

Ae*l 2

= - 128 cos 8a: + 320 sin 8x+ 21.

16

-

128 cos 8a:

because of x

4

320 sin 8x

=

[C

Exercises 4.6

23.

D(D -

2)

25.

D2 + 4

because of cos 2x

27. 28.

D 3 (D 2 + 16) because D 2 (D 2 + l)(D 2 + 25)

29.

(D + 1)(D -

30.

D(D -

31.

D(D 2 - 2D + 5)

because of

1){D

l)

-

2

1

because of

(D - 4D +

2)

33.

1, x,

x2

x4

34.

D 2 + AD = D(D + 4);

35.

e

,

24.

D 2 (D - 6) 2

26.

D(D 2 +

1)

because of x and xe6*

because of

1

and sinx

sin 4a:

because of

2

3

2x

x 2 and

because of

2)

(D + 2D + a:

of

e

and co95x

x, sinx,

because of e _I and x 2 e x

3

32.

,

and

1,

1, e

1

x ,

and

and e

I

e

2x

cos2x e~ x

5} because of

sin

x and

e

2x

cos

x

e" 41

ex e' Sx / 2 ,

36. £> 2

- 9D -

36

= (O -

12)(D

+ 3);

e

12x ,

e~

3x

37. cos\/5x, sin \/5x 38.

D 2 -6Z> + 10 =

39.

D

40.

1,

3

.D

2

-2(3)D +

- 10D + 25D = D(D 2

x, e

5x ,

e

(3

2

5)

;

2

+

l

2 );

5 1, e *,

e

xe

3l

cos:r, e

3l

sin:r

5x

7x

Exercises 4.6 1.

Applying

D to the


we obtain

D(D 2 -

Q)y

=

0.

Then y

and

j/p

=

solution

=

+ c 2 e~ 3x + c3

A. Substituting yp into the differential equation yields

is

y 2.

3x c-je

Applying

D

to the differential equation

=

cie

ix

+

c 2 e-

Sx

-

6.

we obtain

D(2D 2

-7D + 5)y = 0.

—9A =

54 or

.4

=

—6. The genera!

Exercises 4.6

Then y

and yp

=

5A = —29

A. Substituting yp into the differential equation yields

general solution

D

Applying

or

A=

—29/5. The

is

y 3.

= ae 5x/2 + c 2 e* + C3

=


cie

hxl2

+ C2e x - ^.

we obtain

D(D 2 + D)y= D 2 (D +

l)y

=

0.

Then y

=

ci

+ c2 e~ x + C3X Vc

and yv

~

Ax. Substituting yp into the

differential

y 4.

D

Applying


=

+c 2 e

ci

A=

equation yields _31

3.

The

general solution

is

+ 3x.

we obtain

D(D 3 + 2D 2 + D)y = D 2 {D +

l)

2

=

y

0.

Then y

and j/p

=

j4i. Substituting

yp

ci

+ c%e~ x + case"*

4-

C4X

into the differential equation yields

y 5.

=

—

c\

+ C2e~ x + czxe~ x +

A—

10.

The

general solution

is

lOx.

D 2 to the differential equation we obtain

Applying

D 2 (D 2 + 4D + 4)y = D 2 (D + 2)

2

y

=

0.

Then y

=

cie'

21

+ c 2 xe~ 2x +c3 + c4 x He

and yp

= Ax + B.

Equating

Substituting yp into the differential equation yields

coefficients gives

4A = 4A + 4B = Then

A=

1/2,

B—

1,

and the general solution y

=

cie~

2x

2

6.

is

+ c 2 xe~ 2x + ~x + 94

1.

4Ax

+

(4A

+ 4B) =

2x

+ 6.

Exercises 4 a

6.

D3

Applying


we obtain

D 2 (D 2 + ZD)y = D 3 (D + 3)y = 0. Then y

and yp

= Ax 2 + Bx.

Equating

=

c\

+ cae" 3 ^

+ ct x + (2A + 3B) = 4x-

coefficients gives

2,4

A=

2/3,

B=

=

D3

Applying

4

+ 35 =

-5.

—19/9, and the general solution y

7.

2

Substituting yp into the differential equation yields 6 Ax

6/1

Then

c3 x

4-

=

cl


+ c2 e

-1* **

is

2

19

,

+ -x'- —x.

we obtain

D 3 (D 3 + D 2 )y = D 5 {D + l)y = 0. Then

=

y

+ C2X + c3 e~ x + c 4 x 4 + c 5 x 3 + c%x 2

ci

_

= Ax 4 + Bx 3 + Cx 2 Substituting yp into the differential equation UAx 2 + (24A + GB)x + {6B + 2C) = St 2 Equating coefficients gives

and yp

.

.

124

=

24A + 6B

=

6B + 2C = Then

A=

2/3,

B=

-8/3,

C= y

8.

Applying

8,

=

8

0.

and the general solution

8

2

ci

is

+ c 2 + c 3 e _I + -x 4 - -x 3 + 8x 2 :k

D 4 to the differential equation we

.

obtain.

D 4 {D 2 -2D + l)y = D 4 (D ~

ify

=

0.

Then y

—

c\e

x

+ C2xe x + c^x 3 +

C41

2

+ c^x + c$

yields

Exercises 4.6

= Ax 3

Bi 2 + Cx + D. Substituting yp into the differential equation yields Ax + {B - 6A)x 2 + (6A - 4B + C)x + {2B ~2C + D)=x 3 + Ax. Equating coefficients and

j/

p

4-

3

A=

gives

1

B - 6A = 6,4

- 4S + C =

2B - 2C + Then

A=

1,

B = 6, C =

y 9.

Applying

D—4

=

22, £>

=

32

C\e

0.


,

x

D=

4

x

3

2

+ C2xe + x + 6x +

is

22x

+ 32.

we obtain


-D- 12)y = (D - Af(D + 3)y =

(D - 4)(D 2

0.

Then y

=

cie

4x

+ c 2 e~ 3x + c 3 xe ix _

Vc

= Are 4 *. Substituting yp into the differential coefficients gives A — 1/7. The general solution is and yp

y 10. Applying

D—6

=

Cl

e

4x


+ cê~ 3x +

equation yields TAe41

^xe 4x

=

e**.

Equating

.

we obtain

(D-6)(D 2 + 2D +

2)y

=

0.

Then y

and yp

=

j4e

6a .

coefficients gives

D{D —

e

_I (ci

cosx

+ C2sinx) +C3e

6s:


A—

1/10.

The y

11. Applying

=

1)

general solution

—

e~

x

(c\

cos a;


D{D -

1)(£>

2

50Ae 61

is

+ ci sinx) +

e6T

-

^j

we obtain

-2D- 3)y = D{D - 1)(D + 1)(D - 3)y = 0.

Then y

=

cie

3x

+ c2 e~ x + c$e x + c4 _

96

=

5c

61 .

Equating

Exercises 4.6

and

j/p

= Ae x +

Equating

B. Substituting yv into the differential equation yields

coefficients gives

A = —1

=

y

D 2 {D + 2)

12. Applying

B=

and

c\e

3x

The

3.

general solution

— 4Ae x — 3B =

1

4c

—

9.

is

+ c2 e~ x -e x + 3. we obtain


D 2 (D + 2)(D 2 + 6D + &)y = D 2 (D + 2) 2 (D + 4)y =

0.

Then y

=

cie~

+ c2 e~ ix + c$xe~ 2x + c*x + c 5

2x

Vc

and yp

2Ae~

2x

= Axe~

2x

+ Bx + C. Substituting yp into the differential equation + 8Bx + (6S + 8C) = 3e~ 2x + 2x. Equating coefficients gives 2A =

3

85 =

2

6B + 8C = Then A =

3/2,

B=

1/4,

Cy

13. Applying

D +1 2

-3/16

=

cie"

0.

and the general

,

21

yields

solution

is

+ c 2 e- ix + \xe~ 2x + ]x - ~. 2


(D 2 +

4

lb

we obtain

lXD 2 + 25)!/ =

0.

Then y

= pi cos 5x + C2 sin 5x + C3 cos x + a sin x yc

and

j/p

= Acosx +

Ssinx.

24B sin x = Gsiax. Equating

Substituting

coefficients gives ,4

y

14. Applying

D(D 2 +

1)

p into the differential equation yields 24 A cos x

i/

—

ci cos 5a;

+

C2


D{D 2 +

=

and

sin5x

B=

+

1/4.

The

general solution

^ sinx.

we obtain

\){D

2

+ 4)y =

0.

Then y

=d

cqs2t

+ C2Sin 2x + C3 cos x + a sin x + c5 l/e

is

+

Exercises 4.6

= Acosx + Bs'mx + C. Substituting yp 3j4cosi + 3Bsinx + 4(7 = 4cosx + 3 sin x — 8. B = 1, and C — —2. The general solution is


and yp

=

y 15. Applying

(D —

2

A)

ci

+ C2sin2x +

cos2x

2

4)

(D 2 + 6D +

coefficients gives -4

4

- cosx

+ sinx —

=

4/3,

2.

we obtain


(D -

Equating

9)?/

= (D - 4) 2 (£> + 3) 2 = j/

0.

Then y

=

cie~

+ C2Je~

3j:

3l:

+ c3 xe

4:l

+

c4 e

4T

— Axe ix + Be^. Substituting yp into the differential equation AQAxe + (UA + 49B)e 4x = -xe 4*. Equating coefficients gives and yp

yields

ix

49,4

144

Then

A=

B=

-1/49,

D

2

{D —

D

l) 2

2

+ 495 =

0.

2/343, and the general solution

y 16. Applying

= -1

=

Cl e-

to

+

c2 xe~

3x


(D -

2

1}

{D 2

+ 3D -

is

- —xe 4* +

4x .

we obtain

= D 2 (D -

10)?,

ê

2

1)

{D - 2){D + S)y =

0.

Then y

=

cie

2x

+ eye' 5 * + c3 xe x + c\t x + c 5 x + eg

= Axe x + Be x + Cx + D. Substituting yp into the -6Axe x + {5A - 6B)e x - IQCx + (3C - 10D) = xe x + x.

and

j/p

-6-4= 5-4

Then

A=

-1/6,

B-

-5/36, y y

C=

-1/10,

= ae Or + 1

c?e 2

10D

--

D=

—St

Equating coefficients gives

1

- 6B =

-10C = 3C

differential equation yields

-

1

0.

-3/100, and the general solution

1

~

5

xe

98

_

3

1

x

e

36

6

10

.

100

is

Exercises 4.6

D(D —

17. Applying

I}

3

we obtain


D(D -

3

1)

(D 2 -

l)y

- D(D - \f{D +

l)y

=

0.

Then y

= pi^ + <%e~ x + c3 x 3 e x + CiX 2 e x + c 5 xex + ce d

Vc

— Ax 3 e x + Bx 2 e x + Cxe x + D. Substituting yp into the differential equation 6AcV + (6.4 + AB)xe x + (2B + 2C)e x - D — x 2 e x + 5. Equating coefficients gives and yp

&A=

yields

1

6A + AB =

2B + 2C =

-D = Then A =

1/6,

B=

-1/4,

y 18. Applying

(D +

l)

3

C= =

1/4,

c\e

x

D=

5.

-5, and the genera! solution

+ cê~ x + -x 3 e x - -x 2 e x + 6


(D + \f{D 2

4

-xe x -

is

5.

4

we obtain

+ 2D+

l)y

=(D+ ify = 0.

Then y

=

Cje~

x

+ C2xe~* + cjx

4

e~ x

+ c
yc

and yp = Ax i e~ x + Bx 3 e~ x + Cx 2 e~ x Substituting yp into the differential equation yields l2Ax 2 e' x + &Bxe~ x + 2Ce~ x = x 2 eT x Equating coefficients gives A = 1/12, B = 0, and C = 0. .

.

The

general solution

is

y

=

cie~

x

+ c2 xe~ x +

]-x

i

e"

x .

Li

19. Applying

D 2 — 2D + 2


{D 2

we obtain

-2D + 2){D 2 -2D + 5)y = 0.

Then y

—

x e (c\

cos2x

+ C2sin2x) + e*(c3 cosx + C4sinx)

yc

= Ae x cos x + Be x sin x. Substituting yp into the differential equation yields we obtain cosx + 3Be x smx = e x s\ax. Equating coefficients gives .4 = and B = 1/3. The general

and yp

3Ae

x

Exercises 4.6

solution

is

=

y

(D 2

2x

+ C2 sin2x) + — e x sin

a;.

o

D 2 — 2D + 10 to the

20. Applying

x e {ci cos

differential

equation we obtain

-2D+ 10) [p 1 + D +

= (D 2 - 2D +

y

10) (*>

+ 5)

If

=

0.

Then y

=

cie'^ 2

+ C2xe~ a ^ 2

+036^ cos3x

+ c 4 e sm3x a:

yc

= Ae x cos3x + Be x sin3x. Substituting yp into the differential equation yields {95 I x I 27v4/4)e cos3i — (9^4 + 27i?/4)e sin3x = — e^cosSi + e sin3a;. Equating coefficients gives and

i/p

-

-— A + 9B - -1 4

27

-&4

S-

1.

4

Then

jl

=

-4/225,

B= y

21. Applying


=

cie~

x

is

^ + C2xe~*ft - -ê 1 cos 3a: - ^-ex sulZx.

D 2 + 25 to the differential equation we obtain {D 2 + 25)(D 2 +

25)

= (D 2 + 25) 2 =

0.

Then y

and yp

=

,4a;

10-4sin5x

=

cos 5x

+

=

c\

cos 5x

Bxsin5x. Substituting yp

20sin5x. Equating coefficients gives

y 22. Applying

+ C2 sin 5x + C3X cos 5x + c^x cos 5a:

D2 +

1

=

ci

cos 5x

to the differential equation (Z>

2

+

1)(D

2

4-


A — —2

C2 sin

and

B = 0.

5i — 2x cos

The

10Bcos5x -

general solution

is

5x.

we obtain

+

1)

=

(£>

2

+

2

I)

=

0.

Then y

—

ci cos

+ C2 sin x + C3X cos x + C4X cos

x

a:

Vc

= .4xcosx + Bxsinx. Substituting yp into the differential equation yields 2Bcosx 2 A sin x = 4cosx — sinx. Equating coefficients gives A = 1/2 and B = 2. The general solution is

and yp

y

=

c\

cosx

+ C2sinx + 100

^3: cos

a:

— 2xsinx.

Exercises 4.6

(D 2

23. Applying

+

l)

2


(D 2

2

+

1)

we obtain

(D 2 +

D + 1) =

0.

Then y

= e -/a

.4

cos x

c\ cos

—- r +

C2 sin

---

+ C3 cos + C4 sin + c$x cos x + c%x sin x

a:

a;

a:

Vc

and

i/p

=

(B Equating

+ Bs'mx + Cxcos x + Dxs'mx.


+ C + 2D)cosx + Dxcosx + {-A -2C + D)s'mx - Cxsinx =

xsinx.

coefficients gives

B + C + 2D = O= -A - 2C + D =

-C= Then

j4

=

2,

5=

1,

C=

-1, and

ci

24. Writing cos 2 x

=

h(1

+ cos2:r)

D = 0,

— v/3

cos

2:

+

1.


V3 —

C2 sin

and applying

2:

D(D 2 + 4)

+

2 cos

x

is

+ sin x —

x cos x.


D{D 2 + 4)(D S + 4) = D(D 2 +

2

4)

=

we obtain

0.

Then y

and

j/p

=

-4i cos 2x

-4Aain2x and

C=

=

ci

C32 cos 2x

+

C4X sin 2x

1/8.

+ C5

+ Bx sin 2x + C. Substituting yp into the differential equation yields 5 + 5 cos2x. Equating coefficients gives A = 0, B = 1/8,

+ 4£?cos2i + 4C = The general

solution

j/

25. Applying

+ C2 sin 2z +

cos 2z

=

ci

is

cos2i

+ C2 sm2x + ^1 sin 2a: +

D 3 to the differential equation we

8

obtain

D S (D A + 8D = D b (D + 8) = 0. 2

)

Then y

~

c\

-\-

cix

+

C3e

81

+ c4 x 2 + c 5 x 3 + c$x A

Exercises 4.6

= Ax 2 + Bx 3 + Cx 4 Substituting yp into the differential equation yields 16A + &B + (48B + 24C)x + 96Cx 2 = 2 + 9x - 6x 2 Equating coefficients gives

and yp

.

,

+ 6B =

2

+ 24C =

9

16A

485

96C = -6. Then A

=

11/256,

D(D —

26. Applying

B=

2

{D +

1}

D(D -

l)

and

7/32,

2

1)

C=


we obtain


(D + 1)(D 3 -

D +D2

1)

is

= D(D -

3 1)

(D

+

1)(Z>

2

+

1) =

Then

= de + C2 cos x + C3sinx + a + c$e~ x + <%xe x + 31

j/

2 x

c?x e

Vc

and yp

= A + Be~ x + Cxe x + Dx 2 e x

.

Substituting yp into the differential equatioi

Wxe x + {2C + AD)e x - 4Be~ x -A = xe*~ e~ x + 7. Equating

coefficients gives

4£>

=

1

2C + 4D =

-4B = -1

Then

A=

-7,

B=

1/4,

C=

-1/2, and

=

cie*

+ C2Cosi + cgsinx -

y

27. Applying

D 2 (D — 1)

£>

=

1/4,


D 2 (D -

1)(Z>

3

and the general solution 7

+ ê" 1 - ^xe x +

^

1)

= D 2 (D -

l)

4

=

Then y

=

cie

x

2 I e

we obtain

- 3D 2 + 3D -

+ C2xe x + c$x 2 e x + c4 + c 5 x + cex 3 e x

102

is

0.

.

Exercises 4.6

= A + Bx + Cx ex Substituting (—A + 3S) — Bx + 6Ce x = 16 — x + e*.

and

p

yp

.

j/


Equating

coefficients gives

-A + 3B =

16

~B =

-1

6C = Then 4 = -13,

B—

C=

and

1,

y

28. Writing (e*

=


1/6,

c\e

x

+

+ e" 1 )' = 2 + e 2x + e~ 2x

1.

c%xe

x

+ c3 x 2 e x -

and applying

13

is

+ x + -x 3 e x

.

6

D(D-2)(D + 2)


we

obtain

D(D -

2)(D

+

- 3D 2 - 3D + 2) = D(D -

2)(2D 3

2

+

2){D

+ c 2 e 2x + c3 e xf2 +c4 + c$xe 2x +

cee'

2)

{D

+

1)(2D

-

1}

=

0.

Then

=

y

x

cit~

2*

Vc

21

= A + Bxe + Ce~ 2A + 9Be 2x - 2QCe~ 2x = 2 + and yp

and

C=

2x

e

Substituting yp into the differential equation yields 2x _2x Equating coefficients gives = = e

+

=

D(D —

1) to

cie~

x

A

.

—1/20, and the general solution

y 29. Applying

.

+ c2 e 2x + cze x/2 + 1 +

4 1)(D -

B

1/9,

is


D(D -

1,

^xe

2x

-

^

e

~ 2 "-

we obtain

2D 3 + D 2 = D 3 (D - if =

0.

)

Then y

and yp

= Ax 2 + Bx 2 e x

Equating

.

coefficients gives

D 3 (D — 2)

ci

,

+ c3£ x + cê 1 +

+

2

C5X

+ C6i 2 e I


A= y

30. Applying

=

=

1/2 and

ci

+

c2 x

B—

1/2.

The

general solution

+ C3e x + eâre* + ^x 2 4- iUV

is

1


we obtain

D 3 {D - 2){D A - AD = D h (D - 2} 2

)

.

2

(D

+ 2) =

0.

2A + 2Be x =

1

+

3

e

.

.

Exercises 4.6

Then

=

y

+ c2 x + ae 2* + fê"^ + c 5 x 2 + cex 3 +

ci

c7 x

4

+ cgxe 2*

Vc

4

21

= Ax + Bx + Cx + Die Substituting yp 2 (-SA + 24C) 24Bx 4SCx + IQDe 2* = 5z 2 - e 2x 2

and yp

3


.

'.


-8A + 24C = -24B = -48C -

160= Then

A=

-5/16,

B = 0, C = =

S 31. Applying (2D

-

1)

-5/48, and

O=


(2D - 1)(16D 4 -

1)

-1.


+ c2 x + c$e 2x + ae- 2* -

ci

5

^x 2 -

j^x 4 -

is

^xe 2*

we obtain

= (2D -

2 1)

(2D - 1)(4D 2

+

=

1)

0.

Then y

=

c\z

x t2

+ c2 e~ x ^ 2 + C3 cos -x +

c.j

sin

-x +

Csxe

x^2

Vc

and yv

— Axe

x^2

coefficients gives

.


A~

1/8.

y 32. Writing 2cosh x

=

D(D -

e

x

=

The c\e

+ e~ x

1){D

+

general solution

x l2

+

x 2 c2 e~ ^

1)(D 4 -

+ C3 cos \-x + c\ sin \x + \xe x ^ 2 2

D(D —

2

=

cie

x

+

+

5D 2 + 4) = D(D -

1)

c2 e~

x

1) to the differential 2

(D +

2

1)

(D - 2)(D

equation

+ 2) =

+ cse 2x + c4 e~ 2j + c5 + c$xe x + <^xe~ x yc

.

.

=

cie

1

+

c2 e~

x

Equating

.

and yp = A + Bxe x + Cxe~ x Substituting yp into the differential equation yields AA - 6Be x + 6Ce~ x = -6 + e x + e~ x Equating coefficients gives A = -3/2, B = —1/6, and C = 1/6. The general solution is y

.

a

Then y

x 2 e ^

is

1)(D

and applying

8Ae*^ —

+ c%
^ I

104

- \xe x + \xe~ x o

o

.

0.

we

obtain

Exercises 4.6

33.

The complementary function

is

=

yc

c\e

Sx

+

initial

Using

.

we obtain — 64,4

Substituting yp into the differential equation

The

Sx

c2 e

ax

y

=

Cl e

y'

=

Scje

+ c2 e~ 8x -

81

-

8c 2 e~

D

to annihilate 16

=

16.

Thus

we

find

A — —1/4

and

yp

=

A.

\ 4

8x

conditions imply

ci+c 2 = -

8ci

Thus

ci

=

=

C2

The complementary

1/2,

function

is

yc

=

c\

+

x

4

c2 e

.

Using

y' initial

2B)

+ 2Bx —

x.

j/

p

.

- x + -x 2

= -c2 e~ x - 1 + x.

conditions imply

=

2

and

C2

= — 1,

=

l

1.

and j,

35.

+

= Ax + Bx 2 Thus A — —1 and

x we find

1

- ci+c 2 e

-C2 ci

to annihilate

we obtain (A

c\+c 2 =

Thus

D

2

and y

The

0.

8

8


B=

=

5/8 and y

34.

8c 2

=2-

e

_I

- x + -x 2

.

D 2 to annihilate x — 2 we find

= Ax + Bx 2 Substituting yp into the differential equation we obtain (— 5,4+25) — lOBx = —2+x. Thus A = 9/25 and £ = -1/10, and The complementary

function

is

yc

— ci+c2 e bx

.

Using

^9

,

The

initial

,T

9

conditions imply

d+co —

1

1

2

j/

p

.

Exercises 4.6

Thus

ci

=

-41/125 and

c2

=

41/125, and

36.

The complementary yp

= Ae 21

.

function

41

=

u y

41 e

1

125

is

=

yc

9 ^ + —x

c\e

x

x

25

cê~ 6x

+

1

ox

125

.

D—

Using

.

,

10 2 to annihilate lOe 21

we obtain SAe 21

Substituting yp into the difFerential equation

=

lOe 2^.

Thus

we

find

A=

5/4

and

The

initial

y

=

cie

y'

=

cie

x

+ c 2 e~ 6x + ê 2*

1

-

6c2 e-

61

+ \z 2x

.

conditions imply

3

d Thus

37.

ci

= -3/7 and

c2

The complementary

—

8cos2x

4 sin x we

difFerential

A=

2,

-

and

5/28,

function find yp

-6ca = --.

is

i/c

=

c\cosx

+

C2sinx.

= Ax cos x + Bxsinx

Using (D 2

+

1){D 2

Ccos2i + Dsin2x.

-f-

2B cos x — 3C cos 2x — 2A sin x — 3Z) sin 2x = and D = 0, and

equation we obtain

B = 0, C =

-8/3,

y y'

=

cos x

ci

+ C2 sin x + 2x cos — a;

+

4) to annihilate

Substituting yp into the 8 cos 2x

— 4 sin i. Thus

8

- cos 2x o

= — ci sin x + C2 cos x + 2 cos x — 2x sin x +

— 1

sin 2x.

<5

The

initial

conditions imply

-

7T

sin

x

-Cl

Thus

ci

=

-7r

and

c2

=

The complementary xe

x

+5

we

find

yp

0.

-11/3, and

y 38.

=

= —

function

—— 11

jt

cos x

+

8

2x cos x

is

= Ax + Bx 2

yc x e.

— — cos 2i. 3

3

^

= C\ + cê x + csxe x Using D(D - l) 2 to annihilate + Cx 3 e x Substituting yp into the difFerential equation .

.

106

Exercises 4.6

we obtain A + (2B

The

initial

+ QC)e x + 6Cxe x = y

=

ci

y'

=

c2 e

y"

=

c2 e

+

c2 e

x

xe x

+ 5.

Thus

A=

5,

B=

-1/2, and

C=

1/6,

and

+ c3 xe x + 5x~ \x 2 e x + \x 3 e x

x 3 x + c3 (xe x + e x + 5 - xe + ±x e

x

)

x

+ caixt? +

2e

x )

-

e

x

-

xe

x

LV.

+ \x 2 e* +

conditions imply

d +c =2 2

+ C3 + 5 =

C2

+ 2c3-l =

c2

Thus

39.

=

ci

8, C2

=

—6, and C3

—

3,

y

=

8

function

is

yc

The complementary

2

-l.

and

—

=

6e

e

x

+ 3xe x + 5x- JiV + ^zV.

3l

(ci

cos

2a;

+ C2sin2x).

Using

D i to annihilate

x 3 we

- A + Bx + Cx + Dx Substituting yp into the differentia! equation we obtain 2 3 3 {8A -4B + 2C) + (85 - SC + GD)x + (8C - 12D)x + 8Dx = x Thus A = 0, B = 3/32, C - 3/16, and D = 1/8, and find

2

yp

3

.

.

The

initial

y

=

e

j/

=

e

2l

21

(ci

+

cos2x

C2sin2x)

3 + J-x + -^r + ^x 8' 16 32 2

- 2sin2x) +

[ci{2cos2a:

02(2

cos2x

2ci

=

ci

2, C2

=

The complementary x

+

e

x

we

3

3

o2

o

2 .

8

find yp

+ 2c 2 + |; =

2

4.

-3/64, and

y

40.

3 — + ^x + -x

conditions imply

Q=

Thus

+ 2 sin 2x)] +

=

e

2l

function

= Ax 3 +

-

(2cos2x

is

Bx*

yc

=

+ Cxe

ci

x .

A

s in2x)

+ ax +

+

+ ^3

|^ +

C3X 2

+

cê

1 .

Using

D 2 {D -

1) to annihilate


we obtain

.

Exercises 4.6

(-6A + 24B) - 24Bx + Ce x = x + y

The

initial

=

x

e

A=

Thus

.

+ c 2 x + c3 x'2 +

ci

c4 e

+ 2c3 x + Cie x -

y'

=

c2

3/"

=

2c 3

y'"

=

c4 e

+ eye 1 x

-

1

Ze

x

=

+

2, C2

=

1,

The complementary e*{2

+ 3a;cos2a:)

C3

=

0,

and

function

=

Axe x

e~

x

+ yp

+ xe x

.

+ C4 =

+ C4 + 1 =

2c 3

+ c4 + 2 =

C4

=

+ c4 =

is

yc

=

0.

—2, and

24

c\e

x

+

cie~ x

.

—

Using {D

l)(D 2 —

2D +

5)

2

to an

+ Bxe x cos 2x + Cxe x sin 2x + De x cos 2x + Ee x sin 2a:

The complementary function 9-

+ xe x

x

we obtain

Up 42.

x

e

+ xe x

6

41.

2e

and

24

- ^x 3 +

c2

2 ci

o

1,

+ xe x

- ^x 3 2

C=

-1/24, and

conditions imply Cl

Thus

x

]-x

x - \x 2

-x+

B=

-1/6,

is

yc

= c\+ C2t~ x

.

Using

D(D +

1)(D 2

+

I)

3

to annihilate

x 2 sin x we obtain

= Ax + Bxe~ x + Ccosx + D sin x + Ex cos 1 + Fisini + Gi 2 cosx +

43. Applying the operators to the function x

we

{xD - l)(D + 4)i

find

= {xD 2 + 4xD -D-4)x = xD 2 x + 4xDx - Dx-4x =

x(0)

+ 4x(l) - 1 -4x =

and

(D

+ 4)(iZ) -

l)x

= (D + 4)(xDx -

x)

= (D + 4)(x-l-x) = 108

0.

-1

2

/fi sinx.

Exercises

4.

Thus, the operators are not the same. n)

44. Since y^

(x)

=

n >

for

1,

yW + a n_iy( n -V +

an

•

•

+ aiy'p + a

=

yp

+ +

+

+

a

—=

it.

Exercises 4.7 The

particular solution, yp

especially

checked by substituting

1.

The

=

+

«iJ/l

U2i^, in the following problems can take on a variety of forms,

where trigonometric functions are involved. The back into the

it

auxiliary equation

is

=

form can best be

differential equation.

m 2 + = 0, so yc — c\ cosx + C2 sinx and 1

cos

W= Identifying f(x)

validity of a particular

x

sin

— sin x

x

=

1.

cos x

secx we obtain sin

=

ua

x sec x

cos x sec i

— — tanx

=

1.

1

Then ui =

U2

In| cosx|,

=

y for -5r/2

1.

The

< x<

and

x,

=

cosx

c\

+ C2sina: + cos x In

|

cos x\

+ x s'mx

it/2.

auxiliary equation

is

m + = 0, so yc = ci cos x + ci sin x and 2

1

W= Identifying f(x)

=

cos x

- sin x

sin

x

-

1.

cos x

tanx we obtain

= — sin 2 tan

a:

=

cos x

—

1

=

cos x

— sec x

cosx iiq

Then uj

=

sinx

— y

for -tt/2

< x<

hi sec

=

|

c\

tt/2.

x

cosx

—

sin x.

+ tanx|,

«2

= — cosx,

and

+ C2 sinx + cosx (sinx —

In

|

secx

+ tanx|) —

cosx sinx

7 Exercises

The

4.

auxiliary equation

m

is

+

=

1

0,

=

—

x

sin

x

sin

+

cos x

c\

cos

W= Identifying f{x)

—

so yc

x and

C2 sin

x

=

1.

cos x

sinx we obtain

— — sin 2 x

u'j_

—

u2

cos x sin x.

The „

1

1

1

u\

=

U2

= — - cos

— sin 2x — — x — -

1

— ~x

£ cos x

sin

x.

and y

for

—00 < x <

The

—

c\

cos x

+ C2 sin x +

=

ci

cos x

+

C2 sin

-

sin

1 cos 2 x

ro

is

2

+ 1 = 0,

so yc

=

— Identifying /(x)

=x—

=

sec x tan x

tanx, «2

y

for

-tt/2

The

co;

x — -x cos x

ci cos

cos x

u\

-

00.

auxiliary equation

Then

— -2 cos x —

sin

+ C2 sin x

x

x

=

1.

cos x

we obtain

u[

= -sini(secxtanx) = - tan 2 x =

u'2

— cosx(secxtanx) —

——

and

'

-

1

sec

I

ci cos

x

+ C2 sin x + x cos x —

=

c\ cos

1

+

C3sinx

+ xcosx —

m2 + 1 =

0,

so yc

=

ci cos

cos x

cos 2 x

x

sin

x —

sin

x in

sinx In cosx| |

+ C2 sin x and

sin

— sinx =

3

n cosx|, and

=

is

2

tanx.

tt/2.

auxiliary equation

Identifying f(x)

x

sin

x

-

1.

cosx

we obtain til

— — sin 1 cos

u'2

=

3

cos x

=

x

cos x (l

110

—

sin

2

xJ

.

|

c<

7 Exercises

Then

u\

=

3 cos x, y

112

=

sin

x

—

and

5 sin x,

—

ci cos

x

+ C2 sin x + ~ cos 4 x + sin 3 x — ^ sin 4 3;

=

c\ cos

a:

+ ci sin +

- (cos 2 x

—

c\

cos x

+ C2 sin x +

- cos 2 x

a;

1

1

+ sin 2 x)

for 6.

—00 < x <

The

Ci

cos x

+

2

x —

sin

2 x~j

+ sin 2 x

2

— sm 2 x

+

o

•J

=

( cos

1-2 + - + - sin x 1

C2 sin

a;

00.

auxiliary equation

is

m +1= 2

0,

=

so ye

cos x

C]

cos x

Identifying f(x)

=

+

sin

—

sin

u,

=

--

«2

—

secx.

C2 sin

x

=

1 and

1.

cos x

x

sec^x we obtain sinx

Then ui

-

--

1

— — sec x

COS 2

U2

=

In

I

sec

x

+ tanxl

and y

for -ir/2 7.

The

<

x <

=

c\

cosx

+ C2 sinx —

=

ci

cosx

+ C2sinx — 1 + sin x In

cos a: sec x

|

secx

+

tt/2.

auxiliary equation

is

rn 2

—

1

=

0,

so yc

—

cie

x

W= Identifying /(x)

+ sin x In

=

coshx

=

\(e

x

+

e

x )

we obtain

+ C2e~ x =

and

-2.

]

secx

tanx|

+ tanx|

4.

Exercises

4.

Then 1 -2* * + -1 x u^--e

and y

for

8.

—oo < x <

The

=

cje

=

cae

=

c%e

, 1

x

x

+ c2 e

_, 1

+ c4 e

x

+ ae

x

-

i _, -e 1

+

-x(e

+

-x sinh x

,

i

i

_

+ -ze 1 - -e x -

x

-

x

e

)

oo.

auxiliary equation

m 2 — 1 = 0, so yc = cie x + C2e~ x

is

W=

-

=

-2.

—

e

Identifying /(re)

and

sinh2x we obtain „-3x 1

4

4

4

4

Then 1

T

- le-. 12

and 4

=

cic*

=

cie

z

+ c a «-* + ^(e 2l -e+ C2e

I

+

!l11

)

-sinh2x o

for

9.

—oo < i <

The

oo.

auxiliary equation

is

m 2 - 4 = 0, IV

so y c

=

cie

2x

+ cieT 2*

„2z

„-2i

2e^

-2e -2x

=

112

and

7 Exercises

— e^/x we

Identifying f(x)

=

obtain u\

ui

1

=

= —e 4x /4x. Then

l/4x and u'2

u2 =

-ln|x|,

1

r

"'

x I

t

j*

^

--j

and y for

10.

>

x

The

=

cie

21

+ cae"

2*

+

2 (e * In

^

-

\x\

e~

2x

dtj

>0

x

,

0.

auxiliary equation

is

m z - 9 = 0, so yc — cie 3x + C2e" 31 e

W= Identifying f(x)

=

=

Six/e^ we obtain u\

3i

e

61

|a:e

1

111

=

and

-6.

= ~\ x Then

u'2

-

-61

1

= "24 e„Hte

and

-3x

= —J*

«2

and y

=

Cl

24

=

tic Ci e

-

for

11.

—00 < x <

The

e^ + c 2 e~ 3x - ê~ 5x - \xe~ 3x 3x

?xV 3 * 4

1

„„-33x (l-3x) -±xe-

„-3i 3x

+

4

c3 e-r 1,3"

00.

auxiliary equation

is

m 2 + 3m + 2 =

(m + l)(m -1

W= Identifying f{x)

=

1/(1

+

e

T )

e

-e" 1

+ 2) =

so yc

0,

=

cie~ x

+ c 2 e~ 2x

and

-2i

= _ e -s«

-2e" 2a:

we obtain

+ e1

1

.2i

=-

Hi

-

~

1

Then

Ul

=

ln(l

+ ex

),

y

for

12.

-00 < x <

The

u2 =

ln(l

+

e

1 )

2x

-

e

1 ,

+

e*

1

cie -*

+ C2 e-

=

cae~*

+ c 2 e- 2x +

(1

x .

and

+ e~ x ln(l +

=

e

+ e1

e*)

+ e~ 2x ln(l + e x - e" 1 )

+ e- I )e" 1 ln(l + e 1

)

00.

auxiliary equation

is

m 2 - 3m + 2 =

(m - l)(m -

e

x

2e 2x

2)

=

e

=

0,

3x .

so yc

=

cie

1

+ C2e 2x and

4.

7 Exercises

4.

Identifying f(x)

=

e

3l

+

/(l

e

1 )

we obtain 1x e.

t

+

1

4 Then

t*i

=

H

+ ex ~

1

e

)

y

for

13.

—oo <

The

a:

<

x

u2

,

=

=

Cl e

=

cie*

x

ln(l

e

=

e*

+ ex

l

-

e'

x

+e x

1

+ e1

and

),

+ c 2 e 2x + e x ln(l + e x )

+ c3 e 2x +

x

+ e x )e

(1

ln(l

e

+ e 2x ln(l + e x

2x

+ ex

)

)

oo.

auxiliary equation

is

m 2 + 3m + 2 = (m + l)(m + 2) = 0, so yc = c\eT x + c 2 e~ 2x

Identifying f(x)

—

sine 1

e~ 2x

e"

W=

-e~

x

,-3x

-2e~ 2x

we obtain

^ sine 1

e

„-3x

= -e 2x sme x

_f.-3z

Then

in

= — cose 1

,

y

for

14.

-co < x <

The

=

«2

e

x

.

cosx — sine 1 and ,

=

c\e

=

eie

T

+ c2 e

21

-

x

+ c2 e

2x

-e

x

e

cose x

21

sine

+e

x

cose

x

-e

^sine*

1

oo.

auxiliary equation

is

m 2 — 2m + 1 = (m — I) = 0, so y 2

W=

e e

Identifying /(x)

=

e

1 tan

1

1

xe

1

1

xe

c

—

c\e x

1

+

e

r

x we obtain

ul

ti 2

=

^

= e^tan c 2i

:

= — xtan x

=

114

tan

1

x.

1

x

+ c 2 xe x

and

and

7

e

Exercises

4.

Then ui

=

U2

=

tan

—

i

x — -

x tan

x

In

x +—

+

1

(

x'

and y

cie

=

cie

—oo < x <

for

15.

=

The

1

1

+ c2 xe x +

1

tan

ê x [(x 2 -

+ cgxe 1 +

+ |j

a:

l) tan"

m 2 — 2m + 1 =

is

e

=

Identifying f(x)

1 e / (l

+ x2)

x

l)

2

xe x

=

= e

= — 5 In

+

a:

2 J,

y

16.

The

- ^ In

(l

«2

=

=

—

e

c\e x

+ C2ie x

2x

=

tan

cie

1

-1

x,

2

*(l

+ x2

1

)

+ x2

1+x 2

)

'

and

+ c2ie T -

ê

1

In (l

+ x 2 ) + xe x tan

1

x

is

m 2 — 2m + 2 —

[m —

(1

+ j)][m —

(1

— i}} =

0,

so yc

—

and e^cosx

e^sinx e

Identifying f(x)

=

e

I

x

cos x

+ e 1 sin x — e x sin x + e x cos x

seci we obtain I'm TP 1

—

4 u\

=

x,

and

00.

auxiliary equation

Then

+ x 2 )^ xe"

we obtain

u2

—00 < x <

^x tan"" 1 x

so y c

0,

+ ex

1+x 2

for

+

x-\n(l + x 2

1

(m —

W=

u\

1

oo.

auxiliary equation

Then

e

112

=

In cos |

y

=

35

T

=

sinx)(e x secx)

-e 1x

1

— — tanx.

and

2:1,

c\e

(e

.2i

x

ainx

+ C2e I cosx + xe x sinx + e T cosxln

I

cost!

cje x sin x

+ cje 1 cos x

7 Exercises

for -tt/2

17.

The

4.

< x<

tt/2.

auxiliary equation

m 2 + 2m + 1 =

is

IV Identifying /(x)

=

e~ x

In

l)

x

e

=

+

(m

=

2

so yc

0,

cje _T

=

+ C2xe~ x

and

xe

—e~ x

—xe~ x + e~ x

i we obtain ,

x

e

\nx

= - llnl

,-3» x

e

=

,

"2

x

xe

Ul =

x

e

~

In

x

=

21

In

a;.

Then u\

1 1 2 2, = —-x tai + -i

112

=

x In x

—£

and y

for

18.

>

x

The

=

cje

=

ct e

-x X

X

+ C2xe -ix +

C2xe

x

„2„-a: x

1

,

+

x

irx*e

Vax l

+ ,

1

2 --2 x r

^-x T e

x ^2„-n + x*e \nx-x 2 e i

x

x

lnx-^x 2 e

0.

auxiliary equation

is

m

2

+ 10m +

W= Identifying f{x)

=

= (m + 5) 2 =

25

so y c

=

c\e~

hx

+ oixe" hx

„-5i

„-5x

xe

e

-5c-

0,

—

-Sxe-^ + e- 5*

51

„-l(te

e

fx we obtain

e

u,

xe -5x e -lQx

= e

-5* -lta e

e

e

~bx

-S*

u'2

Then u\

,z e -5i

=—

——

I

Jin

U2

=

xq

dt,

/* e~ 5t /

—=- dt,

Jin

>

I

xi)

>

t

and 2/

=

„-5x JJ cie -

31 + c2 a:e -5x

e

-Si 31

/"

5f

X e

j

dt

/

Jzo

116

t

01 + xe -5i

[

X C

/

Jxa

51 ,

—5- cf* t

anc

Exercises

x

for

19.

The

>

0.

auxiliary equation

x

C2e sin 3i

is

3m 2 - 6m + 30 = 3[m -

+ 3i)][m - (1 - 3i)] =

(1

0,

so yc

=

cie* cos3x

and

W= Identifying /(#)

=

\e

x

T

cos3a

~3e x sin

x

e

x+

e

e 1

3c cos

cos 3jt

x

smZx)^ tanZx)

1 sin

9e 2x

2

9 cos

(e^cosSxlfe^tanSa:)

,

1

sin

3a:

+

3a e

=

x

sin

3e

2x .

3x

tan 3a: we obtain (e

3a;

—

1,

-(cos3x

- sec 3x)

9

3a:

1

Then ui

=

^

sin 3a:

—

^

In sec 3a; |

+ tan 3a

|

1

«2

= — ^= cos 3a

and y

=

I

cie cos3x

—

~ for -7r/6

20.

The

4.

< x<

~e x cos3xln

cie^cosSx

|

sec 3a;

+ tan3a| — ê x sin3a;cos3x

+ C2eI sin3a: —

—

e

x

cos

3a: In

|

sec 3a:

+ tan3a|

7t/6.

auxiliary equation

is

4m 2 — 4m +

1

\e

= je^Vl —

2

=

0,

\xe*t 2

+

x 2 e '

= (2m —

W= Identifying f(x)

— e^sinSxcosSa:

+ C2eI sin3i +

xl2

I)

so y c

a 2 we obtain xe x / 2 e^ 2

VT^

1

4c 1

4ex

4

Then 3/2

U2= |\A - x2 + l sin

1

x

—

c\e

x ^2

+ C2ie x / 2

and

+

7 Exercises

4.

and y for

21.

=

-1 < x <

The

c^

2

+ c 2 xe x t 2 +

±-e

x'2

fl

12

^

- x2f

+ \x i fl 2 j\-x* + \xex ' 2 sin" x 1

o

8

1.

auxiliary equation

m 3 + m = m(m + 1) = 0, so yc = ci -f C2 cos x + C3 sin x and 2

is

cos 3;

1

W Identifying /(x)

=

=

tan x

sin

x

—

cos

x

a;

-

cos x

— sin

1.

a:

sin

tan x

1

W

3

=

=

cos x

—

sin

tanx

a:

x

= — sin 1

cos x

—

x

sin

— sin x — cos x

1

=

sin

—

tan a: we obtain cos x

1^

/2

1

sin

x

cos x

cos 2 x

= — sin x tan x

— sin x — cos x

—

1

=

cos x

— sec x.

cosx tan x

Then u\

= — In

112

=

cos x

U3

=

sin

I

cosx|

1 —

In!

secx

+ tanx!

and y

=

+ C2 cos x + C3 sin x —

c\

+ sin 2 x — sin 1 In = for

22.

—00 < x <

The

C4

+ C2 cosx -f

C3

|

sec

sinx

—

ln| cos;r|

x

+ cos 2 x

+ tanx| cosxj

In

—

sin.

|

x In sec x |

+ tanx|

00.

auxiliary equation

is

m 3 + 4m = m (m 2 + 4) = 0, so yc = ci + C2 cos 2x + C3 sin 2x and 1

W=

cos 2x

sin

2x

-2sin2x

2cos2x

—4 cos 2x

—4 sin 2a:

118

Exercises

Identifying f(x)

—

we obtain

sec 2x

«i

4.

=

Vi

= sec 2z

cos 2x

sin 2x

-2 sin 2a: —4 cos 1%

2 cos 2a:

Sin

1

=

- sec 2x 4

—4 sin 2x

2x 1

2cos2z

"4

— 4sin2:r

sec2a;

cos 2#

1

—

—

-2 sin 2x

tan 2x.

4

—4 cos 2a:

sec 2a:

Then u\

=

U2

= --x

-

In sec |

2x

+ tan 2i|

4

=

«3

- In cos2x| |

and y for -tt/4

23.

The

=

ci

+ cscos2x + C3Sin2a; +

In

^

|

sec2z

+ tan2i:| - 7Xcos2x 4- ^ sin 2a: In 4 8

o

cos2a:|

7r/4.

auxiliary equation

is

m 3 — 2m 2 — m+2 =

(m — l)(m — 2)(m+l) =

0,

so yc

= ae x +C2e 2x +C3e~ x

and e e

Identifying f(x)

|

—

e

31

x

2e 2x

1

21

4e

-e~ x e"

= 6^.

1

we obtain e

21

1

6e 21

1

2e*

Se 2* e

e e

31

4c

21

x

e~ _I

-e

e"

1

e~ x

x

-e" 1 3 e *

e" 1

1

-3e 41

~

6e 2 *

2e 3 *

1

6e 21

3

£ 2

2a;

-

7 Exercises

4.

„2x e e

Then

u\

= -ê 2",

U2

-

je*,

y

= =

for

24.

—00 < x <

The

and Cl e

x

cie-^

=

113

jje

41 ,

x

2x 2e'

x

ie

2x

e

'

6e 2 *

3x

6

and

+ c2 e 2x + c 3 e~* + c2 e" + c 3 e

-

\
+

^

+ i-e 3* 24

+ -e J

1

o

00.

auxiliary equation

is

2m 3 — 6m 3 = 2m 2 (m —

3)

x

1

W=

e

=

0,

so yc

=

ci

+ c2 x + C3e 31

31

3e*x

1

ge 3* Identifying }{x)

w'l

= x 2 /2 we

=

—

obtain e

Wi

9e^

2

1

3e

=

g e 3x

W, =

=

9^

9e 3x

3z

1

1

6"

9e 3s

Oe 31

x 2 /2 1

U3

3*

9e 31

/2

e

«2

3x

a;

1

a;

1

9e 3 *

9e 3 *

~

1 2„-3x 6 IS

x 2 /2

Then 1

1

4

111

= 24* "Si 1

U2

=-

1

X

3

3

T8 "3

X 54

8

81

243

and y

1 1 1 4 1 - —x — ~x 4 = ci+ c%x + c$e ^ + —x

120

1

2 x -

^ 1

-

1

55

and

Exercises 4.7

— oo <

for

25.

The

x <

oo.

auxiliary equation

4m 2 —

is

1

= (2m —

l)(2m

+

—

1)

0,

-

=

Identifying f(x)

u2

= -ze 1 /4 + e

it

xe

/4.

x^2

=

/4 we obtain

x/4 and

u'2

=

so y r _

=

c\e

x^ 2

+ C2&~ x ^

and

-1.

—xe^/i. Then ui

=

z 2 /8 and

Thus y

=

1/2 J c.p c l e ->*

^ + -x'e xf2

c2 e + ftf -t-

-J-

and 2

The

initial

16

conditions imply

+

<=3

1

1 --c 2 -

^

3

Thus

26.

The

C3

=

3/4 and C2

=

auxiliary equation

1/4,

is

2m 2 + m —

= (2m —

1

z/2 fi

±e*/

=

(x

+

l)/2

1

i=

0.

and

W= Identifying f[x)

=1

C2

2

e

l)(m

+

1)

=

0,

-x

-e"*

we obtain

4 = -3«*(* +

l).

Then ui

= -e-/ 2

u2

=

=

c1 e

(^-2

-gie*.

Thus !/

and

l/a

+

C3e~

3!

-a:-2

so yc

=

c\e

x/2

+ cê

x

and

27.

7 Exercises

4.

y'

The

initial

Thus

ci

=

=

oCie

l/2

x

-c 2 e

-

1.

conditions imply

8/3 and C2

=

1/3,

ci

-

c2

-

2

=

1

-ci

-

C2

-

1

=

0.

and

= Tho

auxiliary equation

m 2 + 2m — 8 =

is

e

2at

2e Identifying /(i)

=

2e

_21

+ 4) =

{m — 2)(m e

2x

so yc

0,

—

cje

-4r

- -6e -2i

-4e- 4:c

e~ x we obtain

—

-e„-4x — -e -3i 1

=

Ui

2

*

6

3

Then

Thus 1

1

-2=:

-x

18

12

4

6

18

9

and y'

The

initial

Thus

cj

=

=

2 Cl e

21

-

4c 2 e~

ix

+ \e~ 2x - \e~ x

.

conditions imply

25/36 and r2

=

4/9,

ci

+

c2

-

— = 36

2ci

-

4c 2

+

—=

1

0.

and *

=

C

36

+

g

e

9

4

122

+

C

9

-

21

+ C2e

41

and

Exercises

28.

The

auxiliary equation

ism 2 — 4m + 4 = (m — B 2x

W—

(l2x 2

=

so yc

0,

-

cie

21

+ c^xe 2* and

„ D 2x „4x

2e

Identifying f(x)

2

2)

2xe 2x

2x

+ e 2x

6x^ e 2x we obtain

—

u'j

=

6x 2 - 12x 3

u'2

=

12i

ui

=

2x

3

-

3x

u2

=

4x 3

-

3x

2

-

6x.

Then 4

2 .

Thus y

=

Cl e

21

+

tW +

(2x

3

- Zx 4 )

^de^ + axe^ + e 2 * (x 4

e

2x

+

(4x

3

-

3x 2 ) xe 2x

- x3

and y'

The

initial

=

2c ie 2 *

+ c2

+ e 2*) + e 2 *

(2xe 2x

2c\ ci

=

1

and

C2

=

+ 2e 2x

-

3x 2 )

identify f(x)

—

+

c2

=

1

=

0.

4

-z 3 )

-2, and

=

e

=

e

2x

21


and

3

conditions imply ci

Thus

(4x

(4hix)/x.

FVom

- 2xe 2x + (i

(z

2z

+

- i3 -

4

y

—

y

yi

—x

and yi

x

2 e *

l)

.

+ -*y = - Inx x

x

—

we compute

xhix

x 1

xlna:

1

+ lnx

Now u\

—

(lna;)

2

so

u\

= — -(lna;) 3

and v.2

= — Ini x

so

U2

=

2(lni)

2 .

,

(x

4

- x3)

.

4.

7 Exercises

4.

Thus yp

=

-^x(\ax)

3

+ 2x(laxf =

§x(lnz) 3

and

—

y

c\x


+ C2X In i

-x(lnx) 3

4-

.

,,4,6 - zv + jf = 3 1

y

and

identify /(x)

=

1/z 3

.

From

=

?/]

a:

2

and yi

=

2

3

x

x

a;

3

X°

we compute

=

3x 2

2x

Now

I1

X

-— = -?

3x

4

4

=

x4

-

2x

Ul

= 3?'

.

x 3 /x 3

=

"1

80

and x 2 /x 3

u2 — ,

1

—

x4

u2 =

so

x5

1

T

.

4x 4

Thus Vp

~

4x 4

3x 3

~

12x

and y

=

ye

+ yP =

cix

2

+ c2x3 +


and

identify /(x)

=

Prom

x

W{y\,w) = — x

—

yi

1 /2

i 1 /a

x

sinr

l

y2

—

cosx

x

^sinx ^ cosx — ^x 3

tii

=

cosx,

U2

=

sin x.

^cosx and

l

^sinx we

x

cos x

— jx

x

3' 2

2

'

Now uj

=

sin

i

so

ti^

=

cos

x

so

and

Thus j/

=

c\x

=

cix

^ 1 '/2

cos a:

+ C2i ^sinx + x

cosx

+ C2X

1|/2

sinx

124

l

/2

+ x ^2

2

cos x

.

+ x ^2

e

Exercises 4.7

32.

(a)

Write the equation

in

the form „

and

identify f{x)

=

secflnxj/x

1

2

From

.

secllnz}

1

,

=

y\

cos(ln x)

cosflnz)

W=

and

1/2

=

sin(lni)

we compute

sin(lni) _1

sinfins)

cos(lna;)

x

Now tan(lnx)

,

so

«j

= m jco5(lna;jj,

so

U2

=

inx.

+

(In x) sin(ln 2)

and

— —

%L

x Thus, a particular solution yp (b)

The

general solution

=

y for

is

—

cos(ln x) In

(a)

We

ci cos(lna:)

— n/2 < \nx < n/2

have yc

differential

=

c\e~

\

+ C2sin(lna;) + cos(lnx) In |cos(lni)| +

or e -7r / 2

< x <

x

+

C2xe~

x

e"^

2 .

The bounds on

and we assume yp

A=

have y c

due to the presence of

— Ax 2 + Bx +

C.


4,

=

B= cie

-1

—16, and

C=

B=

particular solutionis yp

+C2xe _r and

— =

A

21.

*

xe

-1

— ie _T

e

Identifying f{x)

A

+ 2B + C = -3

2A

We

Ina; are

equation we find

4A +

(b)

(lni)sin(lnar}

.

A=

so that

.

is

sec(ln x) in the differential equation

33.

cos(ln x) |

*

+ e _T

x

e~ /x we obtain u,

,

U2

=

ie --

x

e

x

a -2x

I

I

e

-2x

e~ e" /x

=

jx

= -1

1 " a;

=

ix 2

—

I61

+ 21.

7 Exercises

Then

4.

u\

=

—x, U2

=

lnx and yp

Since (c)

-xe -1

Adding the

a solution of the homogeneous differential equation,

is

results of (a)

we take yp

=

xe~ x kix.

and (b) we have yp

4.

= -xe~ x + xe~ x In x.

=

2 4x -

16a:

+

21

+ xe~ x lnx.

We have yc = c\ cos x + C2 sin x. We use undetermined coefficients to find a particular solution of y" + y = 2x — e 3*. Assuming ypi — Ax + B + Ce 31 and substituting into the differential equation we find A = 0, B = 0, and C = — ^ Thus yPl —2x — jQe 3x Next we use variation of parameters to find a particular solution of y" + y = cot x. We have .

.

cos x

W= Identifying f(x)

=

cot a;

— sin x

sin

x

=

1.

cos x

we obtain ui

sin x cot x = — cos = -

"2

=

2 cos X

cos x cot X

sin

a:

~

cscx

— smx.

a:

Then ui

=

sin

U2

=

In esc x I

a:

— cot x\ + cos x

and j/pj

=

sin

x cos a;

+ sin x In

|

cscx

— cot x\ + sin x cost =

2 sin

x cos x

+ sin x In

Using the superposition principle we have that a particular solution of y" yp

=

ypl

+yP2 = 2x —

—

e

+ 2 sin x cos x + sin x In

126

|

cscx

+y = —

|

cscx

2x

cotx|.

— cot x|.

— e 31 + cot x

is


Chapter 4 Review Exercises =

1.

y

2.

False; see

3.

False; consider

Problem

multiple of

0.

45, Section 4.1.

=

The statement would be

independent on an interval 2

-Sx + (-2)

if

dependent;

5.

(-oo.oo); (0,oo) or (-oo,0)

6.

True

7.

False; see

8.

x

9.

A + Bxe x

10. 11.

Problem

(l

neither

-x

4.

=

=

and fzix)

2

)

+

linearly

"Two

read

dependent even though x

+ x2) =

2

(D -

2 [p - 4D + 5) Identifying P(x) — we

2)

2

have

=

cos 2x

;r—- dx

I


—

cos

=

^

= -2- 2/x we y2

13.

From

m 2 — 2m — 2 =

14.

From

2m 2 + 2m +

3

=

=

sec

I

2

2x dx

J

^ tan 2xj

have /

m=

we obtain

=

e

P dx = -2x - 2 In x

— — dx =

e*

we obtain

m 3 + 10m 2 + 25m =

cos 2x

sin 2x.

J

y

From

2x

—

cos' 2x

J

±

1

e

x

and

2 j x dx

=

Tjê

1 -

\/3 so that

m = -1/2 ± \/5/2 so that

-W '

we obtain y

=

I

ci

m— ci

^

cos -^-x ^

+

0,

m=

C2e"

5*

127

^

+ C2sm -^-a:

—5, and to

+ c3 ie~

is

not a

functions fi(x) and fo{x) are linearly

31, Section 4.1.

y2

15.

if it

a constant multiple of the other."

is

(2

These are

x.

true

51 .

— —5

so that

— Chapter 4 Review Exercises

16.

2m 3 + 9m 2 + 12m + 5 =

From

we obtain y

17.

From 3m 3

+ 10m 2 + 15m + 4 = =

y

18.

4

x

^ + e~ 3x/2

=

e

m=

'.

-1/3 and

m-

^-x +

m-

m = —5/2 so that

— 1, and

+ c^xe~ x

c2 e~*

^c2 cos

we obtain


y

+

m=

we obtain

Cl e~

—1,

1/2,

±

-3/2

-\/7/2 so that

c3 sin

m-

-2, and

m=

±-/2i so that

= c\e^ 2 + C2e~ 2x + C3 cos \/2 x + q sin v^x.

y

D

cie~^ /2

+ 2m 2 + 6m - 4 =

From 2m 4 + 3m 3

19. Applying

=

m=

( 3x/2 ' I

D

we obtain

i

(D 2 - 3D

vTT ^11 \ ^ —^ x + c 2 sin — — 1 + .

ci

cos

J

+

C3

+ 5) =

0.

Then

23

+ C5X + CfjX°

C4I

y=

and

= A + Bx + Cx + Dx 1

i/

p

solution

coefficients gives

A =

is

9


.

35 + 2C) + (55 - 6C + 6D)x + {5C - 9D)x 2 + 5Dx 3 = -2x + 4x 3

(bA Equating

3

=

e

a*/a I

_

vTT

/"

ClCOS

B=

-222/625,

.

x

+ C2Sm

C=

46/125,

36/25, and

_ j__ _ v^l

x

222

\

46

+

36 _ x

D= 2

+

(D - 2D

+

x

+

4/5.

.

The

4

3

1)

= (D -

general

^ 20. Applying

(D -

3

l)


we obtain (D -

1)

3

Then y

=

cie

1

c^xe?

4-

+

2 I

c3 x e

+

3

c4 x e

x

+ csê 1

yc

and

j/

p

= Ax

2

e

x

+ Bx

3 x

+ Cx

e

A x

e

.


\2Cx 2 e x + GBxe 1 + 2Ae x = Equating

coefficients gives

A—

0,

B— y

21. Applying

D(D 2 +

1)

=

0,

and

c\e

x

C—

+ cê 1 +


D{D 2 +

l)(D 3

1/12.

iV.

The general

^ê 1

solution

-

we obtain

- 5D 2 + 6D) = D 2 (D 2 +

1){D - 2){D

- 3) =

Then 1/

=

ci

+ c2 e

+ c3 e

is

+ c 4 x + c 5 cos x + eg sin x

to

128

0.

l)

5

=

0.


and yp

= Ax + B cosx + C sinx. 6,4

+ (5B + 5C) cos x + (-5B + 5C) sinx = A=


y 22. Applying


=

B=

4/3,

c\

+

1r

c2 e

=

24.

The

1

+

.

- sinx.

—

and

m? — 2m + 2 =

is

= —1 we

j/(7r)

auxiliary equation

+ C2X +

c\

is

obtain

m2 — 1 =

so that

-

=

c\

e "

—2A =

equation yields

differential

.

auxiliary equation

j/(tt/2)

- cosx

is

Then

+ C2X + cie x + cix 2

ci

= Ax 1 Substituting yp into the A = —3. The general solution is

=

1

-

solution

}

y

The

4

It + c%e^ +

The general

1/5.

2

and yp

23.

C=

2 sinx.

D to the differentia! equation we obtain D(D 3 - D = D 3 (D - 1) = 0. y

gives

-1/5, and

+

8

cse

x

m—

and

1)

Assuming j/p = .4x + 5 + Csinx and substituting S = 0, and C = — \ Thus j/p = -x - \ sinx and

=

=

y

Thus,

0.

Equating

coefficients

.

±i and

1

=

C2

(m — l)(m +

— 3x 2

6.

so that

1/

e^fci

—

m

e

I-7r

cosx + C2 sin x). Setting cosx.

= ±1 and

3/


=

c\e x

we

find

+ C2e -X A = — 1, .

.

y Setting

j/(0)

=

2 and y'(0)

=

3

=

1

-x--smx.

+C2e

c\e

z

we obtain

+ c2 =

ci

-

ci

Solving this system

we

find C\

=

-^

and

y= 25.

The

auxiliary equation

is

13 —

m 2 — 2m + 2 =

3

x

.

5

- -e

[m —

e e

Identifying f(x)

—

e

x

1

x cosx

sin

x

+

x

e

x

sinx.

—

(1

— i)] —

(1

(e

I

;c

cosx)(e taEx)

_

.

is

.

1 cos x

— e x sin x +

tanx we obtain ,

1

-

e

sin


-x -

and

W=

„ 3.

- -=

c2

= — | The

C2

e

2

0,

so yc

=

= -e 2x cos x

.

cje1 sin x

+ cze? cosx


Then

uj

= — cos x, y

= =

26.

The

x

cie sini c\e

x

— am x —

U2

s\nx

c

+ C2ef

In

cosx

e

x

and

,

=

so y c

c\e

+ C2t~ x

x

W= Identifying f(x)

2e x /(e

=

x

I

+e

)

=

—

Then

Ui

—

tan

-1

e

1

-e2 +

-

112

,

=

j/

27.

The

auxiliary equation

cie

x

er

particular solution

is

ex

+ e -i

tan""

=

-4,

yp

j/'(0)

1

e

i

x

= -4

=

=

=

l/2

cie

c\e

x/i

=

find c\

=

|

,

The auxihary equation

is

e

1

-

m2 + 1 =

0,

\e

*' e2

+ e" 2 tan -1

1

= (2m -

+ c2 e

3!C

+

l)(m

caê

-

5

3)

31 -

is

4.

we obtain

+ C2 — 4 = — 4

+ 3c 2 + c 3 =

a ~ -§

y= 28.

9

1

+ c 2 e 3x + c3xe Sx -

§

ici

we

+

and

,

ci

Solving this system

1


and y"(0)

0,

- -e x +

+ e 2x

+ c 2 e~ x + e 1 tan"

y Setting y(0)

„3x

3sc

e

Vc

A

-2.

+ e2x

1

2m 3 - 13m a + 24m -

is

and

e

1

4

secx

we obtain

+

ex

|

+ tan:c|.

cosx\n secx \

m 2 — 1 = 0,

is

+ tan x\

— e x sin x cos x + e x sinxcosx — e'cosx In

+ C2e T cosa; —

auxiliary equation

x

sec

|

,

and

C3

—

1.

Thus

*n-L 3x + xe 3x -4.

so yc

=

cos x

c\

cos x

—

sin

x

130

+ C2 sin x and

sin

x

cos x

-

1.

e

=

x .

so that

+ tani|

— Chapter 4 Review Exercises

=

Identifying f(x)

sec

3

i we obtain

smx

= — sin x sec i x =

,

,

u,

— Then ui

=

cob x sec x

=

sec i.

11—x— = — 1

2 cos

U2

=

=

tan

x.

C2 sin

#

2

x

sec

2

x

2

Thus y

=

c\

cos £

+

=

ci

cos x

+ C2 sm x

— -1 cos x sec > i + smi tan x 1

sec x

2

=

+

1

— cos 2 x cosx

1

C3 cos

x

+ C2 sin i + — sec i. Li

and ,

yp

The

initial

= — C3 sin x + C2 cos x +

1

- sec x tan x.

conditions imply c3

+

= C2=

Thus

c3

=

c2

=

l

l 1

2-

1/2 and

y

=

- cos x A

+ -sinz + £,

- sec x. z.

5

Applications of Second-Order Differential Equations:

Vibrational Models Exercises 1.

A weight of 4 lb (1/8 slug), attached to a spring, position with an initial

2.

A

upward

released from a point 3 feet above the equilibrium

is

The

velocity of 2 ft/s.

weight of 2 lb (1/16 slug), attached to a spring,

The

equilibrium position. 3.

5.1

Applying the

initial

spring constant

conditions to x(t)

=

x(0)

Then

=

c\

-2, c2

=

=

is

c\

is

3 lb/ft.

released at rest from a point 0.7 feet below the

4 lb/ft.

cos5f

= -2

ci

is

spring constant

+ c2 sin5(

and

x'{0)

and

-

— — 5cj sin 5t + 5c2 cos bt

x'{t)

=

5c2

gives

10.

and

2,

A = ^4+1 = 2V2

and

=

tan^>

—=

-1.

A

Since sin x(t) 4.

=

<

>

and cos

2\/2 sin(5t-

Applying the

initial

0,

is

ci

=

=

c2

1,

conditions to x(()

=

=

ci

c\

=

Thus 5.

>

z(t) «s

and cos$ <

^

Applying the 3/(()

=

— — ?r/4.

cos At

+ c 2 sin At

and

1

=

x'(0)

and

sin(4(

initial

Thus

= — 4ci sin At + 4c2 cos At gives

x'(t)

4c2 - -2.

-1/2, and

A = Jl + 1/4 = Since sin^

4>

jr/4).

z(0)

Then

a fourth quadrant angle and

0,

4>

=

is

^

and

tan

=

—^-

- -2.

a second quadrant angle and

tan

_1

(-2)

+

7T

=

-1.107

+

it ft*

2.034.

+ 2.034).

conditions to x(t)

-\/2cisini/2t

+ V2c2 cosv^t a;(0)

=

ci

=

c\ cos \/2'

+ C2Sin -J2t

and

gives

= -l

and

x'(0)

132

=

\/2c2

=

-2-/2.

Exercises

Then

ci

=

=

—1, C2

—2, and

A= <

Since sin

Thusi(t) 6.

<

and cos v/5sin(\/2i

=s

Applying the

initial

is

0,

c\

=

4, C2

=

2,

+4=

y/l

=

conditions to x(t)

=

^.

tan~ 1 ^

+ = tt

+ tt

0.464

=s 3.605.

=

=

ci

cos 8t + C2 sin 8f and x'(t)

c\

and

4

=

i'(0)

8c 2

=

= — 8ci sin 8t + 8C2 cos 8(

gives

16.

and

and cos0 >

>

—^ =

=

tan

+ 3.605).

A = V16 + 4 = Since sin0

and

1/5

a third quadrant angle and

z(0)

Then

5.

a

is

0,

and

2\/5

=

tan

^

=

2.

=

quadrant angle and

first

tan

-1

2 ss

1.107.

Thus

x(t) ss 2\/5sin(8t+ 1.107). 7.

Applying the

initial

=

conditions to x(t)

cj

+ C2 sinlOt

cos lOi

= — 10c sin lOt + IOC2COS lOi

and

gives

=

x(0)

Then

ci

=

01

1,

=

z(f) 8.

«

i/ToTsintlOf

Applying the

Then

c\

—

Since sin x(t) 9.

=

and cos0 >

>

-4,

+

-

a

first

conditions to x(t)

=

c\

+

a — 3,

0,

=

z'(0)

10e 2

=

1.

and

=

tan

y~ =

10.

=

quadrant angle and

tan

-1

10

«

1.471.

x(0)

=

A=

\/16

ci

cost

+

— — c\ sini + C2Cosf

C2sini and ^(t)

= -4 and

x'(0)

=

c2

=

gives

3.

and

0,

+9=

-4

5

tan0=—

and

=

a fourth quadrant angle and

is

~~

tan

-1

(— g)

—0.927. Thus

0.927).

16x

Thus

1.471).

- Owe

obtain

x

10.

VIM

-

01

is

and cos0 >

<

5 shift

From mx"

initial

and

l

and

1/10,

A = Vl + Since sin0

=

ci

so that the period tt/4

=

Prom mx" + 120x =

we obtain

ns/mft, 2;

=

4

ci

m= =

cos

—7=

.

t

4

+ C2 sin —;= f

1/4 slug, and the weight

ci cos

2^/30/m (

+

is

8

lb.

C2sin y^30/mt so that the period

1

Exercises

5.

m, m=30/ir 2

C2sin2( so that the frequency 0,

x(0)

=

-1/4, and

=

0,

x(0)

-

0,

+ 40x =

0,

x(0)

=

1/2,

From \x"

14.

From fx" +

15.

From fx"

(b) (c)

=

2/2iv

is

=

13.

(a)

and the weight

slugs,

-\-12x

72a:

and

lb.

ar'(0)

=

x'(0)

2

=

=

we obtain x

we obtain x =

=

-jcos4\/6f-

sin4v/ 6*-

we obtain x = £cos8t.

= -1/4, x(n/8) = -1/2, x(tt/6) = -1/4, i(jt/8) = 1/2, — — 4sin8i so that x'(3ir/l6) ~ 4 ft/s directed downward. x' If x = AcosSt = then i = (2n + 1)tt/16 for n = 0, 1, 2

i(9tt/32)

=

v^/4.

-5sin2i and

x'

=

x(tt/12)

16.

From 50x" + 200x =

17.

From

20a:"

+ 20a: =

0,

0,

The 20 kg mass has 20 kg: i'(tt/4)

-5sin2i

=

x(0)

(b)

-

=

x(0)

(a)

(c) If

0,

0,

and

and

x'(0)

x'(0)

= -10 sin (

+

16x

=

0,

x(0)

x'(tt/2)

=

is

is

obtain 2

obtain x

=

m/s;

=

= -10 sin t

=

50 kg: x'(n/4) so that

The 50 kg mass

even and downward when n

-1, and x'(Q)

- -2 we

(

=

w/2 seconds and the amplitude

is

=

From \x"

+x=

0,

x{0)

x'

=

-lOcosf.

nir for

=

1/2.

and

n=

.

.

.,

10

m/s

placing both

is

odd.

is

feet.

3.6).

In 4?r seconds

x!>

sin(2i

134

0, 1, 2,

-

obtain

y/E/2

x

m/s, x'(jt/2)

moving upward; the 20 kg mass

vibrations.

19.

and

-10cos2f.

is

- ^sin4i = —^cos(4f -

x

The period

= -10 we

then 2 sin ((cos (-1)

moving upward when n

From x"

= -10 we

the larger amplitude.

-by/2 m/s,

masses at the equilibrium position.

18.

approximately 97.3

vibrations/second.

x'(0)

and

is

+ 0.588).

it

will

make

8 complete

Exercises

20.

From

1.6a:"

+ 40a; =

0,

=

x(0)

=

x If

21.

x

=

5/24 then

From 2x"

*

=

+ 200x =

0,

= -f cos lOt +

(a)

x

(b)

The amplitude

(c)

37r

=

7rt/5

is

=

i(0)

5/6

and k — 15

and

2jitt)

=

§sin(10t

5/4 we obtain

\ sin5t

-

=

t

=

-2/3, and x'(0)

^ sin lOt ft

+

cos5t

+ 0.927 +

5 (f

=

-1/3, and x'(Q)

=

^-

sin"(5t

+

I

+ 0.927). + 2n?r)

0.927

(e)

for

71

=

0, 1, 2, ...

.

5 we obtain

0.927).

and the period

2-k/IO

is

=

n/5

cycles.

If x — and the weight is moving downward for the second time, t = 0.721 s. If x' = f cos(10t - 0.927) = then 10* - 0.927 = tt/2 + tot or t = = ji 0, 1, 2, ...

(d)

5.

—

then lOf

(2n

+

0.927

=

or

2tt

+ 0.0927

l)ir/20

for

.

= -0.597 ft X '{Z) = -5.814 ft/a

1(3)

(f) (

g)

(h)

=

a/'(3)

59.702 ft/s

2

= thent = ^(0.927 + wr) for n = 0, 1, 2, ... and x'(t) = ±f ft/s. (j) If x = 5/12 thent = ^(tt/6 + 0.927 + 2jijt) and i = ^(5tt/6 + 0.927 + 2rwr) (k) If x = 5/12 and x' < then = ^(5tt/6 + 0.927 + 27iw) for n = 0, 1, 2, From x" + 9x = 0, x(0) = -1, and z'(0) = —/3 we obtain (i)

If

x

t

22.

1

and 2,

23.

x'

=

If x'

=

3 then

t

fci

Now,

m

= =

40 and

20/32 so

obtain x{t) 24. Let

—

first

=

m

=

^

k*i

=

120

k/m =

sin8\/3

we compute the

120(32)/20

=

2 —7= sin

/

y3

\

-7tt/18

+

37

.

n

=

0, 1, 2,

.

.

.

.

.

— 4tt

3

+ 2n7r/3

and

t

=

-?r/2

+ 2h7t/3

n=

for

1,

weight

is

+

192 and x"

192x

0.

=

4(40)(120)/160

Using x(0)

=

and

x'{0)

= =

120. 2

we

first

Let ki and

weight.

be the spring constants and

fo) the effective spring constant of the system. Now, the numerical value of the

W — mg — 32m, so

From these equations we

w=

+

=

1-

denote the mass in slugs of the

Ak\k2/{k\

effective spring constant k

32m = &i(^

so

=

sin 3t

for .

3

From

k

V3 —

— — cos 3f

+ 47r/3).

2V3cos(3t

.

30. Since the

find 2fci

mas

=

3/t2-

and

The given period

of an 8-pound weight

30

2

=

32m =

k

-71/4

=

is

Ak

fc 2

^

of the

combined system

1/4 slug, we have from or

k

=

225.

is

w = k/m 2

2-k/w

—

ir/15,

Exercises

We now

5.

have the system of equations

+ ki

fcl

2ki

=

225

=

3fc 2

.

Solving the second equation for ki and substituting in the 4(3fc2 /2)ft2

+

3fe/2

Thus, hi

=

375/4 and

=

fc]

Prom x =

c\

coswt

+ C2smut,

26. Let /I

amphtude

=

+4

is

x(Q)

A=

=

+

yjxg

cos0

=

c\/A and sin0

= 2V2~[^cos5t- ^sin5f] =

28.

If

x

is

zero; that

29.

=

is,

x'(0)

—

weight

46.88

is

lb.

we obtain

vq

w

sm uit

.

)

then the

when i

=

=A

ci

-c 2

— coswf

— sin

,

A i =

.4cos(uf

2\/2cos(5(

+ 5tt/4).

= —c^/A

27. i

= A cos(w( +

=

xo cos tot H

(wq/w)^

first

o

wo

=

22

«^

and

xq,

we obtain

so that

I If

_

equation,

5

1125/8. Finally, the value of the

x

so that the

12fca

=

hk2

^ = 32m= ^ = 25.

lZfcf

=

k2

first

then

minimum and maximum

,

.

A +

uji

velocities occur

when x"

=

—Auft cos(wt+#)

0.

= — ui 2 Asin(ut + 4>). Differen= ±1, tiating we find a'(t) — — j4cos(wt + 0). Thus a'(t) = when cos(wt + 0) = or am(ut + maximum acceleration occurs when = Thus, the displacement is ±.A. Using T 2tt/w orw= 2jt/jT we see that the magnitude of maximum acceleration is w 2 A = (2it/T) 2 A = 4tt 2 A/T2 From

x(t)

,4sin(wf

4- 4>)

we compute the

acceleration a(t)

=

x"(t)

3

)

.

30.

If

x

and is

= (

Asin(uif-l-i^) the

=

(—n/2

extremes for x occur when

— 4> + 2mr)

for

n

=

0,

1,2,

...

,

x'

=

Auj cos{ut+)

=

0,

or

f

=

(n/2 — +2nn)^

Thus, the time interval between successive

2ir/w.

136

maxima

Exercises 5.2

Exercises 5.2 1.

A

2-lbweight

attached to a spring whose constant

is

is

The system

1 lb/ft.

resisting force numerically equal to 2 times the instantaneous velocity.

equailibrium position with an upward velocity of 1.5 2.

A

16-lbweight

is

downward

velocity of

above

4.

(a)

below

(b) from rest

5.

(a)

below

(b) heading upward

6.

(a)

above

(b)

7.

From \x"

.

+ x' + 2x = 0,

8e" 4t

-

displacement 8.

x

=

1

-

9. (a)

5e~ 2v/3i

(l

5y/2e~ l /4

FVom x"

+

e

x

(

(a)

damped with a

x(Q)

x

=

=

heading downward

-1, and x'(0)

then

f

=

=

we obtain x = 4te~ tt - e" 4' and

8

1/4 second.

If x'

=

then

t

=

1/2 second and the extreme

feet.

=

0,

x(0)

=

=

If x'

2\/2().

0,

and

a/(0)

then

f

=

5

^

we obtain x = 5te~ 2,

= V^/i

t

and

second and the extreme displacement

+

16x

=

0,

i(0)

=

1,

and

i'(O)

=

we obtain x

=

\e~ 2t

-

\e~ st

.

*.

*

x'

= y~ st (5 - 2e 6 ') = when * = $ In § « 0. 153 second; if x = |e" 81 [e 6t = g In 10 = 0.384 second and the extreme displacement is x = —0.232 meter. 1

From O.lx" +

0.4r'

+ 2x =

(b) «

=

2 e-»^ — —7=cos4t -/5

(c)

x

=

If

then 4t

+

4.25

i(0)

0,

=

-1, and x'(0)

1

=

=stn4(

=

2tt, 3tt, 4tt,

^

...

e

- 2i

=

s i n (4t

so that the

we obtain =

e

_2t

lo)

j-cos4i

-

=

then

£sin4(].

+ 4.25). first

time heading upward

is f

=

1.294

seconds.

12.

(a)

From \x"

is

+ 1&e - 0, x(0) = 1, and x'{0) = -12 then x = -fe" 2 + fe" 8 m (ie — l) is never zero; the extreme displacement is x(0) = 1 meter.

(a)

\e~ st

-

I0x'

From i" +

=

is

starts 2 feet above the

feet.

(h)

(b) x

11.

=

x

If

-2

From \x" + -fix' + 2x x

10.

16te" 4 '. is

The weight

(b) heading upward

(a)

=

from the

starts

1 ft/s.

3.

x'

The weight

The system

2-lb/ft.

resisting force numerically equal to the instantaneous velocity.

equilibrium position with a

damped with a

ft/s.

attached to a spring whose constant

is

is

+ x' + 5x = 0,

x(Q)

=

1/2,

and

x'{0)

=

1

we obtain x

=

e^ 21 Qcos4f

+

£sin4().

Exercises 5.2

-*^

(b) x

=

(c)

x=

If

e

then 4t+7r/4

n=

for

^

+

co B 4t

=

= ^""»in (« +

gin*)

it, 2jt, 3tt,

...

J).

so that the times heading

downward

are

(

=

(7+8*1)7^

0, 1, 2,

l

.5

-.5

-1

13.

Prom -^x"+0x'+5x = Owe find that the roots of the auxiliary equation are m (a)

(b) (c)

14.

a

-

25 > If 40 25 = 2 If 4/3 25 < If 4/3

2

From 0.75x"

>

5/2.

then

=

5/2.

then

< p<

then

+ fix + 6x = 1

= — ^0±^402 —

5/2.

>

and

3i/2

we

find that the roots of the auxiliary equation are

m =-f ±|03* -18 and cj

If

15.

x(0)

=

and

Prom 40x"

x'(0)

= -2

then

+ 560x' + 3920x =

Cl

0,

cosh -
=

x(0)

0,

x 16.

From 40i"

+

1120x'

+ 3920x = T X

17.

From x"

+ 6x' + 9 = 0,

x(0)

=

0,

~

18.

x

=

then

t

=

2/3(u

-

=

=

_^

(-H+7V2)t p 6

0,

and

x'{0)

14

From x"+0x'+25x =

we

The

=

quasi-period

is 7r/2

obtain

=

2 we obtain

+ ^,(-14-Tv^)( ,

14

-2/3, and x'(O)

v >

18.

-e " sin 7(.

x(0)

2) so that

+ c 2 sinh-09 2 -18t

c2

=

x--ê- 3t + If

18*

= -3/03 2 and x'{0) = 2 we

and

=

--

v we obtain (t;

-2)(e-

3t .

2 ft/s.

see that the roots of the auxiliary equation are

2tt

h^JlQO-0 2

so that

138

=

6.

m=

—^±.\\J 100

—

Exercises 5.2

underdamped motion

19. For

x for

=

e-

M

d cos \ju A

some constants

and

(?

-

2

A2

B and

some constants

the quasi-period of 20.

The time

interval

7.

+

= Be- Xt sm

interval

\ju 2

-

A2 f

+ 7] -

A2

,

x'.

between successive intercepts

vV-A The time

fl]

The difference in times between two successive maxima is 2wf \/u/2

(w+l)7r-^

21.

- X2 1 = Ae~ u sin j^w 2 - X 2 1 +

c2 sin \jio 2

and x'

for

t

2

between successive values of __

(2n

—

nir ~~

+ 3)w/2 -

yfui

(

2

7T

- X2 ~

for

which

+

1)tt/2

(2n

y/ui 2

- A2

(15) touches the graphs of

-

y

— ±Ae~ xt

tt

22. From equation (17) (2n

+

l)7r/2

i/w

which |ir,

is

^?r,

extrema 23.

The

is

.

.

and

if

x'

is

then

(

=

v'w

2

x

of

2tt, 37t, ....

= Ae~ Xi sin (Vw 2 —

the ratio of the values at

e

24.

tt,

-

nir

4>

-A

=

2

_

tv/2

~~

Vw 2 -A 2

e~' sin((

+ tt/4)

'

is 2?r; if

x

=

then

*

=

|tt,

so that the time interval between intercepts and

tt/4.

quasi-period of x

maxima

=

2

The period

half of the quasi-period. .

-A

2

(a)

If

S

>

(b)

If

x

~ -ê" 21 sin (4t + tt/4)

is

t

and

A2 f

£

+

-I-

^

is 2-k/^/lj 2

— A2

2ir/Vw 2 — A 2 that ,

so that the ratio of consecutive

is

-A((+27r/v^Â !I)

very small then x n

is

slightly larger

then 5

=

than x n +2 and the rate of damping

27rA/\/u> 2

-

A2

=

4tt/4

=

tt.

is

slow,

Exercises 5.3

Exercises 5.3 1.

If

+

Ix"

\x'

+ 6x =

10cos3i, x{0)

=

xc

— ™ (cos 3i + sin 3f)

and x p

-2, and x'(0)

e

^2

4

(a)

x" + 2x' + 5x

If

=

and zp

=

+ C2 sin -^-f1 1

1

64

"

*

2

x

=

1

3/17

=

12cos2t + 3sin2i, z(0)

is

\

SU1 2

10,

+

T

=

5 then x c

)

-1, and x'(0)

3sin2t so that the equation of motion

+ Sm

(C ° S

=

e

_l

{c! cos

2i+c 2 sin2i)

is

+ 3 sin 2t.

l e~ cos 2t

(c)

steady-state

<*»)

then

\/47 ci cos

I

^/if

~3 ° 0S [

2.

=

so that the equation of motion

/

,

=

X

=

transient 3.

From x" + xp

+

8x'

= — jCOs4(

I6x

=

=

8sin4t, x(0)

and

0,

so that the equation of motion

I =ie-'"

x'(0)

xp

+ 8x' +

lfiz

x

As 5.

f

—

>

=

ê" 625

cc the displacement x

From 2x" Xp

=

+ 32x =

—

>

4t

(24

5e" *cos4( — 2e~

sin4t so that

x

= --cos4t +

1

7.

i

=

sin(4t

By Hooke's law

=

2t

+

- 0.219) -

and

lOOt)

te-

4t

-4t

and

-icos4t.

ai'(0)

-

=

we obtain x c

-^-e-'(24cos4( o25

0,

and

9

-sin4t 4

2,

6. Since

c2 *e

+

=

cie

_4i

+ c 2 te _4t

and

7sin4().

0.

68e~ 2! cos4t, x(0)

2

+

+

4

-'

- e sin 4t, i(0) = 0, - -ê~'cos4( - g5g e"' sin4t so that

From x"

we obtain x c = Cie~ At

is

4 4.

=

x'(0)

+

1

-e

=

>.

-!I

we obtain x c = cicos4t + C2sin4t and

cos4t-2e

i, ' !

'sm4(.

t-

ê _2t sin(4( - 2.897), the amplitude approaches ^85/4 as

the external force

is

F(t)

—

kh(t) so that

140

mi" + /3x' + kx —

kh(t).

t

oo.

Exercises 5.3

8.

+

Prom \x"

-

and x p

2x'

+

+

jfcoat

=

ix f|

20cosf, x(0)

=

and

0,

=

e

56 -2t( a [-—

+

From lOOx"

=

and xp

.

13

1600a:

=

1600 sin

St,

C2sin2t)

x(0)

=

.

and

0,

z'(0)

13

- Owe

=

£sm4t(2-2cos4i)

2

=

=

obtain x c

c\

cos At

+ ci sin 4(

then

1

-

- sin4£

-

3

o

=

htt/4 for

sin St.

=

If

(c)

Ifz'

§(l-cos4t)(l+2coB4t)

for

extreme values. Note: There are

= |cos4t-f cosgi = n = 0, 1, 2, ... at the

=

+

32 — -sint.

13

(b)

x1

e~ 2( (cicos2t

-jsinSf so that x

x=

=

obtain x c

72 .A 56 „ cos2(-—- sin2I +Tr:cosf +

13 (a)

we

sin£ so that

x

9.

=

x'(0)

£

ra

=

0, 1, 2, ...

.

Othent = w/3+nn/2 and

many

t

=

other values of

n/G+nir/2 t

for

which

0.

(d) x(tt/6

+ ror/2) - V3/2

cm. and x{ir/3

+ mr/2) -

-1/3/2 cm.

(e)

10.

If

x"

+ 1\x' + u^x =

J*b sin

xc

7* describes

=

e

-At

underdamped motion then

ci cos

\/w 2

-

A2 t

+

~

c2 sin \Jui 2

\2 t

and

F 2

4A 7 If

sin^

cos#

=

=

cj/^/c2

(w 2

+4

— 72 )

,

/\j (w

cos^ 2

(w 2

-

2

- 72 ) 2

-2A7F0

-7

+

(w

c2

/^ +

— 7 2 ) 2 + 4A 2 7 2

2

2

4A 7

)

c

2 ,

,

sin0

2

+

(w 2

-7 2

;

COS ft.

)

= -2A7 / v/(w 2 - 7 2 2 + 4A 2 72 )

then the equation of motion

,

is

F sin(7*

\j{u

11.

(a)

If g'(7)

test

(b)

=

then 7 (y 2

+ 2A 2 - w 2 ) =

shows that g has a maximum value

The maximum

value of g

is

9

so that 7 at

7

=

(V^ 3 - 2A 2 ) =

=

y/u 2

2

- A 2 ) 2 + 4A 2 72 or

7 =

— 2A 2

F /2Xy/u 2

+ 0)-

Vw 2 - 2A 2 The .

.

A2

and

.

first

derivative

u

y

.

Exercises 5.3

12.

(a)

x"

If

+ 0X1 + 3x =

and

m = \ (~0± \J0 2 — 12^; =

7 (b)

yjZ

When

-

Fa

2

=

/2

3,

,

\x"

7

15. (a)

Prom x" xp

=

+ uj 2 x =

shown

is

=

=

J^cost*, x(0)

16.

=

From x" + w 2 x ip

17.

2V(cos7i

—

—

-

7

2

+ 5sin2t =

0,

and

2

= hm

focoswt, x{0)

=

=

^

we obtain

ic

=

.

ci

coswi

+ (vjsinwi

ci

cosu*

+

and

2,

cos 7i.

sin 7* —Jo* —

Jo = ^— tsinwt.

—27 and

0,

2w

=

x'{0)

we obtain x c =

and

_lim

= — 1,

and

x'(0)

x(0)

= — cos2t — =

jt/4).

F

1

- sin2t

and

x'(0)

=

— =

=

— 2u

f

sinwt

1

we obtain i c

3

5

4

-tcos2f. 4

+ -tsm2f +

8 5sin3t, z(0)

=

x'(0)

(Fot/2ui) sinuii

+ 4x = — 5sin2t + 3cos2t, = \t sin 2t + |( cos 2t, and

From x" + 9x =

g

amplitude of xp

5\/2, the

cos iOt

— 72

'

(Jot/2w) sin wt so that x

5v/2sin(2i -

F

= —

coswi)

x 18.

so that

)

From x" xp

<

for various values of 0.

- 80/2.2-^12-4 =

uj

—

<

or

0,

when

given by

is

-5cos2t

x Jb

- 0/2 >

3

in resonance

is

then

2 and g(f)

(Focos7()/ (y 2

00 f-^w 'is 2

underdamped motion. The system

3

+ 2x' + 6x = 40sin2i

= Vu 2 - 2X2 =

is

=

xp 14.

this

1-J% then the roots of the auxiliary equation are

the resonance curve

and the family of graphs

If

<

where we require that

5(7)

13.

<

we obtain x c =

c\ cos3f.

C2Sinu;f

and

(sinwf.

=

cicos2t

+ C2sin3t,

xp

=

+ C2Sin2f,

-|(cos3(,

and

x=

2 cos 3f

—

+

5

sin 3(

—

18

19.

(a)

From cos(u — sin

w sin w =

v)

=

j[cos(u

cos u cos v

—

v)

—

+ sin

cos(u

+

sin v v)].

and cos(u

Letting

follows.

142

u

=

5 -t cos 3f 6

+

v)

=

3(7 —

— sin u sin v we obtain = 5(7 + the result

cosucosw

w)t and w

,

Exercises 5.4

(b)

5(7

lim

(c) v

t-0 2ef

'

20.

=

If e

— w)

sin et sin ft

w

=s

so that

= hm

x — (Fo/2e7)sineisin7f.

27

27

'

=

10cos7i, x(0)

0,

and

=

x'(0)

= — — [cos6( cosf —

5

—

—

coset am ft

(-.0

+ 25x =

From x"

then 7

am ft. '

we obtain

5

x — — ttt(cos It —

cos 5f)

12

sin6( sin

t

—

cos6t cos

(

—

sin 6t sinil

=

12

5

—

sin 6t sin

t.

6

Exercises 5.4 1.

Solving the differential equation q" conditions g(0)

initial

=

=

g(t)

=

initial

q(t)

2

3.

Since

4.

Since i? 2

5.

Solving

ff

g(0)

=

— L/C =

0,

2q'

+

5 and ^(0)

=

^q" +

and q(O.Ol)

«

=

0,

100g

+

—15/4 and

ci

+

—

100g

=

20t

imply

=

imply

we obtain ci

is

=

and

—

c-i

i(t)

we obtain

=

and

and

0.

—

+

c\ cos At

C2sin4t

4- 15/4.

The

Thus,

15 sin

q(t)

=

c%

i(t)

—

q(t)

—

it.

cj cos

2t/5t

+ ci sin2v^t + 1/5.

V5/50. Thus

= ^cos2v5t +

\.

underdamped.

damped.

q(t)

5 and c 2

c\

\t

is critically

=

=

The charge

e

_20t

5/2.

^5cos40t+ ^sin40t)

4.5676 coulombs.

0.0509 second.

=

60 we obtain

=

At

the circuit

the circuit

-201 e

ij'(O)

—

= ^sm2V5t +

— 4L/C = —20 <

q(t)

t as

=

conditions q(0)

16?

—- cos

2. Solving the differential equation bq"

The

4-

imply

is

as

(ci

cos 40i

+ c 2 sin40t). The

initial

conditions

Thus

^25 + 25/4

zero for the

e~ 20t

first

sin(40i

+

1.1071)

time when 40*

+

0.4636

=

jt

or

'

Exercises 5.4

+

6. Solving \q"

and

(f(0)

=

+

20tf

imply

=

300g

=

ci

we obtain

=

6 and oi

=

7.

Solving §g" + 10?' + 30g

=

q(0)

=

g*(0)

=

i(t)

=

imply c 2 q(t)

Solving

=

find e 40*

we

=

we

Be"

1/3 which implies

300 we obtain

=

c2

=

= 10-

cie~

2at

+ C2e~ m The

initial

.

conditions q{0)

=

4

-2. Thus q(t)

Setting q

=

q(t)

£

=

q(t)

201

- 2e-

<

e"

0.

3

m .

Therefore the charge cos3( + c2 sin3t)

'(ci

is

never

+ 10. The

0.

initial

conditions

-10. Thus

10e~

3t

(cos3t

+ sin3i)

maximum

see that the

and

i(t)

=

when

charge occurs

60e

3t

—

t

sin3t.

tt/3

and q(x/3)

«

10.432

coulombs. 8.

Solving q"

=

g(0)

+

lQOtf

and ^(0) q(t)

Solving

iff)

=

+ 2500
we

-0.012e- 50 '

+ 2q' + 4q = = A cos t + B sin

form yp

Thus,

150/13 and

-50*

B=

+

2B) cost

100/13.

The ,

Qp(t)

and the steady-state current

charge occurs (cos v^Sf

z(t)

when

t

=

26- 50 ' - 70te _5Ot

-I-

(35 -

2.4) sin

i

steady-state charge

=

150 -

jg- cos

f

+

— 100

=

.

= 1/35 and g(l/35)

+ sin%/3 1). The

conditions

as 0.01871.

steady-state charge has the

we

find

50 cost.

is

.

sin

t

is

*j>( ( )

=

100 -— smi + —cost. sini+ rT T3 150

.

From

and

Z= \ZX*TW we see that the amplitude of ip(t)

The Eo


=

100 and 7

=

60.

^g"

is

z aV

z*

V 11.

and


t.

,

10.

+ 0.012

we obtain yc = e~ l

(3A

A=

l^te

maximum

see that the

9. Solving q"

+

initial

+ 20g' + 1000g =

100 sin

f.

z To use Example

Then

X = L 7 -1 ^(60) Z=

yjx?

+

-5^"

= /Y2 + 400 w

and

144

13.3333,

24.0370,

3 in the text

we

identify

Exercises 5.4

From Problem

then

10,

i

p

where sin$ angle.

s=b

12. Solving \q"

+ 20^ +

we

1000?

=

has the form qp (t)

,4 sin

=

=

60*

is

4.1603(60*

=

-

is

a fourth quadrant

0.5880).

(a cos40t

p (t)

+

conditions q(0)

10^

—

1

+

{2400A

+

c 2 sin40i).

The

steady-state charge

cos 40*. Substituting into

the differential

-

160OD) sin 40*

16005) cos 60f

A=

-1/26,

13

B=

+ 400Z?) cos 40*

— cos60* + —4 1

4/17, and

sin 40*

+

1/17-

The

steady-

— cos 40* 1

1

I

D=

I

is

Id

q'(0]

(1600C

C=

-3/52,

OCi

=

+

+ 400 cos 40*.

= -^cos60* + ^sin60* +

100g

and

(400C

-

+

= -—sin 60*-

and the steady-state current i

sin 60(

200 sin 60*

qp (t)

IJ

^

cos 40*

-

if

=

imply

c\

^

sin 40*.

Li

= e- 10t (ci cos 10* + = q = -1/2. Thus

150 we obtain q(t)

q{t)

14.

«

+ B cos 60* + Csin40* + D

we obtain

coefficients

state charge

*

-0.6667 and

find

+

As

p (t)

we obtain ^(f)

{- 1600,4 - 2400B)

13. Solving \q"

0)

-0.5880 and i

Equating

+

= -XjZ and cos<£ = RjZ. Thus tan0 = -X/Rk

Now

equation

(t)f« 4.1603(60*

c2 sin 10*)

+ 3/2. The

initial

= -i e -' ot (cos 10* + sin 10*) + \

-+ oo, q(t) -» 3/2.

By Problem 10 the amplitude of the steady-state current is Eo/Z, where Z = y/X 2 + R2 and X = L*i — I/C7. Since Eq is constant the amplitude will be a maximum when Z is a minimum.

R constant, Z will be a minimum when X = = \j-jLC The maximum amplitude will be Eo/R7 Since

is

0.

Solving

£7 — I/C7 =

for

7 we obtain

.

15.

By Problem 10 the amplitude of the steady-state current is Eq/Z, where Z = \/X 2 + R 2 and X ~ Ly — l/Cy. Since Eq is constant the amplitude will be a maximum when Z is a minimum. Since R is constant, Z will be a minimum when X — 0. Solving L7 — I/C7 = for C we obtain

C=

l/i7 2

.

145

Exercises 5.4

16. Solving O.lq" qp (t)

=

+

lOq

+

A&inyt

=

100sin7t we obtain

(100-7 Z )^sm7( +

A=

Equating coefficients we obtain initial

conditions q(0)

=

1

q

(0)

~

(100

100/(100

imply

=

,4sin7(

we

=

vlsin7i

A=

we obtain

-

see that q{t)

.^v

=

'

the circuit

1

—VLC ^=

1

ci

g'(0)

=

is

,

io

i

imply

~

ci

we

3 sin?*. The is

+qP {t)

—^ ).

qa

where

find

Eacosft.

LC7 2 Thus, the EqC + cos

1

c2

VLC

1

H

1

charge

is

— EqC/{\ — LCy 2 ) and

form of qp (t)

coefficients

we

'

C2

=

iij^/LC

A—

q{t)

i(t)

\

= =

qo

=

and ci cos

and ^(0)

—ciks'mkt

5=

fct

=

=

\js/LC

71. '

EoCf

.

t

At cos kt + Bt sin kt where k

find

^cosfct.

EdjlkL. The charge

+ C2 sin io

sin

1

.

—

.

, „ „ 2 - LCf

+ k 2 q = -2kA$mkt + 2kB cos kt =

obtain

conditions g(0)

1

qp (t)

is

we

EoCt

, i

£bC

/

1

in resonance the

1

initial

equation

differential

EqC/(1 —

—VIC ==cos —j== VW

.

,

sin

q

The

-

charge

(t/\fLC)

cicos

+ c2 sin

*

and

90

5=

Substituting qp {t) into the differential equation

Equating

^j

— 7 2 ). The

—107/(100

— £7^ B cos ft =

+

and

cos

e,

1

When

Thus, qp {t)

is

i(f 1 )

18.

=

Bcos7t. Substituting gp (t) into the

conditions q{0)

The current

=

100sin7(.

7 t-cosl0t).

(cos

2

+

coefficients

initial

find

^-7sin Wt)

1 Q sill

in the text

,(*)

The

B = 0.

=

C2

1

— £7 5 ^ Equating

^

^

1

qp (t)

and

)

where

we

is

i(t)=

FVom Example

(


-7 2 )Scos7t =

and

72

+ c2 sin lOt- + qp (t)

cos lOt

1"

10Q

17.

c\

-f 2

—

c\

-

q(t)

and the current

=

q{t)

Bcosft. Substituting gp (t) into the

fct

imply

+

cj

=

—— (&£cos

fcf

+ sinfci)

=

g

- <7o* j sin &t + io cos fct + 146

sin

2k ij

and

ci

+ 02k cos kt +

(

is

—

I

io/fc.

cos kt.

The

current

is


19.

Prom

0"

+

=

160

8(0)

0,

=

1/2,

9

and

—

\ cos4f 2

The amplitude 20.

period

is 1,

is tt/2,

= 2^3

fl'(0)

+

we obtain

-^sin4t = 2

and frequency

U6 = 0then4t+7r/6 = nn or t = j(njr-ir/6) or f = |(?r/3 + mr) for n = 0, 1, 2, ...

for

+ —V

sin (^t \

6/

is 2tt.

n=

1, 2, 3, ...

.

If

^=

then 4*+tt/6

= x/2+nn

.

Chapter 5 Review Exercises since k

=

1.

8

2.

2tt/5, since

3.

5/4 m., since x

4.

True

5.

False; since

6.

False

7.

overdamped

8.

-tt/4

9.

9/2, since x

10.

ft.,

(a)

4.

\x"

+ 6.25x =

0.

= — cos4( +

|sin4f.

an external force may

=

c\

Solving fx"

cos
+

6x

=

+ C2 sin \/2fc

The amplitude

(c)

If

x

(d)

If

x =

=

1

then

then

downward (e)

i

x'{3tt/16)

for

is \/2,

= i

—

period

rnr/2

=

t.

=

subject to x(0)

x (b)

exist.

and

tt/16

+

t

cos4i

—

is tt/2,

=

for

and

sin4i

=

x'(0)

+ w/2

n =

0,

= -4 we

\Z2sin (At

and frequency

-tt/8

nn/4

1

for

1,

2,

n

+

obtain

3tt/4)

.

is 2/ir.

=

1, 2, 3,

The motion

is

upward

13.

From mx"

14.

=

even and

=

12.

11.

If x'

n

n odd.

+ 3ir/4 = 7r/2 + titt or i = 3tt/16 + nn. From \x" + \x' + 2x = 0, x(0) = 1/3, and x'(0) - Owe obtain x = |e _2t - £e _4t From x" + £x' 4- 64x = we see that oscillatory motion results if 2 - 256 < or (f)

for

then 4f

.

<

\0\

<

16.

+ 4x' + 2x = we see that non-oscillatory motion results if 16 — 8m. > or < m < 2. From \x" + x' + x = 0, x(0) = 4, and x'(0) = 2 we obtain x = 4e _2t + 10fe _2( If x'(t) = 0, then -0.2 ^ 4 094 i = 1/10, so that the maximum displacement is x = 5e .


+

15. Writing gi"

§£

=

+ sin 7*

cos yt

w = 64/3. From Example when7 = \/6473 = 8/\/3. 2

xp = A/u 2

16. Clearly

17.

—

jê

',

we

in Section 5.3

8 cos 7*

+ 8 sin yt we

see that the system

is

identify

A

e

_!

x(0)

,

=

2,

and

x'(O)

cos

2^

=

and x

= e ~«

=

restoring forces satisfy k\x\

+

t

The

fci

+

- 2W and £3 = 4W we find


(c)

conditions x(0)

initial

To compute the maximum

=

we obtain x c

^

sin

=

2v^t)

e~ 41

(c\ cos

+ ê~

2\/2t

+ €2 8^2^2*),

l .

=

From qp

=

9"

+

104 ?

y% sin 50f

(a)

9=

(b)

i=

(c)

g

=

,

=

x zj

=

k

=

1

_

fcl

k{x\

+

X2)

=

k\x\

x'(0)

=

+

J_

J_

C2 sin

when

+

=

2/3 imply c\

1

and

C2

=

+ |coB2^t =

0,

and

and g/(t)

= ^4§+| =

+

i-

we obtain qc = cjcoslOOt

+

|a/(t)|

=

\\l*9

C2 8inl00t,

§cos50t, and

L dt 2,

-

is

lfyfig.

+ ^ sin 50t,

sin50t(l

4mg/$.

2^|t.

and

-|cosl00t

The

*2

then becomes x" (

fcj-

we have

= I/2W+1/4W - Z/4W. Then k = 4W/Z = + (4g/3)x = 0. The solution

1/Jfc

ci cos

and

100sin50t, g(0)

sin lOOt

20. Differentiate

1

From

{k\[ki)x\.

speed of the weight we compute

i , (0-2y|aln2^|( 19.

~j^

mx" + kx = x(t)

The

=

^2x2 so xa

k ^xi

From

and

a state of pure resonance

in

18. (a) Let k be the effective spring constant and x\ and 22 the elongation of springs k\ and

(b)

=

suffices.

From lx" + x? + 3x = xp

4

+ ^x =

form x"

in the

cos50t)

+ R ~ + ~g =

= i?(t)

or

t

=

jmt/50 for

and use ^{t)

dt

148

—

71

=

i(t)

0, 1, 2, ...

.

to obtain the desired result.

.

6

Equations with

Differential

Variable Coefficients Exercises 1. 2.

The The

auxiliary equation auxiliary equation

6.1

is

m 2 - m — 2 = {m + l)(m - 2) =

is

4m — Am +

m =

2

2

=

1

(2m —

=

l)

=

2

so that y

=

so that y

=

cix

1 '' 2

+ C2X 2

1

c\x

+ cax

1 '' 2

.

In 1.

+ c 2 lnx.

3.

The

auxiliary equation

is

4.

The

auxiliary equation

is

5.

The

auxiliary equation

is

m 2 — 2m — m(m — 2) = so that y = cj + C2X m 2 + 4 = so that y — c\ cos(21nx) + 02 sin(21nx).

6.

The

auxiliary equation

is

m s + 4m + 3 =

7.

The

auxiliary equation

is

8.

The

auxiliary equation

is

m 2 - 4m - 2 = m 2 + 2m - 4 =

9.

The

auxiliary equation

is

25m 2 +

10.

The

auxiliary equation

is

4m 2 —

11.

The

auxiliary equation

is

12.

The

auxiliary equation

is

13.

The

auxiliary equation

is

m 2 + Am + 4 — {m + 2) 2 — so that y = cix~ 2 + c 2 x~ 2 lnx. m 2 + 7m + 6 = (m + l)(m + 6) = so that y = cix -1 + C2i~ 6 m 2 - 2m + 2 = so that = x cos(lnx) + casinflnx)].

14.

The

auxiliary equation

is

m 2 - 8m + 41 =

so that y

=

15.

The

auxiliary equation

is

3m 2 + 3m + 1 =

so that y

= x~ 1,/2

16.

The

auxiliary equation

is

2m 2 -

17.

Assuming that y

=

so that y

c\

2

.

1

1

=

(m + l)(m

so that y

=

c\x~

l)(2m

+

so that y

x"1 and substituting into the

m(m -

c\ cos

l)(m

-

2)

-

6

=

^

1+ly

(| In x)

=

1)

c\x~ l

+ cax~ 3

4-

c 2 x~

+ C2

so that y

1

"^,

(5 In x)

=

c^x 1 / 2

+ C2X~

=

x4

[ci

x 1 ^4

cos(51nx)

= x m and

m(m -

|ci

[ci

lnx)

cos

cos

(-^lnx)

-f

c^sin

+ qjsin

equation we obtain

m 3 - 3m s + 2m - 6 -

(m - 3)(m 2

+c 3 sin(\/21nx)

substituting into the differential equation

l)(m -

2)

+m-

1

=

ro

3

.

.

+ ca sin (5 lux)].

+ 2) =

0.

Thus

18. Assuming that y

1 ^2

[ci

differential

(\/21nx)

.

"^ + c2 x 2+ A

cjx

Ji

m+1 =

2

=

—

—

so that y

so that y

so that y

= (2m —

+ 3) =

- 3m 2 + 3m -

1

.

we obtain

= (m -

l)

3

=

0.

lnx)].

(-^lnx)].

Exercises

6.

Thus y 19.

Assuming that y

m(m -

l)(m -

— xm

c\x

and substituting

- 2m(m -

2)

=

+ C2x\nx + C3i(lnx) 2

we obtain


m

- 2m + 8 =

1)

.

3

2

- 5m + 2m +

8

= (m +

l)(m - 2)(m

-

4}

=

0.

Thus

20. Assuming that y

m(m -

= xm

l)(m -

2)

and substituting

- 2m(m -

-1

=

y

cix

+ C2X 2 + C3X 4

.

we obtain


+ 4m - 4 = m 3 - 5m 2 + 8m - 4 = (m -

1)

l){m

-

2

2)

=

0.

Thus y 21. Assuming that y

m(m-

= xm and

l)(m- 2){m-

3}

=

c\x

+ C2X 2 + C3X 2 In x. we obtain

substituting into the differential equation

+ 6m(m - l)(m-

2)

^ m* - 7m 2 + 6m = m(m - l)(m- 2)(m + 3) =

0.

Thus

=

J/

22. Assuming that y "

= xm

+ C2l + C3X 2 + C4X

Cl

and substituting

3 .

we obtain


m(m-l)(m-2)(m-3)+6m(m-l)(m-2) + 9m(m-l) + 3m+l =

m 4 +2m 2 +l = (m 2 + l) 2 = 0.

Thus y 23.

The

=

c\ cosfln x)

auxiliary equation

is

initial

=

ci 4-

C2%~ 2

and

24.

The

=

2, C2

=

-2, and y

auxiliary equation

is

m

2

y

The

initial

— — 2cix~ z

y'

.

+ c2 = -2c 2 =

ci

+ C4 In x sin(lnx).

conditions imply C!

Thus,

cos(m x)

a:

m 2 + 2m = m(m + 2) = 0, so that y

The

+ C2 sin(ln x) + C3 In

=

2

-

2x

— 6m + =

cix

2

8

4.

-2 .

= (m — 2)(m —

+ C2X

i

and

4)

y'

=

=

32

conditions imply 4ci

+

4ci

+ 32c2 =

16c 2

150

0.

=

0,

2c\X

so that

+

4c2£

3 .

Exercises

Thus, 25.

The

ci

=

16, c 2

-2, and y

auxiliary equation

y

26.

=

=

is

m2 +

initial

The

auxiliary equation

is

m

y

=

cix

initial

conditions imply

(-oo,0).

c\

-

=

so that

0,

=

1

and

- 4m +

2

2

conditions imply c\

In the next two problems

1

l&x 1

2x A

+ C2Sin(lna:)

C!COs(lni)

The

The

=

C2

and

=

+ C2X 2 lnx

=

we use the

.

X

+

=

2

2)

=

= —x

t

+ 2sin(tna;).

+ C2(x + 2x\nx).

2c\x

=

Thus y

3.

X

cos(lni)

so that

0,

=

1

y

10

substitution

—

Thus y

and

5 and C2

= -ci-sm(lni) + C2- cos(lni).

y'

2.

= (m -

4

5x 2

-

7x 2 ln;z.

since the initial conditions are

on the

Then dy

dy dx

dy

dt

dx dt

dx

and dt 2

27.

The

differential

equation and 4(

The

dt\ dx)

dt\dt)

auxiliary equation

is

2

initial

initial

conditions imply

y 28.

The

=

2t»


a

c\

=

1

2 and

= (2m —

1

y'

=

initial

conditions

_. _„__,_„„_

-4.

[=1 2

l)

\c

=

y'lt)

x

=

0,

so that

t'^ + c 2 (t^ 2 + \t~întj

1

'2

- 5(-x)V 2 ln(-x),

auxiliary equation

is

m 2 — 5m + 6 = y

The

initial

=

c\t

2

=

y'(t)

S,

tfw

Uj

(m — 2)(m —

+ C2t

3

and

y'

3)

=

conditions imply Ar,

*

<

become

t=2

The

dx 2

dt

+ C2 = — 4. Thus

-KÎni = 2(-x}

and

2,

[=1

and

dx 2

dx dt

=

y(t)

t,

4m 2 — 4m +

[y)

become

conditions

y-citW + C2t 1/2 lnt The

dt

^f + = 0;

6.

-\-

Rm —

8

=

0,

2cit

so that

+ 3c2f 2

.

0.

0.

.

'

interval

1

Exercises

6.

from which we find

c\

=

6 and

—

c
=

y 29.

The

auxiliary equation

is

m2 —

—2, Thus 2

6t

-

2t

=

3

so that yc

=

6x 2

In

1

<

a:

,

+ ci In x

c\

=

W(l,lnjs)

2x 3

+

0.

and 1

a:

1/x Identifying /(x)

30.

The

=

1

we obtain

y

—

c\

auxiliary equation

= — xlnx

=

c\

+ C2lnx + ^x 2

—

c\

is

m a — 5m — m(m — 5) =

x 3 we obtain y

31.

The

auxiliary equation

=

c\

Uj

= — 5X 4

+ C2X 5 —

and

so that yc

x

25

is

2m 2 + 3m +

1

j/

=

cix

1

—

S we ol5ta^ n u

'i

=

=

The

x" 1

is

112

—

xeX

—

=

m 2 — 3m + 2 =

x 2 e T we obtain u\

The

u\

—

«2

=

5 lux, and

x -2

_J r

=

so that yc

=

c\x~ l

2

—

x3 ^ 2

cix""

1

~~

I '^2 Then ui -

=

jx 2

+ cax-W - \x + ^-x 2

x

x'

1

2x

2)

and

=

so that yc

=

x2

—

xe x

—

c\x

+ C2X 2

auxiliary equation

=

cix

+ c 2 x 2 - x3 e x + 2x 2 e x -

=

c\x

+ c2 x2 + x 2 ex -

is

to 2

— 2m +

1

= (m —

W(x,xlnx) -

l)

.

Then

uj

=

—xê?

+

2xe x

+ x3ex - x2 ex

x

2

.

=

1

152

so that y c

xlnx

x 1

,

.

+

In

x

and

.

<

2xe

— jx 3

15

(m — l)(m —

— —x 2 e x

C2X~*/ 2 and

-3/2

6

=

4-

eX

y

33.

and

.

Then

1)

x-'/

5

lV(x,x 2 } Identifying /(x)

+

l)(m

+ c 2 x -V2 + I T - It 2 + \x 2 -\x =

auxiliary equation

5x 4

l/5x.

£ — x 2 and u 2

I

32.

.

5

= (2m + ,

;

,

&

5

)

=

=

u'2

+ C2X 5

= 3X 2 and

1x2

—x s + 7X S inx = ci + C3X 5 + ^x 5 lnx.

W(x-\x-V 2 = Identifying f(x)

— ^x 2 lnx,

+ C2lnx + ^x 2 — ^x 2 lnx + ^x 2 lnx =

x.

5x 4

=

jx 2

ui

and u 2

1

Identifying /{x)

=

Then

u'j

=

cjx

+ C2ilnx

and

2xe x

—

2e

x ,

Exercises

—

Identifying f(x)

2/x we obtain u[

= -2lax/x

and

x(lnz)

=

2/x.

+

2x(lna;)

u'2

Then

m

-(lni) 2

=

,

112

=

6.

2\nx,

and y

34.

The

auxiliary equation

Identifying f(x)

U2

— x\nx — y

x,

=

c\x

+ C2x\nx —

=

c\x

+ C2x\vlx + x{lnx) 2

m 2 - 3m + 2 =

is

= xinx we

and

cix

=

+ C2^ 2 +

we use the

i

=

- l)(m -

.

=

2)

X'

1

2z

=

and u 2

- -x 3 lna: + x 3 lnx - x 3 =

3

so that y c

x

11

-rx

2

=

c\x

+ c^x 2

and

.2

X

= — zlnr

obtain u[

*4

In Problems 35-40

(ro

2

2 .

=

cja:

Then

Ins.

3

1

*±

£.

+ cix 2 - -x 3 + -x 3 \nx.

£1

following results derived in

= |x 2 — ji 2 lni,

i/j

Example 6

When

the text:

in

x

=

e*

or

lnz,

dx~

35. Subsituting into the differential equation

The

auxiliary equation

undetermined

is

coefficients

y

m2 + 9 + 8 we

=

try yp

cje"

1

x2

dx 2

x dt

dt

we obtain

= (m +

= Ae

dt 2

2t .

l)(m

+

=

8)

=

so that yc

This leads to 30Ae

=

2t

e

2( ,

cie

so that

+ c2 e- 8( + ^-e 21 = c^" + C2X~ S + -^z 2

-t

+ A=

C2e~ 8t

.

Using

1/30 and

1

.

36. Subsituting into the differential equation we obtain

The

auxiliary equation

undetermined ,4

=

1/3,

B=

is

coefficients

5/18,

m 2 - 5m + 6 = we

try yp

=

At

(m - 2)(m -

+

B.

3)

=

so that yc

This leads to (-5.4

+

=

QB)

and y

=

cje""

+ c 2 e Jt + -(+ J

— = ci^ + 1(5

c 2 :r

J

+

- lnx O

+

— 18

cje

+

21

6At

+

C2e

=

3t

2f,

.

Using so that

1

Exercises

6.

37. Subsituting into the differentia! equation we obtain

The

auxiliary equation

coefficients

we

try yp

m 2 — 4m+13 =

is

= A + Be

l .

so that yc

=

2

e '(ci cos3f+c2sin3i).

+ lOBe = 4 + 3e', 1

This leads to lZA

so that

Using undetermined

A=

4/13,

B = 3/10,

and y

=

e

=

x2

zi

+ C2 sin 3t)

(ci

cos 3t

[ci

cos(3 In x)

4-

—+—

e'

+ C2 sin(31nx)j +

+ io

38. Subsituting into the differential equation

^-x. 1U

we obtain

2m 2 — 5m — 3 = (2m + l)(m — 3) = so that yc — cie - '/ 2 + C2e 3t Using undetermined coefficients we try yp = A+B^+Ce 21 This leads to -3A-6Be -5Ce 2t = l+2e'+e 2t so that A = -1/3, B - -1/3, C = -1/5, and The

auxiliary equation

is

.

t

.

,

y

=

c\e~

t/2

+ c2 e3 '

ê*

- \e 2i =

c\x~

1 ^2

+ c 2 x 3 -\x- ^x 2

.

39. Subsituting into the differential equation we obtain

The

auxiliary equation

undetermined

is

coefficients

m 2 + 8m — 20 = we

try yp

{m + 10)(m —

= Ae~

st .

2)

This leads to

=

so that yc

-35Ae~

st

=

=

5e~

and y

40.

=

Ci e~

m+

c2e

2t

- \e~ 5t = ax' 10 + c2 x 2 - ±x~ 3

From dx 2

x2

l

dt 2

154

dt

c\e~ iat

.

31 ,

so

+ cê 24 Using that A = -1/7 .

Exercises

it

6.

follows that

dx 3

x 2 dx [dt 2

~

x2

dx\

x2

dt 3

dt

dt 2

x 2 dx

J

dt 2

d2y

dt 2

x3

dt

+

x3 dt

x 3 dt

2

dt)'

we obtain

Substituting into the differential equation 3

dt

[ dt J

x 2 dt 2 \x)

\x)

x 3 \dt 3

d y

x* \dt 2

j

dy

dy\

fd^y

dy

or

d3 y

The c$e 3t

auxiliary equation .

A=

-17/12, y

The

m 3 -6m 2 +llm-6 = (m-l)(m-2)(m-3) =

Using undetermined

so that

41.

is

=

BI

cie*

=

coefficients

we

try yp

= A + Bt.

T* !t

m

1

m = m(m +

uq and u(b)

=

m yield the system Cia~

— = /iio-ui\ ab \

,

Thus

b

,

u(r)

=

The

auxiliary equation

and u(b)

—

is

m2

—

ui yield the system c\

Cl

=

u\ In a

^H ~ J t

— «i \

~Tb-a

C2 In

c2

a

=

=

ci

uo, c\

„M

U{r}

~

t

Ul lnq

-"0 lnfc

— uq In 6

ln(a/6)

+ ,

=

+C2

=

u\b

-"l MVfe)

=

cir

uq, cib"

— —

1

-1

+

,

In

a:.

C2-

+C2 =

The boundary

«i- Solving gives

uqo.

a

u ib - «oo

—

b-a ii

+C2lnr. The boundary conditions u(a)

+

C2 In 6 ,

and

ln( a /&)

Th„ hUSB

i

o

so that u(r)

+

and

}

Ziifl

\

so that u(r)

.

,

—a

=

1)

1

9

•>

+

ci

42.

This leads to

17 17 — - -t = cix + c^x* + c%7? - — - -

9*

2

is

= c\e +C2e 2t + {HB-6A)-6Bt = 3+3t, so that yc

-1/2, and

+ c 2 e' + c3 e M -

auxiliary equation

conditions u(a)

f.tPy ... dy

"o

_ 111

=

C2=

u\. Solving gives tto

—

u\

M^}'

_ «oIo(r/6)-ui ln(r/q) ~ " tafc/fr)

=

uo

1

Exercises

43. Letting

6.

iî-lwe obtain dy

=

dx

dy dt

and 2

_

d y dx 2

d dy dx


The

auxiliary equation

is

m 2 — 3m — 4 = y

44. Letting

t

=

3x

+4

=

cii

4

d?y dt

_

_

dt 2 dx

dt

d?y dt 2

we obtain

(m — 4)(m +

+ c 2 f-' = d(x -

I)

=

1) 4

+

so that

c2 {x

-

l)"

1 .

we obtain

^-^

-

s

dt dx

dx

— dt

and dx 2

dx\


dt 2

dt

The

auxiliary equation

is

y

45. Letting

i

= i + 2we

'

we obtain

air

9m + 21m + 9 = 2

dt 2

dx

dt

3

(3m 2 + 7m

+ 3) =

so that

(-T+^)/6 + ^((-7-^1/6

_

cl(

=

ci(3x

+ 4)<- 7+ ^>/ 6 + c2 (3x + 4)(- 7+ ^>/ 6

.

obtain

dy dx

_

dy dt

and

£l2 ~ dx

d2y (d}L\ - d2y — ~ ~ dx \dt ) dx dt

dt 2

dt 2

'

Substituting into the differential equation we obtain

The

auxiliary equation

y

=

c\

is

m2 + =

cos(in t)

1

so that

+ ci sin(ln () = a

cos [ln(i

156

+ 2)] + a sin [ln(x

^ Exercises 6.2

Exercises 6.2 1.

=

lim

n—too

On

x n+1 /{n lim

n—too

+

1)

=

lim

c"/Vi CO

The

= — 1,

on (-1,1). At z

aeries is absolutely convergent

^ -n

the series

j

is

the harmonic series

n=l 00

which diverges. At x

=

1,

given series converges on (—1,

On+1 2.

lim

n—too

=

^ (—1)"

the series

converges by the alternating series

z n+1 /(n+l) 2

=

lim

74—t OO

X = J

lim

on {—1,

series is absolutely convergent

alternating series test.

At x

=

Thus, the

1].

00

The

test.

1,

At x

=

-5

a convergent

1).

the series

is

—1, the series

—— (— 1}"

p-series.

converges by the

Thus, the given

series

rt=l

converges on [—1,

3.

1].

p±l = n—lim too

Urn

2 n+l x n+l /(n+ 2"l n /7l

1)

2n

—

iim

111

=

21*1

00

The

<

series is absolutely convergent for 2\x\

or

1

|x|

<

—

At x

1/2.

—1/2, the

series

^

(—11* fc

fc=l

00

converges

by the

alternating series test.

At 1

=

j

^—

1/2, the series

is

the harmonic series which

fc=i

on [-1/2,

diverges. Thus, the given series converges

On+1 4.

lim

n—too

The

On

lim

rt—too

The

lim

n—too

5n+1 i" +1 /(n + 5 n z n /n!

series is absolutely convergent

On+1 5.

=

On

=

lim

l)!

series is absolutely

(a:

Iim

"-co

on (-00,

(x-3) n+1 /(n +

n—too

—

l)

- 3)V«3

convergent for

1/2).

n+l

\x\

=

00).

3

=

|x

Urn

—

3|

f-^-l <

1

|i-3|

= |x-3| At x

or on (2,4).

=

2,

the series

£ —n -

n=l 00


test.

At x

=

4,

the series ti=i

the given series converges on

[2, 4].

~ j

'

s

a convergent

p-series.

Thus,

Exercises 6.2

On+1 6.

=

11m

n—-oo

(x

On

+

7)" +1

n /Vn+T = lim oc V n + n— + 7)"V"

lim rc—*oo

(x

|3

+

=

7|

+ 7|

|x

1

V"

(-1)" —

p-series.

Thus,

00

The

series

absolutely convergent for

is

\x

+ 7| <

OO


At x

test.

—

=

on (—8, 6). At x

or

1

—6, the series

j —=

^2

is

—8, the

series

a divergent

jL-

the given series converges on [—8, —6).

On+1 7.

lim

n—too

The

—

On

(x

-

5)" +1 /10n+1

(2

-

lim

n—-oo

—

5)"/ 10"

series is absolutely convergent for

^ (_— 00

series

l)

k

-

^(—1)*

(— 10)*

—

—

lim

n-ooio |x

—

5|

—

|x

5|

1

=

<

\x

1,

—

|x

-

51

10'

1

—

'

<

5]

on {—5,

10, or

15).

00

1

diverges

by the n-th term

At x

test.

=

15, the series

—5, the

(— — — = ^— °°

^

=

=

At x

l)*10 fc -j

diverges by the n-th term test. Thus, the series converges on (—5, 15).

it=i

8.

«n+l

lim

=

On

(n

lim

n—>oo

l)(x-4)" +1 /{rc + n(i-4)™/(n + 2) 2

+

3)

2 ..

= hm

(n+l)(n + ,

x(n

+

2)

2

-4 = i-4 111 ,

„,, 2 3)

x

1

£ ^7— 00

The

series is absolutely convergent for \x

—

4|

<

or on (3,5). At

1

x —

00


comparison

On+1 9.

lim

n—»oo

The

with

^—

(n

lim

n—'oo

+

.

—"OO

=

On

n+1 x n+1

=

0.

nx" +1 /{n

+

l)

-

2 <"+ 1 >

—

l)i n /n 2 "

^

—

the series

1)1x1 '

1

1

=

00,

nn In (n-l)(n +

lim

1

'

2n

=

lira

"-«(n-l)(n+l) 2

Vti

+

x|

I/

1

—

1

lim

«—»(u-l)(/i+l) 2

(^)T =

lim n^(„-l){ n+

l)2

[(l

+

l/ n )n]2

Ixl

158

=

-Ax\

=

-r^

diverges by the limit

#

l) 2»+ 2

(-l) fc k

fc

converges on [3,5).

1

x

(n

series

the series

5,

= n— lim 2
n\2 n x n

lim

n—too

Thus, the

l)!2

series converges only at

lim Ti

—

On

fln+1

10.

test

=

At x

test.

3,

" Exercises 6.2

The 11. e

, *

x1

00).

4

x3

x T -+

nx=|l + x +y + ,

S1

on (-00,

series is convergent

3

2

...j

-

4

\ /

.

, w 13. sinxcosx .

—

x ,

„

—— x3

x5

^

/

504Q

12Q

g

X'

5

7

4

xe

3

_ 16. 17

a

1-

x x — +2 3

Î= ^ tan

'

r 4

2

\

x3

6+

=

cosx

~2 e*

+ e- =

(l

+«+

-b

x5

x7

120

5040

1

—X

4

x2

z3

2

o

2

+ e- x

„

x

4

x

1

19-

x2

1

x4

x6

-T + T"T + -

^ \

=x

,

5

+

,

15

I?* 7 315

+

'

_6

.

x5 ~ —

x6

120

'

720

X6

X4

1

ex

—

+

^ + 2*

'

24

6

•

720

X4 x

3

1

\ h

—+

^+^+

+

X6 x

105

3

—5

T4

+

T T s [ 1

—

44x 8

12

~3

+

4

X3 x

2x 6

"

720

24

~

6

+

2

12

1

i2

5x 4

61x 6

2

4

48

1440

24

—_ x3

x5

4

3

H

llx 4

+

+

_...)

— + Xx24

6

—

x^xjT+

+

2

18.

,

= l-x 2 +

'-/

X

X2 x

3

/

+

+

X2 x

4

,

X4

,

+

120

2

2x 23x — + — - — + V =z*- — +— 45 7

x-

15.

\ "

\ X X T 5? —. =!-, + ___ +

—

1

W

*3

^

,

X2

l-

W ^

-r h

X5 x

"

z

x7

h -r

—

+ +

-7T

6

X - X + X -... -„=|l-x + T T s

12. e

X3 x

\ (

+

+

+ '"

2

x4

x6

x

2x3

...

2x 5

4x 7

3

15

315

3x 2

4x 3

1--

1

4

" Exercises 6.2

21. Separating variables we obtain

— = — dx

=4- ln|d

= — x + c =>

y

=

c\e~ x

.

V Substituting y oo y'

+ V=J2 n=l

= ££Lo

W

1

"1

CnX n into the oo

+

£ CnX" «=0

equation leads to

differential oo

oo

+

l)ck+l x

k

+

fc=0

oo

£

CfeX*

= £[(* +

A:=0

ft=0

0, 1, 2,

Iterating

l)Cfc+l

+C

fc

^

k—n

it=n— 1

Thus

{k

+

l)cjt + i

+

Cft

=

and

ck +i

=—

for

+

1

Cl

-

-Q>

C2

- ~2 C1 =

C3

= "31 C2 =

Ci

1 = - c3 =

,

k

1

and

=

Ar

we

find

r>

5

1

1

1

24*

so on. Therefore

y

=

co

- cqx +


1 1 2 i -cqx - -cox 3

I

1 + ^cqx*4

=

cq

l

~

24

6

x+

1

x

1

Q

2

,

~r

+

1

x 2i

we obtain

— = 2dx

=^>

ln|j/|

=

2x

+c

«

=

3 cie *.

1/

Substituting y

= E£L

CnX

n

into the differential equation leads to

*

t/-2y='£

£ c«x" = £

f;[(fc

+

fc=0

k=n

fc=n-l

=

2

n=0

n-l

l)cfc+1

-2c

t fc

]x'

=

*=0

160

0.

(*

+

!)<*+!**

~

2

£ k=Q

ck x

k

— 4

]x

fc

=

0.

— Exercises 6.2

Thus

{k

+

l)c fc+1

-

=

2ck

and Ck+\

= T~T7 ch> ci

=

C2

=

c3

=

=

for

we

Iterating

0, 1, 2,

find

2co

§

=

2°°

c2

=

-co

C3

=

24°°

16

2

«=

4

and so on. Therefore

=

y

co

+ 2cqx + -cox* + 2

=

co [l

+ 2x +


we

^(2x)

2

-cox"

+

6

— cox* + 44

l

+ -{2xf + -^(2x) 4 + --]= o

=

lnjj/|

=

- x 2 y=

=

ci

=

C2

=

3

00

E

+c

0, (fc

ci

+ 2c 2 x +

oo

E ^" +2 = E

-

oo

+ 3)c t+3 x i+2 -

E cfcx^

3

*=n

fc=n-3

Thus

^

3,n ' nto tne differential equation leads to EJJLo cn

oo j/

-

3

1/

i/

2l

obtain

— = x ax Substituting

£ i(2*)» = ^

£

"

+ 3)c* +3 - c^*" 2 = 1

[(*

+ 3)c,fc + 3 — Cj; =

— —

and

C3

=

3~C0

C4

=

C5

=

1

c7

=

eg

-Cf., for

1

=

1

0.

k

=

0, 1, 2,

Iterating

we

find

Exercises 6.2

and so on. Therefore V

=

co

=

e-o

+

J jcox°

1

1

x3

+

T

+

1

* 6 pCoZ"

,

+

^ 1

/x 3 \

2

T

—

1

-^CQX 9

1

+

2

fx 3 \

1

+ 2~3

(t

^-

= -x 3 dx =>

ln\y\

=

y

=

~x 4

+ c =>

4

y

= ce" 1 ^4

.

xn ' nto tne differential equation leads to Y.™=o °n

£ n=l

+ xHy =

=

ci

CO

QO

OO

*/

'

'

we obtain


Substituting y

+

+

nc^"" + 1

£ c^

3

£

=

n=0

00

(*

+ 4)c* +4 x* +3 -

fc=-3

£

ck x

k+3

it=0

+ 2c2 x + 3C3X 2 +£[(* + 4)cfc+4 + c

fc

jx

fc+2

=

0.

it=D

Thus

ci

=

C2

=

C3

=

(k

0,

+ 4)c k +4 + ct =

and c*+4

= ~ fc

^4

&

c fe'

=

0, 1, 2,

find 1

C4

= --CO

C5

=

c

= -8 C * = 2-4^°

C6

=

C7

=

1

8

Cg

=

Cl2

=

1

1

= Cn =

Cio

= -_._<:„

and so on. Therefore 1

1

t

CO

1

~T


s

1

i/4\2

4

=

1

+

2V4

2

3

/

,

3

12

1

V

4

we obtain dy y

_

dx 1

-x

In

|y|

= — In |1 —

162

x\

+c

y

=

l-x'

Iter)

Exercises 6.2

Substituting y

= J^L

cn xn into the differential equation leads to oo

(1

- x)y'

oo

-y="£ nenX*n— 1

oc

E n=l

~

1

_

~

EW n=0

k—n

*=u-l

k—n

oo

00

= £(* +

£

"

lJOfc+lZ*

n

oo

£ cti*

-

fcCfcX*

OO

Thus

=

ci

ci-co

=

+

(ft

0,

iJpfc+i

-

(*

-

+

l)cjt

l)c*+i

(*

+

-

o-

=

a = co

and

=

Cjfc+i

Iterating

and so

EP+

-^+

we

c fc

=

A

,

1,2,3

find ci

=

co,

£>;

=

Cl

=

Co

C3

=

c2

=

Co

on. Therefore

y

=

co

+ cox + cox 2 + cqx 3 +


=

co [l

+ x + z2 +

i

= Eô

+

•

•]

=

co

=>

In|tf|

=

21n|l+a:|

+ c =>

y

=

E

ci(l

~\~

c^x" into the

differential

OO (1

3

= 7^-

we obtain

—y - T-7—d x x Substituting y

•

+ x)y' - 2y =

equation leads to oo

CO

E

tkwe"-

1

n=l ^

+

E

-

^

k—n

oo

l)c fc+1 s*

+

fc=o

C!

£ fi=0

k=n

oo

=

2

n=l

Jc=rc— 1

- E(* + ~

oo

E ***** - E 2

jfc=0

jfc=l

2co

+

+ z) 3

£ K* +

l>«*+i

+

-

2)c*)s*

=

0.

.

Exercises 6.2

Then

-

ci

{k

and

—

ci

+

2co

=

+ (k-2)c k =

l)ck+1

2co

—

k 2 = - jT^J c *>

Cjfc+i

Iterating

we

=

ft

1, 2, 3, ...

.

find

ci

=

2co 1

and

The

=

0c 2

=

c4

=

co (l

+

so on. Therefore

=

27.

cA

auxiliary equation

is

co

+ 2coi + coi 2 =

2z

+ x2 ) =

m 2 + = 0, so y — c\ cos X + C2 sin £. 1

cq(1

+ 1) 2

.

Substituting y

—

^ n=0

differential equation leads to oo

oo

k=0

k=0

oo

oa

n=2

n=0 (

k=n-2

k=n

oo

=

O* +

2 )(*

+

l ) Ck+2

k + Ck\x =

0.

fc=0

Thus

and

(k

Cfc+2

=

+ 2}(fc +

l)cfc +2

___i__

Cic

164

,

+c fc

fc

=

,l,2,.

n

c„3:

int

Exercises 6.2

Iterating

we

find

C2

1

=

Co

_

2 C3

1

—

C

3-2 i

and

=

CQ

=

CO

+ ClX -

~

C2

2^

4-3-

1

1

fc5 L

—

C6

=

-eV

c7

-

"7^ C5 =

5-4.3.2

=

4

<=1

1

-ei* 1

1

o

auxiliary equation

is

1

-cox

2

-

1

,

1

-(agr + «iE

m2 —

1

=

0,

1

— cix J + —^ cox 4 1

+ C1

-**i;

The

4-3

so on. Therefore

!/

28.

1

1

a=

so y

=

(2n

cje 1

Is

1

=coc OS * + Cie n *. i

+

+

,

1)

,

C2e s

Substituting y

.

= ^2

differential equation ieads to

n=2

CO

00

00

n=0

k=0

k=0

fc=n

= £[<* + 2){fc +

l)Cfc+2

"

ck \x

k

=

0.

fc=0

Thus

{*

+

2)(*

+

l)£* +2 -cfc

=

0,

*

=

0,

1,

2,

Cn 1"

'

nto tne

Exercises 6.2

Iterating,

we

find 1

CO

1

ci

C2

1

C3

Ci

—

and

2

=

co

+ c\x +

C2X

=

co

+ c lX +

\cox

1

r

co

=

The

=

3

+

i

-

1

=

Cl

7-6-5-4-3-2

7-6

'

so on. Therefore,

y

29.

Cy

1

*

+ 2

+

1

2

C31

.

^x + ^cox + ^^J^* + 3

4

S» is

1

r

°°

+ Cl

4

1

2

auxiliary equation

.

1

5

x2n+1

m 2 — m = m(m — 1) = 0,

5

i

1

=

00 C03h

=

so y

00

-

v'

=

E

00

n(n -

£

iKx«- 2 -

n=2

ci

1

W

00

= £<* +

,

Substituting

= £[(* +

-1

00

2)(*

2)(*

+

l)c fc+2 z*

+

l)c* +2

-

-

(fc

£(* + l)c fe+1 x +

l)cfc+1 ]z

it=0

Thus

(k

+ 2)(k +

M ~{k +

l}c

166

.

+ Cl 9inh x

+ C2C 1

1

n=l

.

1

into the differential equation leads to

y"

.

l)c k+i

=

fc

=

fc

0.

3/

=

Cn^"

0.

Exercises 6.2

and

c*+2

=

we

Iterating

=

*

cfc+i,

+ 2)

(k

0,1,2

find C2

1

=

C1

2 1

-

C3

C2

=

C3

=

1

3-2

3 1

=

C4

1

Cl

i

5i

and so on. Therefore 1/

=

co

+ c\x +

1

2

-ctx*

cq

-C + The V

t

gj-cia:

=

l

1

+

jjCi^

1

+ x + -x 2i + -x 3i + is

4

+

4

+

+ ci

Co

1

ci

auxiliary equation

— £nô

1

+

1

-x-'

2m 2 + m = m(2m +

.

=

1)

0,

=

so y

+ cue

ci

x^2 .

Substituting

CnI ™ int0 tne differential equation leads to 2y"

+ y' -

2

£ nc^xn-1 £ n(n - lW" + n=l 2

n=2

=

2

£{ft +

2)(fc

+

l)cfc+2 T

+ £)

fc

*=0

=

£

(Jfc

2(k

and Iterating

Ck+2

(2(fc

+ 2){fe +

we

+ 2)(k + =-

\)ck+i x

k

k=Q

+

l)cfc+2

(t

+

l)c i+I ]x

Jb=o

Thus

+

l)ck+2

+

{k

+

1

2{k

+

Cfc

k

+ i,

l)ct + i

=

0,1,2,.

2)

find C2

C3

C4

= =

1

"2

1 C1

2

1

"2

C2

=

3

1 1 - ~--c 3 =

1

1

22

3-2

1

1

2^4!

=

C1

Cl

fc

=

0.

2

Exercises 6.2

and so

on. Therefore 1

1

1

2

Co ci

2

x

= Co + Ci =

-2 +

l

- -CiaT

222

2 2 2

'

/x\ /xv2

1

4

= Co — 2ci = -iCi

1 1 1

1

= C + Ci - -Cii + -

1

3

/x\ J3

1

2U) -3iU)

-Cix 3

+

3!

+

^^ | ^(|)" = c + | I(^)^c + c Cl

l

1(

o

-2/2

o

Exercises 6.3 1.

Substituting y

= E^Lo

^^

Cn^" i nto the

differential

OO

j,"

-

equation we have

CO

n(n

-

n=Z

l)^" 2 - £ cx^

oo 1

n=0

fc=0

f

k=n-2

oo

$> + 2)(* + l)<*+3** ~ £

=

Jt=l

fc=n+l

oo

=

2c 2

+E

\{k

+

2){k

+

- c^Jx* =

l)c k+2

0.

it=l

Thus c2 (fc

and Choosing

ct+2

co

=

1

and

c\

=

we

=

+ 2)(* +

=

l)c fc+a

__l_—

-c _i=0

^

find I

C3

=6

C4

=

C8

=

C5

=

Tio

168

fc

fc

=

1 , 2 ,3,....

c *-

Exercises 6.3

and so on. For

cq

=

and

c\

—

1

we obtain c3

= 1

C5

=

Cg

=

and so on. Thus, two solutions are

2.

Substituting y

— Y%Lq

Cn^" into the differential equation we have

y"

oo

00

CO

+ x 2 y=J:

l)^"" 2 4-

n(n -

n=2

£ n=0

k=n-2

oo

^x" +3 =

+ 2)(fc +

2c 2

jfc=n+2

+ 6c3 x +

+ 2)(* +

+

ljcu-a

c*_ a ]»*

-

k=2

Thus c2

(A

^ Choosing

Cfe+2

Co

=

1

and

ci

=

+

2)(k

=~ (fc +

we

+

2)( fc

=

C3

=

l)ct +

+D

2+e*-2-0

Cfc

=

-2 '

find 1

C5

and so on. For

co

=

and

ci

=

1

=

Cg

=

C7

=

C7

=

Cg

=

we obtain c4

=

C6

=

k

+

£

k=2

fc=0

oo

=

l)ck+2 x

*

2

'

M

0.

C*_ 2 z*

Exercises 6.3

and so

on.

Thus, two solutions are

m= .

Substituting y

+

— Y^=o

y"

-

W

Ci»^

B = , - 1*6 + Jj^ -

and

n into the differential equation

+

If

=

we have

CO

CO

£ n=2

~

2

1

~

2

£ n=l

A=n— 2

= £(fc +

W

00

n

+

£ n=0 k=n

fc=n

+ Oct+ai* ~

2 )<*

£

2

it=0

+

E

c* x

*

fc=0

fc=l

oo

=

2c 2

+ co +

£

[(A

+ 2)(* +

l)c fc+2

-

(2fc

k=l

Thus

(k

+ 2)(k +

l)ck+2

-(2k-l)c k =

Q

and

2k

Ck+2= (k

Choosing

q>

=

1

and

ci

=

we

—

1

=

k

Ck

+ 2)(k +

'

l)

find c2

= --

C3

=

=

C5

C7

=

-

•

•

•

=

1

06

and so on. For

cq

=

and

ci

=

1

7__

~

336

we obtain C2

=

C3

=

Ci

=

Cfi

=

1

6 C5

=

1

24 4

C7=

1

m 170

-

=

1

M--

-

l)ck ]x

k

Exercises 6.3

and

so on. Thus, two solutions are

__x

m = l-- x 4.

Substituting y

=

and

„i

= x + -J +

-

a*

+ -- x r +

Y.™=a Cn^" i nto the differential equation we have

y" -xy'

n(n -

+ 2y=Y,

£ nc„x" + n=Q £ c^"

l)^"" 2 -

2

n=2

n=l r

4=n

fc=n-2 oo

= £(* +

+

2)(*

l)c t+2

£

^-

Jt=0

=

2c2

k=n

oo

fc=l

§

+ 2co +

[{*

+ 2){* +

l)ck+2

Thus

+ 2co =

2c3 (k

+ 2)(k+l)ck+2 -(k-2)c k =

and c2

<

Choosing

For co

=

co

=

and

1

a =

and

ci

=

1

^ we

a

=

-co

" (*+*2)(ft+l)

k=W

'*'

find

ca

= -l

C3

=

C4

=0

Ce

=

C5

c$

=

Ct

=

Cio

=

-

=

=

=

=

=

we obtain C2

=

C3

= "61

C4

=

eg

C5= .^120

0.

oo

+2

£

QfcX*

fc=0

-(k- 2)ck \x k = 0.

Exercises 6.3


=

yi

5.

Substituting y

=

- x2

1

and

y3

=

-

x

\x* -

~x

5

Y^^Lq r-n xn i nt ° tne differential equation we have oo

co

on

/ + *V + xy = £

n(r,

£ nc^ + £ c^ n=l

- l)^"" 2 +

1

n-2

1

n=-0

fc=n-2

it=«+l

fc=n+l

oo

oo

= £(* + 2){* +

oo

+ £(*-

l)c t+2 S*

fc=0

fc=2

l)cfc - 1 X

t

+

^

Cfc.jX

ifc=l

,

oo

=

+

2c2

(6c3

+ co)x +

+

2 )( fe

+ *Cfc_i]x = fc

+

iJcfc+a

Jc=2

Thus c2

+ co =

6c 3 (*

+

2)(*

+

=

l)qt+ 2

+ kcfc-i =0

and C3

<

=

-^

*+ a =

<

-(Jfe

Choosing

co

—

1

and

c\

—

we

+ 2)(Jfc +

'

find 1

<*

= -g

^4

=

=

C5 1

and so on. For

co

=

and

Ci

=

1

we obtain c3

= 1

C5

=

C7=

*-

*- 1

l)

C6

=

5

252

172

2 3 4 '

'

-"

and

so on. Thus, two solutions are 1

,

1

5

« = 1 "6 I + 45 X 6. Substituting

y

=

1

,

fi

5

4

""I

,

+232*

-"

Y.™=o cn I " i nto tne differential equation we have 00

00

00

n-2

n=l

n=0

00

CO

CC

= £<* + 2)(* + l)a +2 ^ + fc

2

A;=0

£

+

lfc=l

2

£ k=0

C *x*

00

=

2c2

+ 2co +

Y, K*

+ 2)(* +

l)c,t +2

+

k=l

Thus 2c 2 (ft

+ 2co =

+ 2)(* +

+ 2(* +

l)Cfc

=

and

2

c*+2

Choosing

and so

co

=

1

and

on. For cq

=

c\

—

and

we

c\

=

= - j-jT^ c

=

fc.

1,2,3,...

find

1

c3

=

-l

C3

=

C5

=

C7

=

1 '

=

we obtain C2

=

C4

=

C6---- =

.

2{k

+

ft

l)ck ]x'

Exercises 6.3


=

y1 7. Substituting

y

l-^ + i^-ix 6 + --

=

and

-

l)y"

x

-l

x>

± xS-JL x

+

?

+

....

Y£?=o Cn 1 " i nto tne differential equation we have 00

(x

=

y2

+ y' =

00

£

n (n -

n=2

00

£

l^x"" 1 -

l^x"" 2 +

n(n -

n=2

.

£ n^x""' n=l it=n-l

it=n-2

fc=n-l oo

oo

= £(* +

l)kck+l x

k

oo

~'£{k + 2)(k + 1)^+2** + E(* +

oo

= -2c2 + ci +

^

[(A

+

k

t=o

Jt=0

fc=l

l)c k+l x

-

l)fcc t+ i

(A

+ 2){fc +

l)c k+2

+

(k

+

\)c

M

\x

k

=

0.

Thus

+ ci =0

-2c2 (k

2

+

-{k + 2)(k +

l) ck+l

l)c k+2

=

and 1

cfc+2

Choosing

co

=

1

and

—

c\

we

=

find C2

I

fc

+

^ 1

=

03

=

C4

1

— =

=

=

C3

=

ft

Cfc +l'

^TjT2

-

-

1)2,3,...

=

For Co

0.

C4

_,

= i

.

=

and

ci

=

1

we obtain

,

and so on. Thus, two solutions are 4" (1/1

8. Substituting

y

=

=

and

1

Cnx" into the

y2

=

a:

differential

00

(1

+ 2)y" + xy' -

y

=

£ n=2

1 1 1 1 1 + -x " + -x 3 + -2 * +

equation we have 00

00

n_1 n(n - l)cnX

+

.

£ n=2

2n(n

-

l)c„x"-

2

+

00

£ nc„x" - n=0 £ n=l t

k=n— 1

fc=n— 2

fc=n

00

00

=

E

(A

=

4ca

-c +

+

ljft^+jx*

+

£

fc=n

00 2(fc

+

2)(fc

+

ljcfc+a**

+

00

E ***** - E

00

EP+

!)ftcfc+l

174

+

2(fc

+ 2)(Jfc + l)c k+2 4-

(ft

-

ljcjtjx*

= 0.

.

Exercises 6.3

Thus

-

4c2

+

{k

l)Jfccfc+i

+ 2{k + 2)(k +

=

co

l)c fc+a

+ (k-

=

l)ck

=

0,

«

=

1, 2, 3,

k

1, 2, 3,

.

.

and

Choosing

co

=

1

and

C2

=

c *+2

=

ci

^

so on. Fbr co

=

=

and

l)Acfc+1

2(k

we

-=1=0,

and

+

(fc

+

(fc-l)c

„w,. 2)(A

+

fc

.

.

.

.

.

1)

find

C2

=

C\

+ ,

1

=

= -^.

<*

\,

=

^4

=

C5

0,

^

we obtain c2

=

c3

=

C4

=

C5

=

eg

=

=

0.

Thus, two solutions are 1

2

4 9.

Substituting

i/

= Z^Lq

Is

If

°n xn into

differential

oo

(x

2

-

l)y"

and

n(n

- lj^x" -

Tt=2

-

l)cfci*

+4

k=n

+4

fc

= -2c2 + 2co + {-6C3 + 6ci)a: +

2

£

2

[(*

"

*

+ 4k + 2 )

c*

Thus -2c2

+ 2co =

— 6C3 +

+ 3A + 2)

k=n

oo

£ ***** + k=0 £

fc=2

2

n

ck x k

k=\

fe=0

(fc

ac

£ ncnx" + 2'£c nx n-l n=0

CO

- £(* + 2 )( +

ifc=2

2

k=n~2

oo

£fc(fc

'

oo

£ nfn - 1)^"~ n=2

k=n

=

cix

equation we have

oo

+ 4xy' + 2y=Y,

=

j/2

480

24

c fc

-

(At

6ci

=

+

2)(k

175

+

l)ch+2

=

~

(

k

+ 2 )( k +

l)c*+a]

=

0.

.

Exercises 6.3

and

Choosing

For Co

=

co

=

and

1

and

c\

—

=

ci

we

C2

=

eo

c3

=

Cj

Ck+2

=

ck

ca

=

l

c3

=

c5

=

c7

=

C4

=

C6

=

Cg

=

C2

=

c4

=

ce

=

C3

=

C5

=

C7

=

=

ft

,

2,3,4,.

..

.

find

=

•

=

1

we obtain

1

-

-

•

•

=

=

1.

Thus, two solutions are j/i

10. Substituting y

= H^Lo

=

1 4-

CnX

n

x2

+ x4 H

and

2

+

l)

y"

+ x3 + x h

n=2

oo

H n n " l)^" n=2

2

(

OO

~

6

£ n=0

+

=2

.

^ oo

CO

= £*(*-

H

we have

oo

- Gy = Yl n ( n ~ Vcn*" +

ft

x


00

(x

=

y2

E t=0

+ 2)(* +

(ft

l)c* +2 x

fc

-

6

£ t=0

cfc:c

*

oo

-

2c2

-

6co

+

(6c 3

-

6ci)x

£

+

[(ft

2

-

ft

-

6) c*

ifc=2

Thus

(ft

-

3)(ft

=

2c 2

-

6C3

— 6ci =

+ 2)c + fc

6co

(ft

+ 2}(ft + l)c k+2 =

and c2

—

3co

c3

=

Cl

=

ft-3 c *'

176

=

2 3 -4,. >

+

(Jfc

+ 2)(ft +

1

Exercises 6.3

Choosing

Co

=

1

and

=

c\

we

find c2

=

3

C3

=

c5

Ci

=

1

=

=

c7

-

-

-

=

1

and so on. For

=

Co

and

cj

=

we obtain

1

C2

=

C4

C3

=

1

C5

=

C7

=

C6

=

=

Cg

=

•

•

-

*

=

=

0.


= l+ 3x 2 +x 4 -~x b +

yi

and

y2

5 11. Substituting y

=

2

+ x*.

00

+ 2)y" + 3xy' -y=-£n(n-

ljî" +

2

n=2

CO

£ n(n - lfax"n=2

oo

oo

£ *(* "

1><***

+3

l)<*+2**

+3

k=0

oo

£ nci" - n=0 £ c„x n=l 4=n

n

k—n

co

co

+ 2 E(* + 2)(* +

fc=2

2

fc=n-2

fc=n

=

x

n T,^Lq c„x into the differential equation we have 00

(x

=

E ***** - k=0 E

cjt**

k=l oo

-

(4c 2

- co) +

(12C3

+

2ci)x

[2(*

+ 2)(fc +

-

=

+ Y,

l) Clt+2

+

2 (fc

k=2

Thus 4c2

12c 3 2(fc

+

2){fc

+

co

+ 2ci =

l)ct+2

+

+ 2fc -

2 (fc

l) cfc

=

and

Cfr+2

=

A:

2

-2(fc

+

2lfc

—

1

+ 2)(i +

*-

Cfc

l)

'

2 >3,4,....

+

2*

-

l)

cJ x*

=

0.

Exercises 6.3

Choosing

cq

-

and so on. For

1

and

co

ci

=

=

and

we

c\

find

=

1

c2

=

-

C3

=

C5

C4

= "96

=

C7

=

"

=

C6

=

-'-

=

we obtain

=

C2

=

C4

1

6 C5

and

so on. Thus,

12. Substituting y

=

= I^

two solutions are

52%Lq Cn^" i Qto tne differential equation we have oo

(x

2

-

l) y"

+ xy' -

J,

=

£

- j>(n - IK*"" 2 +

n=2

£W

n=l

n=2

k=n—2

jt— n

OQ

£

-

(-c 2 -

*(*

-

lJCfcl*

co)

-

£

- &3Z +

(*

£

+

2 )( fc

[-(fc

+

l)cfc+2**

+ 2)(* +

-2c 2 -

co

-6c 3

=

l)c fc+2

+

=

(fc

-

^

=

+ 1)q =

and

C3

=

Cfc+2

=

fe

—

1

^j^

c *>

178

2,3,4

+

l)ck+2

Thus

+ 2)(fc +

£ c^"

n=0

k=n

CO

fc=2

-(*

-

k=n

CO

=

oo

oo

oo

B «(« - l)c„i

£ +

OO

to***

(fc

a

-

-

£

l) ck ]

C

x*

Exercises 6.3

Choosing cq

=

and

1

c\

=

we

find

C2

= --1

C3

=

=

C5

C7

=

•

•

=

1

and

—

so on. For cq

and

—

c\

1

we obtain C2

=

c4

=

eg

=

C3

=

C5

=

C7

=

=

-

=

-

0.

Thus, two solutions are !

13. Substituting

j/

n J^™= o c„x

=

=

"

1

2

i

1


00

f"

-

{x

+

00

-y= £n(n- l)c«* n " 2 - £

l)y'

m=^

j

4

8

2

n=2

n=l

«

we have OS

n

-

1

OO

fc=n— 1

OQ

= £{fc + 2)(fc +

l)cfc+2

^ - £ Ac

fc=0

n

n=l

A=n

h=n-2

OO

£ W" - n=0 £ cn^ fc=n

OO

OO fe

z*

-

fc=l

£(* + ljqt+i** -

£ c*x

it=0

fc=0

fc

oo

=

2c 2

£p +

a - co +

-

2)(k

+

l)ct +2

—

(k

+

Thus 2c2 (k

+ 2){k +

-

l)c k+2

ci

-

co

=

-

(k

-

l)(ct +1

+ Cfc) =

and

Cft+2

Choosing

qj

=

1

and

ci

=

we

=

cq

=

and

ci

=

fc

=

2,3,4

find C2

and so on. For

^±1^^

1

=

2'

we obtain 1

C3=

6'

^6

l)ck+1

-(k +

l) C(t ]x*

=

0.

Exercises 6.3

and so

two solutions are

on. Thus,

VI

= 1 + x2 + \ = ££Lo

14. Substituting y

+

|«*

CnZ" into the

+

•

•

y

>

2)y

=

OO

£

«(n

GO

CO

£ £ nci" - n=0 n=l

- l)^"" 2 -

n=2

t=n

k=n-1 OO

-

2c"*"

k=n

oo

£

-

1)<*+2**

£ n=0

fe=n+l

CO

= £(* + 2 >(* +

•

we have


CO

_ xy _ ( T +

"

= x + \* 2 + \* 3 + \* 1 +

V2

•

E

*^^ fc

oo

c*-i^

fc

"

E 2c*^

oo

=

-

2c2

2co

+

+ 2)(* +

£[(*;

1

)

c *+2

" ck-l]x k =

~

(*

+

*

-

1, 2, 3, ...

2 ) c*

0.

Thus 2c2 (ft

+

2){k

+

-

l)ct+2

(A

+

-

2q,

-

2)c*

= =

c fc _!

0,

,

and

Choosing co

=

and so on. For

1

and

Co

=

C2

=

co

Cft+2

=

m

ci

=

Ci

=0,

we

+ (^

+ 2)(Jc +

A

=

1' 2

=

-,

-

3

---

l)'

find

=

C2

and

ci

=

1

C2

=

CO

=

=

c3

1,

1

oj

1

Cs

=

11

5f

we obtain

=

,

C4

5,

= ^,

C5

= i

and so on. Thus, two solutions axe yi

1 3 1 a i = l+i 29 + -x + -x 4 +

15. Substituting y

= ££Lo

— 11

Cnx" into the

<; 5

+

.

and

---

differential

00

(x

-

l)y"

- xy' + y =

ih

1 3 = x + -x

4

1

5

equation we have 00

00

£ «(« " IK*"" n=2

1

+12+8

00

-£"(«" l)^^"" 2 - E ncn^ + E

1

n=2

n=l

n=0

t

k=n— 2

fc=n— 1

180

k=n

k=r,

Exercises 6.3

00

£

=

OO

(k

+

l)kck+1 X

- -2c 2 +

+

k

-

00

£

E H* +

(Jfc

+ 2)(fc +

+

2)(*

l)Cfc +a i*

+

l)c,t + 2

+

(A:

CO

£

-

fcct Z*

-

l)fcct +1

(k

+

£ c*x*

-

l)c k ]x

k

=

0.

Thus

-2c2 + -(*

+ 2){k + l) CJc+2 +

(k

=

Q)

-

—

l)*Cfc+i

(A

—

l)qt

=

and

Ck+2

Choosing

co

=

and

1

ci

=

we

Co

=

and

=

c\

1'

2 3 '

-"

find

C2

and so on. For

*-

-JT2-(k + 2)(k + l)'

1

ff

=

C4-0

C=l

\,

we obtain

c2

= Ci(l +

^

= 3

=

C3

=

C4

=

-

Thus,

0.

+ i* 3 + ---)+Q

i:

c

and

The

initial

— —2

conditions imply C\

= -2

V 16. Substituting

j/

=

(l

and C%

=

6,

so

+ ^x 2 + ~x 3 +

)+

6x

= 8x -

2e

z .

52%Lo Cn 1 " in*o the differential equation we have

(z+l)y" ~(2-x)y'

+y OO

OO

= j>(n - l)^"" + 1

ti=2

OO

£ n(n - l^x"' n=2

2

-

2

£ n=l

W

1

CO

OO

"1

+

£ ncx" + n=0 £ n=l ^

r

k=n-2

k=n— 1 OO

fc=n-l

OO

= £{* +

l)kc k+l x

k

A;=n

OO

+ £(* + 2)(* +

ljct+ax*

~2^(H

fc=0

fe=l

fc=n OO l)cfc +1 i*

+

t

t=l

fc=0

CO

-

2c2

-

2ci

+ co + Y, P + 2 )( + fc

l)t*+2

-

(*

+

l}cfe+ i

+

(*

+

ljefclar*

OO

£ *c x* + £ -

0.

fc=0

ck x

k

Exercises 6.3

Thus

(k

+

+

2)(fc

2c 2

-

2c t

l)c t+a

-

(fc

+ co =

+

+

ljcfr+l

+

(ft

l)cfc

=

and

Choosing

cq

=

and

1

C)

1

C2

=

ci

Ct+2

=

^

=

we

- -co

"^

Cfc+1

Cfc

*

'

=

1'

2 3 -

find 1

2

and so

=

on. For cq

and

ci

=

we obtain

1

C2

and so

=

C3

1,

=

Q=

0,

on. Thus,

y

= C (l-lx 2 -lx 3 +

±-x 4

1

12

+ --)+C2

(x

+ x 2 -]x 4 +

---

and j/

The

initial

=

2 Ci (-x - ±x

conditions imply C\

!/

=

17. Substituting y"

-

+ ^x 3 +

—2 and

t?2

) + C2

= — 1,

+ 2x - x z +

(l

)

.

so

=

2(l-^-^ + l^ + ...)-(x + ^-^ + -)

_

2

x

.

2r *

^+^+

.-.

xn i nto tne differential equation we have Y^?=o °n 2xy'

+ 8y =

£ n(n - l)^""

2

-

2

n=2

£ "ex" + 8^^" n=0

n=l (

^

^

oo

=

£

k=n

k=n

k=n— 1

oo

(fc

+ 2)(fc +

l)c*+3S*

-

2

£

oo

fccfc**

+8

fc=0

£

c t x*

fc=0 GO

=

2oa

+ 8co +

£

[(*

+ 2)(* +

l)c k+2

fc=l

Thus 2c2 (fe

+

2)(fc

+

+ 8co =

l)c t+2

+

182

(8

-

2A)cfc

=

+

(8

-

2fc)c fc ]x

i:

Exercises 6.3

and C2

C*

=

-4co

0fc

Choosing

co

=

1

and

c\

—

we

—

2k

+2=

+

8 Ct

2)(fc+l)

*

'

=

2 3

1'

'

find C2

= —4

C3

=

C4

=

c5

=

=

c7

=

cio

=

eg

=

•

=

•

4 3

=

eg

For cq

—

and

c\

—

1

eg

=

0.

we obtain C2

=

C3

= -1

C5

=

C4

=

•

=

I5

and so

on. Thus,

s/

= C (l-4x 2 + 1

(x-x 3 + ^x 5 +--)

^) + C

2

and

= The

initial

(-8x

C,

conditions imply Ci

y

18. Substituting y

—

T,%Lo Cn^

=

+ ^x 3 ) + C2

3 and C2

= 3^1 -

n into

Ax 2

=

2

+

l)y"

+ 2xy' =

£

3x a

+ ~x 4 +

12x 2

4x 4

+

.

00

00

n=2

.

equation we have

differential

n(n - l)cnX n +

-)

so

+ ^x 4 ) = 3 -

00

(x

0,

-

(l

n(n - l)c„x"" + £ 2nc x n £ n=2 2

1

2=1

k=n

k—n-2

k=n 00

00

= £*(*fc=2

l)c fc x

fc

+

£

00

(fc

+ 2)(* +

l)c* +2 x

fc=0

fc

+

£ *=1

2fccfc x

fc

00

=

2c 2

+

(6c3

+ 2
J^[k(k fc=2

+

l)c k

+

{k

+ 2)(k +

ljc^+a]**

=

<>

Exercises 6.3

Thus

=

2c 2

+ 2 Cl =

6c 3

k{k

+

l)cfc

+

0,

{k

0,

+ 2)(k +

l)c k+2

=

and e2

=

C3

= "3 C1

a+2 = "-j^2 C^ Choosing

co

=

1

and

=

cj

we

find c3

=

=

c4

A

=

c5

•

=

2, 3, 4, ...

=

0.

.

=

For Co

and

c\

=

1

we

obtain

1

C3

-3

c4

=

C5

= "5

C7

=7

=

ce

c$

=

=

1

1

and

so on.

Thus

and ,/

The

initial

conditions imply qj

=

Cl

=

(l-x 2 + and

=

Ci 1

19. Substituting y

y"

= £Jô

+ (smx)y =

dii" into the

1,

3

£ n(n - l)^' 2c2

+ 6c3 i +

4

-x 6 + --).

so 1

1

5

7


2

12ciX

we have

+ (x- ^x 3 + ^-x 5 *

=

x

2

6

)

120

i

+ 20c5 x' +

+

cox

(co

+ ax + ca x 2 +

+ CiX 2 +

- ^coj x 3

)

H j

=

2c 2

+

(6C3

+ ca)x +

(12C4

+ ci)x 2 + 184

(20c 5

+ c2 -

3 ^co) x

+

=

0.

Exercises 6.3

Thus 2c 2 6c 3

+ cq =

12c4

2flc5

=

+ ci =

+ c3 -

=

\cq 6

and c2

=

C3

- --CO

1

D C4

Choosing

cq

=

1

and

=

ci

we

=

C2

and so on. For

cq

=

and

= "12 C1

find

C3

0,

ci

=

1

c2

=

0,

1

=

=0,

C4

=

C5

120

we obtain c3

=

C4

0,

=

=

c5

"IS'

and so on. Thus, two solutions are Vi

20. Substituting y»

+

j/

=

£J^L

_

gn

=

[2c2

+ = Thus

(2c2

=

1

-

^+

1

+

and

J/2

—

—x

£ —

4

+

•

-

-

.

Cn^" into the differential equation we have („

_ 1)^^-2 +

+ 6c 3 x + CO

+

I2c4 x

+ CiX +

cq)

+

(6c3

\

C2

2

+

_ Ix2 + 20c 5 x

- ^c

+ ci)x +

3

+

2

+

^ x

(l2ci

(c 3

) (cq

-

+ c2 -

x3

+

^cq) x 2

+

^cj*j

-

+ c lX + c2 x 2 +

=

0.

c3 x

3

+

-)

+

+

+

'

Exercises 6.3

12c 4

+ c2 -

\cq o

=

and C2

_ __1

C3

=

-TCi 6

= -12^ +

C4

Choosing

Co

=

1

and

01

=

we

find

= ~.

and

=

so on. For co

and

q=

1

we c2

and

=

l/'

+

-3

=

C4

0,

i

=

obtain

=

c3

0,

= ~,

C4

=

two solutions are

so on. Thus,

21. Substituting y

C°-

72

tne differential equation

£)n=o -n^"

e-^g^-lKx"-

we have

2

n=2

+

1

i

~x

x2

\

=

[2c2

+ 6c3 x +

=

(2C2

+ Co) +

12 Ci x 2

(6C3

hx

~ l x3

*

+ 20c5 x 3 +

—

+ Cix + C2x2 + 03x3

]+

[cq

+

(ci

+ ci - cq)x + {12C4 + c 2 - c\ +

Then 2c2 6C3 12C4

+ co =

+ ci -

+ c2 -

ci

186

'

'

)

co

+

= ^Cfl

=

-

cq)x

^co)x

2

+ +

(c 2

-

-

-

ci

=

0.

+

Exercises 6.3

and

1

Choosing cq

=

1

and

ci

=

we

find

C2

and so on. For

cq

=

and

=

cj

1

=

03

~r

1

we obtain

+

-x*

=

=

c4

6*


yi

2. Substituting

y

= l-

— 2n=o

-x*

1

+ e x y-y='£n(n-l)cn x n n=2

|2c2

+ =

(2c2

y2

= x- -X s +

ci

+

1

+

ci

-

co)

2

+

[ci

+

2c2 i

+ 3c 3

^ci

x

2 a:

+ 4ciX 3 +

•)-

£ c„x n=0

n

+ 20c 5 x 3 +

+ ci)i + +

4

we have

J

12c4 x

(2c2

1

00

\

+ x + -x 2 + -x 3 +

+ 6c 3 x +

—x

-2

(1 1

=

and

cn :c " into tne differential equation

00

y"

1

+

(6c 3

(

3c3

+ 2c2 +

+ 2c 2 )x +

(l2c4

+

)

3c3

Thus 2c 2

+ cj - co =

6c3

Uci

+ 2e 2 =

+ 3c3 + c 2 +

-c

l

=0

2

+

+ c2 +

-

[co

+ cjx + c-zx 2 H

2 ^ci) x

+

•

•

=

0.

]

Exercises 6.3

and

=

02

2°°

~

C1

2

1

1

1

C4—4C3 + -C2--C,. Choosing

and so

co

=

1

and

on. For co

=

c\

=

and

we

c\

find

—

1

1

1

2.

o

we obtain

and so on- Thus, two solutions are

m= + 23. Substituting y

V"

= I^Lo

- xy =

£ n=2

1

2

o

1

X ~ X 6

o

2c 2

n(rc

+

V2

1

X ~ X 2

2

+

1

X

^

1

6

~

X

4

24

+

Cnx" into the differential equation leads to

- l)^"" 2 -

£ k=l

[(*

£ n=0

^

00

=

£

+

(*

2)(*

fc=0

+ 2)(fc +

l)c k+2

-

ck ^)x

k

=

1.

Thus 2c2 (*

+ 2)(* +

=

1

l)c fc+2

-<:*_!=()

and C2

= 2

Cfc+2= (yfc

+ 2)(yt+l)' 188

+

l)c ft+2 x

fc

-

£ fc=t

A=n+1

fc=n-S

=

=

.

+ "'

fc=1

>

2 3 <' >

!4.

Exercises 6.3

Let Co and C\ be arbitrary and iterate to find 1

C5

=

=

C2

55

20

and so on. The solution

=

Substituting

3/

=

co

+

is

2b 1

cix

,

+ —x +

1

,1

-coi-

1

,

+ —cix* + 12

1

c5

+

" Sn^=o Cn 1 ' nto the differential equation leads to OO

y"

— 40

- Axy - Ay = 1

CO

oo

n{n - l)^' - £ 4nc*s* - £ 4c*x B £ n=2 n=0 2

Ti=l

^

k=n~2

^

k=n

=

k—n

oo

oo

£ £=0

(A

2c2

-

+ 2)(i +

l)c* +2 x*

oo

£

-

-

t=l

£4

Cfc

s*

jfc=0

oo

=

4cq

+

£

[(A

+

2)(fc

+

l)ct + 2

- 4(* +

k=l

Thus 2c2

-

4c

=

1

and c2

=

^

+

2co

^^(TÂ^*'

fc

=

l,2,3,..

l)c k ]x

k

Exercises 6.3

Let Co and ci be arbitrary and iterate to find

+

C2

=

2

C3

=

14 +

2c°

=

3

5!

14 +

C!

3

3!

14 11 + C2= + +2C0= 14 4

=

C4

C1

4

4!

2

4! 1

14

and

so on.

The

solution

14 +s

=

C7

1

1

c5

5

3!

4

13

6

4!

4

17

7

5!

13

+2CO 4!

16

17

16

8

261

4

409

64

is

(5 /261

+ a^y

4

\

+

Qi

/409

fi

t^y -

64

\

+

,

15 1

25.

Two power

Substituting

26,

If t

of

13 i

17

4!

5!

*

261 6!

fi

64 105

409 7!

series solutions are

„

n —2

3!

^y

**y

7

16

2"

64

n

=

1

-, +

2

Vf

n(u-2)-..(H-2fc + 2)

,21c

into the second series gives the polynomial solution

—

into the first series gives yi

= L—x

fe

dx

—

—— and de

'

dt

.

1

—

dx*

2i

j/2

=

z>

whereas substituting

2 .

2

—

d e — = and the boundary- value problem becomes at*

t,

dt 2

190

in

terms

Exercises 6.3

Substituting B

=

^ Cnf


we obtain

A2

Choosing

cq

—

1

and

ci

=

we

find c2

=

C3

=

"T

=

C5

A2

<=4

=

A4

and

so on. For co

=

and

ci

=

1

C7

=

Cg

=

C2

=

ca

=

we obtain

A2

C5

=

C6

=

A3

C7=

504 CS

=

C9

=

and so on. Thus

Now

or

the boundary condition S'{0)

—

implies

C2

=

0.

Thus

C2

=

and

Exercises 6.4

Exercises 6.4 1.

Irregular singular point: x

=

0.

2.

Regular singular points: x

=

0,

—3.

x

=

3.

Regular singular point: x

=

—3.

4. Irregular singular point: x

=

1.


=

0.

5.

Regular singular points: £

=

0,

±2i.

6.

Irregular singular point:

x

=

5.


=

0.

7.


=

—3,

8.


=

0,

±i.

9.

Irregular singular point:

x

=

0.


10. Irregular singular point:

x

=

—1. Regular singular points: x

3. Irregular singular point:

11. Substituting y

2xy" -y'

= Xô

+ 2y=

(2r

2

cn^"

-

+r

3r)

2.

1

coz^ + 2r

and

(A indicia! roots are r

=

and

r

2

-

3r

+ r){2fc +

=

+r-

£[2(ft

=

-

2r

3)cfc

=

3/2. For r

ci

For r

=

-

3/2 the recurrence relation Cft

and

The

ci

general solution on

s

(0,

c2

2cq,

collecting terms,

(fc

=

+ r)c + 2ct _ fc

'X -

-

2cfc _i

k

=

we obtain

1

]x*

+r - 1

(

0.

is

w

-2co,

=

c3

4 -co-

is

=

t

-(2TT^'

=

oo)

-

+

0, 3.

the recurrence relation

«-m?*jand

3)

=

+ r)c k -

l)(fc

-

r(2r

±5.

2,

and


which implies

The

=

C2

=

^.

=

1-

2 3 -

Cs

--

=-

9l5

D0

-

is

-Oi( 1 + 2 ,-tf +

| a!

-

+

...)

+p

l

^(l-|, +

192

^-^

+

...).

=

0,

Exercises 6.4

12. Substituting y

= H£Lo

2xy"

CnX

n+r

+ 5y' + xy=

and


(2r

2

+ 3r)

cox

£ P(* +

+

r_1

+

(2r

2

+ r-

r)(k

+

collecting terms,

+ 5) c\x

7r

l) Ck

+

+ 3) =

0,

5{k

we obtain

r

+ r)ck + c

fc

_ 2 ]x

k=2

=

0,

which implies 2r

2

+

(2r

and

The

(k indicia! roots are r

=

and

C2

For t

=

=

and

The

general solution on

,. 13. Substituting

W

=


+

j/

\v'

=

Ci

(0,

=

=

00)

is

=

0,

so c\

a

=

ci

.

=

0,

fc

_2

=

=

For r

0.

—3/2 the recurrence

relation

is

^2,3,4,...,

-(2T3tjP

-^co,

cs

=

0,

—^co,

c3

=

0,

C4

=

^co.

=

gjgCo.

is

...)

+ ft

l

(

.^. +j ^. + ...).

n+r into the differential equation and collecting terms, we obtain

- jr) co^- 1 +

£

+ r)(* + r -

[4(A

l)c fc

0,

which imphes

and

r

I

(*»-

r(2r

+ 7r + 5)

^( _^ + ^ +

ZSJLo Cn£

+V=

c2

2

=

+ r)(2k + 2r + 3)c* + c

—3/2 and C*

3r

4r

i(/c

2

7

— -r —

f r

(

4r

—

7\

=

-J

+ r}(8fc + 8r - 7)q + Cfc_i =

0.

+

+ r)c k + c

fc

_i]x*

+r - 1

Exercises 6.4

The

=

indicial roots are r

and

r

ci

For r

=

=

and

The

ci

general solution on

„.

0i

1

(

14. Substituting y

=

=

(0,

- 1I + I |

2xV - zrf +

(i

2

+

o i

i

2 -co,

2ct-i

=

c3

_

.

rt-

+ 7)*'

(8*

C2=

4

-^ggCO-

s|

„

-.

q '"^'---'

1

C0

=

Ca '

~m5

C0

-

is

.-^. +

l) V

_

is

=

+

3r

l)

l

.

+

...).

equation and collecting terms, we obtain

differential

-

2r2 (

^- 5 s?

+ ^/.(i-i, +

...)

Cn£ B+r into tne

EfJL

=

C2

-^>

oo)

i.

is

Jfc(8fc-7)'

— "

Cfc


2 £fc-l

-2co,

7/8 the recurrence relation

=

7/8. For r

_ "

Cfc

and

=

cox

r

+

+ E[2( + r )( k + fc

(2r

2

+ r) -

-

r

r+l t

CiX*"

+ 0^ +

(*

cfc

+ ck . 2 ]x k+T

k=2

=

o,

which implies 2r

2

-

3r

+

=

1

2r

and

The

+ r) (2fe +

[(A


=

1/2 and r Cfc

and For r

c2

=

1

=


=

=

1,

2

(2r

-

+ r) ci = ~

2r

so ci

3)

=

+

0.

-^h)> c3

-~co,

=

-

l)(r

=


=

=

c*-2 "*(2fc

+

fe

1)'

194

=

2 3 4 *

1

i^g^

is

Cfc

0.

2,3,4,...,

C4

0,

0,

+ ct _ 2 =

For r

=

=

0,

l}c k

t

1)

'

is

Exercises 6.4

and

The

C2

general solution on

(0,

oo)

=

c3

-tttco, 10

=

c4

0,

=

3xy"

+

(2

n+T Z^Lo CnX

-y=*

-

2

-

r)

3r

and

(Ar

—

ci

=

=

r

general solution on

,. ot

1

( 16- Substituting y

+

(0,

^+

= X)ô

ci

-

00)

is

»

+ L«. +

...).

s

and

collecting terms,

r

=

+ r)c k + 2(k + r)c k -

—

l)c k

-

=

=

c2

—

r(Zr

1/3. For r

+ r)^,]^"

(k

1)

(fc

=

+ r)c k -i =

0.


1

=

c3

JqCO'

—

is

1

co.

—

ifc

,

1

C2

-co,

+

+

—

—

co,

1,2,3,...,

1

-

...)

&

C3

1

-

+ ftl ./.( 1 +

"j^co-

J,

+

^

+ 5

^» +

crt x" +r into tne differential equation and collecting terms, / 2

\

=

we obtain

is

^^

(2 x-Q)y=(r which implies

2

-co,

c

The

-

1

=


and

r

+ r)(3fc + 3r -

and

and For r

-i I

0,

which implies


360

co^"*

£ P(* +

+

The

I


(Zr

=

r^Q>.

is

,.cix«»(i-^ + 1 L^ + ...) + a,( 15. Substituting y

=

2\

-r+-)cffL T +

^

r

'52 [(*

+

r)(*

+

r

0,

r

2

-r+\ =

(r

— ^) (r -

~"\

=

-

2

l)c*

...).

we obtain

+ -c* - ck ^

1

,

,

Exercises 6.4

and

The

+ r)(fc + r-l) + -j


=

2/3 and r

Ci

For r

—

=

=

1/3. For r

=

c*

and

=

—

is

CS

=

C3

=

560

OD

-

Co

-

is

and

The

0.

1, 2, 3, ...

^CQ,


=

£*_i


ft

,

+k

3k 2

~

Cfc

l60 general solution

on

{0,

oo)

is

1/3

17. Substituting y

= E^Lo

+r

Cn%"


and

collecting terms,

[2(ft

+ r)(ft + r -

we obtain

00

2xy" -

(3

+

2x)y'

+ y=

(2r

2

-

=

and

The

(ft


=

3{fc

2

r

Cfc

= -

For r

ci

=

Ck

-

5/2. For r

=

(2k

-

ci

=

£

+ 2(fc

+r-

—

(2*

l)ct _!

=

,

5)

C2

-

+

=

2r

-

3)ct_i

=

1,2,3,...

1

-rco,

C3

= - -CO. 6

is

2{ft+l)ct-i

-co,

,

+ 5) C2

4

= ^CO, 196

=

1,2,3,...,

C3

=

32 693°°-

l)c ft

+ ct.jjx^*-

0.


3) Cfc _!

-

5)

o

fc(2fc

and

r(2r

-

= ô,


ft

5)cfc

fc(2ft

and

r-1

+ r)c -

— 5r =

+ r)(2ft + 2r and

5r) eo*

0,

2r

which implies

-

is

1

Exercises 6.4

The

general solution on

,_ Ck

1

+

( 18. Substituting y

2xy"

oo)

is

|,.^_^ +

— ££Lo

+ xy' +

(0,

(x 2

CnX n+T into the

-

=

y

(r

2

...)

+ai w(

f)

q,/ +

+ | I+

+


differential

-

1

(r

2

+ 2r +

+ £[(* + r )( + r ~ fc

!> c *

jj)

+

Cl z

(*

r+1

+r

jt-2

=

)

c*

~ 5* + Cfc-a]^ y

0,

which implies

T

z

+ 2r+-)ci=Q,

and

The


=

—2/3 and

r

~~

3k(3k

and

C2

For r

=

-jCo,


and

The

=

C2

general solution on

19- Substituting y 9a:

V

(0,

oo)

a

2/3, so

- 4)

-- 0.

*

,

=

For r

= —2/3


2,3,4,...,

9 128*°-

C3

=

°'

C4

=

C3

=

0,

C4

= j^<*.

is

= ~C0, is

= J^Lo CnX n+r into the differential + 9x + 2y= for 2 - 9r + 2) cqx t


V

+ T, $( k + ifc=i

r )( k

+r-

+

2cfc

+ 9(* + r -

l)c*_i]*

is

Exercises 6.4

and

The

[9(k indicia] roots are r

=

ci

=

=

ci

general solution on

20. Substituting y 2 2x y"

=

=

(0,

oo)

ZJJLo cn i

Tl+r

+ 3ay + (2x

=

(3r

-

l)(3r

1)

+

2}c k

+

9(k

=

2/3. For r

c2

=

l)cfc _!

=

0.


1

'

2 3 '

is

'"-

'

1

1

--co,

=

2)

+r -

*-

fc(3fc-l)

-

gco,

~

C3

7

~Y20

C°'

is

1

--co,

c 2 ==

—5

=

c3

cq,

-— 1

cq.

is


=

-

2

4-

(3*-2)^

"


and

The

=

1/3 and r

and

9r

+ r)(k + r -

Ct

For t

-

9r 2

which implies

(2r

+r-

2

and

collecting terms,

we obtain

r

l) cox

CO

+

=

and

The

[(jfc


= -1

+r-

2

and

For r

ci

=

1

=

(2r

-

+ r)(2fc + 2r + 1) =

r

_

and

+ r)(fc + r -

[2(fc

l)cfc

+

3(fc

+

r)c k

-ck + 2ck _

0,

2r

which implies

£ k=l

=


1/2.

For r

l)(r

+

Ije*

+ 2c

= -1

<=

2

fc

_i

=


= -2c

C3

,

=

4 qCq.

is

C *~"

*(2*

0.

_

2cfc_i

2co,

=

1)

+ 3)' 198

fc-

1

" "*'---' 2

is

l

]x

k+r

Exercises 6.4

and

c\

-

--co,

=

02

5

The

general solution on

=

21. Substituting y

2x 2 y" - x{x

-

00)

(0,

Yl^=o Cn% l)y'

-y=


-

2

2r

and

\{k


-

= —1/2

2

-

-

r

and r


=

=

=

1

1)

(2r

-

~, 2k Ck-l

c2

=

'

cô,

and

V

solution on

x(x

-

(0,

2)y"

=

we obtain

+

(k

+ r)ck -c k -(k + r-

l)(r

-

1)

=

l) C*_!]x

fc+r

Y1%Lq CnX

n+r

-

=

(A;

+r—

l)2c fc _!

=

—1/2 the recurrence

0.

relation

is

1,2,3...

1

1

-co,

c3

= ^co.

*

=

1,2,3,.

Q-^co,

c3

= ^cq.

is

00)

= C,X^(l + i, +

22. Substituting y

collecting terms,

is

^ = 7^7;, 2k + 3

The general

and

l)ck

k=

'

-co,

+

l]c fc

For r

1.

1

=

ci

1

cq.

t

+ T){k + r -

+ r){2k + 2r -

and

=

l) cqx

(2(*

Ck

For t

-— 945

0,

which implies

The

r

E

+ =

-

C3

Co,

is

n+r [2t

— 35

^+i

I3

+

...)

+CM (

1


+ y'-2y= (-2r 2 +

3r) cox

+ i, +

^

+ 3L I 3 + ...

and

collecting terms,

-

+r+

.

)

we obtain

r~l

CO

+

+ ifc=0

+r-

l)c*

-

2c*

(*

l)(2fc

+ 2r-

l)ck+l ]x

k+r

'

Exercises 6.4

-2r 2 + 3r = -r(2r -

which implies

and

The

[(fc

+ r)(fe + r-

indicia! roots are r

=

-

1)

—

-

2]c t

j

ci

+r+

=

For r

0.

=

=

--co,


and

13.

ci

general solution on

Substituting y

=

xy"

(0,

oo)

Y,%Lo cn x"

+

2y'

+r

-xy=

=0.

l)cfc+1


is

*»0.1,2,....

= _

C2

c°'

03

~

"i^g

00

'

is

Ck+i

The

+ 2r -

1

32

For r

l)(2fc

ill

3/2 and r

2k-

and

(fc

=

3)

=

k-2~

=

^k

_

ct

=

A:

,

0, 1, 2, ...

x

-

c2

2co,

-2co,

c3

=

0.

is

into the differential equation (r

2

+

i

r) c

r"

1

+

(r

2

and

collecting terms,

+ 3r + 2) ax

+ £[{* + 0(* + r -

+

l) CJt

2(fc

we

r

+ r)c* - c

fc

_ 2 ]r

Jfc=2

which implies r

2

r

and

The

and

+r = 2

+

r(r

3r

+

+

=

1)

=

2) ci

0,

0,

+ r)(k + r + l)c k - c k -2 = 0. r 2 = —1, so ci = 0. For n = the

(k indicial roots are ri

=

and

c2

=

C3

=

C2"

=

gjCo C5

=

C7

=

=

CO

(2i^TIj!

200

-

recurrence relatic

Exercises 6.4

For r2

= —1


is

<*= J,* \w *(* - 1)

A-2,3,4,...,

and C3

=

C4

=

=

C5

C7

=

=

OD 4i

=

C2n

(2n)!

The general

solution on

(0,

oo)

is

OO

i

1

„to<2n

+

i)r

= — [Ci sinha: + C^cosha;]. 24. Substituting y

x

= S^=o %£™ +7-

V + V + (V - 1)

y


=

(r

2

- ±)

+

(r

a

+

2,

+ £[(* + r)(fc + r -

and collecting terms, we obtain

+ |) C^+l

l)cfc

+

(A

fe

-c* 4

fc=2

=

+ r)c -

+ c*_ 2 ]z* +r

0,

which implies

r

2

+

2r

and

The

^

+

Ck

indicial roots are ri

=

1/2 and T2

=

—1/2, so

*(*

+

,

1)

ci

=

0,

+ Cfc_ 2 =

c\

£

0.

=

=

0.

=

2,3,4,...,

For

r*i


is

Exercises 6.4

and

c2

=

-jjjco

C3

=

c5

C4

=

=

=

c7

•

•

•

=

C0 ^!

(-1)"

C2n= For T2

= — l/2


C0

-

(2^TI)!

is

°k ~ 2

c= and

-

l

C2

— — ^rco

C3

= C5 = C7 =

C4

=

2 3 4

2!

=

CD

5f

The

general solution on

(0,

oo)

is

x 2n

— 5.

Substituting y

x(x-l)y"

=

x

T.%Lo Cn%

1^2

[Ci sin x

+ C2COSX].

n+T into the

differential

+ 3y'-2y =

(4r

-

r

2 )

cox

"- 1

1

+

£

[(fc

+r -

+ 3(k + r)ck - 2c k ~i}x k+r

=

12)c t _!

-

(k

+ r-2) -2]e

_!

=

+r-

+ r)(k + r -

1

0,

Ar

which implies

and


-(fc

—

+ rJffc + r- ^cjj +

r

2

—

ttAr

r(A

—

+ r202

r)

l)(fc

= fc

0.

l)q

The

=

indicia! roots are ri

=

4 and ti

-fc{fc

For

0.

- 4)ct +

=

r

- 3)cj._i =0,

k(k

k

=

1, 2, 3,

.

.

-

fc

=

1, 2, 3,

.

.

.

or

—

-

4)c*

+

-

(A:

3)e*_i

-

0,

=s-

c3

=

.

Then 3ci

-

2c2 ca

c,=

Taking

Taking

The

cq

cq

^

=

and

and

q=

y

=

ci

=

+ C3 =

(

oj is arbitrary

*-^

=

,

5,6, 7

we obtain c,

=

-co

C3

=

C4

=

C5

=

-

-

=

0.

we obtain

C4 ?^

general solution on

-

=

+ 0c2 =

Ocy

and

2cq

(0,

00)

Ci (l

=C

1

('l

K

ci

=

c2

cs

=

2c4

ce

=

3C4

C7

=

4c4

=

c3

=

.

is

+ ^x + ix 2 ) + Ci +

(x

4

+ 2x 6 + 3x 6 + Ax 1 +

^ + ^ )+C n=l f nx" 2

6

6

2

'

+3 .

.

Exercises 6.4

26. Substituting y

=

y"

+

X)^Lo £n%

-J

n+r into the differential equation and collecting terms, we obtain

-2y=

(r

2

+ 2r) cox'- 2 +

(r

2

+ 4r +

3)

c,^" 1 fe+r-2

fc=2

=

0,

which implies r

2

+ 2r =

(r

{k

The

indicial roots are

n=

and

2

+ 4r + 3)

+ r)(k + r + —

r%

2)c k

—

—2, so c\ 2c fc-2

_

+

r(r

0.

=0

2)

cj

=

-

2cfc _ 2

For

r\

-

=

0.


is

,-234

and

=

-co

C3

=

C5

06

The

A

1

c2

=

C7

^li

00

=

=

-

'

result is

second solution

is

„-/(3/*).fe f

_

dx

(1+1^ + f

^

^

+ r

204

...)'

1

/,

1

9

7

4

19

6

\

Exercises 6.4

f ( 1

The

7

1

general solution on

19

1

7

2x 2

96

(0,

00}

19 -x* 2,304

,

=

xy"

+

(1

£?|Lo

- x)y' -y =

=

which implies

r

2

=

r

I "+r into the

W

c,i

E

+


solution

=

f"2

=

+C2 y2{x).

C\y\{x)

differential


+ r X* + r "

[(*

^

+ r)ck -(k + r)ck -

+

1

]x

k+r - 1

and the recurrence

=

£

Ti

*

-

=

relation

is

1.2,3

is

=

yi

A

+

+ r) a Cfc-(* + r)i*_] =0.

ck

One

4

and (*:

The

19

2

is

y

27. Substituting y

,7

1,

1

3

second solution

cq (l

=

+ x + ^x 2 + ^x 3 +

coe*.

is

,-/(i/*-i>fa »2

=

yi

IS

y

-^/i('-* + =

e

1

Ins

-

x

general solution on

2-2

(0,

00)

3,

28. Substituting

1/

= E^Lo xy"

-)*-'/(j-

+ -±-x 2 - -±-x 3 +

L

The

i*'-s*' +

Cn^"

=

-\

e

x

lnx -

e

x

J

+

T ^

?-s** + -)' K

1

n-nl ,

x

n .

is

=

+r

3-3!

1

Cie

1

+ C2 e x

[lux -

£

J

1

n


+ y={?-r)

^

+r+

+

\){k

J" n\

and

+ r)c k+l +

ifc=0

which implies r

(fc

2

-r=

+ + t-

l)(fc

r(r

-

1)

collecting terms,

=

+ r)c k +i +c k =

Q,

=

we obtain 0,

=

0,

Exercises 6.4

The

indicial roots are

n=

and

1

—

—

For r\

0.

and

ci

= - ^co

C2

=

03

=

C4

=

3T2

result

second solution

_

7

/,

1

/

= S^Lq

+ y' + y =

which implies

r

2

1

1

3

=

_ + ...-

38

1

4

5

19

dx

(

yi

^(1-* +

J

\

,

r

J

^-^

/ 1

7

1

+ ...) 38

\

,

A

2

=C

m +C

c„x" +r into the

V""

+ lnx +

2 yi

differential

1

r

~x + ^x

+ £[(* + r )(* + r ~

^

+

(*

and

indicial roots are

n = r2 =

+ r) 2 c k +

c k -i

=

Q.

and the recurrence

relation

^-^fi1

1. 2,

'

2

+

...).


oo

(k

The

-

is

yi (x)

xy"

,

7

,

general solution

Substituting y

2

^

dx

f

f 1

9.

W

is

-*J J-J + foa*-

The

00

is

1

A

is

C0

'm 5!4!

The


1

*

206

=

3

is

+r

)

c*

+ c t -i]x fc+r

-1

-

Exerc/ses.6.4

One

A

solution

is

second solution

is

=

VI

— =

—

!

V\

—

dx

dx

.(l-^ + H-l^+H**-...)

/

5

Z

1

[,

!riI

yi

—————

yi /

f

VlJ

-/J =

=

23

i

23

2

677

5

„

5

,

_ l2

+ 2l +

677

o

23

,

__ x3

+

.

,

\

3

_^ + 677

+

_,

1

4

...j

[

= The

VI In

general solution on

(0,

*

+ VI

oo)

/„ I

Substituting y

—

xy"-xy' +

$2%Lo Cn x

y=

(r

2

'i+T

-

r) c

=C

r"1 a:

The

indicial roots are

n=

1

23 — x

, i

677

,

+ l^r>x* +

+r+

and

1

y l (x)

+ C2 y2(x).

r%

we obtain

00

+

r

{k

+

tne differential equation and collecting terms,

lr^'

which implies

and

5 ,

is

y

3.

+

2x

£

2

—

[{k

r

+r+

—

r(r

—

l)(fc

1)

+ r)ck +j -

+ r)c k +i -(k + r-

=

For ri

=

1

+ r)ck + ck ]x k+r =

=

l)(k

0.

{k

l)c k

=

0.


is

Exercises 6.4

and one solution

is

=

yj

A

cqx.

second solution

is

dx 1

f ( 1

1

i

The

1

1

1

2

general solution on

1

oo)

(0,

1

3

\

2

= T^Lo

Substituting y

xV +

~

cn ;C

1

1

2

3

is

" +r

+V=

1

,

4

y 1.

1

,

= C\x + C2V2{x).


P+

+

2r

and

collecting terms,

we obtain

l )

£P+

+v-

r)(k

l)c k

-{k + r-l)ck +

(k

+ r-

l)c*_i]z*

k=l

= which implies

r

and

The

=

r%

=

solution

+

- (r(k

second solution

+ r-

e-

-^,

=

=

l)ct _ 1

*=

=

0.

is

1,2,3,....

= cox(\-x + ^x 2 - ^x 3 +

/XG X ^-^dx = y

f(l-l/x)dx

1

+ ^x 3 +

+x+

etc

3!

The

2

-

=

c xe"

is

.

=

l)

is

yi

A

2 l) c k

l

and the recurrence relation

1

ck

One

-2r +

+ r-

[k indicia! roots are r\

2

yi \lnx

+x+

general solution on

(0,

-x* oo)

+

—x 1

J

+

=

f 1

1

x j -e dx

,

j/i

J

>

is

V

= Cm(x) + C2 y2(x). 208

1

+ 1 + 2* +

1

3!*

2

+

dx

+r

Exercises 6.4

32. Substituting y

=

xy"

Y.™=o cnX

n+r mio tae

+ y' ~ 4xy =

2

r cax

r

~1

differential

+

(r

EP+

+

2


+ 2r +

r)(k

l)

+r-

cn r

l)ck

+

(k

+ r)ck - ic^x^-

1

*=2

= which implies r (r

and

The

(ft

indicial roots are

n = ri =

0,

One

A

solution

2r

= +

0,

l) ci

+ r) 2 c k -

— Vl

=

0.

and the recurrence

^,

ft

=

relation

is

2,3, 4

=

ax

5

l(1+2ia+

yi

£

~

I

WK'-^r -?* +-> J

+tl . +

...)

= M /(i-2, + §*»-fx' + ...)« fa = n

The

0,

is

I

W =

=

is

second solution

02

=

+

~

so c\

Cjc

2

2

yi In

x

+ yi [-x +

general solution on

(0,

oo)

5

A

-

8

23

[in

x - x2

^+ u

is

j/

=

Ciyi(x)

+ C2y 2 (x).

+

-

6

gx 6 +

]

Exercises 6.4

—

33. Substituting y

n+T into the differential equation and collecting terms, we obtain Yl^Lo <^x

+ (x- IV -

xy"

2y

=

r

2

^ -

1

£

+

[(fc

+r-

+

+ r)ct + (* + r - 3)^_

(*

l)c fc

I

]^

ft+r - 1

= which implies

r

2

=

and

+ r)(k + t -

(k

The

indicia! roots are ri

=

=

T2

fc(*

+

2)c k

(k

+ r-

and the recurrence

"

2)c fc

+

(*

-

relation

=0,

3)cfc_i

3)ct _i

k

=

0.

is

=

1, 2,

3

Then

=

=>

cj

=

=

=

ci

=

+ 0c 2 =

=

c3

=

-

-ci

—

0c2

3c 3

2co

ci

-2co

and

C2

is

arbitrary

and

Since

ci

Thus, yi

= =

and c%x

2 .

=

A

//

1

have co

second solution

^—

n = ^7 ,

= — 2co, we

ct

dx

=x

1

1

=

Taking

0.

C2

=

we obtain

eg

=

C4

=

C5

=

-

-

is

^

J

1

dx

~ \

1

x

j

^V~ X+ 2 X 2

,

11,

1

T

y

x

+

4i

1

x

a

) 1

2

^ln*-- + *--^ + -*<-....

34. Substituting

j/

=

EJJLq Cn£

xy" -y'

+x

3

n+r

y

=


and

-1

r

(r

2

-

"

2r) cox 2

+

(t

+

£

+ [(A

1

+

2r) c 2 3:

(r

r+l

+ r)(fc +

r

k=4

-0 210

2

-

+ -

l) c : x

(r

2

collecting terms,

+ 4r + 3)

l)cfc

-

(*

we obtain

r+2

c 3 x'"'

+ r)cfc + c k ^)x jfc+r-1

Exercises 6.4

which implies t

-

2

2r

=

r(r

2

-

l) c,

{r

r

2

+ 2r = (r

and

(A

The

indicia! roots are r

C2

arbitrary. For r

is

=

=

2

and r

3

=

cq arbitrary

and

=

C3


C2

=

we

c3

0,

0,

=

+c

2)ck

0.

0,

0,

fc

_4

Also,

=

*

1

One

solution

Taking

co

=

=

Cfi

eg

=

cio

=

C7

=

— en —

0.

is

and

ci arbitrary

we

find C3

=

c$

=

C5

=

_ * = 24 C2

A

second solution

=

C10

=

C8

=

C9

1^

C2

= -

is tin y2

The

C7

general solution on

(0,

oo)

=

fr, c

2

\x /

2

— -±x 6 + -±-x 10 . r

24

1,920

is

y

=

0.

when

4,5,6,....

find

C5

=

is

«--*?Sri)' Taking

=

+ 4r + 3)

so cj

0,

=

2)

+ 2)c2 =

r(r

+ r)(k + r-

=

-

Ciyi{x)

+ C2 V2(x).

r

=

2,

ci

=

0;

but when r

=

0,

Exercises 6.4

=

35. Substituting y

Y,%Lq

^


and

collecting terms,

we obtain

00

It follows

x 3 y"

+y=

=

and

that co

cox

r

+

Ck

The only

solution

= Enô

36. Substituting y

xV -

we obtain

+V=

if

=

+

[c k

(k

+r-

+ r - 2)c

l)(k

lt

_ 1 ]x

fc+r

=

0.

+ r-l)(k + r-2)c k ^.

-(k

=

y(x)

is

£ A=l

0.

Cn x7l+r into the differential equation and collecting terras,

rcox

1

-

we

obtain

CO 1

£

+

+ r)(k + r -

{k

([

+

1)

l]c k

-

{k

+r+

l)ck+1 }x

k+r

-

0.

fc=o

Thus r

=

and the recurrence

relation

is

=k + —

Ck+1

*-

°k

Then and

Q =0,

c2

so on. Therefore, one solution

=

y(x)

37. Substituting y

=

x*y"

JZ^Lo Cn2

+ 3xy' —

n+r

8y

1' 2 -

1

=

'

'

i

1

-co,

= -co,

c3

c4

=

7 -co,

is

Cf>


= ^[{n + r)(n + r -

l)c„

+

and

3(n

collecting terms,

we obtain

+ r)c - 8cn ]x n+r

71=0

00

=

[r(r

-

1)

+

3r

-

£

+

8]c

[("

+ r){n + r + 2) - 8Kx" +r

71=1

= Taking cq

/

and c„

=

for

r{r

The

genera] solution

38. Assume x

==

is

is

y

-

0.

n= 1)

1, 2, 3,

.

.

.

,

we have

+ 3r - 8 = r 2 + 2r - 8 -

= Cjx

-4

+ C2^

(r

+ 4)(r -

2 -

a regular singular point so that oo

CO

n=0 Multiplying both sides of y"

+

P(x)y'

x

+

V+

7i=0

Q(x)y

=

by x 2 we have 1

x (xP(x)) y

212

+

2

(x Q(x)) y

=

2)

=

0.

Exercises 6.5

*y

PnA y' +

(f:

/

\n=0

/

\n=0 Substituting y

~ J^Lq

On,x

n+r


CO

\

/ 00

x 2 Y,(n

+ r)(n + r- l^z"*" 2 + x

n=0

o.

we obtain

/ 00

£ £ p„x" \«=0 \n=0

\

(n

+ r)cn z»+ '-

1

/

/

n+r

\n=0

/ \n=0 CO

\n=0

= The

coefficient of the lowest

indicia! equation is

39. Identifying po

The

=

40. (a) Substituting

w—

/

Vi=0

power of

a:,

obtained when n

-

1)cq

then r(r

5/3 and go

indicial roots are

\n=0

=

—1 and

+ porco + qoco =

-

1)

+ po^ + 50 =

-

is

0, is

1)

+Por +

becomes

aw

4y

=

a singular point at co.

P(w) = 2w and Q(w)

= — 4/u; 2

we

see that oo

is

a regular singular point.

Exercises 6.5 1.

Since v'1

=

1/9 the genera] solution

is

y

= ciJy 3 (x) + c^J^îx).

2.

Since v 2

=

1

the general solution

y

—

c\J\{x)

1

=

25/4 the general solution

3.

Since v

lojco-

1/3.

\/x, the differential equation

see that there

—

indicial equation is

—1/3, the

dw

(b) Identifying

\Tt=0

°-

w 2pLi + 2w %Land we

/

0.

r(r

The

\n=0

/

is

is

y

=

+

C2Vi(x).

c\Jb / 2 ( x )

+ c ^-&/2l x

)

Exercises 6.5

4.

2 Since v

=

5.

Since e2

=

the general solution

is

y

=

ci Jq(x)

+ C2Yo(x).

2

—

4 the general solution

is

y

=

ci

+ C2l^{i).

2

=

2 the general solution is y

=

ci

J2(3x)

2

=


y

=

ci ^1^3(63;)

6. Since

i/

7. Since

i/

8. Since 9.

If

j/

=

f


y

is

is

—

ciJi/ 4 (x)

+

+ c^J-iîx).

C2V2(3x).

+ C2^_jy a(6x). ,

ar-V2 w {a;) then

= x-Wv'(x)-\ X -V%(*),

y'

= x~

y"

l

' 2 v"(x)

- x- 3 ' 2 v'(x) + ~x- b l 2 v{x).

and

x2y" + Multiplying by x

1/2

+ X2x2y =

2xy'

10. FVojn y

=

is

v

=

x n Jn (x) we y'

-

x

=

x

n+1 n+l

J% 2

[x

-x n+1 lx

11.

From y =

x~ n J

n

y'

=

is

=

y

=

x""

=

0.

2

l)x

+

n~ l

(2n

)

v.

0,

c\x~ x ^ 2 Ji/ 2 {Xx)

+ C2X~~

1 /2

J_ 1 / 2 (\x).

-

(1

n 2n){x J'n

- 2n)x n J'n +

1

(n

J£

2 2 + xj'n - n Jn + x Jn

JZ

+ xJ^ + (x 2 -n 2 )Jn

1

(since

*

2nx

n~ l

J'n

+ n(n -

n - 2 J„.

l)x

we have

Jn +

+

= xn J% +

y"

and

a solution of the

we

_1/2

find

+ 2nxn J'n + n(n -

Therefore, x n Jn

3

+ xv'+(\ 2 x 2 -±)v =

Substituting into the differential equation,

x n+l Jl

V + x^v' + (aV - ^

+ 92^-1/2(^3;). Then

c\J-l 2 {^ x ) j

+ ni"" 1 ^

z n j;

=

3/

we obtain 2 x v"

whose solution

a;

Jn

is

2

+ nxn ^Jn + x n+l Jn )

-n + n- 2n 2 )x n ~

l

Jn

+ x n+l Jn

]

]

a solution of Bessel's equation)

original equation.

find

x-"J'n

-nx- n - Jn l

and

y"

=

x~ n J'^ - 2nx~ n

214

-1

J'n

+ n{n +

l)*""" 2 Jn

.

Exercises 6.5

Substituting into the differential equation,

xy"

+

Therefore, x~

12.

From y

=

n

(1

Jn

+

2n)y'

is

+ xy =

aT"" 1

—

x~

n~ l

=

0.

2

{x

we have J%

+ x J'n + (since

(x

Jn

- n 2 Jn )

2

\


is

a solution of the original equation.

we

yfx J„(\x)

find

l

= X^Jl(Xx) + -x-

y'

l'2

Ju {Xx)

and

=

y"

2

+ Xx~

X Vx~J'J(Xx)

'2

l

Jl{Xx)

l - -x-V 2 MXx).

Substituting into the differential equation, we have

x 2 y"

Therefore,

+ (aV -

-JxJ^Xx)

13.

From Problem

14.

From Problem y

15.

=

xj-i{x)

=

From Problem y

=

A) „ =

^

=

-Jx

=

0-

a solution

=

n

1/2

=

n

10 with

1

we

find

we

10 with l

n

=

-1 we

y

=

find y

+ XxJl(Xx) + Jn

is

2

(X x

2

- u2 )

J„(At)]


x x l 2 J1/2 (i). From Problem

=

From Problem

xJ\{x).

with

n = -1/2 we

find

11 with

n — —1 we

find

11

— x~

1

J-i(x).

From Problem

n =

we

find

n — — 3 we

find

11 with

1

J_i(x).

12 with A

=

2

and v

=

17.

From Problem

12 with A

=

1

and e

=

18.

From Problem 10 with n x J-$(x)

x 2 J'J(Xx) (since

find y

From Problem

3

2

of the original equation.

16.

y =

[X

-xJi(x).

x~ J\{x) = -x~ l

is

10 with

+

„2

= -x 3 J3 (x).

=

3

we

we

find y

±3/2 we

find y

=

=

find

*Jx Jo(2x).

y

=

x s Js{x).

yfx Jyz{x)

and y

From Problem

=

yfx J-z/2( x )'

11 with

Exercises 6.5

19.

The recurrence

relation follows from

-vju {x) + xJv -i{x) = -J2 00

2n+i>

(-1)""

2n+f-l

(-1)" n)

(x\ 2n+ »

(-l)V

=

-E

=

E + ^Jn!r(l

g, (-l)"(y

,

^nirfl +

(-l)"{2n y

+

y

(2)

+ n) + n)

V2

/x\ 2n+(

'

f)

+ n)

V2

20. Using

•M*)

= =

(-D

E {2n

n

'x\ 2" +

+ v)(-l) n

"- 1

E ^ 2n!r(l + c + n)V2, (

1

(_l)n

W*) = E

/

x n 2n+f-l

we obtain ^[x"7,(x)j

i^-VvM

=

z"Jj(s)

=

+ i>)(-l) n / x\ 2 " +t -1 *"E n)V2 ntj2n!r(l + +

+

'

(2n

i/

2n+u

(-1)"

+

(2n

„tl2nl(y

f)(-l) n

/asN 2

^""

1

+ n)r(i/ + n)V2

-

2n+i/

„(-l)»2-i

(I)" (I)

E ~

+

(2n

i/)(-l)

n

/x\ 2 " +

"- 1

+ n)rff + n)V2

2n!(e

E ^

+

f(-l) n 2n!( I/

1

216

+ rt)r( + n)V2 l/

Exercises 6.5

_

Alternatively, in

JJx).

An

integrating factor for this equation

The recurrence

22.

The

pf"^-

x", so

is

is

1

a linear first-order differential equation

— ax

v

\x J„(x)}

— x"J

ir

-i(x).

relation follows from

recurrence relation follows from Example 3 in the text and Problem 21:

2Jl(x)

24.

(-1)"

that the formula in Problem 19

we can note

21.

23.

,y

r

--

\2vJ*(x)

By Problem 20 By Problem

19

- 2J„ +1 {x)\ = -

-^-fxJi(x)]

dX

we obtain

= xJ

J'n {x)

2J so that J'q(x)

=

(x) so that [* JO

=

+ xJ„-i(x)] - 2J„ +1 (x) =

\xJu+l {x)

rJ

(r)

dr

=

J„_i(x)

= xMx).

rJi(r)

r=0

J-\{x) and by Problem 22

(x)

=

J-t(x)

- Mx) =

JSC*)

- Mx)

—Ji(x).

25. Using Problem 20 and 24 and integration by parts

Jx

n

J {x)dx =

26. Using Problem 25 with

n

jx

n n j x -\xJa {x))dx = j x ^~{xJ^(x))dx n-

- (n-

=

x

=

x n h{x) - (n -

-

xVi(x) +

— 3

we have

l

xJi{x)

(n

-

1)

I)

ji

jx n-1

Jb(a=)

and Problem 23 we have

J

(x)

xJ]{x)dx

(-Jo(*))

-1 lja:"

3

dx

n-2

-

(«

=

x 3 yi(x)

+

=

x 3 Ji(x)

2 + 2x Ja (x) -

2a;

2

Jo(i)

-

4

-

J

I)

2

/ i""V

xJ (x)dx

4xJi(x)

+ c.

(x) <*c

-

J„+i(x).

—

J

Exercises 6.5

27. Since 2

we

obtain

= xJ3 / 2 (x) + xj_y 2 (%)

28.

By Problem

21

we obtain

29.

By Problem

21

we obtain - J_ 1 / 2 {x)

Ji/ 2 (a:)

= xJ

1(/ 2

.

j 30.

By Problem

21

we obtain

By Problem

21

By Problem

21

21

xJ^j 2 {x)

+

,

/

.

— 2

,

—

,

,

3cosx

\

a:

7)/ 2

=

(x)

/15sina: /15sina;

5— -

2

—

s

6sinz

+ xJ_ij 2 {x)

—

/— 15cosx

\

so that

15cosa;

-

so that

3sinx

+ xj3 / 2 (x)

15sinx

feh /

+ xJ_ 5 / 2 (x)

/3cos:r

2

we obtain — 5.7_ 5 / 2 (:e) = xJ_y 2 (x) !(l)

\

.

2^1/2(2;) so that

/3sini

2

/

we obtain 5J5 / 2 {x) =

^

so that

(;z:)

pZ /cosx

.

we obtain -SJ_y 2 (x) = xJ„i/ 2 {x)

,

By Problem

=

.

,

hiiW 33.

+ xJ_ 3/ 2

^H™(^

.

32.

(^)

so that

-*t*w--i™{— +sm v-

ZJj,ji{x)

JT 31.

(n-l)!22"-l

J

+

+COSX

the function 35. If yi

=

Iv{x)

\

-

-^-^

{-l)"i 2n

fx^+"

6cosa:

^- +

is real.

=

i~"J„{ix) then

S/i=i-"

+1

218

^(

l x),

^

.

so that

34. Since

-viC

\

1

\

SiQX )-

Exercises 6.5

and

*V + aVi " (* Similarly,

y 36.

If

= t/i

+

VI

= I-v{%) — + C2l~v(x).

t/2

ciIl,(x)

=

2

=

i~" [(w)

i"J- y (ix)

= Mx)

= Mx)

satisfies

+

{«)

a

((«)

2 - " ) M*c)] =

i

"

=

0.

the differential equation, and the general solution

is

Jrn^ dx dx dx

j f

,

,

.

,

,

J,

x2

=

T

,

,

a:

,

Jo(a:) In

a:

hx 3

x2

5x 4

2

+—

4

37. Using (8) with v 7

=m

)

T

+

3x* — — 128

23a:

5

23a:

, ,

6

--~ + x6

x*

11a:

=

-£ d r

\

2

fx + ^_

_+_+ 5a:

23x 6

.

6

13824

flN*"", "

(-1)"

(-l)

/S^'"

(-1)"

n^n!r(l-m +

fl)

W

m J (a;). m

m we have '

(1

+ i + ,) (-5)

i-

i

r£

^

+ i t ,| (-i )

Using the formulas on Page 315 in the text we obtain

P6 (x) = and

4

4-

,Jj»!r(l-m + n ) V2>

= 38. Using (7) with f

x2

xe

we have

^- V»

mK

x4

x

(I

= M*)lnx + (, (l-

(a)

^(w) +

Jo(x) then using equation (35) on Page 299 in the text gives

V2

39.

2

— 16

(231a:

8

- 315x 4 + 105x 2 -

5)'

=(-')•«.

t

Exercises 6.5

Pj( x )

=

A

{429a;

-

7

693a;

5

+

315a;

3

-

35a;)

.

16

P6 (x)

(b)

40.

We

-

satisfies (l

2 a;

y"

-

2xy'

)

+ 42y =

d_

If

satisfies (l

- x 2 ) y" -

2xy'

+ 56y = 0.

use the product rule for differentiation:

+ n{n +

dx

41.

and Pi{x)

=

x

=

(1

- z 2 )^| + (~ 2x )^. + n ( n +

=

(1

-

l)y

x 2 )y"

-

2xy'

+ n{n +

l)y

=

0.

cos# then dy

Sm0a .

d9^-

(fiy

9

sin

dy

Tx'

Q-r-z

—

cos

dx*

de*

dy — dx

,

and

sin#S( +

cos(?^(

do

That

do

+

n(n

l)(sin0)y

=

sin#

-cos 2 S)

1

^ - 2cos$^ +n(n+

l)y

=

0.

is,

42.

The polynomials

43.

By

are

shown

the binomial theorem

[l

+

+

44. Letting

(t

x

2

-2xt)y

=

1 in (1

(1

-

1/2

2t

on Page 316

in (18)

in

the text.

we have 2

=l-±(t 2 -2xt)+^(t2 -2xt) + --- = l + xt + ±(3x 2 -l)t 2 + ---.

2xt

+ t 2 )"

1'2

+ t 2 )- ,/2 =

we have

,

-

(1

t)"

1

= jl— = l+t + t 2 + t 3 +

...

(\t\

<

1)

t

00

=

£'" 71=0

From Problem 43 we have n f;pn (i)t =

Equating the letting

i

(i-2*

+t

a

coefficients of corresponding terms in the

— — 1 we (i

r l/2 = £t n two

series,

we

-

see that

Pn (l) =

have

+2t +

2

y

1

'2

=

(i

+1)-!

= i - t+t 2 -

= L

=

"T*

i

£(-i)V=£pn (-i) n=0

n=0

220

t ",

3t

3

+

.

.

.

(\t\
1.

Similarly,

-

.

,

Exercises 6.5

so that

45.

The

P„(-l)

=

(-1)".

recurrence relation can be wrtten

ft+iW = 3

k

=

1:

k

=

2:

k

=

3:

w

=

k

= =

4:

n—

0:

=

1:

n—

2:

71

"-

^

C

1

=

-?*•-!)-! 35

ort

3

1

Pl(x)

=

1^/2 (X — l) = z

ft(x)

=

0, 1, 2,

8

35

1

Sdx 2

^

^

2 " J

(l2

8 &x*

30

4

" »' "

-2x 2 +

(

-

(1** - 1-)

l)

=

T

f 2

T

"fjUT

- 35

=

n #»

,

5 /35

\

15

3

=

3:

47. For n

3

"2*

P6 (x) =

5:

.

1

?)-!(?"-

W W

...

4,

3,

"2

MkFT

k

2

2*

= 2,

15 x x+ T ,

3^

231

8,'

16

,

^(12x 2

a\ -- 4]

"

3 *'

+ 312 " 3 = »

48<

16

= -x 2

3

1

2

2

'

5 5»

12to3

and 3 we obtain

^P

a

2 (x)

dx

=

(9x

4

- 6x 2 +

l)

dx

=

J

|

and

j

P${x) dx ^

=j i

(25x

6

-

30x

4

+ 9x 2 )

=

In general,

£ 48. All integrals of the form

P*(x)

=

2^

Pa (x)Pm (x) dx

are

for

for

n

n=

^

0, 1, 2,

m.

-

_315

6

2

.

X4 +

" 72l > =

105

IF

?

3

1

" 5*

X

Exercises 6.5

49. Let

=

y2

+ x)-ln(l-x)}-l

-x$n(l

that

=

J/2

1 t:

x

1

+

l+x

+

I-

+ x)-ln(l-x)}

-\ln(l

and 1

2

(l

+

(l

+ x) 2

x)

1

1

r

+

+

(1-x) 2 1

+

+

(l-x) 2

1

1

2

l+x

1

+

1

+

l-x

1

1

2

1

+x

1

+ 1

1

-x\

1

l+x + l-x

Then (1

- x)(l + x)y'i -

2xy'2

+ 2y2 =

0.

Chapter 6 Review Exercises 1

.

The

auxiliary equation

is

m— =

6m —

=

y 2.

The

auxiliary equation

is

The yp

auxiliary equation

= x4 —

x2

In

is

m2

The

=

+ c 2 x~ lli

4-

C2X~ 3 In x

= (m —

6

.

3)

2

(m +

+ ^x 3

2)(m —

3)

=

1/2)

so that

.

=

and a

particular solution

is

x so that

auxiliary equation

is

ro

2

—

c\x

2

+ C2X 3 + x A —

— 2m + 1 — (m — l) 2 = y

5.

cix~ 3

— 5m +

y 4.

1/2

cix

2m 3 + 13m 2 + 24m + 9 = (m + y

3.

so that

1

=

c\x

x

2

In x.

and a particular solution

+ c 2 x In x + -x

is

yp

—

jx3 so that

.

Since

P(x)

=

the singular points are

-2x (x-2)(x 2 + 2x

x=

0,

x

=

—1 +

+ 4) \^3i,

— Q(x) ^ v '

and and x

— —1 —

1 (at

-

2)(x

=

and

Q(x)

=

222

(x 2

-

+ 2x + 4)

^/Si. All others are ordinary points.

6. Since

P(x)

2

2 4) (x

+

4)


=

the singular points are x 7.

2,

—

x

—

2i,

afo

-

—2, x

the regular singular point

is

x=

and the

Alt others are ordinary points.

^

5)'

Q

^ =0

irregular singular point is

=

£

5.

Since

P(,, the regular singular points are x 9.

= — 2i.

Since

=

8.

and x

=

=

and

and i

=

=

5.

=

—3. The irregular singular point

There are no irregular singular points.

Since

the regular singular points are x

=

3 and

a:

is

i

=

0.

10. Since

ifa^+l) the regular singular point 11. Since P(x)

= —x

12. Since P{x)

= —2x/

13. Substituting

?/

=

and (x 2

= S^=o

x

is

-

4)

cn x

'i

=

0.

The

[X^

irregular singular points are

6 the interval of convergence

and Q(x)

mto * ne

=

9/

(a:

2

-

differential

4)

an

is

+ xy =

2c2

C2

=

+Y, [*(*

-

1)<*

Choosing

co

=

1

and

ci

=

we

find

,

= and x —

-co < x <

,k-2 k~2

+ c k ^\x

and c k-3

x

equation we obtain

it=3

which implies

lj

i

.

c

=

—i.

oo.

interval of convergence is

oo

y"

+

-2 < x <

2.


and so on. For

co

=

and

=

ci

1

we obtain c3

=0

Ca

=

1

12

=

C5

C7=

=

fk


and

14. Substituting y

= ££Lo

1

dj^' into the differential equation

y"

-

4y

=

£

[*(*

- Ik* -

we obtain

4ct _ 2 ]x

fc

-2

k=2

which implies Ck

Choosing

Co

=

1

and

c\

=

wc

=

777

rr

fc

i

-D'

=

2, 3, 4,

find C2

=

2

C3

=

C5

C4

=

=

=

...

=

2 3

C6

4

=

45

and so on. For

cq

=

and

ci

=

1

we obtain c2

=

C3=

C4

=

ce

=

2 3j

C5=

2

15 5 4 C7

=3i5

224

=

.

.

.

.

=


and


and 2

I » = C ( I+

15. Substituting

=

?/

,

2

+

X

»

4

*

+

X

YH

3


Y5£=$

7 a:

3i5

+

\

---)'

we obtain

oo

-

(x

l)y"

which implies

+

C2

3y

=

=

(-2c2

cq

=

1

3q,)

and

c\

=

_ -

(fc

-

1)(*

en

we

" 2 H-l ~

k{k

-

+ 3ck _ 2 }x k

l)ck

~2

=

and

-

ci

+

2)0^1

3ck „ 2

_

fc-3,4,5,...

,

.

1}

find

=

C3=

=2'

2'

^=

8

=

-

we obtain

1

c2

and

~

(*

k{k _

C2

and so on. For

£

+

3co/2 and ck

Choosing

+

=

1

=

c3

0,

-,

c4

1


and

= G2

V2

=

16. Substituting y

Cn*™ ' nt0

{

I

+ ^x 3 + ^x 4 + equation we obtain

diiferential CO

y"

-x

V + xy =

2c 2

+

(6c3

£

+ co)* +

[(*

+ 3)(* + 2)c*+ 3 -

it=i

which implies

C2

=

0, C3

= — cq/6,

and

— C * +3

Choosing

co

=

1

and

c\

=

we

=(A +

1

3)0t

+

*

C *'

=1

2)

find 1

Ci

=

C7

=

C5

=

c8

= en =

C10

=" =

225

•

-

--

=

.

2 .3.----

(*

-

l)c*]i

fc+1

=

=


obtain C3

=

ce

=

Cg

=

C4

=

c7

=

Cio

C5

=

eg

= cn =

-

•

=

= •

=

..


=

yi

17. Substituting y

2x 2 y"

=

T,%Lo Cn%

- ^x 3 -

Co (l

n+r into the

~x 6

and

)

y2

=

c Y x.

equation we obtain

differential

+ xy'-(x+l)y =

for 2

-T-ijcox'+Y, |2(* +

r)(A

+r-

l)c k

+

(k

+ r)c k -ck -

it=i

which implies 2r

2

- r-

1

=

(2r

+

l)(r-

1)

-

ck

=

and [(A

The


=

1

and

+ r)(2* + 2r r

=

1)

—1/2. For r

-

=

^mrry SO

For r

CI

= —1/2

=

c2

\co,

the recurrence relation Ck

Two

c\

=

1

fc

^ = 0.


=

1

'

^co,

2 3 '

c3

=

^co.

is

= fc(2*

so

=

\\ck

-

3)

*

'

=

1

1

2 3 ---> >

1

-eg,

C2 c

= -~2 C

linearly independent solutions are

and

226

<>'

c3

=

"18°°*

is

c ft _i]x*


18. Substituting y

—

n+T into the differential equation we obtain EJJLo cn x CO

2xy"

+ y' + y =

(2r

2

-

r) cqx'-

1

+ £[2(* + r){k + r -

\)c k

+

=

0.

+ r)ck + ck ~i}z k+r

(k

fc=l

which implies 2r 2

-

=

r

-

1)

\)c k

+

r(2r

=

and

+ r){2k + 2t-

{k

The


=

=

and r

For r

Ct

=

=


Two

=

=1

>

2 3 >

is

-" 1

= -—c

c3

^co,

b

.

90

is

= ~*(2fc

Cl

*

1

=

c2

-co,

Cfe


= ~k(2k~-1)'

Ck

so

=

1/2. For r

-loo,

+

C2

fc

=

1)'

=

2 3

1'

Aco,

'

C3

= -^CO.

linearly independent solutions are

yi

= c ,(i-, + 1

-^-~, 1

1

,

3

+

...)

and ?/2

V

19. Substituting y

i(l

-

=

630

30

3

/

n+r into the differential equation we obtain EJ^Lo Cna:

x)y" -2y'

+ y=

(r

2

-

3r) coaf'

1

+

£

[(*

+ r)(fc + r -

l)c*

- 2{fc + r)c k

it=i

-

(*

+r-

l)(fc

+r-

2) Cfc _!

+ c^]^"

1

= which implies r

2

- 3r =

r(r

-

3)

=

and (*

+ r)(k + r-

3)c fc

-

[{k

+r-

l)(k

+t-

2)

-

l]cfc _!

=

0.

-1

=


The

indicia! roots are

n—

3

and

=

t-i

2

+ 3* +

(*

_

For

0.

n=

3 the recurrence relation

is

l)c -i fc

so Cl

One

solution

=

209

11

5

= -co,

c2

-co,

c3

= —co-

is

5

, /

11

209

o

\

3

SI

A

second solution

=

V1 -/

is

^(l-2x + ^)(l + |x + f§^ +

/ / 1

=m J =

1

1

3/

=

yi

Cn^"

xV -xy'+(x

2

+r

...)

\

1

(r

2

-

2r

£

+

+

[{ft

l) cox

we obtain

+ r 2 cjx r+1

r

+ r)(* + r -

l)cfc

-

(A

+

k=2

= which implies r

2

-2r +

l

= (r-l) 2 =

r 2 cj [(ft

The

indicial roots are

n=

=

=

+ r){k + r - 2) + so

1,

C*

c\

=

=

1

..-).


+ l)y=

1

91

^+^+

(-^3 +

+

_ 1,lni V^" fl r3? + 4? + lfa 36 + ""J

1

l?-2?-16?-3fe—

-^ lnz +

20. Substituting

1

|i| 3 x

1]^

+c

and

-^T'

*

228

=

2,3,4

fc

_2

=

0.

+ c + ^_ 2 1* fc


Thus

-

C2

-4* 1-5

and one solution

A

Ci

-

C6

=

—

—

l*T (-7

1

61°°

2,304°°

is

second solution

is

e^/ 1

xdx

f

vi

— -

y

^

f

V

/•

x(l

_l x 2 + ^__ _ l6 + i f

/1

5

1

23

5

2

21. Substituting y

xy" - (2x

= ££Lo

-

l)y'

+

+

E

c„:r

(x

[(*

-

n+r into the l)y

-

2

r cox

+ r " 1)^ +

+

{(r

(*

-\ ,

23

4

+

)

=

2

6

equation we obtain

differential T- 1

...

+

2r

+

+ r)cfc -

l) ci

2(k

-

(2r

+r-

+

l)ck

ifc=2

= which implies r (r

+

2

l)

2

=

0,

ci-(2r

+

l)co

=

0,

and (fc

+ r) 2 c fc

(2k

+ 2r-

!)<;*_!

+ ck_ 2 =

0.

I)co]

xT

^

ck . x

-

+ ct - 2 ]i fc+r

-1

— Chapter 6 Review Exercises The

indicia! roots are

n = r2 = 0,

so c\

=

-Co,

—

and

cq

Thus C'2

and one

solution

second solution

i

i/

=

V - xV +

—Co,

= —Co

C4

=

co (l

+ x + -x 2 + -x 3 + ii 4 +

=

Co e

dx

=

x .

is

y>

22. Substituting

1

1

=

C3

is 1ft

A

1

=

£™=0 2

(*

-

y—^

e

cn x"

2) y

+r

=

=

dx

;^

e

=

e

2

+

-

-

r

£

r

2) CoX

+

+r-

2

+r-

+

[(*

[(r

2) ci

-

l)c fc

lnx

e

-

we obtain

into the differential equation (r

Jx

2cfc

-

-

rco]

(A:

+

x

r

r+l r4

-

Ijcj^

+c

fc

fc=2

= which implies r

2

-

-

t-

r

2

=

2

+r-

-

(r

2)(r

-

2) ci

+

rco

-

1)

=

0,

0,

and [(*

The


-2ci

+ Co —

+ r)(k + r -

=

=

2 and r2

and

1)

fc(fc

-

-

-(k + r- ljc^ + c k _ 2 =

2}c k

=

-1. For V2

-

3)c ft

(k

-

0.

-1,

2)c fc _i

+ ct _ 2 =

0,

k

=

2, 3, 4, ...

Thus

-2c 2 cs

and

-

— c% + c\ = cfc

=

c\

1

=

ci

^co

+ co =

=?

(ft-2)cfc_i

C2

=

-c

fc

=>

=

Ci

_2 ,

230

C2

=

^co and

A

=

|co

C3 is arbitrary

4,5,6,....

.

_ 2 ]z fc + r

Chapter 6 Review Exercises Taking

cq

=

1

and

=

C3

we have

=

Cl

and

so on. Choosing co

1

2'

=

and

1

=

C2= C2

-,

=

-

= -^,

CB

2

=

03

we have

1

1

ci=c2 =

c

0,

4=2>

C5

3

=

20'

and so on. Two solutions are

and Vi

!3.

Substituting y

xy" + y' + xy

—

=

= Qx-

J^nLo cnX

n+r ' nto the

2

2

r

r CQX

+

(r

+ 2r +

1

(x 3

1

+ -x 4 +

3

~x

r+i

+

£

+

equation we obtain

differential

l) Cl x

5

[{k

+ r)(k + r -

l)c k

+ {k + r)ck + ck - 2 ]x k+r =

which implies r (t

+

2

=

0,

2 l) ci

=

0,

and (k

The

indicia! roots are r\

=

ri

=

so ci

0,

c*

=

+ r) 2 ck + cic-2 = =

0.

and the recurrence

-^'

^

=

relation

2,3,4

Thus c2

= --co

C3

=

C4

=

=

C7

=

=

1

C6

=

C0

64

-2^4 C0

and ?/i

3/1

=

rs\ co

1 (

1 1

— \x 2 + ^-x_ i .

4

64

l—x — 2,304

6

+

is

e

7

Laplace Transform Exercises

1.

#{/(*)}

-

3.

S

=

#{/(()}

=

et

te-

J*

-7 e 4.

\

dt

"s

/'(2t

+

-

S

—

+

l)e-

3t

"s )

dt

=

J*

e

st

-

=

dt

(°

(~l

-i)

te

'

s

~ St

-

-| e £

- flt

¥{f{t)}

r

=

(cosi)e-

st

=

= (--^ye-^sinj-

dt

=

(

\

<

t

(>

1

0, f,

<

—

f(t)

=

—-i—e^'cost + + S

£

1

J?{/(t)}

1

2t

-

2,

-

2

Hit J\

t

'.

s

>

,r '

S

+

s>0

l),

l

;

+

/2 .

s

e~

st

smt

1

*/2

>0

1

1

= -e

= J™te- St dt = (-\te~ at - ~e~ at 0,

e_s

e

(e-"

+

2

1

8.

-

-ji-ye-^cost

00

#{/(*)}

=

>

s

J 0-(o + ê-"' )=-Vrre" 2 s + l' «a + l

/(()

"s

e

e

e

Jn/2

7.

~ St s

s

s

6.

e

^2

- ^° -

- (--t e ~ s -

;e

{sint)e- st dt

S

s

o

J™ e-

ie

3

-St

s

o

e" st

\

—

= #{/(*)}

—

- e~ at

)

5

JO

5.

=

s

St 2{f(t)}= f Ae- dt = JO

%{f(t)}

e' st dt

~e~ st dt +

S

2.

7.1

e~ a

s>0

,

s

1

l)e~

st

dt

=

2

st (--{t - l)e~ -

\

3

\ e -st S

232

—

=

2

-je"',

s>0

Exercises

9 -' (

f 1

-

Ho,

*
t<

o <

1,

1

>0

t

/V <

f 0,

-

<

f

-

t)a-'

+

^

=

«

"2 e

a t

-st

10 0,

11.

12.

13.

{/(*)}

=

#{/(*)}= 1 WJ

#{/(()} 1

-

t>

e'

+ V'dt

[°°

e-

2t

[°°

te

4t

e-

st

fe-t'+^dt = _- £ll e -(»+2)t /o s + 2

= rte^'dt =

dt

JO

s

,

-

>

" 8 '*

-tV 3

(

-

2

e(

3-s

_JL~f e

^(3

(s+ 16.

t

e- (smt)e-" dt

+

l)

1)

= £° e

t

2

2

s

,

(s-3)3

t

J™

+

+

1

-

s2

t

s) 2

+

3

(s+l) 2 +

(cmt)e-* dt

+ -s

2

3 -')'

>

l

1

(l-s) 2

<

= Jîsmt^+^dt

+ 2s + 2' =

J^°(cost)e^-'^dt

1-5

(1

-

-

si so -e-

+

cosi l*

(l-s) 2

+

s>

1

a-1 1

s2

-

2s

+

,

2

l

l

e P-)*

S

s

,

e 3

o

(4

-

X

°°

s) 2

—

S

>

,

s

>

1

-5

+

a

,

2

—^ w)

"
(3-5)

2

=

1

4

= jT t 2 e 3t e~ st dt =

(3-s) 3 ^{/(*)}

-^—te^ \4- s !

s)*

1

—

oo

e-*

Jo

3-s

15.

= 0-

e^^'df =

7

-h- st dt =

1,

#{/(«)}

e

1

(4

14.

=

jo

'

-(e" sa

b

°°

/

-

e

/

00

> -2

1

7.

1

Exercises

17.

Jf{/(t)}

=

7.

t(coat)e-

J

st

st

s

+ s

1

(s

-l

2

-5

2{f{t)}

=

t(8ini)e" 5C

jf

-1

2

2

+

1)

>

S

,

/ao

18.

dt

* 2s (c08t)e

+

1

1

(s

2s

7

(P +

if{4(-10}

23.

^{(

25.

^{t 3

27.

^{l + e -"} = i +

}

=

4 10 — 4-

+ 4-^ s s s

+ 6(-3} =

J

2

+ 3f + 3( +

s

33.

^{l+2e 2t +

e

4(

a

if{4t -5sin3[}

y{sinhA:(}

i

^ + 3-| + 4 + -

=

l}

—-!-

2

-l + 1)

{s 2

1

(sint)e ;

=

=

20.

*{t B } =

22.

^{7( +

24.

y{-4f 2 +

26

^{8i 3 -12( 2 +6i-l} =

-

s

+

^ s

—

2

2

4

+ ^-r s - 4 -

2

^} -*

16f

2 + 9} = -44 +

16

8^-124 + 4-"

y

32.

i? ( c „ s_5, + sin5, ) =

34.

%{coshkt}

- i} -

234

{

9

^+^

so.

-fc 2

^.,i„ h <) - if {e-

3}-^ + -s

^-.- + 5}.^-^ + ^

3

3 -5-2s + 9

5

28.

4

s

= s

35.

}

*

+

2

s>0

,

= 2^

21.

31.

5

1)'

s

+

'

4!

4

if{2t

29.

+

st 2

l)

19.

2

2

-*_

«»- a + .-.,_-i -2 + -i

^

5

-

i

=

3 ,

-

5

-^_ + ?A-5

a

Exercises

37.

i?{sin2tcos2f-}

38.

i?{co S

39.

^{costcos2r-}

40.

41.

42.

2

t}=i?

= if

43. Let for

3

u=

}

=

Ys

s

+

l 2^ +4

=if j^cos3f + ^cosfj =

^{ S m(cos2t} =i?

y {sin

cos2t

l

=J? {^cost -

if{ Sin( S in2[}

=

j^sin4fj

+

{l

=

Is ^

du

s dt

2 s2

9

+

1

\

-

±

^

(

=

Is

+

:

|

(^l C0S 2 )}=^{i

f}= ^{ sin( st so that

cos3f}

\

- în(} =

{|sin3t

and %{t"}

=

Si n

/

e~

t-i(ism3 i -i S mt)} = a

st t

=

dt

(

In

V

s

r{3/2)

47.

If

r(5/2)

3^

s 5/2

4s 5/2

we attempt

to

compute the Laplace transform

*^ If s

=

=

for

of 1/t

2

atdt+

we obtain

rk

stdt

then

which diverges.

If s

<

then

which diverges.

If s

>

then

which diverges. Thus, the Laplace transform of

1/f

2

does not

exist.

5^_l^

= - ) - du

a > — 1. r(i/2)

7.

-^r{a + 1)

i

1

Exercises

,

7.

48. Since / and g are of exponential order there exist numbers

and

< Ne

\g{t)\

dt

for

>

t

T.

t

>T,

and

is

M, and

N such that

Then

= for

c, d,

< Me<*Ne dt = A/TVe'^ 1

\f(t)\W)\

of exponential order.

Exercises 7.2

24

s5

'

3!

.

8.

if-lL.i +

9.

se-

Us +

xto 1

ii.

12.

5s

-

i*— + 4

s

s&

s

\4

S+1/4J

2

\5

s

J

'{?T49}

= ^ _1

+ At + 2t 2

8_

lJ

-

2/5

4

5

J

=

{? '

I

si sin7t

7

2 236

3,1,

|/(t)|

< Mea

Exercises 7.2

13.

g-i/-il-} = ,y-iJ 2 I4s

16.

i?" 1 !— I

17.

iy '

"

^ 26

1 25

=

l/4j

2

10cosh5f

h I

s2

U

2

+ 2j

\s 2

+2

U

2

+ 3§J

U

2

+ *-

+

a2)

\( s _2)( 5 -3)( 5

-

2

9

3

S

s2

+

1

+

l)(s-2)J

\(s-2)(s 2

+ 4s + 3)J

S

3

+ 5j"9

9

+

s-2

ft

(

s-3

+ 2

1

1

1

12

s

s-1

3

1

15

2 sin 3*

s/2

1-jf-if 1

*

-

-^ ^-^}" ^

12

fl2

2cos3i

+ 2j

9

}-^"4 (a3)

\s(s-l)(s

=

+ 3/~3

5-4

-6)J

+ 9)}

-

s2

\/2

*

\9

20 J

^1 (,-a^ + X

+

J

2

1*3

23

—

\s 2i

lj

U|^|| + 9J

i?- 1 /

X

a*

+

S

5

s-2

3

s

+

1

5

s

+

2

1

6

1

6

1

1

+

+ a8 '^'

-6/~2 e

1

s

M

s-2j

+ 3J

G

15

3

C

5

— Exercises 7.2

s+1

28.

2

(s

-4s)(.s

+ 5)J

U ... (11 MIs s2

4) 30.

J£ s

s-l 1 {s + 1} J

i,i 2 2

s

s

W

32.

U

S-

33.

1

!

-9J

,,

l)(s 2

l)(s 2

+

35.

36.

^

=

#{/(*)}=

s

13

+ 4)J

e~ st dt

=

7

+

+

s

1

2

+

) > 1 J

12

2 cos

-

1

r e^- ^dt=-^~ — 3

s

jo

-

2

s 6

—

s2

1

+ 4j

+ sin f — fact that

45

1

„

1

1

1

s

3

1

2 cos 2t

sin

1

S1Il2i

=

^ (s

3.

5

-

^{(*

3

e-

-

10)

2

=

-

-sinf

- sin2t

3

6

3 s

2

2

1

+4

2

s

2

+4}

— — sin 2t

/™/(x)di = Jo°g{x)dx

if

both integrals

values of x.

>

fors>3

3

2.

jf/ te -«}

s (

y { etsin3( } =

+ 2)4

4.

(73^

^{(

1

238

=

10

e-

_J +

fs s 4- 6) 2

J

I

3!

21

_„

hV3t- -Lsin^f

Exercises 7.3

1.

-4 e

6v3

3/

t

many for a

s

1

,-5t

36

= 1 — f — cos t + sin t

I

1

+

20

5J

r^-4'^T2) = 4 COs2i+ 4

6i/3

-t^-t 2 s + l

+

4

~5 s

1

45

2-^ + ^-2 + +

and 36 we use the

J

+

5

-i

2

1=^-1/

g(z) for at most finitely

2{f(t)}

s

5 2

l4"^?4

Û

.J =

= In Problems 35

+ 4j

s

1 2

1

f

s2

8

5-4

36

'

s

_^L_ J_.^L} = J_

*

+ 4)j

+3

6*

\(s 2

f{x)

„_i

\6\/3

*

+

\(s 2

3? -if

34.

r

Utf-J

* 4

1

+ 4 )( s + 2 )}=^

(s 2

20

(

7(

}

=

>

10!

(s

+ 2) 2 +

16

exist

and

Exercises 7.3

7.

^

-

14

.

it.

18.

t

coa

W

.

ls 2

W

2

'

4s

+ 6s + 34 J 1

_

=

s 2 (s

l(s

1

„ 1

, l

s

+

I)

1

1

1 ,

+4

2

s-l + 36

1

2

7^

"

1

2s-l +

ê cos6t}

1)

, \(s

W

v-J^

J

J

+ 2)^-^

2)

(

't, + 2) 2 +

(

(S

2 (s

+

"

1

,

[

J-lJLi} = + ^

, t

=

_} + 5J

^, }

*

"

(s-l) 2

* l = ±tV \6(s-l) 4 J 6

— V

+

^

=

Win

5

U

[

22

1

*

(

+

J

3t}=y|ê +

i

y-i

9

,

s-l) 4 J

-J

21

2

_

+ V' - le-'costt) = 1 J- - 1 12 2 2 + 2 (s+

y|L e -' sin 2 t} = y{e

5)2

- *{*»«» -

I)*}

J

12.

_

(fi

10.

11.

=

S ioh3t}

5 { [

\(s

1

-

I_

s

+

2)3

-

2

—^2—^1

t

(s

l

+ 2) 2 +

l

2

=

e_2t cos

_

5 + 5. 2 "I? + 3) 2 + {.s

v-J*

5

_

s+l

(s

+ 2)3

_

(s

+

+

1)*

3 2 (s

6(s + 2)V

239

-

1

52 j

1

4

*

-

2e_2t 9in(

/

3)

ot2 2 3)

!

l22

2e- 3t coS

5

I

-»-t_

I

2

+

5t- ê- 3t sin5t

1)

fe

5

J

t „-t

_ 1 _ 5e _,_ 4te _,

3

(2e

2

J

(C

+

6*

6

_«

Exercises 7.3

23.

a?{(t-i)qi(t_i)}

= i^ S

24.

if{e

2 -'

«K (t

-

=

2)1

2 J? /e-f" '

(t

<3i

-

1

25.

26.

# {tK ^{(3i

-

(f

+

=

2)}

'X {(t

;

-

2)

m It -

- 3)} = 3 ^{

lj'tt (t

if{cos 2t

K (f -

tt) }

-

3? {cos 2(i

-

-

10e-. -

+

s

+

1

+ 2°U (t - 2)} =

2)

+

3e~ 3a

27.

—

=

2))

3)}

=

~

+

sl

3

s

- 3) * (t -

3)

+

y*

(t

-

3)}

3s

tt) It (t

- tt)} =

4f=P

» *H*(«-!)}-*M'-i)«K)}-j£ 29.

30.

1 ^{( ( -l)V- ^(t-l)} =

^/(e

1

"5 (*

33.

36

yj

^" -

37.

-

1

f

e" 2a

5)}

=

1

{ivrTyj

y{icos2(}

=

if

f

(t

-^ -

? l)<

(S

-

SJe

1

"5

« (i -

5)

+ 5e*" 5 * (i -

5)}

=

6 ,

** _

+

sin(i-7r)%((-7r)

= ^-

1

f

e~ 25

e

{-— --

^ = --f ' d$W + 4-J

-28

55

p~ 2s

+

)

r^}

S

(s2+4)

2

240

= -'«(*-2)-{*-2)K(*-2)+«*->*(*-2)

Exercises 7,3

38.

gft B mh3*} =

2

/

6s

^

1

-^(p^)-

39.

1

ds 2 Vs 2

'

^

-j.(-J—'U

+

(s

2

+

(s-2) 2 +

ds \{s

'

\(s 2

44

+

46. (e)

2{2-4
2

2

1)}

=

y {[(( / 2

55.

>

V

2

\2

J

[(5

+

3 l)

+ 3) a + 9] 2

- 2(3 + 1 )„

1

+

_2) 2 + 36f

[( 5

l)

<- 2

12 [( s

+

(s 2

y

2

1)

+

2 1]

'dsVs 2 + l//

48. (b)

57.

t-smt
if{/(i)}

2tt)}

+

i

49. (a)

50.

(d)

''

— —

l-

+

s

l)

2

2

+ 1\

2t

-

s

l]

K

(*

-

1)}

-

^{[(f -

l)

2

-

+ 2(t -

1)

+

l]

_s

J?{t-*qf(t-2)} = #{t-(t-2)«(t-2)-2'W(t-2)}= s -^2 i?{sin

1

1

l- if ^ -.f\2 -'f /

s

56.

2

s

2{l-
2

^ 36,/

s

53.

l)

47. (f)

s

52.

(s

5 + 1 j-cf-if »-'f 1(^ + 2^ + 2)2/ ^ 1

45. (c) 51.

2

I

J

+

+ 3) 2 + 9 J

2

l)

1)

ds\(s 2

l/ 6

41.

2 3

-

2

a

i?{sin

t

-

= -^{^(i-a)-^((-6)} =

sin(f

p-as

-

2tt)

K (t -

— ,-6s

2tt)}

2e

e s2

- -^-y -

a

J-^-j-

*

(t

-

1)}

1

Exercises 7.4

4.

X{y" -

V+

by}

= 5.

We

solve

= s

2

Y(s)

%{y" -2y' + S

-

if {y"}

2

-

y}

4if{y'}

-

sy(0)

=

Y( S ) -

*/(0)

=

if {0}

-

sy(0)

+ 5if{y} -

4\aY{a)

-

y(0}}

+ 5Y(s) -

-

2{sY(.s)

-

y(0)\

+ Y(s) =

2

(s

-

4s

+ 5)Y(s) -s + 5

0.

y'(0)

(s

2

-2s + l)K(s)-2s +

=

l

2s- 1 - I)*

Y(s)

(s

6.

We

solve

%{y" +

y}

=

if {1} s

2

=

1/s.

Y( S )- S y(0)-y'(0) + Y(s) (s

2

+ l)Y(s)-2s-3 VI r(S} ~

1

7.

if{/V«fT} = -if{ e [Jo

8.

(

}

s[s

X{ tco&rdr) = -¥{cost} =

10.

if|^ e-r cosrrfrj =

X\ fTsiRTdA =

s(s 2

+

l)

ifje-'cosf}

=

J

- if{tsint}

=

- (a \

11.

2[j\e^dr} =

12.

if{

13.

s

2

+

s+l

1

J (s+l) 2 d

s+ +

r* 'o

-t, T 1 Te-
s ( s2

l

1

+

1

2s

+ 2)

2s

1

^ cfes 2 +l,/

s (s 2

+

2 2

l)

(s

^ ++

l)

2

+

=

if {sin (} if {cosi}

=

^

ifn/'sinrÛîf/rsinrdxU-A^. [Jo o ds \s s 2 /

1)

m^} -^

£ sinTcos(f-7-)dr[ =

ifU

+

+3 +l

l

J

14.

S (s*

sl

1)

t

9.

,

—^~ -

=

a

)

2s

+

>

= - — -f.

/

d«

Jo

<*

\

Te-

T

—d

(1 -

1

+

X

1/

s

1

^

ds\s(s +

l)2)

=-

2

(s 2

3s 2

a (s

2

+

l

+

l)

3

2 l)

Exercises 7.4

15.

•

^l^}-!*

^{i*

16.

5-1 I

21.

y-

22

if- l {-j^—F(s)}

.

{^F(

23

+

(s(s

24.

25.

26.

27.

y -1

j

,

=

5 )}

+

e

=

- 5l

l]

t

*/(t)

=

/(r) e

|o

cos2t*/(()=

- 5(1 - T

=

1

*sini

=

/Or) cos 2{(

J

-

/

sinft

-rWr = cos(t- t) =

if-if—l-.}=e\(s + l) 2 f i?

-1

S

j

,

(s

„ 2

=

+ 4) 2 J

= -/

t

/V

* e -'=

'

r

e-<*-

T

>d7-

Jo

cos2f

*

2

1

=

I /'cos2rsin2(t 2/0

— cos 2t sin 2r)
1

rl

1

l

+ -sm4r

=

-sin2i f-f 2 \2 -( sin 2i

+

4

=

^(sin2i

— 16

+

= ^sin2i + 4

+

8

+

— cos 16

[sin2i(2sin2*cos2f)

-^cos2i 16

2

[2

+

l)

cost

2t ~ 3T

dr

= -±

1

1

o

3

,2t-3r (

3

^

tWt

- sin2(

cos 2r dr

/

—

cos 2t

Jo

4

sin 2t

2t cos 4t

1)

——

cos 2t

16

+ cos2t

+ cos 2 2t -

sin

L

(cos

2

2t

2

-

-

2f

ll J

244

sin

=

2

2t)

-

-tsin2f 4

/

-sin4r(ir

Jo 2

16

/

sin 2t sin 4t

.

— cos 2((cos4t -

- sin 4^ +

-

— - cos At

cos 2t

4

o

=

(s-2)(s 2

lo

2

-sin2f y-T

—J

f dr^te"'

e"'

2 Jo

=

-I

Jo

isin2t

cos2r(sin2( cos2r

=

1

I

v

'

2i

*

,

=

r) dr

\=e-Ue = f eê^dr = f e \(s + l)(s-2)J Jo Jo

g-*J

jif

o

JO

l)J

= -ri s(s + 2)

)

W

Jo

I

- 3t

*{ e «*

20.

+

l)[(s-l) 2

l)J

}

\s(s 2

+

(s

>

1

e

I

cos2f]

'

}

Exercises 7.4

28.

x f

=

e"

j£

2r

8inTe" s Tl 1

sint / Jo 2

=

e

* '{[(TT^TTf

i* + 5) 2 } + 4s

«2

-2(

''" r

- t) dr =

sin(t

'

=

6

-2i

— - cos t

—

sini(cos2i

— ^ sint

2

(cos

2

^sinf (-cos 29. Let

u

—t~

i

—

t so that du

h}

e

2t

sin t

- cost sinr)dT

cos2t) dr

sin

2t

2

—

-

t

+ sin 2 t + l + 2cos 2 () -tcosf = ê" 2 '(sint -

2

sin

-

^ sin2t

t

t

-

+ ^ cos t(2 sin t cost)

tcost

(cost)

dr and rt

f*{g +

~

\

2

t *

'

~

[t

— ^cos t

1)

f*9=f 30.

-

-(1 V

/

sin

sinr(smtcosr

Jo 2

o

^

-2 '

/' 1

sin2TdT - cost

— -sin ((cos 2r) —

e

2t

e

- T )dr=-

f{r)g(t

= tf(rMt-r)+h(t-r)]dr= JQ

= f f(t- u)g{u)du g*f.

f f(r)g(t - r) dr + JQ/' f(r)h(t - r)dr

JQ

= [f(TMt-T) + h(t-r)}dT = f*g + f*h 31-

32.

s

*{/(()}

33.

2{f{t)}

34.

#{/(()}

/"

/ e~
#{/(()}

JO

-

-r-

dt

at

T te

/

~

s(l

dt

=

_

+

1

fff

e

+

_ ee(

3t

Bint

^

A-

(2

-

t)e

_s(

1

#{/(*)}

s— / -e- 2"Jo

e

"sint rft

=

1 „

s2

1

dt

"

s(l

-e"

+ e-<")

-e"

s 2 (l-e- 2s )

W+ ^——^ e

J

o

36.

1 ~~

e~ aa )

2

jf te~

-^y

35.

(1-e-"8 ) 2 s(l-e- 2as )

dt

r

1

_-2a

8t

Jq

6

e" 2 "8 Jo

e-

2

e

~"s/2

1 -

+

l

l~e-' s

j

coth

— 2

Exercises 7.4

38-

2{f{t))

=

T^f\-«costdt~ 1zLri Exercises 7.5

1.

The Laplace transform

Solving for

%{y] we

of the differential equation

is

obtain

s

Thus, 2.

y


s

—

1

= -1 + e'.


is

3

Solving for

X {y} we obtain „.

1-s 2 2 s {s + 2)

,

^_J +

Thus, 3.


we obtain t£{y} =

1

-75

-.

(s

Thus, 4.

+ 4) 2 y


11

3

As

2s

As + 2'

=

2

(

is

1

+ 4i?{s} = s

+4

2

+

+ 4'

u+


2e

_4t

:

.

is

s

Solving for f£{y}

s

2

+

tete~

1

_3 e -a.

l


*tf{»}-|f(0)

Solving for ?£{y}

11

'

l

we obtain

vr

1

1 (

52

+

l)( s

1

_l)

246

s+1 1 1 + 13j-1

2('

.

Exercises 7.5

Thus, 5.

y

The Laplace transform a

Solving for

2

%{y} we

=

X{y) -

s+b + 5s + 4

2

4

2{y) -

-

sy(0)

13

X{y) =

0.

obtain 3

-6s +

13


*{y} -

*s(0)

-

2

(s-3) 2

2

+ 22

= --e 3t sin2(.

T/


Solving for ^£{y)

39 + 4'

- 6 \s2{y) - 2,(0)] +

y'(0)

Thus,

2

11

1

is

3

s

0.

1 -4t - -e


s2

7.

+ 42{y] =

y(0)}

3s+l

4 -t

=

j/

%{y) we

is

obtain


Solving for

+ sin().

- y'(0) + 5 [sX{y} -

sy(0)

Thus,

s

^(cosf


a2

6.

—

~e*

-

j/(0)

6 [s

is

%{y] -

j,(0)]

1

+ 9 X{y} =

we obtain 1

,

+ s2

2

s (s

~

1112

2

3)

2

27 5

10

1

27 s

9

-

3

9

1 (s

-

3)

2

'

Thus, 2

2

1

^27 + 9 8.

The Laplace transform s

Solving for

2

f

-27 e +


X{y} -

sy(0)

-

y'(0)

3

1

-

4 [s

10

T

te

is

%{y) -

y(0)}

+ 4 %{y}

6_ »4

we obtain s

5

-4s 4 + 6

3 4 (s

-

2)

2

4 s

9

1

8 s2

3 2

1

4 s3

4

Thus,

247

3!

1

1

4 s

_

13

2

8

(s

-

2)

2

'

Exercises 7.5

9.


Solving for 10.

2

2{y} -

%{y) we

Solving for !£{y}

2

~

ay(0)

i/(0)

is

- 4 [s 2{y} - y(0)] +

1

=

Z{y}

obtain

The Laplace transform a


sy(0)

-

y'(0)

Thus, y

.


X{y} -

-

_

= jê 21

.

is

- y(0)} + 5 X{y} = ~ + ^.

2 \sJ£{y}

we obtain iVi

_ ~

2

4s

s 2 (s 2

7

1

25

s

+s+1 _ ~ 2s + 5)

-

7_ 1

25 5

7

1

1 t

5 s2

25 (s

+

1 _1_

-7s/25

+

109/25

-

2s

+5

+

2

5 s

s

s- I) 2 + 2 2

2

51

2

-

|

25 (s

l)

2

+

22

Thus,

y=

+

2l 11.


^^

e

'

COs2t

+

l

'

e Sin2 *-

of the differential equation is a

Solving for ££{y}

2

X {y} - sy(0) - y'(Q) + X {y}

=

.

we obtain a

2{v} =

3

(

-

s

2

+a

s2

+

1 )2

y

=

cos t

s

a2

1

+

s2

1

+

1 ( a2

1

+

1

)2'

Thus,

12.


— ^ sin t —

of the differential equation s

2

2{y) -

sy(0)

-

cos

t.

is

y'(0)

+

16

=

2{y}

-

.

3

Solving for

%{y} we

obtain

2 W _~

s

2

s(

+ 2a + 1 _ ~ s 2 + 16)

Thus, V 13.

^

{3-2) 1

5!

^

6

=

4


2

=

J_

1

16 a

115 00344 +

16

16


2{y} -

sy{0)

+

- y'(O) -

[s

15

a

16 a

2

+4

+

2

4

1

2 a

2

+ 42

'

1

+

sin4(

"

2

is

X{y) -

3/(0)]

3

= (s

248

-

"

1

l) 2

+

1

'

.

Exercises 7.5

Solving for

% {y}

we obtain ,

_

111

1

„ s

-

2

v

+

2s

2 s

2)

s-1

2 (*

-

1j

1

1

2

+

-

2 (a

1

l) a

+

'

1

Thus,

=

y 14.


z

^

-

r 4-

ê* cos


X{y} -

-

53/(0)

-

j/(0)

2

ê' sin

t.

is

=

[s^{y} - S (0)]

-

(s

Solving for Z£{y} we obtain

v/

s 2 (s

-

4 s

2)2

Thus, !/

15.

The Laplace transform 2 \s

3

X{y] -

11

=

(s

+

s

1

it

4 a

-

1

2t

+

te 7 4

2

-

2

1

1

4 (s

-

'

2)2

is

2 y"(0)]+3[ S i?{t/}-sj/(0)- !/'(0)l-3[ S if{j/ }- y (0)]-2^{y}

115 s-1

+3

2s

,

2

+ 74 ( - 74 e 7 4

we obtain

Solving for

w

4


- V(0) "

2

s (0)

111

1111

1

,

l)

l)(s-l)(2s

8

1

+

+

l)(s

2s +

2)

1

9s +

18

l

1

,

l

= 7

T

.

1

9s + 2'

l/2

Thus,

16.

The Laplace transform 5

3

£{y) -

Solving for

2

s (0)


is

- V(0) - y"(0) + 2\s 2 X{y} - sy(0) - y'(0)) -

[s

2{y} - y(0)] -

2 <£{y)

we obtain a

2

+

12

>-l)(s+l)(s + 13

60

13

1

s

-

1

2)(s2

16

1

+

20 s

+ 9) 1

39 s

1

+

+ 2

3

13

s

130 s 3

+

9

65 s 2

Thus, 13

^60 e

,

13

-20 e

,

+

„

16 C

+

39

249

3

„

CO83(

l30

1

-

Sin3t

65

-

+9

'

=

3

^+

.

fl

Exercises 7.5

17.



s*X{y} Solving for 18.

%{y} we

obtain


Solving for

4

%{y) we

- *V(0) -

3

s y(0)

is

g

%{y) =


2{y} -

A

s y(0)

=

Thus, y

.

2

-

sy"(0)

- 2{y} =

y"'{0)

0.

cost.

is

- sV(0) - V(0) -

f"'(0)

- 2{y) = \. 3

obtain

~

sV

-

1)

111 +

~

4 s

11

-

+

4 s

1

+ 1

11 2 a2

+

T

Thus, y 19.


%{y} we

-t

It 1 1 + -e --e -t + -smi. £

£

of the differential equation S

Solving for

=

.

is

-€-*.

X{y}-y(0)+2{y} =

obtain bests

=

5e"

+1)

S

3

+

1-

1]

Thus,

y 20.


= 5%t-l)-5e-^°U{t-l).


is

s2{y}-y(0)+Z{y} = \3

Solving for

%{y} we

V«. 3

obtain

*{y} =

2e" s($

+

l)

s(s

+

1

l)

-2e -a

1

8

s

+

1

1

.3

1

3

+ l]

Thus, y 21.


= l- e -'-2[l-e-('-


1

£

']

W((-l).

is

a

o

we obtain

Solving for !£{y) 1

l

"

_ ~

1

s 2( s

+

S

2)

_s

s+ 1 _ sHs + l)~

ii.ii 4 s

1

_J 2s !+ 4s + 2

250

6

-t

1 1

4i

+

1

2

Exercises 7.5

Thus,

»=-i4 + k '-H<'- >-k 2

22.


i

of the differential equation .2 2

s

X{y} -

-

sy(0)

y'(0)

2<1

-"

is

+ 4 X{y} = l-tl s

Solving for

a

%{y} we obtain

2{y} =

1

s(a

2

s

Ills

1

—

e

+ 4)

s(s

+

2

4s

4)

4 s

2

12

+4

+4

2 s2

_, ri

—

e c

Thus, y 23.

=


~ cos 2t ^

sin It

- r i-icos2(i-l) «(t-l).


s

Solving for ?£{y}

-

i

2{y) -

2

sy(0)

-

j/(0)

is

+ 42 {y} =

e"

2* 8

s2

+

I

we obtain i .3

2

1

i

s2

+

6s 2 + 4

1

Thus,

y 24.


Solving for

2

%{y} we

=

coe 2f

+

\

Bin(f

-

2jt)

-


^{y} -

-

aj,(0)

i/(0)

-

5

sin 2(f

-

27r)

(t

*tl

-

2tt),

is

-

[s

y(0)]

+

=

6 <£{y}

obtain

X{y}=e =

e

a(s-2)(s-3)

+

1111 6s

2 s

-2 +

1

(s-2)(s-3)

11 ^ 3

1 .

3-2

s~3

+

„1

s

Thus, »

=

+ .6

25.


2

-

6

1)

+

-

2(

e

is

g-jre 2

31

3


s

e

2{y}-sy(0)-y'(Q)+X{y} =

i

Us

g— 2irs

.

3

i

s

4 s2

+4

Exercises 7.5

we obtain

Solving for ?£{y}

'1

3

ITS

S

.3

+

2

- e~ 2™

1

-

s

rl

s

.s

l.

2

1

+

+

s

l.

+

2

l

Thus, y 26.

=

- cos(t - n)\%t -

(1


2

Solving for

2{y) -

-

-

[1


-

S y(Q)

^{y} we

it)

y'(0)

cos(t

-

-

2ic)]°tt{t

+ sint.

2tt)

is

1

+ 4[s %{y) - 2,(0)] + 3 X{y] =

e~ 2a

-

-

e~ is

+

e

_6s

obtain 1

%{y} = =

tj

3 s

2 s

—

e

-4a

+

1

65 +

l

— e -2s 3

1111 2s+l

3s

+

11

11

3 s

2 s

11

-

6 s

+

+

+

,-6s

e~

3

111

+

+ 3.

6 s

l

1111 +

2 s

3 s

+ 1

11

T

+3

6 s

Thus, y

1

1

3

2

1

-i

i-A e -('-» + Ae-"('-«

-st

6

A

+

3

2

"!/((-

D

Z

+

2)

A _ A--(t-B) + A P -3f'-8)
27. Taking the Laplace transform of both sides of the differential equation and letting c obtain

2{y"} 8*2{y} -

sy{0)

s

2

-

y'{0)

+ X{2y'}+X{y} =

+ 2s 2{y) -

%{y} -cs-2 +

2s

2y{0)

X{y} (s

2

2c

+ 2s +

+ X{y} = + %{y) = l)%{y}

=

+ 2c + 2

cs

+2 (a + l)3 2c

(

+

5

l)2

+ 1-1 (s + l) 2 c

s

+

(a

c

+ l

'

+2 + 1)3

2c

s

(s

+2 + l) 2

'

Therefore,

y{t)

= c^" 1 { -1^} + (c + 2) if"

1

1

252

(J^p}

=

OB

'

+

(c

+ 2)te-

( .

=

y(0)

we

Exercises 7.5

To

find c

we

let

y(l)

=

2.

Then

2

=

ce

y(f)

_1

+

(c

+

2)e

_l

= (e-l)e^ +

=

(e

2(c

+

lje^ 1 and c

+

=e-

1.

Thus,

l)(e-'.

28. Taking the Laplace transform of both sides of the differential equation and letting c

=

y'(0)

we

obtain

%{y") s

2

X{y} -

ay (0)

- y'(0) s

2

%{y} +

9s

X{y]

+

r-

c

9y(0)

2{2Qy)

= ^{1}

+ 20if {y} =

- 9s ^{*,} + 20if{y} =

J s

(s

2

-9s + 20).S?{i/} = - + c s

-

s(s 2

+

9s 1

s(s-4)(s-5) 1/20 a

s2

20)

+

- 9s +

(s-4){s-5)

c_

1/4 a

—4

20

c

s

—

5

s

—

4

c

+ — 5 a

Therefore,

To

find c

we compute B '(i)

and

let y'{l)

=

0.

= _ e * + e «- c

(4 e

*-5e Bt )

Then

0=-e 4 + e 5

-c(4e 4 -5e 5

)

and C

~

g

4e

5

4

~ -

e

4

5e

5

_ e- 1 ~ 4- 5e

'

Thus,

W) 29.

6

20

4


+ 5

4-5e^

I

20

+

4(4-5e)

of the given equation is

y{/}+if{t}i?{/} = y{(}.

5(4

-

'

5e)

Exercises 7.5

Solving for Jf{/}

30.

we obtain if{/} =


of the given equation

=

Thus, /(*)

.

smt.

is

#{/} = 2{2t} - 4%{smt}Z{f}. Solving for

%{f] we

obtain 2s 2

yrf1 _ XUJ

Thus,

31.

2

(s

%{f) we

M

obtain

(s

-

l)3( s

+

8 s

i)

/(t)


2

+

|*

l

.

e*


5%/5s 2

sin \/5

t.

4

1

te *

|

1

(s

-

l)

2

+

(S

Thus,

11 l)

3

8 a

is

+

1)

=

/(*)


3

5

4e~'


+

(s

1

%{f} we

obtain


#{/} =

.

S

+2 {sinf},

l)

2+

%

(

is

=

Thus, /(()

if{l}.

=

e"'.

1


2{f} = Solving for if {/}

T

+

- lte + 4iV

if{/}+if{l}y{/}

34.

-

we obtain

Solving for

Solving for

12 4 (a

+ J(V -

_t 2{f} + 2X{cost}%{f} -4J?{e }

33.

+ 5'

is

-

+

V5

8

1

5 a2

113 +

s

Thus, 32.

+2 _ + 5)


,2

^

2

=

/(()


Solving for

s

is

if{cosi}

+if{e-

we obtain

254

t

}i!?{/}.

(s

+

1)

3

'

+

1

Exercises 7.5

Thus, 35.

=

f{t)


+ sint.

cost

of the given equation is

= #{1} +%{t} + Solving for

X {/} we obtain s^s +

Thus,

l)

-

16

/(f)

=

s

4

8s-2


Solving for if {/}

a

2

^

f1

~

f{t)



+

\ sin2t. 4

11

1

~2s

=

±t

-

1

2

3!

12 5 4

^t 3

'

.

is

{/} we obtain

v/„l

j3

~

g2

+s

2

+

l)

s(s

Thus, 38.

2t

is

2s"

Thus,

Ji?

\ cos

y


=

^

2

fl*

sin*


—

^

,

+ (

l

1

2s

2 (s 2

+

Solving for if{/}

we obtain %{y)

=

—+

7

(s

Prom equation

(3) in

0.005


—+

i

+ 50

of this equation

0.005[sif{i}

'

.

Thus, y

=

te~ 3t

Jtf

{1}

.

3)

ft /

3

is

the text the differential equation di

l)

sin*.

sX{y} - 3/(0) + 6%{y} + 9y{l}^{j,} =

39.

2

4s 2 + 4'

4

we obtain

U'

Solving for

2s 2 +

2

2


jfr

37.

8s +

+ \e~ 2t +

a 36.

2{t 3 }2{f}.

1111,1

3

|e 2t

|

i(r)

dr

=

100[1

is

-°U(t-

i(0)

1)],

=

is

- i(0)] +if{i} +

50 5

=

1

1

s

3

100

e

0.

Exercises 7.5

Solving for !£ {i}

we obtain

*»-îU-.-U. + 100) 2

(s

Thus,

i(t)

40. From equation

(3) in

=

20,000fe-

100i

-

20,000(t

-

the text the differential equation

rf»

005 dt

r*

1

+ + *

i{T)

O02

lje

'

K(t

-

1).

is

= m[t -

dT

Jo

-100 " -1

"

1)

(t

-

" * (* "

*(0)

1)1.

=

or

— + 200i + 10,000 The Laplace transform

/

i(T)rfr

of this equation

=

-

20,000[£

i(0)

1)],

=

0.

is

+ 200 y {i} + 10,000 -

s

-

=

1

1

20,000

we obtain

Solving for

^W

200

"s(s+100)

2(1

6

J

+

s

s

100

(s

+

(l-e" s ).

100) 2

Thus,

=

2

-

2e-

100t

-

200(e-

41. The differential equation

10Ot

-

2K(t -

+

2e

_100 < t - l >

-

+ 200(t -

1)

lle"

100 ''- 1

'^ -

is

fl^ + The Laplace transform

1)

of this equation

lg=

£be-

ta

fl(O)

1

-0.

is

we obtain

Solving for

E /R

EC

'

> + k)(RCt +

When 1/RC

1/fiC

(*

+

Jfc)(*

+

1/ftC)

^we have by partial fractions

W=^ When

1)

=

( V

l/(l/RC-k) _ l/(l/RC-k) \ _ Eo s + a + l/RC R } fe

A we have

£b

R

1 (s

256

+

k) 2

1

(J-

l/RC-k\a + k

a

+ l/RCJ'

1

Exercises 7.5

Thus, 42.

The

R

R


is

10^ +

IO5

=

-

30e'

30e'^/(f

-

1.5).

OX


of this equation

is

3e

3

Solving for if {q}

ls

_ liBa

we obtain

^-(-'i)-^T + i^"(^^)--Thus,

43.

The


is

2.5^ + The Laplace transform

Solving for %*{q}

of this equation

12.5 9

=

5 s!/(t-3).

is

we obtain

X{q}

2

=

+ 5)

s(a

e

-3.

_

(

.

5

1

1 _ H s

5

s

+ 5)

e" 3 *.

Thus, g(t)

44.

The


=

—

f 5

3)

—

e | 5

-B(t-3)^^

_

3 )_

is

or

50^ + The Laplace transform

of this equation

50s 2{q} Solving for

%{q) we

=

100$

+

Eo[^(( -

1)

-H(t -

3)],

g(0)

=

0.

is

100

^{5}

=E

(-e~ s - -e" 3a \s

3

obtain „-3 S

50 [s(s

+ 2)

s(s

+

2)

50 12

s

+ 2j

2 Is

s

+ 2J

'

J

Exercises 7.5

Thus, ?

45.

The

W=

1 e ll K -

" 2(t " 1,

)*(*

-

!)

-

(!

^

-

2(£

- 3)

)^C "

3)]


I + lOi The Laplace transform

=

sin

+ cos

t

of this equation

(t

- ~j^(t -

t(0)

,

=

0.

is

se- 3 ™/ 2

1

we obtain

Solving for

1

y)

(a 2

'

+

l)(a

+

+

10

+ 10)

(s

2

'

101 Vs

s2

+

+

l)(a

+ a

1

+

10)

W

N

+

1/

2

1

101

/_^ +

Vs

+

10

W. s

2

+

1

+ 1

s

2

+

1

Thus,

i(t)

=

rjrr

101

(e~

10t

^

- cos t +

10 sin t

+ _1 (.uw-^ffl + 46.

The

differential

equation


(<-|) + *. (« -

is

of this equation

8<£{i}

Prom Problem

10 c„s

31, Exercise 7.4,

is

+ 22{i} =

j-X{E(t)}.

we have #{£(()}

1-e - '

=

s(l

+

e

Thus,

I_e

(•+!)*«>-£t a(l + e" 258

s )

-

*)

.

Exercises 7.5

and

1-8-

1 = -

L

s{s

+

i L/R L 1

1

R

s

+ e"

L/R + R/L

s

1

s )

(1

-

1

e-")(l

-

e"

8

1 1

+ e~*

+ e~ 2a - e~ 3s +

-3s is {\-2e~ 3 + 2e~ 2s -2e-' +

+ R/L

s

-e h s(s + R/L) 1

R/L){1

)

).

Therefore

=

i(t)

-[1

-1%t-

I

e

R

«

1)

+ 2%t-

-Rl/L

+

2e

- 2%t -

2)

-R(t-l)/L

_

3)

+

1}

_

2e

-R('-2)/t

^( (

I (l - e -«/£) + £(-1)" (l - e -«t-«)/£)^( t _ „). | Tl=l '

47.

The

differential

equation

is

f + fi=^( The Laplace transform

of this equation

i(0)-0.

=

-iy{£(()}.

is

s^{i} + |y{i} From Problem

t ),

we have

33, Exercise 7.4,

=

1

-=

+

1

1

Thus,

R\ „

/

f .,

11

Jl/L)

1

111

and

+

L

s*{s

1

fL/rt

1 +-

L

rt/L)

s(s

L 2 /fi2 |

L

A R

s

\

2

1

_ L/R +

s

s

+

L 2 /R2 \ s + R/L)

L/R s

+ R/L

1

e

s

(L/R

|

L

s

L/R + R/L /

1

-

e

_2

)

Exercises 7.5

Thus, i(t)

<

For

f

<

2

-

5 (' " 5

+

e_nt/L

I

+

)

=

s

+

-

('

e_i!t/L

i

^

+ )

U('-*+* a_ * /L )+A( The

e

~ fl!t " n)/L

)^ " »)

we have t(t)

48.

"

1 t

differential

equation

1

of this equation s

Solving for !£{q}

-

)^ -

- ,ltt - 1)/£

e

1

).

!)

^*<

2

-

is

^+20^ + 200? = 150, The Laplace transform

~ fl( '~ 1,/L

"e

2

Z{q]

g(0)

=

=

?'(0)

0.

is

+ 20s^{9} + 200^f{g} =

— 150

.

we obtain 3

150 s ( a2

+ 20s + 200)

Thus,

î = q(t)

3

1

3

4(s+

4 a

LV

+

10

10) 2

+

10 10 2

3 - luc —lot .nx - 3__ 10t 1K cos r _ r3 e -i°' cail0tlot e sin ^ 4, 4 4 .

4 (8

+

10)2

+

102'

lOt

and i(f) Hcj

If

E{t)

=

150

-

150 %t

-

2)

,

=

-10t 10t

=

(t; 9 '(t) 5

iur I5e-

sinl0f.

then 150 s(s 2

9 (i)

=

| _

C o S lot

_| e 49. The differential equation

-lO(t-2)

+ 20a + 200)

- ê" 10

silll0(

j_

'

2)

-

3

_3 e -io(

t

-2)

is

+ 20$- + ^| at* at The Laplace transform

sin lOt

(1-.-*)

of this equation

**X{q}

+

1009

=

120 sin

10*.

is

20 S #{9}

+

100i?{9}

260

=

120 -

c03lo((

_

2}

Exercises 7.5

Solving for f£{q}

we

vj

obtain

1200

\

(a

+

3

+

2 2 10) {s

1

+

5 a

100)

3s

1

10

+

(a

2

10)

5 s2

+

10 2

'

Thus,

5

5

and

=

50.

The

steady-state current

is

The

differential

equation

is

=

q '(t)

-60ie- 1O(

6sinl0(.

|f + 2A§W 9 =f, The Laplace transform

+ 6 sin lOt.

of this equation

S

2

5 (0)

=

=

,'(0)

0.

is

^b -

+ uj 2 %{q} =

2{q} + 2XsX{q}

s

or (

s

2

+ 2As + W 2 )^{g} = we obtain

Solving for if {q} and using partial fractions

2^ = Eo(l/u L \

For A

> w we

write

s

2

2

(l/w 2 )s s

s

+ 2As + J1 =

X{q} = for

<

we write

= Ek,c(ls

2

for

A

<

e-

Ai

-

J1

)

,

s + 2X + 2As + w 2 ;*

so (recalling that

w2 =

1/LC,)

X

-(A 2

-^

2 )

(

s

+

(s

{a

+

A) 2

—

- w2 1 -

cosh V'A 2

+ 2Xs + J1 =

,a

Thus

+

A) 2

(A 2

s2

+X

S (s

-

A) 2

fl

ImP \s

J

A)

2

-(A2-w2);'

X > u, q(t)

For A

+

1

E c{ -,s

Thus

(s

+ 2A/w 2 ^ _ E

+ 2As + w 2

2

f i.

+

(u>

+ A) 2 +

(w 2

cos i/w 2

-

2

-

A

- A2)

2 ]

,

f

sinh

Â 2 - w 2 ^

so

(s

+

A) 2

+

(u 2

—

X2 ) J

\/w 2

-

A2 j

'

w,

= E C (l - e _A

*

A2 t

-

^

sin

1

.

.

Exercises 7.5

For A

= w,

s

2

+

2A

+ w2 =

+

(s

A)

E s(s

Z,

Thus

for

=

A

+

A)

2

and

L\

2

s

The



N

£b /l

A) 2 /

LA 2 ^s

1/A

X

+

(s

1 s

+

A

(s

+

A) 2 /'

%{q} we

= -EoC(l-e- A( -A(e- At ).

is

of this equation

S

Solving for

+

s

«, 9(i)

51.

1/A 2

fl/X 2

is

^{ q } + —X{ q

= E

}

1

L

+

s

k

obtain

E

E

1

fl/(k 2

+ l/LC)

s/(k

2

+ 1/LC)

+

*/(*

2

+ 1/LC)\

,

L

+

(s

k){s

2

+

L

1/LC)

s

\

+

k

s

2

+ 1/LC

1/LC

)•

and 32

=

Thus,

Eq

=

?(*)

L(k 2

52. Recall from Chapter 5 that so that k i(0)

=

0,

=

16 lb/ft.

x'(0)

=

[e~

+ l/LC)

kt

-

mx" = -kx +

Thus, the

cos [t/^/LC)

f(t).

sin

£

=

sin(i

—

2tt)

we can

= W/g = 32/32 = + 16x — /([). The

sin


=

sinf

-

sm(i


s

2

<

t,

t

<

2?r

write /(f)

-

a

2ir) il(t

-

2tt).

is

1

%{x} + 16X{x} = ,2

Solving for !£{x}

x"

is

sin (t/y/l~C)]

0. Also, since (

and

m

Now

equation

differential

+ k^LC

„-2tT3

+

J

s

2

+1

we obtain „-2tts

y+

2 16) (s

-1/15 s

2

+

16

+

+

1)

1/15 S2

+

l

262

(s

2

+

16) (s

-1/15 s

2

+

16

2

+

1)

1/15 s2

+

1

_-2jts

.

1 slug,

initial

2k

conditions are

.

Exercises 7,5

Thus, x(t)

= -^r sin At + t>U

\

x(0)

+

^-

and

x'(Q)

=

-

sin 4(t

2jt)^

-

-

2tt)

<(< >

m

Now

0x'.

equation

is

+

x"

- W/g = 4/32 —

+

7x'

=

I61

'

5 slug,

The

0.

initial


21

,

f

{X

27r) 1/ ( t

-

2tt)

2tt

_ ,„ 2r „, ,3 ^{z} +-s + 7si?{z} + — + \&%{x)

Solving for !£{x)

-

2-jt.

mx" = -kx differential

sid(*

lo


0.

(t

bU

t

Thus, the

2 Ib./ft.

= —3/2

t

0,

53. Recall from Chapter 5 that that k

sin

-^sin4f + ^sint,

f

=

~ lo

=

and 4

=

2k so

conditions are

is

0.

we obtain

_ -3s/2 - 21/2 _ _3 ~ s 2 + 7s + !6 2

+ 7/2

a

VlE/2

7^*15

_

'

(s

+

7/2)2

+

10

(715/2)2

(

s

+

7 /2)2

+

(v^5/2)»

Thus,

2 54. Recall from Chapter 5 that so the differential equation

^x"

The

initial

Solving for Z£{x]

f(t).

Now

2

m

= W/g =

x"

+ 9x =

16/32

=

1/2 slug, and k

a:(0)

+

4 sin 3t

=

x'(Q)

2 cos

=

&

or

8 sin 3£

^


0.


\

-

+ 24 _ 2 + 9)2-3

4s

~

{s 2

2(3)a

2(3)

12

+

3

(77^5

27

Thus,

55.

The

=

2

-ts'mZt


4.5,

+ 4 cos 3f

we obtain

VI

x(t)

=

is

+ A.bx =

conditions are

10

2

mx" = -kx +

+

4 -(sin3t y

~

3tcos3t)

=

2

-t sin 3t o

+

4

4

y

o

- sin 3t - -tcos3(.

is

dx 4 Taking the Laplace transform of both

sides

dx 4

and using y(0)

El

=

^M-V'(0)-/'{0) =

j/(0)

|^

=

we obtain

is

Exercises 7.5

=

Letting y"(0)

c\

and

=

y"'(0)

we have

C2

v/.

c\1

i

c2^

,

i

,

]

wou

1

1

w

so that i

To

and

find ci

=

Solving for

From 56.

The

this

ci

we

=

=

and

we obtain

find

C2

y(^)

+ c2 +

ci

y"'(L)

differential

1

1

^î

2

and

4

we compute

C2

y"(a:)

Then

?

1

i

a:

=

equation

^v>oL

^

-

c2

+

^

x.

system

yields the

c\

=

2

/EI and

Umax

=

y{L)

e2

= —woL/EI.

=

Thus,

.

is

EI^ = vjo[H{x - L/3) -%x - 2L/3)]. Taking the Laplace transform

of

both

sides

s*X{y] - sy"(0) Letting y"(0)

=

c\

and

y"'(Q)

—

c%

and using y(0)

y"'(0)

=

|J

±

=

y'{0)

(e~^

—

we obtain

- «-»•/»)

.

we have

so that 1 3 — = -ax 2 + -c 6~*2 x + 1

1

y(x)

1

24

'

To

find ci

and

C2

/i,

y

,

(a:)

WQ — EI

L\ 4 m ,/ 3/

L\

(

2L\\,f

2L

3 y

we compute

=

ci

+ c2 x +

1

-

— wo

*-!)

*H)-(*-?) *(-t)

and

El

264

Exercises 7.5

Then y"(L)

=

y"'{L)

=

w

1

i

^ 57.

The

and

c\

C2

2i-j

<>

2

,

c2

Solving for

system

yields the

we obtain

cj

2L

DO

+

_ ^L

= \wqL 2 /EI

w L2

6

£/

w L

1

and

= — ^woL/EI.

c2

Thus,

4

,

21, «B[a:

I

24

18

equation

1

r

Ll

/

differential

2

£7

x- -

~

£7 ^12"

1

3

is

E/^ = ti*o[l-
Letting y"{0)

=

c\

and

4

sides

- y"'(Q) =

^{y} -

sy"(0)

=

we have

y"'{0)

C2

=

and using y(0)

w

^

1

\

(l

y'(0)

-

e"

=

we obtain

Ls / 2 )

.

so that ,

=

.

y{x)

To

find ci

and

C2

1

-c,x

2

1

+ -c2 x

1

1

— 1«0

we compute (x)

=

ci

y"'(x)

=

C2

y

+

c2 x

+

-

—

and

Then

=

y"'(L)

=

yields the

r

1 ,

w

system 21

L2

-f4

w (L\ Solving for ci and c 2

we obtain

c\

= %woL 2 /EI

= Cl+C2Lr + 1 «j

and c2

2

£ ___ =0 3 iu

£

= — ^wqL/EI.

Thus,

Exercises 7.5

58.

The


and using

j/(0)

=

y"(0)

=

is

d^y El-r—r dx*

=

— =— El d^y

mjo or

r

Taking the Laplace transform of both sides

.

dx*

we obtain

s^{y}~ S y^)-y"'{ Q = )

f

1

-.

I

Letting j/(0)

=

C]

and

y"'(0)

—

C2

wc have

" £/

+

55

+

S2

S4

Then

To

find c\

and

C2

we compute

Then y{L) =

=

yields the system

^ = -w Solving for cj and C2

59.

we obtain


c\

= wqL s /24EI

and


C2

= —wqL/2EI.

Thus,

is

-|[^M- y'(0)]-^M = ~ Then

and

= -4-

+ This

is

3 a first-order linear differential equation with integrating factor e/'

s

3

^{y}

= - (\ds = J

so 5?{y}

=

—2 + —c

s

and

266

s

+

c,

^

ds

=

a

3 .

Thus,

J

Exercises 7.5

60.



~

\s

2

Z{y} -

-

2/(0)]

2

is

^[sX{y}\

+ 2%{y} =

0.

Tlien

-S 2

- 2{y}\ -

(

2 S X{y}

-

2s

- Z{y}

}

-

22>{y}

+ 2%{y} =

and

This

is

a separable differential equation so

^§jjjy

61.

=

=

ln^}

-~l

-21n( S

+ 2) +

c

=>

{»}

= ce" 2

^

= ci^ + 2)"

y(£)

(a)


.


is

(as

2

+ bs + c)l£{y} =

a.

Solving for !£{y}

we obtain

^ {V} = as

2

+ bs + c-

Thus, Vl

(b)

Now

if

=

%?{g(t)}

=i?_1

{as 2

+ fe +

C

}-

G(s), the Laplace transform of equation (13) gives

\as 2

+

&s

+ cj

a

I

w

as 2

+ 6s + cj

a

by the convolution theorem. 62.

From

part (b) of Problem 61 the solution of the given initial-value problem

-

sec ry\ (t

/

t) dr

where

y\=!£

Jo

1

1

1 L

+

,

}

is

=

sin

t.

1

Thus, V2

1

-21

and

cife

=

1

f'

—

(sin

I

t

cos t

— cos t sin t) dT —

Jo cos t

—

t

sin

t

-+-

(sin i) /

Jo

cos (In cost!. I

dr—

sin t

(cost)

/

jo cos T

dt

Exercises 7.6

Exercises 7.6 1.



is


is

so that

2.


p~ s

2

so that

S 3.


=

2e-

1

+

e-<

i_1


^(t-l).

is

" ?TT( + e

*<»> =

1

a

")

so that

=

i/

4.


sinf

+

sini^((-27r).

of the differential equation is

*<»>so that 3/

5.


=

^sm4((-27r)

fl

l/((-27r).

of the differential equation is

S

-\"

1

bo that

J/

6.

=

sin( t

-^(t-f) + sin( -f)^( -f)=-cos^(t-0 + co ^( -0


(

S

i


is

so that

y

=

cost

+ sint^fi. 268

2?r)

+*U(t

-

4w)].

(

Exercises 7.6

7.



2 i

is

+ 2s

11

11

2 s

2 s

+

(l

2}

+

O

so that

_

1 8.


2

i e -2£*-i)

-2 8= 3

e

(s-2)

s(s-2)

is

3

1

4 s

1).

-

11

1

4s

2

2 s2

+

r

1

_1

|.2s-2

so that J

9.


O

Of

J.

-I]*(t-2).


is

„-2tts

so that

=

!/

10.


e

-2(t-2,)

sini 9/(f-27r).


%{y} =

is

+

{s

1)=

so that

y 11.


=

(t-l)e- {t


+ 4s +

52

3 (s

s2

13

+ 2)1 + 3

2

+ 4s + s

+ (s

~ 1)a U(t-l). is

13

+2

+ 2) + 3 2

2

+

1

3 (s

+

2)

2

+ 32

so that 2

y

1

= -e~ a sin 3t + e _2£ cos 3t + -e _2((_,r) sin 3(f + ie- 2 *'- 3 "! S in3{t -

12.


X^

of the differential equation 1

(s-l) 2 (s-6)

+

-2s e

5-1

5 (a

+

-

3jt).

is

-is e

(a-l)(s-6)

11 25

c

37r) «(£

-

11 l) 2

25

s-6

1

1

5

s-

tt)^ (t

- tt)

111 2s\

Exercises 7.6

so that

_A e t-4 + 5

13.


l

e

6(t-4)

m{t

-

4).

5


is

13

12,,,,

£/

6

s"

so that 3

.

Using y"(L)

=

=

and V

14.

FVom Problem

we obtain

EI

A

X

6EI

+

X

=

=

and

Assume

io

>

X

2)

2]

\

and

g(t)

=

i§(x-f)\(,-f).

we obtain

!^(lV

15.

X

we know that

13

îv"(o), 2 + i/'(o)^ + Using y{L)

6EI\

3 I2

0<

x3 )'

a;

< 4

^ so that

{ ^

JfOO '

FOB

rO

dt

=

/

9(t)d(t-to)dt

=

g {t

)

fV

3l

e-

st

S{t

-

)

g(t)6(t

=

e-

st

-

t

)

dt

-

6(t

/

°.

-co

16.

If

/(0

=

f 0,

{

(

j

*

e

_3, )

<

(>0

then from

/°° /(*)*(* J-oo

-

(7),

4) dt

=

f°°

J0

270

<5(f

-4)dt

= We' 12

.

Exercises 7.6

17.


2


X{y} -

sy(0)

-

y'(0)

is

+ 2{sX{y} -

y(0)}

+ 2%{y} = -e" 3 "

so that

*M- (s+ + 8

1

1)2

+

1

(

S+

l)2

+

1

and y 18.

=

e"'coB(

-


e~

(t

~ 3w)

sin((

- 3ir)%t -

3jt)


=

(

e" cost

is

* <»> = i so that

y

Note that 19.

y'{0)

=

— — eiiiwt. w

1.



is

so that

Li

Note that 20.

i{0)

=

1/L

^


0.


s

+

is

s

5

so that

V

=

S(t)

-

5e" 5t

.

+5

+

e

- (t - 3,,)

sin t°lt(t

-

3tt).


Chapter 7 Review Exercises =

te'

, 6t

+

1.

2{f{t)}

2.

%{f(t)}=

3.

False; consider f(t)

4.

False, since /(()

5.

True, since lim e _, co F{s)

6.

False; consider /(£)

jo

dt

r°°

^

,12

(2-t)e~ et dt=-^-ê- s

[

10.

f e-

if(e- 3 'sin2(} 1

1

11.

et

=

=

7

dt

= -U- 2a -e- i3 )

= (e

(s

t

-1 ' 2

10

( )

—

+

1

=

.

e

10i .

= 1^0. and

3)

2

g{t)

—

17.

if

J Is

- tt)} =

if {sin2((

Utf-i ^ -103 + 29/ £

2

7.4 in the text.)

1.

4s S2 (

.£{flm2t V(t

Theorem

+4

^{tsin^^-l^

12.

(See

f

+ 4)2

—

- jr)H(t - tt)} =

s

~

5

5

l(s-5) +

22

2

-™

?

|

2

-e

(s-5) 2 +

272

1

22 /


18

= 20

.

cos7r(t

-

-

+sin7r{(

1)

-

-

1)

= H-^(t-^1 = ^^^t ^{htÎ L + nV + tmr i 2s 2

I

21.

y{e~ 5( }

22.

^{(e 84 /^)}

23.

^{e°<^*>/((

24.

1*1=

2

Z/ mr

J

exists for s

>

2

I s

(n 2 7T 2 ) /L 2

J

-5.

"^(s " »)

-

- k)%t -

=

k)}

e~

ka

%{e at f(t)} = e- k *F(s -

a)

l

f dr

=

t

Jo

25. (a) /{()

=t-

26.

=

+

=

- jt) -

sin(

=

1)

-
_

=t-

4)

_

-(a-l)

(t

-

l)*{t

-

1}

- *(t - 4)

_J_ e ~4t<-l) s

%t - 3tt) =

—

-sin((

1

-

jr)<3f(t

[(t

-

+ 2]*(t - 2) = 2 +

2)

-

(i

-

jt)

2)
5

*{* »,}.^ +

* l

(

(

/(£}

-

-2s

2'

S

(a)

!)+<«((

+

sin(t

--^- + -^-3-

= 2 - 2"tf(t - 2) +

(b)

28.

-

i]H(t

_ I e -«-

sint"U(*

*{/(,)}

27. (a) /{()

(C)

l)

^/w}-(rh?-(rriji e

(a) /(*)

(b)

-

=

(b)

<«>

[((

=t-

t°U{t

=i-

2(t

-

-

1)

S

-l)2

+ {2 - t)K(( -

l)
-!)

+

(*-

1)

2)««

(t

(2

-

- 2)

-

2)

-

2)

-

37r)K(t

-

3jt)


-

(b)

+

we obtain

29, Taking the Laplace transform of the differentia! equation

(s-l) 2 so that

=

y

l) 3

(a-

2

+ ^tV.

5fe*

30. Taking the Laplace transform of the differential equation we obtain

^ {y]

=

+ 20)

(s-l) 2 (s 2 -8s

111

6 169 a

-

s-4

6

-

13 (a

1

169

1)2

(s

-

so that

y y

= —e< +

-

-I( e (

169

13

—e 169

4t

co S 2t

5

+ 22

4)2

—

+

e

338

4t

{a

-

2

4)

+

22

sin2(.

338

31. Taking the Laplace transform of the given differential equation we obtain

s(s 2

-4s + 6)

(s-2) 2

(s-2)2 + 2

«

V

+2

so that y

= 5W(t-*)- Be 2 *'"*) cos ^2 (( -

jt)

« (f -

tt)

+ 5 V2 e 2

32. Taking the Laplace transform of the given differential equation

My) ws =

s3 s 2 (s

6

+ fo2 + 1 + l)(s + 5) 1

1

1

5

s2

'

+

3 2

+

s

1

1

s

5

s2

1

1

2

s+1

+ +

+ 5)

e

50

1

1

4" 1

H>

+

l e -(t~2)^( t

_

2)

-

A

?

+

i)( s

+5

s

1

W 1

+5

1

ie-

100

e -5('-2)^((

274

V2 (t -

2s

_

tt)

%{i

we obtain

so that

+

-*> sin

S (s

1

s

(

-2*

13

1

1

-+

>

I

s 2 (s+l)(s

<

2).

2B

e )

2s

Chapter 7 Review Exercises we

33. Taking the Laplace transform of the differential equation

s

3

s 3 (s-5)

(s-5) 2

obtain

2

1

1

25 s 2

125 a

H

127

1

5 s3

125 s

-

_37_1_12_1__1_2_ '

25 s 2

125 s

5

5 s3

+

_37

125 s

so that

y

2

2

125

25

=

(

_

-f

, 2

+

27 e

,, 5t

-

125

5

34. Taking the Laplace transform of the integral equation we obtain

+

s(s

5)

3 .

5s 2

35. Taking the Laplace transform of the integral equation

„(0

=

i

+t+

36. Taking the Laplace transform of the integral equation

(#{/}) so that /(*) 37.

The

=

2

= 6-^

or

+5

„-«

that

so that

5 s

we obtain

|t

a .

we obtain

#{/} = :M

*

±6(.

integral equation

is

10i

+2

2F + 2t.

Taking the Laplace transform we obtain

^

A

(s 3

2

s

^Wl0s + 2

The

differential

+2 = + 2)

—+ 9 s

'

2 s

2

+

= -9 + 2t + 9e~' /5

Thus, 38.

s 2

s (5s

equation

'

45 5.s

+

= 1

.

is

+ 10^ +

1009

= 10- 10*(i-5).

— +" s

-* s

2

+ "

'

a

+

1/5


Taking the Laplace transform we obtain 20

+ 20s + 200)

2(s 2 1

3+

1

10 s

10 (s

+

10j

10

10

2

+

~

M sin 10(

10 2

+

10 (s

lO) 2

+

(l-.-)

102

so that

=

q{t)

To~

e

~ l0t

cos 10(

"

To

e

ô

39. Taking the Laplace transform of the given differential equation we obtain

2{y} =

(L

4!

15!

EIL \48

5

6

2w

s

120

s

15! 120

s3

2

s

s4

6

so that

V

where y"{0)

=

=

cx

2wq

and

L

1

EIL 48" y'"(Q)

120"

=

c2

Using

.

=

ci

u)

j/"(Z.)

=

c2

2

,

Q2

3

2

= we find

and y"'(L)

L 2 /24£7,

ci

--)

~\~

2/

120 V

'

= -woL/4EI.

Hence

wo

1

5

YIEIL ~H

+

L

i2

4

~

2

T

3

L3

+

2

T

L\*

/

1

/

L\

5^2) n*-*)

40. Taking the Laplace transform of the given differential equation we obtain 4 s4

2

+4

+

+4

s4

4

wq

4EJ

s4

+4

so that

y

=

sin

+ where

j/"(0}

—

cj

a:

sinhx

4El and

+

sin

j/"'(0)

^(sinxcoshx —

-

=

^

C2-

Cl

cosh

Using

_ wq ~ El

j/(tt)

cos a: sinh x)

^=

cos

and

-

y'(7r)

sinh

=

we

find

WO cosh j

siring 2

sinhTr

276

-

sinhn

~ ^)]^ x — (

Chapter 7 Review Exercise

Hence,

wo smh f*

y - —=77

2£7

.

,

.

sib

x smh j -

sinliTr

+

8in [

(*

-

—wo— cosh 5 (smxcoshx - coszsinhx)

cosh

1)

,

4£7 smh ir

x

cos

(

" I) siIl!l (* "

0] *

x" I)

(

8

Systems

of Linear

Equations

Differential Exercises 1.

Prom Dx

= 2x-y

8.1

Dy = x we obtain y = 2x-Dx, Dy =

and

2Dx-D 2 x,

and (D 2

-2D + l)x = 0.

Then

=

x 2.

c\e

+ C2te

Dx = 4x + 7y and Dy = 2 (D 2D - 15)x = 0. Then

From

Dx = -y + 1

and

cie

Dy =

t

and

y

=

-

(ci

+ c2 e~ 3

5i

x -

and

'

we obtain y

t

y

c2 )e*

+ C2ie'.

= ^Dx -

x - 2y we obtain y

From

x= 3.

l

=

- c2 e~ 3t

^c\e ht

— t - Dx, Dy =

Dy = ^D 2 x - ^Dx, and

fx,

-

1

.

D 3 x,

and (D 2

+

l)x

=1+

1.

Then x

=

ci

y

=

ci sinf

cos t

+ C2 sin f + 1 +

i

—

1.

and _^

4.

From

Dx - 4y =

1

and x 4- Dy

=

2

x

—

c\

V

=

-7 c

5.

From (D 2 + 5)x-2y = and (D 2

+

1

2Cos(

—

1.1 =

0.

and (D 2

+ \)x =

2.

Then

2

-ci suit 4

and -2x + (D 2 + 2)y

+ 6)x =

l)(D 2

,

+ C2 sin t +

and 4

+t—

- \Dx - | Dy = \D2 x,

we obtain y cos t

C2cosf

—

1

ci

sin

t

4

-

.

4

we obtain y

= £(D 2 +5)x, D2 y = £(D 4 +5D 2 )x,

Then

i

=

ci

y

=

2c\ cost

cost

+ C2 sinf + C300S VEt + C4 sin
and

6.

From (D +

l)x

Dx = -\{D + 2

+ (D =

ci

-C3C0S y/6t

—

-C4 sin \/6t.

—

l)y

=

2D)y, and (D

y

+ 2c2sint —

2

2 and 3x

+ 5)y

cos v^5

f

=

+ (D +

-7.

Then

+ C2 sin V5 — i

\

and

278

2)y

= -1 we

obtain x

=

- £{D +

2)y,

Exercises

=

1

7.

Prom

(P

2

D2 x =

4y

+

^

2

(

l~3

e l and

~3~ C2

D2y =

+ 4){D - 2)(D + 2)x = x

~

Cl

-3e

=

\

4x -

(

v5t+

cos

I

I

—

we obtain y =

e'

2 \ -C2 J

-

Ci

3

i-

.

smV5i + -

|D 2 x -

ê 1

ZJ 2 y

,

8.

.

= jP 4 a:

\e l

4)x

=

,

and

Then

.

ci cos 2t 4- c 2 ain 2t

+ cê 21 + Cie~ 2t + -e

l

5

and y 8.

=

-Ci cos2t- C2sin2f

+ 5)x + Dy = and (D-5)(D 2 + 4)y = 0. Then From (D 2

(£>

+

+ {D -

l)x

a:

=

c\e

y

=

C4e

+ c3 e 2c + C4e~ 21 4)y

=

-e'.

we obtain (D - 5)(D 2

bt

+ C2 cos 2f + C3 sin 2t

5t

+ C5 cos 2t + eg sin 2t.

+

and

and

(6cj

so that C4

+

(D

Substituting into

l)x

+ (D — 4)y =

+ c4 )e + {c 2 + 2c 3 - 4c5 + 2ce) cos 2t +

= — 6ci,

From £)(D 2

(-2c2

+c3 -

2c 5

-

4ce) sin 2t

=

— 5C2, and

C5

=

1/

9.

gives

st

Dx + D 2 y = e 3t and (D + l)y = -8e 3 Then

-6cie l)i

4-

5t

+

cos 2t

+ (D -

=

l)y

- ^C2 sin 2f. 4e

3t

we obtain

D(D 2 +

l)x

=

346 31 and

'.

y

=

ci

+ C2 sin f + C3 cos t —

_4_ 3i

1

=

C4

+ C5 sin ( + ce cos f

— 15

-re 15'

and

Substituting into

=

ci, C5

l)x

+ {D -

- Cl) +

{C4

so that Ci

+

{D

=

C3, ce

(C 5

-

= — C3,

from

D 2 x - Dy =

D{D +

1)(D

t

and (D

+ 3)y = -1 -

3t.

+

4e 3t gives

-

C3

-

C6

C2)sin(

+

(C6

e

.

+ C5 + C2 ~

C3)cost

=

and ci

— C2Cos( + C3S111 +

3)x

+

(£>

ci

+ C2e

1

+

3)y

=

2

17 — 15

e

we obtain

.

D(D +

Then x

and

=

=

x 10.

l)y

-+-

=

'

+ c3 e

3(

- + t

^t

2

+

3)x

=

1

+

3t

and

1

Exercises

8.

=

y

(D + 3)x + (D +

Substituting into

+ c5 e" + cee~ 3i + 1 !

ci

=

Z)y

D

2 and

3(ci+c4 ) +

2

- Dy =

x

-t

.

gives

t

+ c 5 )e- = l

2(c2

2

2

and

+ c s )«- + 3(3 C3 + ce)e- 3t = 1

(ca

— — c\,

so that C4

= — C2,

C5

ce

= — 3^3,

=

-ci

-

-l)x-y = and (D - 1)(D 2 + L> + l)x = 0. Then

l)x

+ Dy =

y

11.

Prom (D 2 {Z?

and c 2 e'

i

-

3c3 e~

3!

+t-

we obtain

^t

=

t/

2 .

(£>

2

-

ljz, Zfy

= (D 3 -

D)x, and

'

x

=

4/2

+e

l

cie

>/3

C2 cos

-—t +

\/3 C3Sin -—-t .

and 3

/

12.

\

\/3

_ tj2

3

From (2D 2 -D-\)x-{2D+l)y = land (D-l)x+Z)i/ and

(2£>

+

l)(D

+

l)y

=

\

_ (/2

.

v/5

= -1 we obtain (2D+l)(D~l)(D+l)x =

-1

-2. Then

x

=

c\e~

y

=

c4 e"

t/2

+

c 2 e~

l

+ c3 e' +

1

and

Substituting into

(U —

\)x

+ Dy = —1 (-|c 1

so that C4

— — 3ci,

C5

= — 2c2,

Prom (2D~5}x+Dy =

e

1

and

2

+ C5e" - 2. 1

gives

-^ )e' 4

1/2

+

(-2c 2

-c 5

-

-3cie

_t/2

{D-l)x+Dy =

5e t

=

Cl e

-

2c 2 e

_!

-

we obtain

^

4t

+

U

l

•J

Dy =

-3cie

4t

+ 5e

(

e

-1

=

so that j,

= --cie 4t + c2 + 280

5e

2.

Dj/

Then

and

)

and y

13.

l ''

l -

= (5-2D)z+e and l

(4-D)a:

=

4e*.

1

Exercises

14.

FVom

Dx+Dy = e* and (-D 2 + D+l)x+y = Owe obtain y=

andD 2 (D-l)x =

e

(

(D 2 -D-l)x, Dy

8.

= (D3 — D 2 — D)x,

Then

.

x

=

ci 4- c2 t 4- C3e*

y

=

-c\

+ te'

and

15.

+ (D 2 + l)y = 1 and D 2 (D - 1)(D + l)y = 1. Then

-

C2

(D 2 -

From (D - l)x and

—

c
l)y

=

_1

-

^(

+ cge"' -

-f

+ {D +

l)x

x = ci+C2t

— C3e' - te* +

+ c3 e* + c 4 e

e

l .

we obtain

2

D 2 (D - 1)(D + l)z «

1

2

and

Substituting into

y

=

(D - \}x + (D 2

+

=

+

C2

1,

)

=

eg

C4, C7

=

V 16.

=

=

-

c2

+

D2 x -

2(D 2 + D)y - sin* and x D(D 2 + 2D + 2)y = — sint. Then Prom

«

-

C2

c\

2 .

gives

1

-

(ce

0, C5

(ci

=

\)y

- ci - 1 + c 5 +

(c2

so that eg

c 5 4- cet 4- c 7 e'

-

2)

l)i

C2

+

+

+ 2,

(ca

+

(2c g

_(

-

2c4)e

+

e

(2c7)e

=

1

and l)t

+ Dy =

+ Qe" 1

^t

.

= -Dy, D 2 x = -D 3 y,

we obtain z

= ci+C2e -t cos( + C3e -i sml +

2

and

2

1

7 cos

t 4-

5

- suit 5

and t 17.

From

Dx =

y,

Dy =

=

z.

t

(c2 4- C3)e~ sint

and

Dz = x we

x^de' + e"'/ 2

y

=

Cl e

z

=

cie

(

+

îc

C2 sin

2

—

t

- ~c^j

+

(c2

-

obtain x

c

C3)e~ cost

+

= D2 = D 3x t,

1

2

5

5

-sint - - cost.

(D - 1)(D 2 +

so that

D + l)x = 0,

+ C3 cos — -t 2

e~'/ sin

^t + (^~c

2

-

^

2

e"'/ cos

^t,

and

18.

!

+ f-^c2 + ^cs^

e~

t/2

sin

= e (D - l)i + Dy + Dz - 0, Dz = -D 2 x + e', and the system (-D 2 + D From

Dx +

z

l

,

™t + and x l)x

(-^-c 2 ~

+

2y

\c-&

+ Dz =

+ Dy = -e

1

e

l

\

e~

t/2

cos

^t.

= -Dx 4- e 4- 2y = 0. Then

we obtain 2

and (-D

2

4-

l)x

1

,

1

Exercises

y

8.

Dy = \D{D 2 -

= \{D 2 -

x—

-

and [D

l)x,

C\e

2t

^

+

=

l)x

-2e* so that

+ C2 cos t + C3 sin( + e',

3

—

2){D 2

c l e2t

— c 2C0St — cgsin t,

and

= — 1c\e 2t — a cost + C2sin(.

2 19.

From

Dx —

so that

6y

=

x

0,

— Dy + 2 =

{D + 1)(D - 3)(D +

2)x

+ y — Dz =

and x

0,

we obtain

0-6

D

-6

1

-D

1

—D

1

1

1

-D

1

-D

=

Then

0.

x

=

c\e

y

=

--cie

z

= --ae -t + -c 2 e,3t - -c3 e -2t

t

+ c2 e 3f + c3 e +

2( ,

-c 2 e dl - -C3e

and

20.

From (D +

l)x

-

=

z

0,

(D

+

l)y

-

=

z

0,

and z - y

D+l

D{D +

l)

2

z

=

0.

D+

-1 X —

D

-1

1

we

obtain

0-1

-1

D+l ao that

+ Dz =

l

-1

D

-1

Then X=

C\

+ C2&

y

=

c\

+

z

=

ci

+ C3e

*

{c2-

+

C3te

c3 )e

_t

+ c3 te~

( ,

and

21.

Prom 2Da; + (D -

= ( and Dx +

l)y

Dj/

y

and

Dx =

cie"'

+ t2 -

4f

+5

no

-

2Dj/

=

t

2

and (D

i1 we obtain (D

cie-

t

+

l)y

=

2f

=

It

2

-

t.

Then

+ 2( 2 -5( + 5

so that

x 22. Prom Da;

=

=

+

-

l)x

-cie"

c

1

+ c2 + -r - 2r + 5t.

- 2(D +

l)y

=

solution.

282

1

we obtain

+ f2

so that the system has

e

1

Exercises

23.

Prom (D + 5)x + y = Then Ax

and Ax- {D + l)y

+ (D + 5D)x + (D + 5)x = 2

=

we obtain y = -(D + 5)x so that

and (D

+ Z)

2

x=

x

= ae~ 3t + c 2 Ie~ 3t

y

=

0.

8.

Dy = -(D 2 + 5D)x.

Thus

and

Using z(l)

=

=

and y(l)

-(2ci

+ c2 )e~ 3! - 2c2 te~ M

we obtain

1

3

=

-(2c,+C2)c- 3 -2c2 e- 3

=

cie

ci

2ci

Thus

24.

ci

=

e

3

and

c2

3

= —

.

The

3

+ c2 e

Then -\{D - 2D)y = y y

=

x

= -e

e

+3c 2 = -e 3

.

solution of the initial value problem

_

x

=

y

= _ e -3 t+ 3 + 2ie -3 i+ 3

1

and {D

[c\

co3\^2*

l

l

+ c2 =

e

-3t+3

2

is

,-3t+3

t(

= -\{D-2)y

From Di-j/ = -1 and 3x + (D-2)y = Owe obtain x 2

.

— 2D + 3)y =

Dx = -\{D 2 -2D)y.

Thus

3.

+ c2 sin\/2t) +

so that

1

and 1

Using x(0)

=

j/(0)

=

l

[(ci

-

\/2c 2 ) cosV2i

+ (v^ci + c 2 )

we obtain C!

(

ci

= —1

and

c2

=

Cl

=

e

t

=

e'

initial

f-^cosv^f 3

?/

1

=

- V2c 2 ) + - =

V5/2. The solution of the

x

+

2

1

-

Thus

sin\/2(]

0.

value problem

—sinvÂ + 6

/

^-cosv^f + ^sin vîj +1.

3

is

+

2 -

Exercises

8.

Dx = 2c2e 2t and Dy = = 2ae 2t = Dx. A system is

25. Differentiating the equation we obtain equations we have x

-t-

y

Thus

2c2«

Dx =

Dy. Adding the

Dx - Dy =

{D-\)x-y =

Q.

Exercises 8.2 1.

Taking the Laplace transform of the system gives

= -2{x) + X{y}

sie{x}

s%{y}-\ = 2%{x} so that

1111 + 111 +

1

>-l)(s +

3s-l

2)

and

s(s-

s

2

2

2

1

,

3s

+ 2)

l)(s

3 s

-

3 s

1

'

2

Then a;

=

-e s

2.

^

e

3

and

y y

=

3

+ -e

-e' 3

.

3

Taking the Laplace transform of the system gives 1

s
s-1

s%{y}-l = 8Z{x}-^ so that eBI

,

2\Vt ~

s

3

+ -s + — — -

777 s{s

7s 2

777752 \){s

1

l

T7T7 -16)

=

1

"

16 7 s

77i

8

—1

7 77 15 s

7 1

+

173

96

53

1

s

-

4

and y

Then

—

1

8

16

15

t

173

e* H

e

53

4i «

e

96

1,1

_ *\ 4(

160

1

1

,

284

173

«

53

_ it

160 s

1

+4

3.

Taking the Laplace transform of the system gives a

X{x} + l-X{x}-2X{y}

sX{y}-2 = 5X{x}-X{y} so that

-s-5 s2 + 9

s

5

3

+9

3 s2

+9

s2

and

Then

11

= -x —

y 4.

5

= — cos 3t —

x

-x'

=

— sin 3i.

7

— - sin 3t.

2 cos 3(

Taking the Laplace transform of the system gives (s

+ 3)X{x} + sX{y} =

± s 1

(s-l)X{x} + (s-l)X{y} =

-

s

1

so that

„.

-

5s

.

3s(s-

1111

1

l)

2

s-

3

3 s

1

4

1

3

(s-

l) 2

and

l-2s -

3s{3

1 1

~

l) 2

_

_1

1

1

3s-l

3 s

1 '

3(s-l) 2

Then 1

1

1

I= 3"3 e -3 ,te 5.

t

t

j

and

V

-+r + 1

=

1

,

3

4

t

t&

'

3


{2s-2)%{x} + sX{y} = 3

(s-3)2{x} + (s-3)X{y} = Z so that

X{x}

-s-3

\L 1

s(s-2)(s-3)

2

5

1

2

2

s-2

s-3

h

s'

and 3s s(s

-

~

1

2)(s

-3)

1 1

5

6a

2 s

8

1

-

2

1

3s-3'

Exercises 8.2

Then 2 6.

y

2

6

2


+ \)2{x}-{8-\)2{y) = -\

{8

sX{x} +

+ 2)2{y} =

(s

l

so that

+

s

w„i = and

1

/2

s

=

+ s+l

S2

(

S

+

11

s

2

+s+

/2

1

+

(V5/2)

:

~ 3/ 2

~W

<*r x]

+

l/2) 2

l

(s

+

1/2)2

+

(^3/2):

Then y 7.

2

2

Taking the Laplace transform of the system gives 2

(s

+ l)2{x}-X[y} = -2

-X{x} +

+ l}X{y} =

2

(s

l

so that

*

'

y

=

113

-2^-1

X lx]

S4

+ 2s 2

2 s2

2 s2

1

+2

and

Then

8.

x"

1

—

2

2y2

+ x = --( +

3 y= sin

V2t.


+ l)2ix}+
l

4%ix}-(s + l)X{y} =

l

{s

so that

|X|

~~

s2

+ 2s + 5 "

(s

+

l) 2

and

286

+ 22

+

2 (s

+

1)

:

Exercises 8.2

+3 + 2s + 5

-s

,

s2

+

s

+

(s

l

+ 22

l) 2

(s

+

l)

2

+ 22

'

Then a;

9.

=

e~'cos2t

+ -e _t sin2(

and

Adding the equations and then subtracting them

=

y

-e~* cos2(

+ 2e"

f

sin21.

gives


^> = r„r

.

n l

8

;

1

13!

4!

2i? + 3^

+

and

v,

14!

,

13!

so that 1 3

10. Taking the Laplace transform of the system gives (s

(s

-

4)

and

w

„\

-

2)(s 2

23

+

1)

+4

s 3 (s-2)(s

2

+

^

+ 2)if{x}-2s ^{?/}=0

4 (a

=

3

4

14,81

5 s

-

so that

=

3

X{ X ] +

l)

2

+

5

5i 2 + l

1

1

2

211

s

s2

s3

5

s-

2

6

*81

5 s2

+

Then 4 = -e 2t —-cost 4

x

5

8

5

.

.

smt

5

and 1 = 1 - 2t - 2ro + -e >,

y

5

6

- cos i 5

2

X{x}+3(s + l)X{y} =

8

- sint. 5

11. Taking the Laplace transform of the system gives s

+

2

T

1

5 s2

+

'

1

Exercises 8.2

so that

1112

sj + i)

1_

Then

=

i

+*+

1

^f

2

-

e~*

and 1

»=3 te

t

1

1

»

-r =r

-/

1

+

r

-t

1

-3-

12. Taking the Laplace transform of the system gives 2e"

-3tf{x}

+

(*

+

l)tf{»}

=

i

+

^

so that

-1/2

(s-l)( S -2)

1

(s-l)(s-2)

11

11 2

+ e"

1

~7^T

2a-2

s-1

+

1

I

s-2]

and

_^ s

_

3

_i

(s

a/4-1 - \)(s i

2)

+ ^

-s/2 e" "

_ s ri

i

+2

(s-l)(*-2)

_

3

i

i i

Then

and

«(*-!).

288

Exercises 8.2

13.

The system

is

=

Z"

+

-3ii

-xi)

2(x 2

= -2{xi-xi)

H (0) =0 *',(0)

=

1

*a(0)

=

1

4(0)

=

0.


-2%{x

}

1

+

(s

+ 2)X{x 2 } =

2

s

so that s

2

s4

+ 2s + 2 _ + 7s 2 + 6

2

s

5 s2

+

1

1

2 5 s

+

1

and [0[

,

s

=

,

+6

2 V6 5VSs 2 + 6"

5s 2 + 6

l

y/6

5\/6s2

5

1

5s 2 +

IT

,

4

s

21

+ 5s + 2 4 s = 7 T5-nr l)( s 2 + n + iv,2 5s 2 + l 6)

3

r„2 2 (s

2

5s 2 +6

1

Then 11

=

% cos *

+

5

\

sin

(

— ? cos s/6

5

and ^2

=

4

- cos t

+

14. In this system xi

sin

f

+

5

—kixi

f

5V6 mi and

of masses

m2

from their equilibrium

m 2 are —

and

k 2 (x 2

—

x\)

— k$x 2l

Newton's second law of motion gives

mix" — — k\x\ 7712x2

Using

= sin v 6

v61

5

£1)

V6

1^2^-

- cos

on mi and

+ £2(^2 —

sin

5\/6

and x 2 represent displacements

positions. Since the net forces acting

respectively,

2

-

5

~

H

5

ki=kt =

ks

=

1,

mi

=

m2

=

= 1,

+ k 2 (x 2 — x\)

-^2(^2 xi(0)

=

~ 0,

*i)

~

zi(0)

k3 x 2

=

.

-1, x 2 (0)

=

0,

and

x'2 (0)

=

1,

and

Exercises 8.2

we obtain

taking the Laplace transform of the system, (2

+ a 2 )X{x }-X{x 2 } = -l 1

^{x }-(2 + S 2 )^{z 2 } = -l 1

so that

Then xi

15.

(a)

By

Kirchoff's

-

E(t)

first

= — -^sinVSt

law we have

Rij + Lii'2 and E(t)

=

ii

and

+ 83. By

22

= Jfci+Z^

=

12

-^sini/St.

Kirchoff's second law,

=

or Lii'2 +Ri 2 +Ri 3

E(t) and

on each loop we have

L 2 i'3 +Ri 2 +Ri 3 =

(b) Taking the Laplace transform of the system 0.01i 2

+ 5i 2 + 5z 3 =

100

0.0125i 3

+ 5i 2 4- 5i 3 =

100

gives

+ 500) #{i 2 } + 5ooy{i 3 } =

(s

400y{i 2 }

+

+

(s

,000

=

400)i?{i 3 }

s

so that

vr

1

_ ~

_

8.000 s2

80

+ 900s

1

80

_

9 3

1

9 s

+ 900

'

Then 13

16.

=

(c)

ii

(a)

By

i2

+ i3 =

Kirchoff's

have E(t) i3

—

=

tf

=

=

80 -3-

-

9 20

c -

- 20e- 9001

first

Li[

80 .™ _BMt — g

+

— ,

and

.

12

=

,

20

- »-™«»3 0.0025i 3 -

13

=

100 — 9

100 _»„, 90m e 9

—

.

.

law we have

ii

Rii 2 and E(t)

=

=

i2

Lij

+

CR\i'2 — CR-iH so that the system Li'2

-flii 2

+

13-

+

Li'3

By

R213

Kirchoff's second law,

+ £q

so that q

is

+ Rii 2 =

+ fi2i3 + î 3 = 290

E(t)

0.

on each loop we

= CR\i 2 -

CR2 i3-

Then

Exercises 8.2

(b) Taking the Laplace transform of the system i'

2

-10i'2

+

+

i'

z

=

10i 2

120

-

120

+

-

2)

+ 5i'3 + 5i 3 =

gives (s

°U(t

-10s2{i 2 }

™

+ sX{i 3 } =

2{i 2 }

10)

+ 5(s +

1)

%{i 3 }

(l

- e" 2s

)

-

so that

120(s (3s 2

+

+

lls

1)

+

48

(i-„-)-

10)s

s

60

+ 5/3

s

+2

+

12 s

and 240 3s 2

+

lis

240

-.-*)-

(i

+

10

s

240

+ 5/3

s

+2

(i-.-).

Then

+ 48e- 6 "3 - 60e" 2£ -

*3

=

12

i3

=

240e" 5i /3

[l2

+ 48 e

- 5(i~ 2V3

-

60e" 2 < l

-2

V(t

>]

- 2)

and

(c)

ij

=

i2

+ 13 =

12

+

-

240e- a

288e" 5 '/ 3

-

-

300e

[240 e

-2 '

-

- 5(t - 2 " 3

[l2

- 240e- 2 «~ 2 >] °ll(t -

+ 288e- 5 " _2 ^3 - 300 e - 2

<

(

2).

-2 >]

* (* -

2).

17. Taking the Laplace transform of the system i'<2

+ 613 =

50sint

+ 612 + 613 =

50 sin f

+

13

lli2

gives (

S

+ll)if{i 2 }

6^{i2}

+

(s

50

+ 6^{i3} =

+

s2

l

50

+ 6)^{i 3 } =

s2

+

l

so that

20

50s (s

+ 2)(s +

15)(s a

+

1)

— 1

= -tz -+ 13 3 + 2

375

1

1469 s

+

Then 20 _,, t2

and

^-T3

e

+

375 e

I469

+

145 113

COSt+

85 Slll(

Tl3

145 s — + TTT7 ^r-^r + 15 113 s + 2

1

85 113 s 2

1

+

1

Exercises 8.2

I3

18.

By

Kirchoff's

E(t)

=

Li[

+

=

first

25

TS m

1 --* 2 ,

(

law we have

Rt2 and E(t)

=

—

ii

Li\

30 -

11

V2 =

+

i2

+

250

_, (

e

+j^-9 e

By

'3-

+

CRi'^

Then

-

=

E{t)

-

=

0.

i2

280 cos t

IT§

Ri 2

+

Li'

-

+

— 810

.

mot.

KirchofF's second law, on each loop

= CRi 2

so that q

_ 15t

ii

=

i3

1

q

=

we

19. Taking the Laplace transform of the system

+ 50i 2 =

0.5i[

0.005i 2

60

+ i2 - h =

gives

-200if{i,

+

}

+ 200)

(s

— 120

+ 100^{t 2 } =

«J?{il}

=

^{13}

so that * 12 '

_ "

24,000 s(s 2

+

200s

6

+

20,000)

1

6

"5s

5

3 (s

+

+

100

100) 2

+

100

6

100 2

5

(a

+

100) 2

+

Then

h=

5 _ 6,-iow cog 1M( _ 6 e _ 100i 5

5

1Q0(

5

and ti

=

0.005i'2

+

i2

=

I

- |e- 100t cos

100t.

20. Taking the Laplace transform of the system 2i[

0.005i 2

+

+ 50i 2 = i2

60

-h =

gives

2si?{*i}

+ 50^{i 2 } =

— s

-200^{ii} +

(s

+ 200) %{i 2 } =

so that

Xin} =

s{s 2

+

200s

6

1

6

5

s

5

+ 5,000) 5

(a

+

+

100) 2

100

6\/2

- (50V^) 2

292

5

(3

+

50V2 - (50^2 } 2

100) 2

havt

CRi'2 so that system

'

100 2

'

is

Exercises 8.2

Then

5

5

5

and

h = (a)

0.005i 2

Using KirchofF's

first

+ j 2 = | - ê- 100*cosh50\/2t -

law we write

Kirchoff's second law

=

ii

12

+ ^3.

Since

E{t) E(t)

=

=

ii

and summing the voltage drops

ê-

100

'sinh50v^(.

dq/dt we have

ii

— ^3 =

dqjdt. Using

across the shorter loop gives

1

ifl, iRi

+ -5,

(1)

so that

Then dq

1

„,

1

,

and

Summing

the voltage drops across the longer loop gives

E
this with (1)

L^ + Rii*.

we obtain

+ R2 3 - hRi +

hR\ +

i

or

£§ + *2i3-^ = (b) Using

L = R\ = R 2 = C =

1,

E(t)

=

50e-"W(t

-

0.

bOeê-^^t -

=

1)

1),

and taking the Laplace transform of the system we obtain

oil (

S

+l)J£{i 3 }-^{9}

=

0,

so that

2{l} =

50e (a

+

_1 l)

e- s 2

+

1

and

=

50e

_1

e"

(t_1)

sin(i

-

1)

293

=

50e

_l

sin(*

-

l)<3f(t

-

1).

q(0)

=

i 3 (0)

= 0,

1

Exercises 8.2

22. Taking the Laplace transform of the system

+ 02 + 80i =

40?

+

0?

+ 202 =

'

2

gives 2

4 (s

+ 2)

+s

}

2

+

s i?{0,}

(s

2

2

i?{02 }

=

3s

+ 2)^{02} =

O

so that

(3s

2

+ 4)

or

Then 02

S

(

2

+ 4)^{02 } =

Is

3

s

2 s2

2 s2

+4'

12

3

2

2

+ 4/3

= -cos-7=t--cos2t %/3

=

0i

- cos 4

0"

and

12—

so that

-3s 3

;=(

+

= -^'-20 2

3

- cos

2f.

4

V3

Exercises 8.3 1.

Let xi

2. Let

3.

n

Let xi

=

=

=

y,

y,

j,,

X2

x2

x2

-

=

=

y',

y',

y',

and y"

and y"

x3 =

=

=

y",

- 4j/ + sin3t

3j/'

—2y'

and

x\

=

xi

x'2

=

3x2

+

j/"'

so that

—

+ sin 3t.

4tj

|y so that

=

x\

=

x'2

= -2x 2 +

3y"

x'j

=

x2

x'2

=

£3

-

xi

Qy'

x 3 = 3x3 - 6x2

+

+

10y

10xi

294

+ f2 +

+

2

+

1

1.

so that

Exercises 8.3

4. Let ii

5.

Let xi

=y,X2 =

=

y,X2

=

y*,

y',

x3

x3

=

=

y",

and

y", Xi

y"'

=

= -\y +

|e' so that

x[

=

x2

x'2

=

£3

=

y"r and j/ 4 ' ,

=

x\

-

1y"

Ay'

-y+1

so that

i2

4 = ^3

6.

Let ii

=

y,

x2

=

j/,

x3

=

y",

x4 =

X3

—

3:4

x\

=

2X3

and

j/

x\

=

x2

x'2

=

Xi

x's

=

X4

j/",

Let xi

=

j/,

x2 —

y',

and y"

=

Let xi

=

y,

x2 =

j/',

and y"

=

+ Ay +

10.

y so that

(-l/t)j/

+

[(4

=

=

-

X2

(

2

2

)/( ]y so that

X2

4~f 2

1

9.

-\y'"

1

x[

8.

=

4>

— Xi+t.

-ix 1 + 4x +

x4 = 7.

4x2

-

Prom x'

+

4x

-

u'

=

7t

10 so that

Exercises 8.3

we obtain 2x'

+ 4x -

2y

=

lOf

2y'

- 4x -

2y

=

-At

SO that

10. This 11.

is

x'

= -2x + y + 5t

y'

=

2x

+ y-

a degenerate system.

Adding equations, we obtain

Dx =

2 t

+ 5£ - 2

Dy = -x + 12.

It.

From x" — 2y"

Let xi

=

x, i2

= smt

=

i',

and x"

£3

=

y,

+ y" =

and 24

3x"

=

2cost

3j/'

=

cost

y'.

Then

xi

=

x2

x2

=

-cost*

x3

=

X4

=

-cost

<

2

2.

+ sint

—

sint.

+

-sint„

1

1

1

/

Xi

-

we obtain

cost

=

5f

— -sm(.

13. Since

Dx - Dy = the system 14. Let yi

=

is

xi,

2

- -x

and

Si -

degenerate.

w — x\,

y3

=

£2,

and

j/4

=

y'\

x 3 so that

=

Wl

mi

'

3/3

=

3/4

T712

296

mi

Exercises 8.3

15. Let z\

=

x,

z2

=

x', 23

=

x", z4

= y,

and

=

z'i

=^

25

so that

22

A = 23 =

23

-

4zi

323

+

425

4 = 25 4= 16. Let 2]

=

x, 22

=

Dx,

Z3

=

10i

2

-42 2 + 3^5

-

and

y,

2£> 2 x

2Dy

= y - 6t 2 + 4t -

10

=-y + 6f 2 + U + 10

so that

f 21 = 22 £*3 = £>23

^3 -

3f

2

+ 2t -

5

= -^23 + 6t 2 + 4i +

10.

17. Taking the Laplace transform of the system gives

so that

„,

625(25s

,

+ 2) +

25

(25s+l)(25s

2

3)

25

1

a

+

1/25

1

2

3

Then 2

2

and x2 18.

The system

=

50X1

+ 4a:i =

25e-'/

25

-

25e

_3t/25 .

is

I1 = '

X2

=

2 .3

+ lx 2

-lx

I

-4

^ -^ "5T '

1

4

2

X2

-

3

-

+ 3/25'

Exercises 8.3

19. Let xi, X2, and #3 be the amounts of salt in tanks A, B, and C, respectively, so that x[

1

=

1

2-

100

4= *3

6

3

T^ x l

-"

50

50

1

+

X3

-160*- 2 "Too 12

100 1

5-

=

- 7^X 2

6

1

100

100

=

13

Too

~

X2

h>

h X3

12

+

X3 ibo

-

20. Let

a\Dx + a^Dy =

b\X

+ foy

+ aiDy =

b3 x

+ b\y

a 3 Dx so that

if

then the system

Dx + Dy = —x - y y = — 1c\e~

and

(a)

(b)

a4

A+B =

B-A=

4-2

5+6

+

9-10

-6

8

—

a2i3

obtain y

8

+6

8.4,

A-B =

+4

2

6

-10-9

f-2-3 4-0 1-2 7

11

-1

f" \u

X

2A + 3B-

2

6-5

^-2-4

^

3

+-

2 y

19

-6

18 >

24

-30; -5

(

+ 2 -1-0^ B-A = 0-4 2-1 [-4-7 — 2 — 3 >

=

28

G

1\ --1

4 11

3

f

(b)

0.10,4

.

+ l\

(a)

=

l

^

(c)

«3

Dx + Dy = —\y we

Exercises 1.

a2

=

0,

degenerate.

is

21. Since

and

ai

5 f

-4

i\ i

5,

298

— —2x and Dx =

—a:.

Then x

=

c\e~ l

Exercises 8.4

-2\

(2 (c)

2(A + B) = 2|

4

\3 3.

(a)

AB =

'-2-9

+

,5

(b)

BA =

(c)

A2 =

(d)

B2 =

(a)

(a)

BC-

6

2)

-6\

-30 +

12

-11

6

17

-22

8)

,6-10

-9 + §J

+

-6-12

19

-18

15+16

-30

31

4

15

-10 - 20 1

+

+ 24^1

18

6

+

12

9

~

30

-

48

-36

-4 + 30-24

- 15+

4 )

27

-4 -1

M9

6

3

22

\

6-12

-4 + 4

+

-32

-6 + 12\ _

18

-3 +

1

5.

1/ 12

AB = -20+ 10

BA =

6

3

,-32

(b)

8

'-2 -30

/ 4.

3

-3 + -15

30

8

-24 +

-10 ,-20

24 J

-16 + 60-36\ 4

16

-

30

+

-6

/

>

+ 20

24,/

2

8

2

-2

24'

3

(b)

A(BC) =

(c)

C(BA) =

(d)

3

/0

0^

U

o,

A(B + C) = /

(b)

BA

8

( -6

-16

6

5\

-4

-5

5

5J

8

10

=

(-16)

3\

5^ 5

12

0)

Exercises 8.4

18

21

24

28

AB is

x

1

and

1

C

7.

(a)

—

(

4

=

104

72

-18

(AB)C

(180)

-10

/2\ T (b) B B

=

/

4

4

(2

5)

=

V5/ 4\ (c)

A+

BT =

8.

(a)

A + Br =

10\

8

16

20

20

25 /

(

6^

+

12

\5/

G :)

8

\ 10

(2\

10/

4

V-5/

/-2

5

-1

7

3

7

5

11

\

-1

4

-( ::) =

)

9.

(a)

(b)

(AB) T

/l

2^

3

U

4y

(- 3

7

10\

,38

75 J

'

=

(A

1

+ B) T =

-14

-u

20

-11

'-4'

-7

T

:*)(

2

1

-3 -6

5

BT A T =

10

(b)

"(1

-6

-14

9

1

7

38

10

75

2

-11

20

8

2

-11

20

8

+

11. 16 /

V

300

54

{-26

4 \

-10)

8

u

2

3 x 3 the product

is

/

AT A

u

78

f

-1

1

-1}

6 (d) Since

2

(1

^

is

99 \ 132

-33,

not defined.

Exercises 8.4

-t

6f

12.

|

2

3f

+

|

-* a

-a,

+ l\ +t

3t-3

et ^

(

=

2*

|

V-iot/

/

2

+t-

10*

-

i

3

-38

-2

+3 -5 \-6f + 4 /-9*

14.

13*

(

+

4/

+ 1^ - 12

-10*

\

8

1

,

2

13*

+

,-6*

{,-(>)

15. Since det

A = 0, A

is

singular.

16. Since det

A = 3, A

is

nonsingular.

14/

1/4-5 3

17. Since det

A=

4,

l-l

2

A is nonsingular. 1/-5 -8 4

18. Since det

A = —6, A

is

I

A=

2,

A

is

4

2

-10

nonsingular.

A-i - -I

19. Since det

3

The

nonsingular.

-2

7

cof&ctors are

-411=0

Ai2

A21

= -1

A 22

^31

=

^32

1

=2 =2 = -2

A u = -4

= -3 A33 = 5.

^23

Then -1

2

-4^ -3

1

-2

5 /

2

/

20. Since det

A = 27, A

is

The

cofactors are

A u = -1 421 = 7 A Z - -1

Au = 4 -4 22 = -l A 32 = 4

nonsingular.

\

-1

(

2

-4

Xi3

n

2-2 -3

=

22

4 23 = -19 ,433 = -5.

5

\ /

Exercises 8.4

Then -1

22

7

-1

.-1

4

nonsingular.

The

f-\

\

27

21. Since det

A=

—9,

A

is

-5

-1>

-19

\22

/

-5

J

cofactors are

= -13 Ai3 = S =5 A 23 = -1 = -1 A i2 = 7 A 33 = -5.

An = -2 A 2l = -2 ,4.3!

7

4-14

-19

Ai2 Ai-i

Then

1-1 -S

S

,-1 22.

Since det

8\ T

-13

7

-I

=

3t

2e

^

0,

A

is

24. Since det A{t)

=

2t

2e

#

0,

-e 4t

3e 4i _(

e* sin t

- ie"»

—

e*

cos

t

!

+ 8sin2t

\

:

X=

-6cos2(- 10sin2t /

+ 8e- 3t \ ., -2e 2( + 4e" 3t / 2e*

so that

dX —— —

A

4e 2(

-

24e" 3i

-4e 2t - 12e- 3 '

»«t

3t cos 3i

4e 4t 2

+ sin 3t ^ — irsinirA 6f

J

302

2e' cos

f

(

IT cos2(

2e-'

A is nonsingular.

A-

5

-1

-4e

27.

-13

nonsingular. 1

IF

-2

-5/

A" = ±
26.

-2

A = 0, A is singular.

23. Since det A{f )

25.

/

2e sin

t

-1\ 7

5y

Exercises 8.4 (=2

-i

t=o s=t (c)

£m*)**=

=2

-2(/(t 2 30.

+

-

.3 s°

l)

2

s J

(

s=0

(

3

-

f

3

(a) ~dt

2f

ri

1

/tan" 1

,

1-1 t-1

a

*t \

*

E

3 1

/ i 1

3

t=0

/ 3t 2

/2

i=2

2(

(

6s/(s 2

+

/3t

" /I

1

l)

3 ln(t

2

(3/2) t

V

-2

+

/l

4

9

2

In 2

6

+3

2/(s

2

-1

,6

3

1

4

0-3

5

6

1

,0

14

/l

|

1

Thus x

=

3,

y

=

1,

and

2

=

-5.

+

l)

^

+

12s 2

3

4(

+

2

tan' 1 2t

fl

-28

-

1

3

l)

1

3^

j

1

I

-5,

,0

*

-

-

4

-

jr/2

2

-1/3

14/3 \

-5/3

28/3

|

1

,0

2

+ 1) - 3 - 3 In 2 + ( - (5/2) -2

1

1

2

J

31.

2

11

-55

)

.

Exercises 8.4

/5 32.

-2

4

1

1

un

1

l\

1

6/7

fl

j

9

1

9\

1

-7

-1

-7

-1

-35/

ijt,

and x

=4—

|

-35

1/7

1

j

4^ 5

|

I

V4

-3

=

Letting a

fl

-1

VO

1,

3

we

t

find

-5

j/

=

5

—

7^

|

-1

/l

0/

VO |t.

-5

-4

fl

7 \

2\

|

33.

-16

4

5

-10

1

-5

1

1

1

-5

j |

,0

1-5/

1

\

9

ij

-45 J

9

0,

|

= -5 —

Letting 2

(1

/l

6^

1

f,

=

and x

-3

1

4

2

--2

7

/l

6 -

11

-5/2 \

5/2

)

j

34.

+ 4f

2

-17

17/2

1

j

\3

4/

1

--2

,0

=

2

1

(

35.

5,

=

y

-3,

1

—8, and ^

—

—3.

/i

1

-8

j

^0

1/2

-2

-1

2

-2

6

-3

-7 -5

8j

4

|

\0

fi 36.

1

-1/2, y

2 -

2

1

-2

=

3/2,

8^

and z

HO

2

2

-4

4

-

5

Letting z

-6

—

-1/2^

7/2 J

H

8 \

|

-4

we

\

find y

5

-10

= — 2 + 2(,

8\

2 |

1

-2

-2 |

j

6

t

22/7

,

7/2.

j

V2

3/7

1

3/2

1

=

4

1

10 =

-21

3

|

fl

Thus x

2/7

(1

|

|

k

2 \

1/2

|

10

3,

1

5\

|

1

-

,0

|

/l

Thus x

-14,

10

-10

and 2

=

\0 8

—

2(.

304

11,

Exercises 8.4

n

-1

1

n

-1

-1

1

-1\

-1

-1

/l

i

1

1

_1

\4

_1

o

—2

2

1

\o

3

2

-2

1

4

1

—1

A

2

9

4

-1

5

-2/

n u

4

I 1

4^

5

\0

|

/l

-

o

1\

1

1\

/l

1

[

1

1

1 1 j

1

2

1

2

1

|

U ii

=

n

3

1

2

1

1

7

1

3

Thus

x2

1,

1

-

0,

6

i3

=

n

o\

1

1

o)

I

and 14

2,

3

=

,0

1

0.

o\

1

2/5

1

-5

-1

-20

-4

0/

1

1

1

1

o\

1/5 f

Letting 13

/l

2

I

\Q

07

=

we

t,

find X2

/I

2^

4

4 j

j

39.

2

4

3

1

- —5* and 2

—5

1

\Q

0/

y

J

=

x\

2^

(\

-2/5 ^

2

—3

3/5

1

j

I

VI

-1

2

-5

7/

I

\0

5 7

7

I

There /I

is

no solution.

-1

1

1\

3

/l

1

-1

3

1

-1

-4

1

-1

-4

3

-3

-12

n

1\

1\

7

I

-1

1

-4

1

-4

-1

I

40. 1

-2

2

-1

6

5

5

I

V4

-7

7

9

>o

5/

^0

5,

I

There 41.

We

is

no solution.

solve

d et (A - AI) For Ai

=

—2

=*.

=!• i

2

2

1

37.

38.

1\

|

i

=

-1 - A

2

-7

8-A

= (A-6)(A-

6 we have

-7

2

1 I

-7

2

-2/7 I

1)

-

0.

Exercises 8.4

= %k2

so that ki

For A2

=

1

.

If

k2

=

7 then

we have

-2

2

1

-1 j

I

-7 so that

ft]

=

fo- If Aa

=

7

I

then

1

K2 = 42.

We

solve

det(A

-

AI)

2-

=

A

2

For Ai

=

1 1

-

= A(A-3) = A

we have 2

1/2

1

1 j

2 so that ki

=

-\ki.

If

k2

=

10

I

2 then

Ki For A2

=

3

j

=

-1

we have

-1

1

1

-1 j

I

2 so that

fci

=

^2- If £2

=

-2

then

1

K2 = 43.

We

solve

det(A For Ai

=

A2

= —4 we

-

AI)

=

—8 — A

-1

16

-A

0\

f\

0/

U

-1 I

16

=

-3*2-

If

(A

+ 4) 2 =

have

-4

so that k\

=

Jfc2

=

4

1/4 I

4 then

Ki =

-1

306

I

Exercises 8.4

44.

We

solve

=

det(A - AI)

—

1

A

1

1/4 For Xi

=

1

1

so that k\

=

-

3/2)(A

-

1/2}

=

0.

=

2&2. If

ho,

=

1

-2)0

1

|

-1/2

1/4

A2

(A

A

3/2 we have

-1/2

If

=

-

then

1/2 then 1/2

1

2

1 |

1/4

1/2

|

I I

so that k\

= — 2&2-

=

If Afc

1

then

-2

K2 = 45.

We

solve

5-A det(A

-

AI)

=

-5-A

9

-1

-A

5

If

Ai

=

4- A -1 4- A —5-A 9 = A(4-A)(A + 4) 4- A -1 -A

-1

then

/5

-1

-5 s5 so that k\

=

and k%

=

01 9

§£3. If £3

-9/5

1

-1

0\

-9/25

(\

o)

=

0/

25 then

9\

K, =

45 I

.25 If

A2

=

4 then

fl

-I

0\

-1

(I

01 j

I

0-9

9

1

-1 ]

I

\5

-1

-4

0/

0/

Exercises 8.4

=

so that kj

and k^

—

k$.

=

k$

If

1

then

K2 = If

A3

= -4

1

then

/9

U so that k\

k$

and k 2 — 9&3

-

1

-

1

9

-

1

4

*3

If

ON

1

-1

l\

0\

-9

1

0/

o)

then

/1\

K3 =

46.

We

3-

AI)

A

=

2

-

A

4 Ai

=

1

=

A)(2

-

-A

=

and

kt

=

0. If

&3

=

n

0\

|

0\

|

1

|

j

0,

0,

,4

A2

-

then

1

so that k\

(3 1

(2

If

w

solve

det(A

If

9

1

then

2 then

/l o

=

and k%

=

0. If

—

(\

0\

1 j

\o

j

1

|

1

-1

,4

so that &i

0\

|

then

K2 =

308

o

o

o;

A

Exercises 8.4

If

A3

=

3 then

/O

0-1

1

md

=

it i

*2

=

*3

j

o

\0

~2

0. If

0\

I

|

\4 so that

-1/2

fl

fl\

I

|

= 2 then

K3 =

47.

We

solve

det(A

-

=

AI)

-A -1

4

=

-A

-4

-2 For A!

=

=

A2

A3

- -2 we

2

4

-1

-2

|

-2fe.

If

=

*a

-A

1

0\

|

V

=

0.

have /

so that ki

+ 2) 3 =

-(A

0J

|

and

fc3

=

then

1

/-2

48.

We

solve 1

det{A

-

AI)

-A

1

6

2-

=

A 2

For A

=

3

6

A 3

1

1

-

-

-A 1

A

we have j

-2

6

|

0^

.

that ki

=

3*3 and k 2

=

Ag. If A3

—

=

1

1

^

0>

—1 j

oy 1

2-

I

1

i

-A =

—3

(1

j

o

3

n

,0

I

(3

-

A}(1

-

A)

2

=

A

Exercises 8.4

For A 2

=

=

A3

1

we have 0\

6

.0 so that

k2 =

and

=0.

£3

If k\

=

(0

0\

1 1

1

1

1

1

oj

^0

0,

then

1

/1\

K2 = /

49.

We

solve

det(A

For Ai

=

3z

-

AI)

-1 - A

-

2

-5

-A

1

=

+ 9=(A-3i)(A + 3i) =

A2

we have

-1-3*

2

-{1/5)

1

+

(3/5)i I

I

-5

=

so that ki

(5

-

If

|i)

—

1

=

fc 2

3*

!

I

5 then

=

K, For A2

=

0.

1

—

3t

—Si we have

-1 +3i

2

-5-1*1°

1 J

-5 so that

=

fci

(5

+

|i)

h-

If

k2

=

1

+ 3i

I

5 then

K2 = 50.

We

l

+

3i

solve

det(A - AI)

=

2-A

-1

5

2-A 1

= -A 3 + 6A 2 -

4

13A

+

10

=

(A

-

2)(-A

2= (A-2)(A-(2 + i))(A-(2-i)) =

For Ai

=

2

we have to

-1

0\

4/5

(\

4

5

0^ I

I

1

j

I

V0

a

1

I

\0

0/

310

0/

0-

+

Exercises 8.4

so that kj

= -1*3

and k2

=

=

0- If £3

5 then

K,

= 5/

For A2

=2+

i

we have -1

f-i

/l

0>

-i

j

0>

I

5

-i

4

1

-i

—*

1

I

I

\

oy

Vo

/ I

so that

fcj

=

ife

=

and &2

*&3- If

=

fca

*

then

K2 = For A3

— 2—i

we have

—10

/i

so that

fe]

=

-1

~ik
and

A:2

=

5

i

4

,0

1

i

-i&3-

If

A3

=

/

0)

0\

i

i

1

{0

0,

i

0/

then

K3 =

1

51. Let

A=

an

112

tt2i

a 22

Then x\\

d f a\Xi

:)( x2j

dt\ (13X1

'

>>*«»^C a-i

52.

Assume

det

A^

and

K

+ a2%2\ _ + 0.4x2 J

(

+ a'i%i + &2X2 + a 2 x,2 \

\ a3x\

+ a'3 x\ + 0.4x2 + a'4 X2 J

0,4

AB =

I,

so that

'an

ai 2 \ /fen

hi

a2 i

022/ \ ^1

^2

(

1

1

Exercises 8.4

Then Oll6ll

+ Oi2&21 =

1

ttll6l2

+ 112&22 —

121612

+ 021622 =

and 021611

and by Cramer's

+ ^21621 =

1

rule

_ "

611

=

,

021

022

det

A

6l2

_a 2i ,

.

det

,

"22

A .

= -au deTA

= an det

A

Thus

B= A-_j = „

1

1

det

53. Since

A

/

022

\

-a21

-ai2

an J

A is nonsingular, AB = AC implies A _1 AB - A _1 AC.

Then IB = IC and

54. Since

(AB)(B _I A _l ) = A(BB

_1

)A

_1

= ALA -1 = AA" = I 1

and

(B

_1

A _1 )(AB) - B _1 (A _1 A)B = B'lB = B _1 B = I

we have

(abj-^b-'a-

1 .

55. No; consider

A= .

1

°\

Vo

0/

f

andj

ll

Exercises 8.5 1

1.

Let

X=

I

2.

LetX=

[

1.

Then

\

Then

X .

B = (° t,

312

B=

C.

Exercises 8.5

3. Let

X= -9\

f-Z

4

6

-1

10

4

X

/x\ 4.

LetX =

y

.

W

Then

-1

1

X'

=

0\

5.

LetX =

1,

Then

o

/

X'

6. Let

X=

.

=

X+

4

3

dx

_ ~

4x

+ 2y + e

= -x + 3y ^ at

( ;

~kt

dx 8.

di

dx 9.

= 7x + 5y - 9z ~_

x

- y + 2z +

e

dx

_ ~

3i

-

7i/

e

9/)

+

I

e

x+

-

dv Zt;

=

-f

3i - Ay

+ z + 2e

(t-

4)e*

~ = -2y + 3z + - Se' dz — = ~2x + 5y + 6z + 2e~' + 5t

^=

( ;

dt

a;

± c

t

t;

+ y + 8sin + t

(2f

+

l)e

41

11. Since

X'

we

=

-5 -10

e

_5t

-5t

and

see that

3

-4

21

dt

dt

at

+ 4sinf +

2t

e

';

dt

_t,

r

f Ue-'cos2W)

l

^-=4x + y + z + 2e5

dt

10.

-3t 2

Then

x'=r5 '

X

2

1

-1

t

Exercises 8.5

12. Since

X'

/ 5 cos t

=

^ 2 cos f

we

— —

5 sin t \

,

and

e<

4 sin f /

-2

5

-2

cos t X - (5

4

2 cost

— 5 sin A — 4sint /

(

see that

13. Since

3/2

X' = we

-3

- - n ^y-a

-3J/2

see that

14. Since

X'

we

4

5

=

2

-4

-1

and

te*

1

-1

X=

4

-4

-1

see that

2

1

15. Since

we

1

2

6

-1

Ul

-2

/

(°\ x'

=

and

1

\

(0

X= -1,

,0

see that i

1

6-10

=

X'

-1

-2

X.

-1

16. Since

cost

— 5 cos ( — cos t — sin ( J

i sin ^

we

cost

\

X=

and

*

\

see that f

X'

=

i

o

1

1

-2

19.

n X.

-1 /

= -2e~ sl / and Xi and X2 are linearly independent on -00 < t < 00 2t and Xi and X2 are linearly independent on —00 < t < 00. Yes, since W(Xi,X2) = 8e ^ and Xi, X2, and X3 are linearly dependent on —00 < t < 00. No, since W(Xi,X2,X3) =

17. Yes, since 18.

— 5 cos t — cos t — sin t J

^ sin t

IVfXi.Xa)

314

Exercises 8.5

W(Xi,X2,X3) = —Me

20. Yes, since

—oo <

t

<

^

1

and Xi, X2, and X3 are

linearly independent

00.

21. Since 2

-1 we

see that

*-G

i)

22. Since

we

2

1

1

-1

-5

X„ +

see that

23. Since 2

^

e*

+

*

I

)

2

1

3

4

and

te*

=

e

ie

1

thatt

/n

1

K

=

c

l

24. Since

3cos3t\

/

f

we

— 3sin3t /

2

-4

2

V-6

1

and ,

1

/

3\

/

Xp +

4

[

sin 3t

3^

-3 sin 3i /

3

V

3 cos

=

see that 1

2

-4

2

,-6

1

1

/-1\

3^

xp + oj

4 I

sin 3t.

3/

25. Let /

X = 1

/2\

6

-1

V-5

I

e-\

Xa =

1

V

1/

1

VI/

/O e

3t ,

and

A=

6

1

Vi

0\ 1

1

0;

on

Exercises 8.5

Then

/-6\ xi =

p

1

— AX,

_t

^

\

6

(

-2

)

- AX,

p" 2t

V-2J 3i

F =

and W(Xi,X2, X3)

—00 <

t

<

20

^

=

AX 3

,

and X3 form a fundamental

so that Xi, Xa,

set for

X'

=

AX

on

00.

26. Let

Xi =

-1-

y/2

X2 = x„ =

1 I

U2 +

2

)t+

(

1

.0,

and

-1

-1

-1

1

Then

x; =

x2 =

set

on —00 < /

e

t

2t

= <

2^/2

^

AX

lt

-2-y/2j 2\

and W(X],X2)

e^' =

-2- V2

/-2\

so that

l\

Xp is a particular solution and

00.

e

7t\

j

/3 e "

316

- e "'\

Xi and X2 form a fundamental

Exercises 8.5

-e'

3te

3e'

We

-

-e 5t

/3te -e'

1

"

l

e }

K

-2sint

2 cost

/

\ 3 cost 31.

(

5'

[

-fe<

29. #(t)

30. #(f)

e

1

28. *(()

"

!

e 2t \

"\

# -1 (t)

and

+ sint — 3sin( + cost /

fe

—

-3e*

-

1

2

(— 3sint + cost \ -3cosf - sinf

2sint\ '

2cosi

have

x(*)=

CI

„7t

;2

[

so that

and

which give

cj

=

3/5, c2

Vl =

=

2/5,

e

5

and

M+

c\

=

—1/5,

=C}e

C2

=

1/5.

Then

^d V 2 = -i

7*

*

(

]

e»

+

I

]

e «,

5 \,3

so

2„2(,3 = 7l

6 a 2t,fi„7t

32.

We

have X(t)

=

Cj

e

'

+ c2

I

j

e

5t

1

so that

X(0)

=

ci

+ C2

X(0)

-

ci

+ C2

and

c\

=

and

which give

c\

=

-1/2, c 2

Vl =

1

=

/-I

1/2,

1

-;r 2

1/2, c2

/I

W

e

5t

=

and

1/2.

Then

V2 =

1

£

/-l e

5t ,

I

Exercises 8.5

H,(t\

33.

We

i fi -t

—

2

f

_I eP -t _l + 2

lJ>t + ^2

Ie P 5t 2

have 3 /

3

V

V-l

/

so that '

X{0)

=d

X(0)

=

and '

Cl

3

—

which give

—1, c 2

=

—3, and ci

=

J

c%

0,

=

+e

'Sie' '

34.

We

—1.

Then

1

fe'

+ e'

-3ie (

9te<

have X(i)

2cosi

=

ci ,

3 cos i

\

-2sinf

/

+

+ sin t J

c2 \ cos

(

—

3 sin t

so that

and

which give

ci

=

—3/2,

C2

=

—1/2, and

*(*)

35. Since X(t D )

= »(to)C = Xo we

=

ci

'— 3cost

=

have

37. Since is

=

*(to)X = IXo

X = *(t)*

arbitrary

it

-1

=

=

C=

0.

Then

+ sinf

2cos( 3 cos

(

+ sin t

ff-^toJXo and X(t)

=

= AX

X=

we know

that

*(«)*

-1

(to)Xo.

4"(t)Xo solves X'

= AX,

and

Xo.

(to)Xo and

follows that

ci

—5 cos t

,

36. Since the column vectors of 9(t) solve X'

X(to)

1,

X=

S^)*"

1

*(t)X we

^

-

=

see that 0.

318

[*(t)*~ l (to) - «(()]

Xo =

0.

Since

Xo

Exercises 8.6 1.

The system

is

and det(A - AI)

=

(A

-

+

5)(A

'-4

2

4

-2

1)

I

=

0.

=

For Ai

ON

/l

0/

V0

5 we obtain

-1/2

I

so that

For A2

= —1

Ki

0,

we obtain '2

2

4

4

0\

/l

10/

V0

I

1

I

so that

K2 =

' [

Then

X = Cl 2.

The system

^ I

-t M + c..Hi. e a

e

5!

,

is

and det(A - AI)

=

(A

- 4) (A + 4) =

/-4

2

8

—4

\

I

0\

0.

=>

0/

For Aj

/l

=

4 we obtain

-1/2

I

ON so that

Vo

j

Ki

0/

I

For Aj

= —4 we

obtain 'A

2

I

0\

/l

0/

Vo

1/2

I

so that

4

K2 —

0.

Then

3.

The system

is

2J

V-5/2 and det(A

-

AI)

=

(A

-

1)(A

-5 -5/2

+ 3) = 2

I

0\

10/

0.

For Ai

/-5

= 2

we obtain

1

( |

so that

V

0/

Ki

Exercises 8.6

For A2

= —3

we obtain -1

2

-5/2

5

I

0\

/-l

07

V

2

I

so that

Ko

0,

Then

x=o m,. + c ! r],1

4.

The system

is

9

1/2

x'=[

IX 2,

.1/2

and det(A

-

AI)

=

(A

-

7/2)(A

-3

9

+ I

1)

=

0.

For Ai

0\

/I

07

Vo

=

7/2 we obtain

-3

I

0' so that

-1/2

1/2 For A2

= —1

Kr

0/

we obtain '3/2

9

1/2

3

I

0\

fl

07

V0

6

I

so that

K2 —

Then

5.

The system

is

and det(A

For A2

AI)

= -10

=

(A

-

8)(A

+

'2

-5

I

8

-20

I

'20

-5

I

8

-2

10)

=

0.

For Ai

0\

(\

07

V0

0\

/l

07

V0

-12 J

8

V

-

=

8

-5/2

we obtain

I

I

so that

Kj

so that

Kq

0/

we obtain

I

-1/4

I

I

0,'

Then

6.

The system

is

x'=[320

2

|x

Exercises 8.6

and det(A

-

AI)

=

+ 5) =

A(A

'—6

2

=

For Ai

0.

ON

/l

o)

\Q

we obtain -1/3

j

Ki =

|

so that

,-3

1 |

For A 2

= —5 we

|

obtain 2

I

0\

/l

I

Oj

V0

—2

I

so that -3

6

K2 —

,

VI

0/

Then

7.

The system

is

/l

- AI) =

(A

-

-

1){2

A)(A

+

1)

=

0.

1

f

OX

2

Vo and det(A

-V

1

=

X'

-1/

1

=

For Ai

A2

1,

=

2,

and A3

we obtain

/1\

(2)

)

Ki =

= -1

3

K3 =

and

,

,V so that

X

=

(A

(') e*

ci

+ c2

/1\

3

e

2t

+

c3

,2/ 8.

The system

is

X'

and det(A - AI)

=

(2

-

A)(A

(

-

5)(A

4

7)

12

-1

0\

5

10

4

^0

5

= =

0.

2)

=

For Ai

2,

A2

=

5,

and A3

(~ 7 )

) ,

K2 =

3

and

,

K3 =

,-5 J /

X = ci

f-7\ 5 I

so that

4\

/

_ 7\ 3

f-7 e

5t

+ c3

5

=

5/

7

we obtain

Exercises 8.6

= -(AH

-3)(A

+

=

2)

/l

K = L

K2 =

,

-i ,3

3/

\

so that

/-1>

X = ci

4

We

have det(A

-

AI)

U/

iy

I

10.

i\ -1

/

= -A(A -

1)(A

/

1^

-

=

2)

For Aj

0.

=

0,

=

\2

=

and A 3

1,

/0\

2

we obtain

fl\

1

W

and

,

K3 =

U/

so that /

X

11.

We

have det(A

-

AI)

=

-(A +

=

/0\

1^

+ C2

ci

+

1)(A

1/2)(A

1

f

+

e'

1

+ 3/2) =

For Ai

0.

^

c3

=

-1, A 2

=

-1/2, and A 3

= -3/2

we obtain 4

(

f-12^

^

K[ =

K =

6

2

,

1-1/

\

2

1-1/

5^

I

4

f

K3 =

and

,

that

4\

X = ci

j

*

+

We

have det(A

-

AI)

=

(A

-

3) (A 1

(

+

6

c2

-1/ 12.

4\

/ e

\

-

A)

=

0.

-1

,0/

For Ai

=

3,

=

A2

^

(

1

V-i/

5/

\

5)(6

2

( ,

and

K3 =

0/

^

-5, and A 3 2 \

-2 I

11/

so that

/1\

X = ci

1

W

( e

3t

+ c2

M -1

2'

e-

5(

+ c3

-2 11

322

„6f

=

6

we obtain

t

Exercises 8.6

13.

We

-

have det(A

=

XI)

(A

+

1/2)(A

-

=

1/2)

r

Kj=

For Xi

0.

= -1/2 and

K2 =

and

A2

=

1/2 we obtain

f j

J

that

X = C!

„t/2

If

=

X(0) then 14.

We

ci

=

2

and

have det(A

C2

-

=

3.

=

XI)

(2

-

-

A)(A 5

3)(A

+

1)

-3

2,

A2

=

3,

and A3

= -1

we obtain

-2\ and

,

*)

{

=

For Ai

0.

/2\

^

f

=

K3 =

,

|

1/

,1/

so that

l-T

X = ci

-3

e

2t

+ c2

e

|

3t

+ c3

|e-»

2/

\ If

X(0)

then

ci

=

-1, C2

=

5/2,

and

C3

=

=

3

-1/2.

In Problems 15-28 the form of the answer will vary according to the choice of eigenvector. For example, in

Problem

15, if

Ki

is

chosen to be

_

|

cost

X = c\

2 cos

15.

We

have det(A

the solution has the form

|

.

-2t,

-

=

AI)

A2

-

8A

+

t

+ sin

=

17

0.

e

sine

4t

+ c2 2 sin

f

For Ai

=4+

i

t

—

e

4t .

cos t

we obtain

so that

x = 1

2

+i

l '

5

\ J

e 6

<4+0*

= /2eost-8int\ \

5cos(

J

/ cosi

4

+ 2 sin i

l

\

5sin(

Exercises 8.6

Then .'2

X = ci

—

cost

sint\

5 cos*

16.

We

have det(A

-

=

AI)

A2

+

1

=

+ c2

=

.,

4(

e

/

For X\

0.

+ cosf\

/2sint

.,

4t

e

.

5sint

V

we obtain

i

'-l-i Kl 2

so that

—1 —

I

i\ 8

1

\

2

—

/sini

it

\

J

cos*\

2cos(

/

+

j

—

—

cost

H

sin*

2sini

Then (sin f

— cos f \

/

— cos ( — sin t

\

2sin(

+C2 2cos( 17.

We

have det(A

-

AI)

=

A

—

2

8A

+

17

=

/

-

For Ai

0.

Ki-.

+ i we

4

obtain

'-l-i' 2

so that

X,

1

=

f'" 1

"'^^'^

2

y

- cost

siRt

(

) e

2cosi

I

/-sint-cosf

«

2sin*

\

J

Then

X = ci 18.

We

have det(A

-

AI)

=

.'

sin

X 2 - 10A

—

t

cost \

+ 34-0.

U

,, 4i

+ C2 =

For A]

— sint —

{

cosf\ e

5

+

3i

,, 4t .

we obtain

-3T

'1

so that

X

1

=

1-3 [

2

^J

e <5+30<

= /«»3t + 3Bin3t\ a I

2cos3(

/'sin3(-3cos .

/

I,

2cos3t

Then

X = ci

cos 3t I

V 19.

We

have det(A

-

AI)

=

A2

+9=

+ 3 sin 31 ^

}e

'

2cos3t

0-

For A t

ot 5(

+ c2

J

=

( siti3t \

I,

3i

we obtain '4

+ 3i'

324

— 3cos3t^

2cos3t

5 '. )e S( '

Exercises 8.6

so that

4cos3f - 3sin3t\

.

/4sin3i

+ 3cos3t

(

1

5cos3(

/

5sin3r.

\

Then

X = Cj

- 3sin3t\

,'4cos3i

5cos3( 20.

We

have det(A

-

AI)

=

\2

+ 3 cos3(

/ 4 sin3f.

+ c2

+ 2A +

5

_ 1+M)(

=

=

0-

5sin3i

\

I

= -1 +

For Ai

2i

we obtain

so that

Xi

„/2 + 2t\

(

2 cos 2f

-

2 sin 2t e

'

+

cos2t

1

Then

X = ci

2cos2(

-

2sin2i

_t

e

We

have det(A

-

AI)

= -A (A 2 +

=

l)

2cos2t

+ c2

+

2sin2f

sin2t

0.

=

For A]

we obtain

/1\ \°/ For Aj

=i

we obtain

f-i\

K2 =

i

so that

x2 =

— sinf

i

/

sini \

/

cos

t

sini

(

,

-

+i j

cos

(

cosi sint

\

)

Then /

(1)

X = cj

+ c2

We

have det(A

-

AI)

=

(A

+ 3)(A 2 -

2A

+ 5) =

0.

/

'

t j

For Ai n

— cos t

+ c3

-sini COS

22.

+ 2sin2C

sin 2t

cos 2f

21.

2 cos 2t i

\

1

cost ,

= -3

sin

(

,

we obtain

-t

\

Exercises 8.6

For A2

= 1 + 2i

we obtain

-2-i

(

K2 =

-

-2

3i

2

V that (

X2 =

-2cos2(

+ sin2t

\

-2 cos 2t + 3sin2t

e

2 cos 2t

y

-cos2t- 2sin2f > -3 cos It - 2 sin 2t

I l

+i

I

\

2 sin 2t

+ sin2£

\

I

Then

X=a

I

0\ -2 e" 3( +

-2cos2f c2

I

-2 cos 21 +

We

have det(A

-

AI)

=

-3cos2t

2 cos 2f

1/ 23.

e +c 3

3 sin 2t

(1

-

A)(A

2

- 2A +

2)

=

Ki

-cos2i - 2sin2(

For Ai

=

=

1

we obtain

2

U/ For A2

=1+

i

we obtain

m

so that

x2 =

e<

i

=

1

cos f

'

1+i ><

faint \

— sint (

— sin f

e

(

+i

,

cost s

cos t

j

Then cost

{

X

=

ci

2

e'

+

-

c2

V 24.

We

have det(A

-

AI)

=

-(A - 6)(A 2 - 8A +

e

20)

=

+

For Ai

0.

Vo/ For A2

[

we obtain

K2 = V

326

2^

cost

c3

— sin f /

1

= 4 + 2i

/ sint \

>

sint

\ cos

=

6

2 sin 2t

2sin2t

\

0.

-

t

/

we obtain

Exercises 8.6

so that sin 2* ^

/

,2cos2i /

(

— cos 2t y

X3 = 2/

\

2 sin 2i

Then '-cos 2t'

sin2f \

X = Cl

e

1

6£

+ c2

e

]

2 cos

25.

We

have det(A

-

=

AI)

(2

-

A)(A 2

+ 4A +

=

13)

2sin2f

2*7 0.

7 28

K =

=

For Ai

2

we

4t .

,

obtain

\

-5

1

\25j For A2

= — 2 + 3t we

obtain

/4

K2 =

+ 3A -5 )

so that

X2 =

p (-2+3i)t

-5

=

McosSt-Ssin 3^ — 5cos3t

f4sin3i

+ 3cos3t\

-5sin3(

e"

/

/

Then

X = c,

/4cos3f - 3sin3t^

,

/ 28

\

-5

e

2t

+ c2

V 25 / 26.

We

have det(A

-

AI)

=

-5 cos 3f

e"

+ 2)(A 2 + 4) =

0.

For Ai t

=

K,

For A 2

=

2i

+

+ 3cos 3^ )

\

= -2

--2t

—5 sin 3t

c3

/

V

-(A

/4sin3( 2(

we obtain

0\ -1

we obtain

(~2-2i\

K =

1

2

1

/

so that

f-2-2i\

X

2

=

'-2cos 2(

\

1

J

/-2cos2(-2sin 2^

+ 2sin 2t\

cos 2t cos 2t

+

i

sin 2i sin 2t

2t .

"

,

Exercises 8.6

Then (

X=a We

have det(A

-

=

XI)

-

A)(A 2

1 /'-2cos2t-2sin

+ C3

cos 2(

s

(1

+ 2sin2f\

cos 2t

Oz

I)

,

27.

'-2cos2(

°^ -1 e- 2t +

+

25)

=

=

1

sin2f

\

,

For Ai

0.

sin 2t

we obtain

25 ^

K, =

-7

|

6 /

For A2

=

5i

we

obtain

K =

1

3

V

J

1

so that ( 1

+ 5A

x2 =

'

e

1

I

1

5,(

cos5(

=

—

5sin5i^

+i

cos 5(

J

cos 5t

4

sin5f

'

+ 5cos5f

s

sin5t

)

sin 5t

(

,

Then s

/25

1

cos 5t

—

5 sin5t

N

'

sin 5t

,

X = ci

-7

e

l

+ c2

6)

,

+ C3

cos 5t cos 5t

K

+

5 cos 5t

sin 5t sin 5i

„

,

,

If

4\

(

X(0)

=

6

V-7/ then ci 28.

We

=

C2

= —1

have det(A

-

and AI)

C3

=

=

6.

-

A2

10A

+

29

=

=

For Ai

0.

Ki

5

+ 2i we

obtain

1

= 1

-2i

that 1

1

e

- 2ij

X = ci

Uos2( + 2sin2f cos2t

1

,

cos 2t

+

\

2 sin

2M

ct

e

+c 3

=

328

-2

sin2t\_

+i

sin 2(

/

-

sin2t

V sin 2t

If

X(0)

5!

- 2 cos 2*

2 cos 2t

„

1

e

5i

.

then 29.

We

ci

= —2

and

have det(A

-

c-i

XI)

—

5.

=

A2

=

For Ai

0.

=

we obtain '1

K= A

solution of

(A - AiI)P

K

=

is

P= so that

x = Cl 30.

We

have det(A

-

AI)

=

(A

+

l)

2

=

0.

(

3

i

+

ca

= -1 we

For Ai

obtain

K 1

A

solution of

K is

(A - AiI)P =

P= 1/5 so that

X = c,

le-'

|

+ ca

te-*

+

x

31.

We

have det(A

-

AI)

=

(A

-

2

2)

=

0.

=

For Ai

2

we obtain

K 1

A

solution of

(A - A,I)P =

K

is

-1/3

P= so that

X = ci

)c

[

a+

c2

(e

2'

+

1

32.

We

have det(A

-

AI)

=

(A

-

6)

2

=

0.

For A]

=

6 we obtain

KA

solution of

(A - AiI)P =

K

is

_

/l/2\

Exercises 8.6

so that

X = ci 33.

We

have det(A

-

AI)

=

(1

-

A)(A

" 1

e

I

-

6(

=

2

2)

+ c2 0-

te

,

6t

2/

For A]

=

1

+

1/2 I

]

e

we obtain

/1\

Ki = \1/ For X2

=

2

we obtain

K2 =

K3 =

and

VI/

1

V0

m

Then

X = ci

/1\

1

+ c2

e'

e

2t

+

c3

1

W

uJ 34.

We

have det(A

-

AI)

=

(A

-

8)(A

+

=

2

I)

0.

=

For Aj

we obtain

8

(2\ 1

\2/ For X2

= —1

we obtain /

K2 =

0\ -2 1

V

1\

(

K3 =

and

-2 V

0/

Then

X = c\

^\ 1

( e

Si

U, 35.

We

have det(A

-

AI)

= -A(5 -

A)

2

=

+

c2

®\ -2

For Ai

=

we obtain

-5 \

=

+

c3 V

f-4\ Ki = For A2

!

1/

I

0.

f

e

2/

5 we obtain

/-2\

K= V

330

1/

1\ -2

0/

A

solution of

(A - AiI)P

=

K

is

P

2

T„ so that

'-2'

X-d -5)+^^

Je

j

36.

We

have det(A

-

=

AI)

(1

-

A)(A

- if =

.

5(

U

+ C3

For A,

=

l

5l

+

w obtain

'1'

K. = For \ 2

=

2

I

we obtain

0\

K=|-l 1/

A

solution of

(A -

A 2 I)p

=

K

is

/

P=

0\ -1 0,

ao that '1^

X = ci

( e'

I

+

c2

°^ -1

°' e

+ c?

-1

We

have det(A

-

AI)

=

—(A -

1)3

=a

For Al

=

1

te

1/

^

37.

j

)

( 2t

we obtain

-(!) Solutions of

(A - AiI)P =

K and (A - AiI)Q = P

are

/0\ 1

I

\0/

(1/2"

and

Q= I

2c

+

Exercises 8.6

so that

(Q\

/0\

1

1

X=q 38.

We

have det(A

-

AI)

=

(A

0< f

l

i

- 4) 3 =

0-

2

+ r ay w

1

t

1

=

For Ai

we obtain

4

/1\

K Solutions of

(A - AiI)P =

=

K and (A - AjIJQ = P

are

/0\

/0\

P=

Q=

and

1

\0/ so that

\1/

m

X = ci

(0\ e

4t

+

c3

r.

e

4t

+

1

1

/

39.

We

have det(A

-

AI)

—

(A

—

4)

a

=

0.

For Ai

=

4

we obtain

K= A

solution of

(A - AiI)P =

K is F.1,

so that

X = Cl

|

(

e

4t

+

c2

T;

i

te,,+

,e

'

ii

If

then 40.

We

ci

= —7

and

have det{A

-

ci

AI)

—

X(0)

=

=

For Ai

13,

= -(A +

1)(A

-

l)

2

0.

K,

= V

332

i

= -1 we

obtain

te

4t

+

Exercises 8.6

For A2

=

1

we obtain

/0\

K2 =

K3 —

and

1

so that

/1\

(-1} e"

1

+

c2

e

+ c3

(

1

uJ If

=

X(0)

then ci

41.

We

=

2, C2

have det(A

=

3,

-

AI)

C3

—

(A

-

and

=

2

W

2.

5){A

+

5)

K = 1

=

0.

=

For Ai

n

K2

and

= -5 we

5 and A 2

obtain

=(~l

so that

*(*)

=

3e

5t 5t

E

_ e -5t^

___

3e- 5 '

__

* _1 (t)=io

_

and

]

1

/

3e -5t

e«

g

-5( 5t

3e

Then

x = *( 42.

We

have det(A

-

AI)

=

(A

=

)*- (o)X(o) l

f

+ 3/25) (A +

1/25)

=

0.

^

e5[

For Aj

+ ^_ 5[

=

j.

-3/25 and A 2

=

-1/25 we obtain

so that

/_.-31/25

p -t/25\

/

1

9 P -t/25

Then 25

X = *(t)*" 43. Using

1

(0)X(0)=

2

e -3t/25

ow™

(

X = x K in t

i

'

1

tX =l l

333

IX 5

,

25 2

-(/25

.»«,.

„-(/25

\

,

Exercises 8.6

we obtain

For nonzero solutions

K

we must have

(A

—

4)(A

—

=

2)

0. If

Ai

=

2 then

Ai

=

3 and A3

'3'

=

Ki and

A2

if

=

4 then

K2 = so that

X = at* 44. Using

X = x K in t

we obtain

2-X

-2N

For nonzero solutions

K

(0

7-X

2

we must have (A - 3)(A -

K,H

and

\0 6)

=

0.

If

K =( 2

^

so that

Exercises 8.7 1.

Solving

2-

A

-1 we obtain

eigenvalues Ai

= —1

3

-2

and A2

Ki=

=

-A =

1.

A

2

-1 = (A-1)(A +

1)

=

Corresponding eigenvectors are

and |

J J

Thus

334

K2 =

(

J

=

6 then

Substituting

into the system yields

2ai

+ 3h =

- ai -

—

from which we obtain ai

2.

—1 and

b\

—

9

=

26i

=

7

-5,

Then

3.

Solving 5

A

-1 we obtain the eigenvalue A

=

8.

A

=

=

(A

corresponding eigenvector

is

-A

11

\ 2 - 16A

+

64

-

2

8)

-

'3'

K= Solving

{A - 8I)P

=K

we obtain

Thus

Substituting



—

1/2 and

X(t)

=

ei

bi

]e |

B

5oj

+

9&i

-ai

+

llfr 1

=

= -2 =

-6,

—1/2. Then

'+c 2

+

1

3.

Solving 1

3

A

3 1

—

= A

A2

-2A-8 = 335

(A-4)(A

+ 2)

7 Exercises

we obtain

8.

eigenvalues Ai

= —2

=

and A2


|_M

=

K,

4.

K 2 =(|

and

Thus

Substituting


03

+ 363 =

2

(12

+ b3 =

3a 3

from which we obtain 03

=

^3

=

1

-

X

3&2

+ b2 +

3a 2

—1/4,

+

3/4, 02

1

=

2a3

=

2& 3

=

1/4, 63

=

A

+

ai

3o!

=

+

fci

—1/4, 01

=

02

+5=

62

3f>i

=

—2, and

(>i

=

4. Solving

-4 1

—

A

and A 2

=

1

4

we obtain

eigenvalues Ai

= 1 + Ai

Ki=

(J)

—

,

+ 2A —

17

=


Ai.

K2 =

and

(~*

Thus cos 4f

+

[

^

I

1

sin 4f

e

+ c2

cos 4f

—

—

sin 4£

= d|

..

cos 4t

+c 2

|« /

V

sin At

\l

^ J

cos 4t

— sin At

Substituting


as 4
— 4&3 = — 4 +

63

=

1

—

4fcj

=

(13

4d2

+

62

=

i>3

336

— 5ai —

46i

= —9

—

56i

= —1

4ai

3/4.

Then

Exercises

from which we obtain 03 '

X(*)

=

=

=

0, 63

— sin 4( e'

ci

—

aa - 4/17, 62

1,

+ ^-'

mi ')e-

— sin At

cos4(

=

1/17, 01

+

(

)

and

1,

t+

f»i

i/17

_)

(

=

1.

Then

+ (]) C -

1/17

5. Solving

4

-

A

1/3

9

6

we obtain the eigenvalues Ai

=

—

=

-

2

-

10A

=

7.


A

+

21

-

(A

-

3)(A

7)

=

A

3 and A2

K '=L

Ka=

and 3)

(g

Thus

xe =c 1(

e

_;)

-+

ri^.

c2

Substituting


+ ^1 =

3«i 9ai


6.

=

55/36 and

61

=

+

56j

3

= -10

—19/4. Then

Solving

-1 - A

5

-1 we obtain the eigenvalues Ai

=

and A2

2i

Ki =

1

=

-A

)

1

A

+4=

a

—2i. Corresponding eigenvectors are

5 I '

=

+ 2i J

and

K2 =

5 [

\l -

2i

Thus

X = Ci

5cos2t

c

cos2(

-

2sin2f

\

5sin2t

/

+ ^2 /

\

2 cos 2f

Substituting

^ = C) cost+ {Z

]sint

+ sin 2t

8.

Exercises

8.


— 02 + 562 —

-

2

=

+ 56i + a 2 +

1

=

-oa -oi

+

62

X(t)

=

=

5cos2f

/

=

-3, 62 \

-1/3, and

5sin2i

/

+C2

ci

Vcos2(-2sin2t/ 7.

—

-2/3, ai

61

+ bi + b2 =

-ai from which we obtain 02

-

=

aj

\

+ sin 2t /

V 2 cos 2t

bi

— /

+

Then

1/3.

-3

\

cost

,

\-2/3 J

+

/-l/3\ '

\

1/3

Solving 1

-A

1

1

2-

A

we obtain the

eigenvalues Ai

=

1,

A2

=

= (1-A)(2-A)(5-A) =

3 5 2,

-A and A3

=


5.

(1}

Ki =

and

1

K3

m = 2

,2,

Thus

/1\

X c = ci

e'

+ c2

f

1\ 2

1

W

-St

Substituting

x„ =

e

61

41

\ c l/ into the

system yields

-3ai +bi

+ ci = -1

+ 3c! =

-26i

- -2

ci

from which we obtain

=

cj

—2,

61 1

X(t)

=d

f

—

—7/2, and ai

=

—3/2. Then

fl\

^

e

(

+ c2

1

e

2l

1

+ c3

m 2

/-3/2\ e

5<

+ I

338

„4(

-7/2

-2

,

sint.

7

A

Exercises

8. Solving

-A

5

-

5

= -(A-5) 2 (A + 5) =0

A

-A

5

we obtain the

eigenvalues Ai

=

5,

=

Aj

5,

and A3

=

—5. Corresponding eigenvectors are 1 /

Ki =

I

K2 =

,

and

1

1

K3 =

U)

\

Thus

/1\

( e

0l

+ c2

1

\

e

5t

1

+ c3

I

e~ 5t

Substituting

bi

\ c i/


from which we obtain

ci

=

=

-1

2,

and

5ci

- -5

56i

=

5ai

= -40

10

01

/I X(t)

=

5(

e

Cl

+ C2

1

e

5t

+ C3

e

- 5t

f-8\

+

u 9.

2

V-l/

Solving

-1-A

-2

we obtain the eigenvalues

Aj

=

4-

3

=

1

and A2

=

A 2 - 3A + 2 = 2.

K,= [_M

-

(A

1)(A

and

K = 3

*

(

-4 Ci

-1

2)

=


Thus

Xc =

-

e'

+ C2

339

„2i

8.

7 Exercises

8.

Substituting


-ai -

— —9

and

X«)^

61

=

Then

6. 1

I

(_ i

= -3

+ 46i = -3

3ai

from which we obtain a\

2f>i

)e«

4

+ ra (-

e

» + (- 9

6 )

6

Setting

we obtain

- 4c2 -

c\

c\

=

and

13

c%

=

= -4

+ 602 + 6 =

-c\

Then

9

5.

2 so 1

X(()

10.

(a)

By KirchofFs

E=

ij

Ri

first

+

Let I

=

t

and second (on each

+ L2

i'

dt \i 3 j

(b)

= 13^_ ê + 2 i

3

loop) laws,

we obtain

]

-(Ri + R2)/L 2 ) \i 3 )

{-ft/Li

so that

'.-2

-5

1

60

and

X = ci[ c

If

XD =

(

^j

then

=

is

+ 13, E =

so that

12 (

i

Xp = ^ ^

)

2

.)^ + ^(t)et

U

-1/

so that

340

6t

+

\E/L 2

.

Exercises

For I{0)

(c)

=

h{t)

=

i2

^

(t)

we

j

find c\

= -12

+ i 3 (() - ~12e- (

and

C2

=

lSe"61

+

30.

8.

-6.

11. Solving

_ \ 1-A

_i -1

i

-1 we obtain

the eigenvalues Aj

=

1

=

-A

and A2

=

=

M

Ki

I

=

2 A - 2A

-

A(A

=

2}


2.

K 2 -=

and

(

Thus 2t

Substituting

into the

system yields

From the

If

we

first

let b\

~

system we have

1,

then 01

=

02

-

=

11

-

&1

+

2

=

(X 2

-a 2 + b'i =

-01

+

61

-

5

=

62

«2

= 62-

0,2

—

&2

1

and

-

The second system becomes oj

—

61

=

a2

ai

—

61

=

—02

a\

= —02 —

—

4.

2

—

5.

This gives 02

=

(>2

=

Thus X(t)

=

,2*

-3/2

-5/2

Cl

-3/2

1

—3/2 and

ai

=

-5/2.

Exercises 8.8

Exercises 1.

8.8.

From

we obtain

Then

*=

,'1

3e'\

,

and

,

1

=

l

2e j

(-2

3

U"'

-e" (

so that

and

2.

Prom ,

X'

=

j'2

3

-n X +

/o\

-2

It

we obtain

Then 2

C

so that

-2(e~^

(2te~ t

t

u=/*- F*=/(-^; )*-( i

and

Xp = 3.

*u=[*) +(_° 4 t

FVom 3/4

-\j

we obtain

342

\-l

2te*

+ 2e-

2e

l

Exercises 8.8 Then /

2e t/2v

10e 3t/2

J

_l -3t/2 "2 e

e -3t/2

_3 e - ( /2

5

5

so that

U

Fdt

=

13

/

*-

\z t

|

and

4.

From x'

=

['

2

.4

we

sin2t -Mx + f\2cos2f/ ) 2

obtain

\2sm2t)

V2cos2f/ Then /

-e 2t sin 2t

Ue

2

'cos2t

e

2t

cos2i \

2^3^12(7

~ 2t

-if ~2 " e

^

~V

le

2f

3 e_a 008 2t

cos 2*

£ B -»aiii2t

9111 2(

so that

U

5 cos 4t ^

Fdf =

^^

^

^ sin 4t ^

and

X P = *U=(

^ cos 2t sin 4! 5.

—

S sin2icos4f

-

I cos2f cos4i \ 8 e

5 sin 2t cos 4t /

From

we obtain

Xc =

et

cl

(l)

+ C2

,e

21

(l

Then 2e

(

e

2t

and so that

and

#

1

=

_ e -2<

2e

-2(

2 '.

Exercises 8.8

6.

From

we obtain

Then

so that

2e- (

4 /-2e-* + Je-** -ee \ at _ 2„-5t + r-)*-( '

u -/*" ip *-/(-^«" and

e" 3£

1

e

20

7.

From

-

3 A

H

1

we obtain

*—

(!)«" + «-("iV-

Then

so that

"-/•-"-/(

)*-(

61e 3,

2(e

3,_i,

and 12

x' = * u ='"o 8.

4

,+ )

(: 4/3r

From

we

obtain 4^

,3t

,

344

„

f" 2

^

)

Exercises 8.8

Then 4>

4e

-

-2eat

3'

36

6

*-!=f

and

„-3(

„3i

,

so that

U=

/#-

1

6

Frf(=

3

24

=

df

/

t

c lit'

fi^^

T3

+

e

S

12

te

2I

and

-

-te*

9.

±e<

From X'

3

2

-2

-1

=

X+i

" \--t

we obtain

Xe =

Ci|

_

Je

l

+ c2

te'

-1

1/2

Then e

*-

l

te*

*

and

1

e~*

=

2e

2ie" 1

-2te _l

_t

2e

-'

so that

PA -

U

=

di

6e -2t

and 1/2

-2 10.

Prom

we obtain -1

e'

1

+ c2

te*

-1

+ 1/2

Then

$

and

-e

(

Ae'-te'

1

e

=

_t

-

2ie~'

-2(e

2e"'

_(

2e~*

so that

U=

Fdt =

e

_i

- 4te _t df

2e

345

-

3e-'

+ 4ie _( \

-2e -i

e

Exercises 8.8

and

11.

From

x

-1\

,'0

,

='i

/sect

x+

o

we obtain ,'cosi\

/

sin t

\

I

sint

— cos

t

Then

#=

cos

i

sin

£

\

t

# _1 = ,

and

|

—cost!

sint

/ cos (

sin

\sint

—cost

t

so that

U=

1

/

*-'Fd£= /(

)dt

t

=

lnj sectl

and

X„ =

,'t

*U =

s

12.

cost

+ sintln

sin

— cos t m

|

t

i

sectl

I

sec

|

t]

From

we obtain

— sint \

,

/

,

cos t J

cost

V sin

i

.

t

Then

—

sin

t

cos

t

\

(

,

e'

cost

* _1 = ,

and

sint /

/

— sin t

cos t \

.

e~*

\

cost

sint /

so that

U= J

t

/

*" F(ft=

r (

,

1

/ J

V

t

\

/ 3 cos

t

+ 3 sin t \

+ 3 sint

J

V 3 sint

- 3 cost/

-3 sin f + 3 cost

3 cos

H* =

and

13.

From X'

=

1

— 1\

11/

X

346

/ cost \ sin t

Exercises 8.8

we obtain .'-sint\ cos

t

/cost',

,

t

\ sin. i

i

Then

—

*-

sin

cos t

t

l

e

cos t

sin

*

and

_]

/

—

'

— sinf cos

t

cosi sin

t

-t

f

so that

and .

cost sin

14.

(

i

t

From

we obtain ,

te"

2'

1/2

+|

|

e

-2t

U/2

2/ Then 1

2t+

1 1"\

e

2

4f

,

lj

2t+n

,-i

^

-2t

*

=

(

2t

+

-1

2

so that

u=y *-

i

Ptft=

j

2 In

t

dt

-lnt

and

+ lnt - 2tln! 4t + 3lnf-4tlnt 2t

15.

„-2(

Prom

X--I

Mx +

— 10/

f \ sec

t

tan

t

we obtain cos t \

,

.

- sin t

/

/ sin

(

cos

t

V

Then cos t

sin

t

\

-sini

* _1 = ,

t

and

cost/

/ cos t

Vsint

347

—

sin

t

cost

J

,2t

(

Exercises 8.8

so that /

,

J

— tan t\

[I —

tan 2 £\

/(

\

tant J

Vln|9eci|7

J

and

X„ =

16.

cos(\

#U =

.— sinf

— sinf

/ £

|

+ \ sin

/

/sinf.

\

tant/

£

+

ln|sec£|.

]

\cost.

From

-1

Wotf

we obtain

X = ci

cos t \

,

c

—

sin

/ sin t

+c 2

.

\

£ /

cos t

Then sin£\

cost

i

*=

I

—

*

and sin

cost

t

_,

=

V sin

I

— sint

/cost I

cos

£

so that

J

./

\ In csci

\csc £/

|

— cot

t\

j

and '

sin

t

|

\ cos £ In

17.

— cot esc £ — cot

In esc t

£|

£|

I

From 2\

1

-1/2

1

/csct

_

\sect

J

we obtain

X = C1

2sint \

i

,

,K +

C

cos £

/

2cos£

V

- sin Je* £

C2

/

,

.

Then

*=

/ 2 sin £ \

cos (

2 cos £ \

— sin t

,

le* }

* -1 = ,

and

cos £ \

/ k sin £ ? \ j cos

£

— sin £ /

,

e~<

so that

U= J/#"

1

Fd£= /f, * J \%cost-tant

W(l,

,

?*, -

\ £ In sintj

J

j

,)

,

In sect| J |

and

X p = #U=

/3sint\ , 1

\te

|cos£y

, t

cos£\

/

+

, ,

.

\-±smt) 348

In

sint

+

/-2cost\,

,

In sect \

sin£

'

.

1

Exercises 8.8

18.

From

we obtain

-

cos t

.

Xc-Cl

sin

/ cos

\

f

+ C2

1

|

+ sin t \

£

1

I

cosf

sint

/

Then

*=

cos (

—

sin

cos f

t

+ sin ( \

cos t

sin

(

*

and

)

=

1

— sin (

cos

£

+ sin f \

cos t

sin

f

— cos t J

I

I

\

J

so that

^ _ y ^-lp^ _

^2cosi y"

+ sini -

2 sin

t

sect

^ 2 sin

^_

^

—

t

- cos! -

In

—2 cos f —

cos t

|

secf

+ tani| \

sin t

and 3 sin (cost

X p = *U = 19.

- cos 2 ( sin

2

—

1

2 sin

cos

2

f

2

—

*

+

- cost) In

(sini

cos ((In [sect

|

sec t

+ tanf|

+ tan(|)

From /

x'

=

/

1 1

e

X+

1

e

(

\

2t

3^

,o

we obtain

^

(

-1

+

C2

H W 1

(0\ e

2t

+ c3

3t

e

.

Then I

*=

1

e

-1

e

2t

0\

4

,2i

and

3>

1

—

-2( lp e 2

o

5e

e- 3t /

,3* e*/

so that

/ie

u = j * _1 Fd£ =

^

l„-2t

c

-|e 2 '\ dt

j i

=

¥

/

and

xp = *u =

-e + \e 2t + (

2 3t

ht e

\te 2t

>

e

Exercises 8.8

20.

From /S

-1

1

1

=

X'

-1\

I o \

-1

t

2e'7

we

obtain

i\\

Xc =

1

ci

e<

+ c2

1

W

e

2(

+ c3

a)

Then e*

e

2*

e

2(

-(

\

and

$

=

1

a

-2t

—

- e -»

e- 2t

v

2"' \

-2f

so that

/-te- t -e~ t -2e~

(ft

+ 2t\

=

and

-3/4 \

-1/2 ^

X B = *U

=

-1

-1

"1/2^ 21.

2 f

+

)

f

2

\ 4

2

2

ie

+* ^Jj

e- ai

+ 2i

.

-3/4,

From

we obtain

_ e 4t

c

e 4(

£

2t 2t

i

J

and

X = *$-

1

(0)X{0)+*^'*- Fds = *1

-2

„4f

.4;

fe

22. Prom

MI

:l)

x+ i/t

we obtain i i

i

+t

-t i

t

350

i

+ -l

t

^

e

2t

-

+ 2t-

J

Exercises 8.8

and

x . --.( 23.

w

,

+ ./r'F*

-

ln

+•

(-j)

•

;) - G)

(

'

-

(l) + (!)

^

From X'

=

/4

1\

«

s)

/50e 7t \

X+

o

(

)

we obtain

and

x=*x(o w)+ */A

24.

t

2ot+6e

5

6

*-^=*.f^-5/u*.fl,60t-12e:;-+ I2/) 5t

Prom

-2\_.

3

/2

we obtain

*=

/sin

(

—

—5 cost

^

2cosf

3 cost

sinf

+

\

3cosi/

/sint

.

*

>

=

+

5 cost

\

— 2cosi

3cost

sinf

—

\

3 cost/

.

and

X = *X(tt/2) + * h/2 f 9~

l

Fda =

*

°

(

)

+ *-

\0J

l)-{l) 25.

(a)

By and

Kirchoff's first

E=

i\R\

+ d_

+ L 2 i[

fh\

dt(i 2 j

(b) Let

I

=

V *2

)

3cos(

-{°i) i

—

12+13,

8[Rt

R2 /L2

f-(Ri + Ra)/L 2 {

\

-R2/L )\i 2 )

Ri/Lj

l

so that

/ ,

/-ll

3\

351

ZlOOsinA

(E/L 2 {

-

\ 1

-

E = iiR\ +13/^2+ £2''i>

so that

11 f

~ 2c03t

-

COSt

and second (on each loop) laws, we obtain

L\i'2

( \sint

Exercises 8.8

and

Then

*=

I

-

,<«

...

J

J

\3Qe

I

*

,

12t

=

,

ant J

\

12t

&e

(12sin(

and T

=*Ti=

^

sin

*-I cos(

™ fat

I

I

coat

so that

If

1(0)

= (

„

]

then

ci

=

2 and c 2

Exercises 1.

For

A=

;o |

^

= ^.

8.9.

r. ^

|

we have

Aa =

'0

°

Ji

i

a3 = aa = 2 )

=

i

;

>-= ra :

i

1

(A 2

=

i

=

1

i

(o

2

A4 =

s

:

;

i

1

1

1 ' i

I

A 5 = AA 4 = AI = A and so on. In general

,A, A*k '.I,

*=1,3,5, fc

=

352

2, 4,

...

6,...

=

a

-

cost) J

Exercises 8,9

Thus

A

(A

A

A3

2 2

,

= I + At+ilt 2 + |A* 3 + --=

I

=

I cosh

(l

^+^+

+ t

*, + l, + ...)

+A ( t+

...)

/ cosh t

sinh

\ sinh

cosh f /

+ A sinh t =

t

(

and /cosh(-I)

sinh(-f) \

V sinh(-t)

2.

For

A=

1 (

^

_

coshi

{

-sinhfA

\ — sinh t

cosh(-t) /

cosh

t

J

0'

^

|

we have

A2 =

I)

a 3 = aa 2 = A 4 = AA 3 =

u

1

=

i

(o

ir "Uf'J (i 4

°

°)

{

1

"

'

Wi ^

dG

8

for

A=l

° 16

and so on. In general

A*=[J 2°J

(

2,3

Thus,

1

0\

1

(l

1/1 0\

0\

+^+i+

,

1

0\

/l

}

and e" 2t

'

353

,

_(

et

Exercises 8.9

3.

Using the result of Problem

1

(cosh ( sinh

4.

t

sinh

_

f

cosh f

cosh t\

I sinh

sinht /

\ cosht

Using the result of Problem 2

X=

e"

I

=

+ «=2

Ci

e

5.

To

solve 1

=

X'

x+

1

we

identify to

=

=

e

X(()

0,

tA

=

F(s)

and use the

t

sinh t

\ sinh

t

cosh t

+

+ cjsinhf sinh t + C2 cosh t

cosh t

sinht

sinh t

cosh t sinh

cosh

/ cosh t

t

cosh t

(cosht

sinht

/ sinh

\ sinh

cosh

2 (

2 1

sinht

1

cosh

1

/

'

t

t

\ sinh

—

a

— -

'o

cosh

2

J

354

cosh s

—

sinh s

cosh s

cosh s

+ sinh s

— cosh t

- cosh 1 4- sinh t

t

2

+ 1

cosh s

sinh t

cosh

1\

=

sinh s

- cosh s

1 1

solve

X'

— sinh s

sinh s

—

+ c2 t

cosh a

V - sinh 3

sinh t

sinh f

t

(cosht

sinh t

+

cosht

ci

t

sinh f

+ C2sinh( + c\ sinh f + C2 cosh c\

cosh

+

cosh t

+ C2 sinh £ ci sinh t + c2 cosh t

ci

f'

(cosh t

cosht

+ C2 sinh t + ci sinht + C2 cosht c\

To

(3) in the text.

/

I cosh

c\

6.

and equation

1

C + e tA /' e-* A F{s)ds

ci

=

Problem

results of

/cosht 1

sinh

t

'

Exercises 8.9

we

identify tn

=

0,

F(s)

=

(

coshf \ |

\ sinh

X(()

=

e

tA

C + e tA

f*

(

,

and use the

Problem

results of

1

and equation

(3) in the text.

j

e" sA F(s) ds

Jtn

(coshf

sraht\/ci\

/coshf

sinhf\

sinht

coshi/ \C2/

\ sinht

cosh(/ Jo \-sinhs

rt

(

coshs

—

sinhs

coshs\

coshs

sinhs /

ds

/cosht

sinht

\ c\ sinh t

+ C2sinh ^ + C2 cosh f /

\ sinht

coshi/JO \0

I c\ cosh t

+ C2 sinh t\

I cosh i

sinh

ci sinh f

+ C2 cosh t /

\ sinh t

cosh t / \

(

+ C2 sinh t \ ^

f c°sh *

sinh f ^ (

t

+ C2 cosh t J

\ sinh t

cosh t J \

+ C2 sinht \

/i cosh

/cicoshi

=

{ c\ cosh V ci

sinh

(ci cosh t ci

7.

To

sinh

t

da

+ C2 cosh ( /

i

I,

t

\ /s

A

(

i

/cosh

sinht/

t

\

A

sinht/

2

A

/sinht*\

/cosh

\cosht/

\ sinht/

solve At

2

we

1

identify fo

=

X(t)

0,

=

F(s)

e

(A

e

=

M

C + e tA

and use the

results of

I'

/'

e~

sA

Problem

2

and equation

F(s) ds

l

e

cie

2i

o

1

,2t

se

+

C2e

cie'

e

2f

+

it

+

„2s

2t

cie

e

2t

,2i

C2e

+

-t I„4(

2(

2

.

1

_

-

e

-

e"'

e 2

-ie"* e

1

3

-se

e'

e'

c2e

,4s

1

+

e

e

£

l„2t

355

2

+

1

ds

(3) in

the text.

A

Exercises 8.9

8.

To

solve

x-i we

identify to

=

0,

=

F{s)

X(t)

=

e

tA

and use the

,

)

^

(

V

1

C + e tA f e~ sA F( S

e'

e

21

C1Z

ds

e

2t

JJo

e

2t

\

e21

+

2

e

e

-3 _1 2

2

-3 -

3e*

+

= 5

9.

-2( IP e

,2t

l

,2i

lis

-e -2s

-3e~*

c\e

-1

-3e- a

C2e

C2C

„-2s

\

t

+

Cie'

2t

( tr

1

C2

E

Problem 2 and equation

results of

' I e

/ c\ \

\

2*

)

+1

2

«3

e'

+Q

e

2<

-3

+

l

e

2

Solving

2-

A

-3 we

find eigenvalues Ai

=

1

6

—

3 and A2

-

A2

-

8A

+

15

=

(A

-

3)(A

-

6)

=

A

=

5.


K,

Ka =

and

u

u

Then

P=

1

1

P"l=f

- 1/2

PDP" =

3

V 1/2/'

V -1/2

3

1

3/2

2

1

-3

6

1

and

D=f\0

10. Solving 2

-A 1

1

2-

=

A 2 - 4A

+3=

356

(A

-

1)(A

-

3)

=

° 5

(3) in the text.

)

Exercises 8.9

we

=

find eigenvalues Aj

1

and A2

=


3.

Ki =

M

I

K3 =

and

' J

Then

so

PDP 11.

From equation e

(2) in

iA

1

=

2

1

1

2

the text

,-1 = e ePDP- = j + ((PD p-l) + I f 2( PD p-l)2 +

i.(3(p

2!

= P H. £D + i(iD) 2 + i((D) 3 + 12.

From equation

lD

-

1

=

,

P -i = p e 'D P -i.

.

(2) in the text

n e

D p-l)3 +

A2

•

+ ,0

-

1,

\

-

+

1

\

•

xl)

x\

7

2!<

K)

I

•

I

/A?

\

A|

+

+ 1" VO /l

A^y

+ Ai( + ^(Ait) 2 + l

+ A 2 ( + ^(A 2 i) 2 + 1

>

•

e

<

-

A2t

.

e

Xnt

j

357

+

Xn t

+ ^{X n tf +

Exercises 8.9

13.

Prom Problems

9, 11,

and

12,

and equation

the text

(1) in

X = e fA C = Pe^P^C

3„3t 3

14.

Prom Problems

e

_

1

Q 5I

3t_| e 5(

_I e 3'

4.

_l e 3( +

Ig5t 3 e 5 t/

y C2

10-12 and equation (1) in the text

X = e tA C = Pe iD p- C J

e

1

Z)(.o

\ e

3t

J

U

4*

)•


True

3.

A -1 = -A

4

-2N/-2

-3

1

j

~

\3/2

1

-1/2,

5.

AB = AC, A _1 AB = A _! AC, and B = C. True, since X' = X' + XJ, = AXi + AX 2 + F = A(Xj + X 2 )F = AX + F.

6.

True, by

7.

False; they are the zero

8.

True, since

9.

True, by the definition of an eigenvector.

4.

True, since

:

Theorem

if

8.8.

and nonzero solutions of det(A — AI)

AK = AK then A(cK)

= cAK =

c(AK)

10. True, since complex roots occur in conjugate pairs.

11.

False; consider

A=

V ,

\-l

2

358

=

=

A(cK).

0.


12. False;

n

i

2\

i

(1

1

3

i

1

-1\ 3

1 J

U 13.

Prom (D - 2)x + {D - 2)y = Dx = 3 - (2D- l)y. Then

=

y Substituting into

14.

From (D - 2)x

(D — 2)x

+ C2e -

21

c

+ (P —

-y=t—2 and

Dx + (2D —

and

1

cie

0,

1

=

2)y

-3a;

and

3

x

* !

= -c 2 e -

+

(D - 4)y

=

=

1 gives C3

x

=

l)y

-

-C2e l

-cje

2t

and

+ 03.

5/2 so that

3

-cje

=

= -6

we obtain (D - 1){D - 2)y

3

ft zt

+

5

-

.

-At we obtain (D

-

1)(D - 5)x

=

9

-

St.

Then

3

and y

15.

From

(D-2)x-y =

= (D -

-3x

-e* and

-t+

2)x

+

x

=

cie

y

=

(£>

=

sin2f and

1

(£)

2

=

-cie*

- 4)?/ =

+ 3c 2 e 5( + _ +

-7e' we obtain

(£>

—*.

- 1)(D - 5)x = -4e*

so that

+ c2 e 5 + ie' '

Then

16.

From

(D4-2)a: + {D + l)y

-

2)x

+ e* =

+

-cje'

5x+(D+3)j/ =

3c2 e

51

cos 2t

-

te*

+ 2e

( .

we obtain (D 2 +5)y

=

Then y

=

2

ci cos

t

+ ci sin ( — - cos 2i + J

7

-

sin 2i

o

and

x= =

-i(Z>

+

3)j/

/13\ — -

\5

ci

-

5

c2

+

^cos2(

sin

i

/

/

1

V

5

+ — -C2 —

3 \ -ci cos 5 / I

i

17. Taking the Laplace transform of the system gives S

y{x}+^{y} =

12{x} + s2{y} =

359

1

^+ 2

5

1

3

3

— - sin 2f — - cos 2t.

l

2 cos 2i - 7 sin 2t.


that

^{x} =

s

-

2

1111 +

+1 + 2)

2s

s{s-2)(s

8s-2 +

~4s~

9

1

+2

8 s

Then x

= -] + \e 2t + 4

o

%"

2t

and

=

2/

9

+1 =

-x'

1

2t

4

o

18. Taking the Laplace transform of the system gives s

2

X{x} +

2 S X{x}

+

2

s 2'{y}

s

2

s(s-2) 2

and

s-2

%{y} = -

so that

2{x}

=

11

11

2 s

2

+

s-2

-s-2 = _3 1_lJ_ + 2 - 2

3 2 {s

4 s

2)

2 s

1

s-2 1

(s-2) 3

1

-

4 3

2

{s

-

2)

2

Then x

= l-le 2t + te 2t 2 2

and

y

= -\ 4

2 3t

(

19.

(a)

X=

|

,

20. Let xi

-f -( + 2 + 12f 2 + 8t + 3

4f

3

=y, x 2 =

y',

x3

=

y",

(b) 1

\t

X' =

and 14

=

2

ê 2t

2

-

y'" so that

Dx 3 = x 4

21. Let xi

=

x, X2

=

y, 2:3

=

£>£,

- ^3 -

and X4

= Dy

Dxi =

X3

DX2 =

X4

Dx 3 =

x4

£>X4

-

+

2e'

-

|f.

so that

2x 3

= -X3 -

3xi

Xi

-

2x!

- lnt +

+ 5* 360

2.

te

2t

1

+ 24f + 8 }

Dx\ = X2

£>X4

-

+ 6t + 5

-Zt 2 12(

}

+

lOf

-4

.

'


22.

If

x=

c\e

l

+ C2te + sint and t

y

=

c\e

+ c2 (te + e') +cos(

l

l

then

=

x'

cie*

+

+ e') + cos t =

c 2 (te'

y

and j/'

=

cie'

=

~(cie*

+ C2(ie + 2e - sin t l

)

+ c 2 te' + sini) +

= —x + 2y — 23.

We

have det(A

-

=

AI)

(A

f

-

l)

2(cie'

+ c 2 (ie + e*) l

-

+cosi)

2cosi

2 cost.

2

=

K=

and

A

solution to

(A - AI)P

=

j.

^ so that

r/

n

K

/o

.

1

24.

We

have det{A

-

=

AI)

(A

+ 6) (A + 2) =

x

25.

We

have det(A

-

=

AI)

A 2 - 2A

+

so that

-«(-i)^ + «(0 e 5

=

0.

For A

=

1

+

2i

""

we obtain Ki

1

= ^

Xl

-fV

+a0f

\i J

j and

^WJÛ

-(\-sin2tJ

\cos2t)

Then

v X = c,

coa2(\

t

,

e'

+ c2

26.

We

have det(A

-

=

AI)

A 2 - 2A

+2=

0.

For A

=

/sin2(\

,

e*.

\cos2t/

V,-sin2t/

1

+i we

3

Ki =

obtain

*

^

^

^3 1

"

(

1+f j (

_ ^3 cost + sint ^

'

V

2cost

j

(

^

^

-

cost

+

3 sint

2 sint

\

+ sint\

2cost

,

/

/

V

361

— cost +

3 sin (A

2sinf

J

^ j

Then (3cosf

j

,

and

t

is

P=

^


27.

We

have det(A

-

AI)

=

A 2 (3 - A)

=

so that

/1\ 1

1)

\

28.

We

have det(A

-

AI)

=

-(A -

2)(A

-

+ 3) =

4)(A

f~2\

X

=

,1,

so that

(o\

3

Cl

e

2'

+ c2

1

We

7 \

f e

4t

+ c3

-3t

12

{-is)

u, 29.

,31

1

have

Then e

2t

4g 4( e

_4 e -2t

;~2(

4t

and 2e~ 2t

- 64te~ 2t \

—

dt

Wte -it

( 15 e

- 2<

+ 32te~ 2t

-e- 4i - Me

-4t

so that p

30.

We

-1 -

1

4t

have

Xc =

/2cosf\ ci

e

\ — $\tit}

, 1

+ c2

/2sin(\ {

\ cost J

Then

*=

/

2cosi

2sint

A cos

t

sin

f

—

sin t

I

sin V-si

cos I

t

;

cos t

and cos t

—

sec

t

sin

f

sin

=

dt J

t

-

In sec

V

|

t

—

cos f

so that

— 2 cos (In \

31.

We

|

— 1 + sint In

sect |

+ tant| \ + tan(| /

sect

have / cos t

^""H

+ sin t \

2cos

(

I sin t

— cos f

+t

)

362

*{

2sint

+ tan

i|

\

e


'

cos

t

+ sin t

sin

—

t

cos

sin

t

-cosi

—

2 cost

5 cos

t

+

5 esc

— Asini -

A cos

t

+

A.

^ sinf

+

^ sinf

+

5 In

—

2t

te

2i

sin

)e

+

e

2t

2t

+ c2

\

*

,

,2t

,2£

esc t

xP = *u=|

f|

—

t|

I

—

j sin

i

+ cos

'

cot

In esc

2t

1/2 1

/2

—

t

cot

f|.

„2i

+

( -te -2t

-te

-« _

-21

^ -1- U=^ t

/

I

te

=

csci

/

f

-I .

t

\

-te

u = /*-^ ss

— cot

I

sint

-1

e

(

1 1" csc(

-1

X c = c,|

t

t

\

A,cos(+ A,sint /

—

2 sin

—5 cost —

j cos

t

'

2sin(

2 cost

1

)tV' + (

„-2t

i2

-

t

e

-2t

9

Numerical Methods for Ordinary Differential Equations Exercises

5.

+4=

Setting x

6. Setting

+

2x

7.

Setting x 2

8.

Setting y

we obtain x

c

=

?/

—y =

—

x2

9. Setting \Jx 2

—

c

c

we obtain a family

we obtain y

+ y 2 + 2y + 1 =

a family

c

=

x2

c;

of vertical lines.

a family of lines with slope -2.

of hyperbolas.

+ c;

a family of parabolas with veritces on the

we obtain x 2

+

+

(y

l)

2

=

c

2 ;

j/-axis.

a family of circles centered at

(0,

— 1).

+ y 2 )~ = c we obtain £ 2 + 2 = 1/c; a family of circles centered at the origin. y(x + y) = c we obtain 2 + xy — c = or y = — ^x ± ^Vx 2 + 4c; a family of hyperbolas.

10. Setting (x 2 11. Setting

= c — 4,

we obtain y = -2x +

c

2

9.1

12. Setting y

l

j/

j/

1

+e =

c

we obtain y —

13. Setting (y

—

l)j{x

—

14. Setting (x

-

y)j{x

+ y) =

2}

=

c c

c

—

e

x \

a family of exponential curves.

we obtain y = cx + we obtain y =

(1

-

1

—

2c;

c)x/(l

origin.

364

a family of

+ c);

lines.

a family of lines passing through the

1

Exercises

15. Setting

=

x

we

c

form a family of

see that the isoclines vertical lines.

y

\

+ y = ewe obtain the isoclines

16. Setting x

y

9.

= —x + c. V,

/ /

\ .X. **

\

V

41

V

\

17. Setting

y

=

/

s

—x/y =

c

we obtain the

isoclines

18. Setting l/y

=

c

we obtain the

20. Setting xe y

—

c

we

isoclines y

=

-x/c. y

19. Setting 0.2x 2

+y =c

we obtain the

y

21. Setting y isoclines

— cos |x = cwe

y =

cos

^x

+

c.

obtain the

=

In c

22. Setting

y

=

(1

—

1

-

obtain the isoclines

In x.

—

y/x

=

c

c)x.

V

we obtain the

isoclines

1/c.

Exercises

9.

23. Solving

ax + By T~ = yx + oy for

y we obtain cy

a family of 24. y

=

cx

is

lines

through the

a solution of the


and only

if

_

_

'

and only

oix

+

jx

+ 5cx

and only

2

and only

Bex

+

(7

- 0}c -

a]x

=

if

6c if

if

if

[6c if

—a

origin.

^ if

c

2

+

(7

-

-a=

0)c

if

8-7±^{3- 1 )l + 4a6 26 if

and only

if

25.

The

isoclines of y'

=

3i

+ 2y

are 3x

2

-

{0

+ 2y =

+ 4aS >

f) c or

3

If

26.

we choose

The

c

=

—3/2 then 3k

isoclines of y'

=

6i

- 2y

+ 2y — —3/2

are 6x

-

2y

=

is

27.

we choose c

The

=

isoclines of

3 then 6i 1

y

=

—

2y

=

3 is

2x/y are 2x/y

c

a solution of the

differential equs

c or

» If

0.

= -3* +

|.

a solution of the


.

— cor y

=

2

~x. c

Setting 2/c

28.

The

=

c

=

we obtain c

isoclines of y'

=

2yj{x

±y/2. Thus y

+ y)

are 2yj{x

=

±\/2 £ are solutions of the

+ y) = y y

=

c or c

2-c 366

x.

differ*

Exercises 9.2

Setting c/(2

equation 29.

The

— c) =

c

we obtain

c

=

or c

=

1

—

Thus y

.

— x are

and y

solutions of the differential

.

=

isoclines of y'

(4i

+ 3j/)/j/

are (4x

+ 3y)/y =

c or

4 c

-

Setting 4/(c-3)

— —x

and y 30.

The

c

we obtain

c

2

—

- 3c- 4 = (c-4)(c + l) =

are solutions of the differential equation

isoclines of

1

y

=

(5x

+

l(h/)/(-4:r

+ 3j/)

and

—

c

5)/(3c

5 and

3/

-

10)

= — ^x

=

and

=

3/

=

we obtain c

c

10j/)/(-4a;

3c

-

10

3c 2

-

14c

5x are solutions

Thus

=

c

4 and c

= -1

and y

=

Ax

=

+

3y)

+

l)(c

c or

+5

4c

+

0.

.

+

are (5x

a

Setting (4c

3

-

5

=

(3c

-

=

5)


0.

Thus c

= -1/3

.

Exercises 9.2 a spreadsheet program which does not support subscripts.

All tables in this chapter were constructed in

Consequently, x n and yn will be indicated as x(n) and y(n), respectively.

1.

—

Let u

x+y—

so that y'

1

=

(x

+y—

l)

2

becomes

u2

and tan

-1

u=x+

c.

Then

y

EXACT 00 .10 .

0.20 0.30 0.40 0.50

2

.0000

2.1230 2 2 3 3

.3085 .5958 0650 .9082 .

1

= 1 — x + tan(i + c) j/

2.

— du — dx

+

and

j/(0)

= 1 — x + tan(x + —

=

2 gives c

=

7r/4, so

J.

EULER h=0.1 h=0.05 2 0000 2 .0000 2 1000 2 .1105

IMPROVED EULER h=0.1 h=0.05 2 0000 2 .0000 2

.

.2440 2 .4525 2 .7596 3 .2261

2

.

.

.

2

2 2 2

.2727 .5142 .8845

3.4823

367

.

1220 3049

2

2.S858 3 0378 3.8254

2

.

.1228

2.3075 .5931

3.0574 3

.8840

Exercises 9.2

n = xln) 1

.

i

UU in

1.20 1.30 1 .40

1:50

1

.

1

.

05 vfnJ 3 UUUU 4 4UUU

.

J

.

isy

bU

j

.

4

lis

.

.

vln) D uuuu .

2 2 2 1

.9800 .4260 .0582 .8207

xln) UU UD 1 1U 1 1 c 1.1b 1.2V

1

.

.

h = 0.1 xln) 0.00

YW 0. 0000

10 .20

0.1000 0.2010 0.3050 0.4143 0.5315

.

0.30 0.40 0.50

.

/

C

1 1

"J

1

Zzs

1.30 1.35 1.40 1.45 1.50

h = 0.1 x(n) vln) 2 0000 0.00 .10 1. 6000 .20 1 .3200 0.30 1 1360 1 .0288 0.40 0.9830 0.50

.

2

.5702

2.3647 2.1950 2.0557 1.9424

h = 0.05 xln) Wn> .00 2 0000 .05 1 .8000 0.10 1.6300 0. 15 1. 4870 0.20 1. 3683 0.25 1.2715 0.30 1.1943 0.35 1.1349 0.40 1. 0914 0.45 1.0623 .50 1 .0460 .

h = xln) .00

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

.05

Vln)

0.0000 0.0500 0.1001 0.1506 0.2018 0.2538 0.3070 0. 3617 0.4183 0. 4770 0. 5384

368

Exercises 9.2

h = x(n) .

.

00 10

0.20 0.30 0.40 0.50

h -

.

v(n)

1.0000 1 10UO 1.2220 1.3753 1. 5735 1.8371 .

.

uu

0.05 0.10 .15

0.20 .25

0.30 0.35 0.40 0.45 0.50

h = 0.1 x(n) 0.00 0.10 0.20 0.30 0.40 0.50

v(n)

0.0000 0.1000 0.1905 0.2731 0.3492 0.4198

h = 0.1 x(n) 0.00 0.10 0.20 0.30 0.40 0.50

YW 0000

0.

0.0000 0. 0100 0. 0300 0.0601 0.1005

05 V(n)

.

x(n)

h = x(n) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

A f\ f\ A

1

1.0500 1053 1668

1 1 1 1

.

.

.2360 .3144

1.4039 1.5070 6267 .7670 1.9332

1

.

1

0. 05

V(n>

0.0000 .0500

0.0976 0.1429 0.

1863

0.2278 0.2676 0.3058 .3427 .3782 .4124

h = 0.05 xln) 0.00

V(n) 0000

0.

.05

0.0000

0.10 0.15 0.20 0.25 0-30 0.35 0.40 0.45 0.50

.0025 .0075 .0150 .0250

0.0375 .0526

0.0703 0.0905 0.1134

1

Exercises 9.2

h xln) 0.00 u

.

1U

0.20 0.30 0.40 0.50

.

vln)

0.5000 1CA U.c bzbu 0. 5431 0. 5548 f\

.

5613

0.5639

h xln) 0.00 0.05 0.10 15 0.20 0.25 .

.30

0.35 0.40 0.45 0. 50

h = 0.1 xln) Y(n} 0.00 1. 0000 0.10 1. 1000 .20 1.2159 0.30 1 .3505 1.5072 0.40 0.

50

1 .6902

h = .00

0.05 .10

0.15 0.20 0.25 0.30 0.35 .

1 1

.00 .10

1.20 1.30 1. 40 1. 50

0.1 vln) 1 .0000

1.0000 1

.0191

1.0588 1.1231 1.2194

0.5000 5125 0.5232 0.5322 0.53 95 0.5452 5496 0.5527 .

.

.5547

0.5559 0.5565 0. 05

xln)

40

0.45 0.50 h = x(n)

05 Vin)

.

vln)

1.0000 1.0500 1

.1039

1.1619 1.2245 1.2921 1.3651 1.4440 1

.5293

1.6217 1.7219

h = 0.05 xlni 1. 00

1.05 1.10 1.15 1

Vln)

1.0000 1. 0000 1.0049 1

.20

1

1.25

1

1. 30

1

35

1

1.40 1.45

1

1. 50

1

1.

1

.0147 .0298 .0506 .0775 .1115 .1538 .2057 .2696

)

Exercises 9.2

h =

12.

v(n) 5000 5250

.00 1 . 10 1

1 -1-

.

^0 V

-J

1.40 1

50

x in

" n> <

1

00 05 in 15 20

1

?5

.

1

.

1

0.5499 0.5747 0.5991 0.6231

1.20

-

h

1

i 1

.

(a)

h - 0.1 X V 1.00 5.0000 1.10 3 .9900 1.20 3.2546 1.30 2.7236 1.40 2 .3451 1

.50

2

.0801

fl .

j

^^^n ^^75

u

J *± ^ ^

M

~>

5

i _>

5746 5868 0.5989 0. 6109 .6228 .

.35

0.

1.40 1.45 1.50 13.

5000

n

1.30 1

.

h = 0.05 x(n) V(n) 1.00 5.0000 1. 05 4 .4475 1.10 3.9763 1 .15 3.5751 1.20 3.2342 1.25 2 .9452 1.30 2 .7009 1.35 2 .4952 1 .40 2 3226 1.45 2.1786 1.50 2 0592 .

.

(b)

h = 0.1 xln) V(n) .00 .0000 0.10 0.1005 0.20 0.2030 0.30 0.3098 0.40 0.4234 .50 .5470

h = 0.05 x(n) .00 .05

0.10 0.15 0.20 0.25 0.30 .35 .40 .45 .50

V(n)

0.0000 0.0501 0. 1004 0. 1512 .2028 .2554

0.3095 0.3652 0.4230 0.4832 0.5465

1

Exercises 9.2

h =

x In) .

n

00 in

0.20 0.30 0.40 0.50

h = 0.05

.

vln)

xln)

0.0000 1822 0.2622 0.3363 0.4053 0.

.

U

.

U

.

00 Uo ±U

0.15 a ^ ZU 0.25 0.30 0.35 r\

.

.40 .45 .50

h = 0.1 xln) vln) 0. 5000 0.00 0.5215 0.10 .5362 0.20 0.30 0.5449 0.40 0.5490 0.5503 0.50

h = 0. 00 0. 05 0. 10

0.15 0.20

h = 0.1 1. 00 1. 10

1.20 1.30 1

.40

1.50

V(n) 1. 0000 1. 0095 1. 0404

1.0967 1 .1866 1.3260

.

uysj 0.13 97 1823

h = xln) 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1 1

.45 .50

.

.

0.2231 .2623

3001 0.3364 0.3715 0.4054 0.

.05

Vln)

0.5000 0. 5116 0. 5214 0.5294 0.5359 .5408 .5444 .5469 .5484

.25

0. 50

f\ fl

u

xln)

0.30 0.35 0.40 0.45

x(n)

^ 0.0000 U U400 /I

0.5492 0.5495 0. 05

Y(n> 0000 0024 0100

1. 1. 1.

1.0228 .0414 1.0663 1 .0984 1 1389 1 1895 1.2526 1.3315 1

.

.

372

Exercises 9.2

h = xln) .

.

00 10 30

(1

0.40 0.50

x In

v In

1

1.6600 1.4172 1.2541 .1564 1.1122

0.20 ?5

X

0.35

1.1916 1. 1499 1.1217 1. 1056

000(1

1

o

X

-

YW 1.0000

0.30

1

40 SO

.0488

n

7

i

11 '

0000 1.0526 1 1113 1 1 77S — X X .

.

1

1.6880 2

X

& *±

** J.

3? 1

1

.

.4361

.

y

)

.00 . 05 10 15

1.1110 1.2515

10 .20 .

I

OS

(1

x In

—J

7

1

0.45 0.50 0.1

U

.

1

?

.40

h xln) 0.00

>

on n os 10 U.J

2

0.20

ns

fl

1

v In

'

X £ J^ u 1 7 IRQ X J J oo X o/ 1. 5556 1. 6939

?s

.

.

0.35 0.

dO

1.8598 2.0619

0.45 0. 50

h a

A fl \J

11

1

1

.

00

.

1 _L

fl \t

0.20 .30 .40 50 .

-

0.1 vln) 0.0000 0.0050 0. 0200 0.0451 0. 0805 0.1266

n illili

on

n O

-

.

.

V

.

U

\>

Vu

VJ

10 15

0.20 0.25 .30

0.35 0.40 0.45 0.50

0050 0113 0.0200 0313 U V J XJ 0451 0.0615 .0805 .1022 0.1266 .

.

.

.

373

Exercises 9.2

12

A

—

AA

U 10 A 1A 0.20 A TA .

a

n .

h =

0.1 V(n) 1 0000 1 1079 1 2337 1 3806 1 5529 1 7557

en 5U

00 05 10 15 20 25 30 35 40 45 50

= 0.1 Yin) aa i 5000 1.10 5250 i nn 5498 1. JU 0. 5744 1 1 4a 0. 5986 1 1.50A 0. 6224

(e)

1 1

1 1 1 1 1 1

>i

IT

1

1

YW

1.0000 1.2000 1.4938 1.9711 2.9060

1.20 1.30 1.40 16. Integrating y'

=

f(x, y)

ft

7>

.

1

J

.

OU

/

1.4634 1.5530 1

.6503

1.7560 AC

A

3UOU 0.5125 U

.

cTc

f\

r\ U

f\

a /4

*i .

0.5498 0.5622 CI A

A

A

.5866 0.5987 .6106

0.6224

IMPROVED EULER

EULER

1. 00 1. 10

.

00 05 10 15 20 25 30 35 40 45 50

1

.

.

Ob v(n) 1 .0000

x(n)

.

h=0 x(n)

rtc

.

1.0519 1 1079 1 1 C A 1.1684 1.2337 1 AA O 1 3043

h =

xl

15.

a U

x(n>

Vin) 0000

1.

1.2469 1. 6668 2.6427 8.7988

we obtain y

/

Jin

dx=

I

Jx„

f(x, y)

dx

so that

y(*n+l

-

y(zn)

=

f{Xn,yn){Xn+l

~

%n)

or

Vn+i

=Vn +

hf{x n ,yn ).

374

=

hf{x n ,yn )

Exercises 9.3

17.

FVom y1

=

f(x, y)

and f(x, y)

m

\[f{x„, yn )

,

y'dx^ I

we obtain

Vn+i)]

^Tl+l n+l 1,

n+>

{

+ f(x n +l

\[f(x n} yn )

+

f{x n +i,yn+1 )]dx

•'In

'it.

so that

-y{x n ) «

y(x„+i

(x n+i

-

x n )^[f{x„,y„)

+ f{x n+ i,yn+ i)\

or

Vn+l

-

Vn

+

^[f{x n ,yn}

Vn+l

=

yn

+

hf(x n ,yn ).

+

f{Xn+l,Vn+l)],

where

Exercises 9.3 1.

We

h = 0.1

use

y"

=

2

-

xln) 1.00 1.10 1.20 1.30 1.40 1.50

3y'

= 2-3(2x-3y+l)

= 9y-6x-l

Vlnl 5 0000 3 .9900 .

3.2546 2

.7236

2.3451 2

.0801

so that

lfn+i

2.

We

=

yn

+

(2x n

-

3yn

+

l)h

+

{9yn

use

- 6x n - l)^h 2

.

h = 0.1 y"

=4-

V

xln)

= 4 - 2(4a; =

4y

-

8x

2t/)

+4

so that

J/n+l

=Vn +

(ten

-

2yn

+

l)k

+

(4yn

0. 00 0. 10

Vin) 2 0000 1. 6600

0.20 0.30 0.40 0.50

1.4172 1.2541 1. 1564 1. 1122

.

- 8x„ + 4)-h 2

.

h = 0.05 x(n) vln) 1.00 5. 0000 4.4475 1.05 1.10 3 .9763 1. 15 3 .5751 1.20 3 .2342 1.25 2.9452 1.30 2.7009 1.35 2.4952 1.40 2.3226 1.45 2.1786 1.50 2.0592 h = 0.05 x(ni v(n) 00 2.0000 1.8150 0. 05 10 1 .6571 15 1.5237 0.20 1.4124 1.3212 0.25 1.2482 0.30 0.35 1.1916 0.40 1.1499 0.45 1.1217 0.50 1.1056 .

.

.

Exercises 9.3

3.

We

h

—

h =

1

ft

use x.

=

y"

= =

2yy'

+

2y(l

0.10 0.20 0.30 0.40

2 !/

+ 2y 3

2y

x In

\ni

0. 50

o

noon 0.1000 .2020 0.3082

00 05 10 V 15 X -J

0.4211 0.5438

0.20

0.35

so that

-

yn+l

4.

We

+ (l+yl)h+

yn

(2y n

+ 2yl)~h 2

=

2x

+ 2yy'

=

2x

+

=

2x

2

2y(x

+ 2x

2

y

+y

2

+ 2y

.40 .45 .50

)

3

We

h = xln) .00 .05 .10

0.15 .20 .25 .30

0.35 0.40 0.45

so that

yn+i

=

Vn

+ (4 + Vl)h +

use

y

= e- y (e- y )

=—

-2y

(2x n

+

2

2x n yn

+ 2yl)\h 2

0. 50

.

.05

0.10 0.15 0.20 .25

.40

yn+l

=y n + e-y^h-{e-^)\h i

0.45 0.50 .

376

1

V

.

1 J-

510 J -L U

0.2025 0.2551 T090 0.3647 0.4223 0.4825 0.5456

0.05 vtn) 1.0000 1.0525 1.1111 1.1770 1.2519 1.3378 1.4372 1.5535 1 .6910 1.8557 2.0561

h = 0.05 xln) Vln) 0.0000 0.00

0.30 0.35

so that

v(n 0000 0.0500 1003

.

h = 0.1 xln) Vln) .0000 0.00 .0950 .10 0.20 1818 .2617 0.30 0.40 0.3357 50 0.4046 .

05

.

h = 0.1 xln) vln) 1.0000 0.00 1 .1100 .10 1.2490 0.20 0.30 1.4310 1.6783 .40 .50 2 .0300

use

y"

5.

75

.

0.0488 0.0952 0.1397 0. 1822 0.2230 0.2622 0.2999 0.3363 0.3714 0.4053

J

Exercises 9.3

6.

We

n

use

—

n

^f^J y"

=

v n U

+ 2yy'

1

= l + 2y(x + y 2

.

,

YAHl uuuu u UUJU

uu n n J. u

\j

= 1 + 2xy + 2y 3

*

.

0.0200 0.0451 0. 0804 0.1264

0.20 0.30 0.40

)

nh ——

1

.50

that

=

Ifh+i

7.

We

yn

+

(in

+

+

(1

+ 2a:„jfr, + 2yl)-h 2

y"

=

-

2(x

y)(l

-

y')

= 2(x-y)[l-(x-y) 2

}

= 2(x~y)-2{x~yf so that

8.

We

h = 0.1

use

y"

=

xy'

x(n) 0.00 0.10

+y+ ^

V(n) 1 .0000

1.1075

.20

= =

x{xy

+

y/y)

+y+

x 2 y+ -Xy/y

+ y+

xy

+

1 1

0.30

y/y

.40 .50

2^y

1 1

-

that

1/n+l

=yn +

(XnVn

+

\/V n )h

+

{

x nV«.

+ Xi\/V» + ^

Vn

+

U

U U

>

.

uu UD

.

±U

U

ISc 1

-wr

U u n

.

.

/ ri

1

UUUU UU1J nn £ n

Ju

U . U11J U . UÛU U . VJ 1 A AAC1 0.0451

0.35 0.40 0.45 0.50

0.0615 0.0805 0.1021 0.1265

.

U

.

A\J

U

.

ZD

U

.

-

h = 0.1 x(n) V(n) 0.00 0.5000 0.10 .5213 .20 .5355 0.30 0.5438 0.40 0.5475 0. 50 0.5482

use

x (n

.2327 .3790 .5504 .7522

h = 0.05 x(n) 00 0.5000 0. 05 0.5116 0. 10 .5213 0.15 0.5293 .20 0.5357 .25 .5406 0.30 0.5441 0.35 0.5466 0.40 0.5480 .45 0.5487 0.50 0.5490

YW

.

h = x(n)

.05

0. 00

1.

0.05 0.10

1.0519 1. 1078 1.1682 1.2334 1.3039 1.3802 1.4628 1.5524 1.6495 1.7551

0. 15

0.20 0.25 0.30 0.35 0.40 0.45 0.50

V(n) 0000

+ Exercises 9.3

9.

We

— —

h 11

h = 0.1 y<"> 1 00 1. 0000 1 .10 1.0100 1 20 1. 0410 1.30 1. 0969 1.1857 1.40 1. 50 1. 3226

use

y"

=

+y -

2xyy'

-V

xy

2

=

2xy (xy 2

~~)+P

X)

X \

=

2zV-2y 2 +

1 J-

X 1 1 1 1

X*

%

10.

We

M

+

y-

_

£)

h +

2xlvl - 2vl

= ri-

YW

= y-y 2 - 2y(y - y 2

)

= y-3y 2 + 2y 3 so that

yn+i

11.

We

=Vn +

(v»

-

vl)h

+ 2(y„ - Zyn2 + 2y 3 )-h 2

.

l

use y"

= = =

2x 2x 2x

10 X V J. J

+ 3y 2 y' + 3y 2 (x 2 + y 3

)

+ 3xV + 3y 5

1

J.

\J\J KJKJ

J. »

Uua j

i

K on m

j. .

in

J. *

mm

1 J. » \J i 1

£*£aJ

nd t t; ncci y ddj

VjOJ 1.13 87 1. 1891

1.2518 1.3301

h = 0.05 x(n> WnJ 1.00 0.5000 1.05 0.5125 1.10 0.5250 0.5374 1.15 1 .20 0.5498 1.25 0.5622 1.30 0.5745 1.35 0.5866 1.40 0.5987 1.45 0. 6107 1.50 0.6225

h = 0.1 x(n) 0.5000 1.00 1.10 0.5250 1.20 0.5499 0.5745 1.30 1.40 0. 5988 1.50 0.6226

W

Uj

If)

{

use

y"

.

1.35 1.40 1.45 1.50

so that

=

fTi

1

.

2

nn

i

.

x2

\J .

—YSRl ir/n

x(n)

h=0 .1 x(n)

EULER

1. 00 1. 10

1.0000

1.20 1.30 1.40

1.4938

Vln) .2000

1

.9711 .9060

1

2

IMPROVED EULER Vln) 0000

1.

1.2469 1.6668 2 6427 8.7988 .

3 -TERM TAYLOR

V(n)

1.0000 1.2400 1.6345 2

.4600

5.6353

so that

yn+\

12. Let f(x, y)

=

1

yn

+

= ax + 0y

[*t

+ vt)h +

so that fx

(2x n

=

+ 3a£i£ +

a,fv

=

0,

and

378

3i£)£ft

all

a .

higher derivatives are

0.

Using the Taylor

Exercises 9.3

series

expansion

we have

for f(x, y)

- f(x n: yn ) +fxixn,yn)h + fy (x n ,yn )hf(x n ,yn ) - f(x n ,yn ) + ah + @hf{x n ,y n ). Since f(x n yn ) ,

=

+ 0y'n = y£

n w»d a

y'

Vn+l

-

J/n

=

13.

To

solve the initial-value

x integrating factor e~

we have

+ ^M/fan, !/n) +

+

^[/(^n.fn) \h[2y'n

=

yn

+

=

ft.

+ V* + ^»

fn+1

)]

+ fi^Vn) + ah + f}flf(x n

+ h{a + 0y'n

,

y„)]

)\

2

i£-

problem analytically we note that the

differential

equation

linear with

is

Then

.

and y

From

=

5

we

find c

=

6e~

x

=

ce

1

-

x.

so that

y

=

&e

x- 1

-

x.

For Taylor's method we use

y"

h=0.1 x(n)

= l + y'

1. 00

=\+x+y -\ =

x

=

yn

1

+y

so that

I/n+i

+

.

10

1.20 1.30 1.40 1.50 {x„

+ yn -

l)h

+

{x n

+ yn )-h 2

.

IMPROVED EULER V(n) 5 .0000

3 -TERM TAYLOR

V(n)

5.

5.0000 5.5300

6.

6

5300 1262 6.7954 7.5454 8 .3847

.1262

6.7954 7.5454 8.3847

EXACT V(n) 0000

5.

5.5310 6

.1284

6.7992 7.5509 8.3923

Exercises 9.4

Exercises 9.4

X

x{n) 1 1 1

.00 .10 .20

1.30 1.40 1.50

5.0000 3.9724 3.2284 2 6945 2.3163 .

x(n)

V(n) .0000 1003 .2027 .3093

.00

0.10

.

.20

0.30 0.40 0. 50

0.4228 0.5463

xln) .00

.0533

2

Y(n) 0000 0953 1823

0.4055

x(n) 0.00 0.10 0.20 0.30 0.40 0.50

0.5000 0.5213 0. 5358 0. 5443 0. 5482 0.5493

x(n) 1. 00 1.10 1.20 1.30 1.40 1

.50

X

00 10 20 30 40 50

0.2624 .3365

V(n)

X 00 10 20 30 40 50

X

1.0000 1

1

.0101

1.0417 .0989 .1905 1.3333 1 1

380

.4110

1.2465 1.1480 1

.1037

1

V(n) .0000

1.1115 1.2530 1

.4397

1.6961 2.0670

X (n)

.

Yin)

2 .0000 1. 6562

n)

00 10 20 30 40 0. 50

0. 0.

0.10 0.20 0.30 0.40 0.50

V(n) 00 10 20 30 40 50

V(n)

0.0000 .0050 .0200 0. 0451

0.0805 0.1266

V(n) .0000 1 .1079 1 .2337 1

1.3807 1 .5531 1.7561

n)

Y(n)

00 10 20 30 40 50

0.5000 0.5250 0.5498 0. 5744 0. 5987 0.6225

Exercises 9.4


1 n)

— = 32 -0.025v

o 2

=

1 2 3 4 5

f{t, v).

OA

12. Separating variables and using partial fractions

1/1 ^v^-

and iD

2^32^0.025 Since v(0)

=

we

find c

—

0.

(

1

^ ^^ +

1

dv

+

v \

-

h

W5( and v{5)

a=

-1)

a 35.7678. 2.128 and

=

we obtain

0.0432. Separating variables

dA A(a -

a\A

a- 0A

-[\iiA- \n(a-

In

-

dt

dA =

dt

(3

A)

A)}

=t+c a(t

a-pA A

a- 0A

—

A= 1

+ c)

,a(t+c)

e

ae a < i+c >

- pAt a

^

+ 0e a(t+c )} A = ae al t+cl <

Thus A(t)

a

Q

= 1

+ /?e a t+c <

>

+

e-*(t+0

=

dt

- VO025 v\) = t +

13. See the table in the following problem. 14. Let

25 .0 32 .0 34 .0 35 .0 35. .0

Vfr025

we obtain

Solving for v

lill

o

we have

+

VomBv

2^32

.

/3

+

e" ac e

c.

OOOC 2570 9390 9772 5503 7128

Exercises 9.4

From A(0)

=

0.24

we obtain 0.24

so that e~ ac

=

«

-

a/0.24

= +

e-

8.8235 and 2.128 A{t)

0.0432 1

2

3

2. 78

13. 53

A approximated A (exact)

1.93 1.95

12.50 12.64

36.30 36.46 36.63

(dava) (

X

'

4 47. 50

49. 40

47.23 47.32

49.00 49.02

5

v(n) 1. 0000

(n)

00 10 1 20 1 30 1 40

8.8235e- 2128 '

A (observed)

t

15.

+

1 1

1.2511 1. 6934 2.9425 903 .0282

16. Simpson's rule on

[i n > r

in

=

is

x„+h

Jx n

For f(x,y)

+ h\

f(x n )+Af[x n

O

+ -h)+f(xn + h)

f(x) the Runge-Kutta method gives ki

=

k2

= hf[x n + -h

fc3

=

hf[

fc4

=

hf(x n

hf(x n ),

)

,

+ -h]

n

+

k),

and f(x n )

+ Af(x n +

382

+

f{x n

+ ft)]

.

Exercises 9.5

Exercises 9.5 1.

For

—

y'

y

=

x —

1


=

is e

e

x ,

so that

and j/

From

y(Q)

=

1

we

find c

obtained in Example

1,

=

we

1.09182470 compared to y 2 j/(0.8)

2.

fts

= e^-ze" 1 + c) = -x + = —x +

and y

1

find y(0.2)

=

e

1.42554093 compared to y4

.

Comparing exact values with approximation!

1.02140276 compared to yi

1.09181796, y(0.6)

=

x

ce*.

«

1.42552788.

REM ADAMS-BASHFORTH/ADAMS-MOULTON REM METHOD TO SOLVE Y'=FNF(X,Y) 120 REM DEFINE FNF(X,Y) HERE 130 REM GET INPUTS 110

140

170 IF

N<4

GOTO

200 210

REM SET UP TABLE

190

(AT LEAST 4)=",N

160

INPUT INPUT PRINT

180

"X0 =",X "Y0 =",Y

240

PRINT "X","Y" PRINT REM COMPUTE 3 ITERATES USING RUNGE-KUTTA

250

DIM

220 230

Z(4)

260 Z(1)=Y 270

FOR 1=1 TO 3

280

K1=H*FNF(X,Y)

290

K2=H*FNF(X+H/2,Y+Kl/2) K3=H*FNF(X+H/2,Y+K2/2)

300 310 320

K4=H*FNF(X+H,Y+K3) Y=Y+(Kl+2*K2+2*K3+K4)/6

330 Z(I+1)+Y

1.02140000, y(0A) a

1.22211880 compared to y3

100

PRINT 150 INPUT "STEP SIZE=", H 160 INPUT "NUMBER OF STEPS

=

=

1.22210646, anc

Exercises 9.5

350

X=X+H PRINT X,Y

360

NEXT

340

I

370 REM COMPUTE REMAINING X AND Y VALUES

380

FOR 1=4 TO N

390 YP=Y+H*(55*FNF(X,Z(4))-59*FNF(X-H,Z(3))+37*FNF(X-2*H,Z(2))

-9*FNF(X-3*H,Z(l)))/24 400 Y=Y+H*(9*FNF(X+H,YP)+19*FNF(X,Z(4))-5*FNF(X-H,Z(3))+FNF(X-2*H,Z(2}))/24

410 420

X=X+H PRINT X,Y

430 Z(1)=Z(2} 440 Z(2)=Z(3)

450 Z{3)=Z(4) 460 Z(4)=Y

470 480 3.

NEXT END

x(a) 0.00 .20

0.40 O.GO O.BO 5.

x(n) 0.00 0.20 0.40 0.60 0.80

1.00

I

vinl 0000

1

.

0.7328 0.6461 0.65B5 0. 7332 0.7232

4.

.40

xfn)

Yin)

0.0000 0.2027 0.4228 0.6841 1 0234 1.0297 1.5376 1.5569

x(a) 0.00 0.20

initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector

initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector predictor corrector

0.60

1

0.80

1.2109 1.2049

0000 .1003 .2027 3093

10

20 30

4227 40

4228

50

54 62 5463

6S40 6842 8420

60 70

.8423 1

1 1

90 1

00

384

.1039


Yin)

00

80

Yin)

2.0000 1.4112 1.1483

0292 0297

2552

1

2603

1

5555

1

5576

initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector predictor corrector predictor corrector predictor corrector predictor corrector predictor corrector predictor corrector

Exercises 9.5

Wn)

x{n) 00 .20 Aft

f\

0.60 0.80 1.00

1

.

4414

1

.

3

1 J.

WnJ

x(n)

1 0000 3

2 .6028

3.3433 3.34B6 4.2276 4.2230

Runge-Kutta pr edict or corrector predictor corrector

00 10 20

1

30

1

1

40 50

80

9719

1

90 1

.

80

1.00

.

.

9719

2 .2740 2 6028 2 .6028 2 9603 2 .9603 3

3486

3

3

3486 7703 7703

4

2280

3

0.0026 0.0201 0.0630 0.1362 1360 0.23 79 0.2385

.

2 2 740

70

v(n) .0000

.4414 6949

1

60

x(n> 0.00 0.20 0.40 0.60

.0000

1 .2102

00

xfn)

initial condition Runge-Kutta Runge-Kutta Runge-Kutta predi c t or corrector

0000

10

0003 0026 0087 0201 0200 0379 0379

30 40 50 60

0630 0629 0956

70

.0956

B0

1359 1360 1837

90

.1837

00

.2384

2384 1

predictor-

corrector predictor corrector predictor corrector predictor corrector

y(n)

00 20

predictor corrector

i .2280

initial condition Runge-Kutta Runge-Kutta Runge-Kutta predi c t or corrector predictor corrector predictor corrector

initial condition Runge-Kutta Runge-Kutta Runge-Kutta predi ctor corrector predictor corrector predictor corrector predictor corrector predictor corrector predictor corrector predictor corrector

t

Exercises 9.5

x(n) 0.00 0.20 0.40 0.60 0.80 1.00

WnJ 1.0000 1.2337 1.5531 1.9961 2.6180 2.6214 3.5151 3

.5208

WnJ

x(a) 0.00 0.10 0.20 0.30

initial condition Runge-Kutta Runge-Kutta Runge-Kutta

predictor corrector predictor corrector

1.0000 1.1079 1.2337 .3807

1

1.5530 1.5531 1. 7560 1.7561 1.9960 1.9961 2.2811

0.40 0.50 0.60 0.70

2

2.

0.80

.2312 6211

2.6213 3.0289

0.90

3

.0291

3.5203 1.00

3

.5207

initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector predictor corrector predictor corrector predi c t or corrector predictor corrector predi c t or corrector predi c or corrector

V
0.00 0.10 0.20 0.30 0.40


1.0000 1.0052 1.0214 1. 0499 1. 0918 1.0918

Exercises 1.

9.6,

Since the improved Euler method the Taylor polynomial through k

a second-order Runge-Kutta method, the formula agrees with

is

=

The

2.

V "'^3\ 2.

Since the three-term Taylor formula error

is

x

n
thus

i.

the Taylor polynomial through k

=

2,

the local truncation

-gj-

where

xn <

c

< x n+1

.

Since the fourth-order Runge-Kutta formula agrees with the Taylor polynomial through k local truncation error

is

y 4.

Wher6

is

is

y"'(c)

3.

local truncation error

(a) Using the Euler

(5| (

c)

method we obtain

where j/(0.1) =s yi

386

xn

—

=

4,

the

Exercises 9.6

(b) Using y"

=

Ae

21

we

see that the local truncation error

4^^=0.02^.

s/'(c)y= Since e

2x

an increasing function,

is

for the local truncation error

(c)

=

Since y(0.1)

2

e

=

is

The

error in (d)

step size

is

is

0.02e

1.2214

=

0.05

-

1.21

=

for

is

=

2 ' 01 '

=

2

e

we obtain

y(0.1)

=

(c) Since y{0.1)

=

2

e

^y2 —

<

0.001333e

Thus an upper bound

0.1.

e

error 0(h),

is

the case.

=

2'

02

1.21.

0.05 to be one-half the error

=

y(0.1) ?w y\

=

2

e

than 0.0244.

is less

=

when the

when h =

0.1.

1.22.

is

2c

2c

<

0.0214, which

= 0.001333e ^6 = 8e M! 6 e

c

With global truncation

0.0114.

see that this

an increasing function, is

=

-yi

Se 21 we see that the local truncation error

the local truncation error

<

for

0.0244.

method we obtain

y"'(c)

Since e 2*

e

halved we expect the error for h

(a) Using the improved Euler 1 (b) Using y "

<

2

=

Comparing 0.0114 with 0.214 we 5.

2e

1.2214, the actual error isy(O.l)

(d) Using the Euler method with h (e)

e

is

2c

<

for

c

<

0.1.

Thus an upper bound

0.001628.

1.221403, the actual error

-

is j/(0.1)

=

j/i

0.001403 which

is

less

than

0.001628.

(d) Using the improved Euler method with h (e)

The

error in (d)

the step size

is

is

1.221403

-

(a)

With

0.000378.

=

=

Se

21

we

Since e 2x

is


for the local truncation error is

Since y(0.1)

^6 = 8e

=

'

e

2

=

e

20

2c

<

0.001333e

3/2

=

1.221025.

global truncation error

0{h 2 ), when

«

y\

—

=

0.1.

1.22.

is

= COOlê 2 ". 6 e '

2' 2

,1

=

'

=

e

'

2

<

for

c

<

0.1.

Thus an upper bound

0.001628.


is

y(0.1)

-

yl

=

0.001403 which

is less

0.001628.

(d)

h

the case.

is


y "\c)

y(0.1) s=

0.05 to be one-fourth the error for

see that this

Using the three-term Taylor method we obtain y(0.1)

(b) Using y"'

(c)

we obtain

0.05



=

1.221025

—

Using the three-term Taylor method with h

=

0.05

we obtain

y(0.1) as

y2

=

1.221025.

than

Exercises 9.6

(e)

The

error in (d)

the step size

is

-

1.221403

=

1.221025


(b)

=

Using j/ '(x)

32e

2ir

we

32e

)

is


for the local truncation error

(c)

=

Since y(0.1)

02 e

=

see that this

is

e

2c

2e

<

e

2 ( 01

0.000002667e

)

02

y(0.1)

=

h—

0.1.

0.000002667e

=

e

02

=

y\

1.2214.

is

=

=

when

the case.

is


pW( c J£- = Since e 21

global truncation error 0(/i 2 ),

0.05 to be one-fourth the error for

method we obtain

(a) Using the fourth-order Runge-Kutta 5

=


is

With

0.000378.

<

for

2c .

<

c

0.1.

Thus an upper bound

0.000003257.


is

y(0.l)

-

yi

=

0.000002758 which

is less

than 0.000003257. (d) Using the fourth-order Runge-Kutta formula with h

=

y(0.1)«tti (e)

The 0(/i

),

when h

error for 8.

(a)

(b)

is

1.221402758

the step size

=

0.1.

-

—

5e

_2x

2x

is

we

is

0.025(1)

Since y(0.1)

(d)

Using the Euler method with h

(e)

The

(a)

_2c

=


(c)

error in (d)

is

0.8234

-

=

0.05

0.8125

<

.025e-

=

e°

=

-10e

-21

we

is

the case.

3c

<

for

1

-

is t/(0.1)

we obtain

=

for

h

yi

=

<

c

0.1.

Thus an upper bound

=

0.0234, which

y(0.1)

With

0.0109.

see that this

-2c

see that this

is

«

=

y%

is less

for

than 0.025.

0.8125.

global truncation error O(h),

0.05 to be one-half the error

is

when

when h =

the 0.1.

the case.

«

y\

=

0.825.


10e

0.05 to be one-sixteenth the

0.8.

Using the improved Euler method we obtain y(0.1)

(b) Using y'"

=


0.025.

halved we expect the error

is

=

yi

-acM! =

a decreasing function, e

=

~



j/(0.1)

With

0.000000187.

Comparing 0.000000187 with 0.000002758 we


step size

we obtain


is

5e

Since e~

=

1.221402571

Using the Euler method we obtain Using y"

0.05

1.221402571.

error in (d) 4

—

(0^L = 0.001667e- 2c

is

.

o Since e~ 2x

is

a decreasing function, e~


is

2c

0.001667(1)

<

=

e°

=

1

for

0.001667.

388

<

c

<

0.1.

Thus an upper bound

for

Exercises 9.6


Since j/(0.1)

(d)

Using the improved Euler method with h

(e)

The

error in (d)

when h 10.

=

(c)

(a)

(b)

=

the step size

0.1.

|0.823413

is

is

-

y(0.1)

is

=

=

0.000305.



y'"

Using

= — 10e

we

0.001587, which

—

With

y2

is less

=

than 0.001667.

0.823781.


0(h 2 ),

0.05 to be one-fourth the error

=; y\


l 0e

g/(0.1) as

see that this

Using the three-terra Taylor method we obtain ^(0,1) -21

=

we obtain

0.05

0.8237181|

-yi

=

when

the case.

is

0.825.

is

-^M!=0.001667e- 2c 6

Since e _2x

is

a decreasing function, e~ 2c <


0.001667(1)


=

Since y(0.1)

(d)


(e)

The

error in (d)

h (a)

(b)

=

is

the step size

0.1.

|0.823413 is

-

1

<

for

c

<

is j/(0.1)

0.8237181|

— j/i =

=

0.05

=

0.000305.

halved we expect the error


0.001587, which

we obtain

for

h

=

y(0.1)

Using y^(x)

=

we

- 2c

M! =

is less

« y2 =

than 0.001667.

0.823781.

0.05 to be one-fourth the error

see that this

when

the case.

is

=

r;


40e

for

global truncation error 0(/i 2 ),

With

Using the fourth-order Runge-Kutta method we obtain y(0.l)

— 40e -2T

Thus an upper bound

0.1.

0.001667.

(c)

when

11.

=

is

=

e°

0.823416667.

is

0.000003333.

120 Since e _2x

is

the local truncation error (c)

=

Since y(0.1)

is

_2c

<

e°

0.000003333(1)

=

a decreasing function,

e


is

=

<

1 for

<

c

0.1.

Thus an upper bound

for

0.000003333. |y(0.1)

-

=

yi\

0.000003225, which

is less

than

0.000003333. (d)

Using the fourth-order Runge-Kutta method with h

y(0.1)m y 2 (e)

The

0(h

(a)

),

when the

when h =

Using y"

=

0.05

we obtain

0.823413627.

error in (d) 4

error

12.

=

—

38e

is

|0.823413441

step size

0.1.

is

-

0.823413627|

=

0.000000185.


—

With


0.05 to be one-sixteenth the

Comparing 0.000000185 with 0.000003225 we see that

_3 ^ I_1 '

we "

y


{c )

% = 38e-^-') % =

is

this is the case.

Exercises 9.6

(b) Since

e^ -1

is

'

a decreasing function

for

<

1

x <

1.5, e

-3 < c-1 >

<

e

-3 ' 1-1 )

=

<

1 for 1

c

<

1.5

and */"(c)y <19(0.1) 2 (1) (c)

Using the Euler method with h 1,(1.5)

« =

Eo.i/£ 13.

(a)

Using

=

2.0532, the error for h

With

0.1109.

«

y(1.5)

=

0.1

£

is

=

.l

=

With h

1.8207.

we obtain

0.05

h =

0.2325, while the error for

we expect

global truncation error 0(h)

= = — lHe -3 ^ -1

.05

y'"

e

we obtain

i?ci.l/£b.05

5:3

We

2.

0.05

is

actually have

2.10.

'

we


h3

(b) Since

0.1

0.19.

1.9424.

(d) Since y(l.5) #0.05

=

=

-3 * 1-1

'

is

= IHe" 3 ^"


^6 =

1 '

< x<

for 1

19ft

1.5, e

3

is

e- 3 ( c -'».

" 3 < c_1)

<

e

_3(1_1)

=

1 for 1

<

c

<

1.5

and 3

v'"(c)

(c)

h -

<

Using the improved Euler method with h

we obtain

—

=


With

0.005950.

=

have Uo.l/ô.05

0.1

=

0.1 is


we obtain

^

0(h 2 ) we expect

=

i

14e

mV

-3(-i>^ 6

3

2.080108.

With k

—

0.05

Em = 0.026892, while the error for h = 0.05 is


h3 y"'{c)

(b) Since e"

=

y(1.5)

^"

1 '

is

So.l/^0.05

* 4. We actually

4.52.

= — 114e _3 ^ -1 we

14. (a) Using y'"

=

0.019.

y(1.5) =a 2.059166.

(d) Since y(l.5) ^).05

=

3

19(0.1) (1)

a decreasing function for

1

<

x <

1.5, e

is

3 (c-

-3 ^ -1

*

,

>.

<

e" 3 ' 1

"" 1 '

=

1 for 1

<

c

<

1.5

and y"'(c)

(c)

~o

<

3

19(0.1) (1)


we obtain

y(\.$)

(d) Since y(1.5) £fl,05

=

=

w

0.005950.

=

0.1

0.019.

we obtain

y(1.5)

w 2.080108.

With h

=

0.05

2.059166.


have .Eb.i/£b.05

=

=

With

=

0.1 is


4 -52.

390

Eq A

0(h

=


2

we expect Eq.\/Eq.q$

)

fa 4.

h

We

=

0.05

is

actually

Exercises 9.6

15. (a) Using j/

5

3<- x

= -1026e

'

we


w 120

y

3 (b) Since e~ <

1_1


is

>

and y (c)

(d) Since

h

=

16.

(a)

(c)

for

—<

8.55fe

<

1

B

<

x

e-

3
1.5, e

=

5

8.55(0.1) (1)

=

0.05

=

j/(1.5)

0.05

We

is

we obtain

£b.05

0.000006757.

(x

=

1 '

<

e^ 3 ! 1

we obtain

0.1

=

0.1

is

£b.i

=

* 1'

=

1 for 1

<

c

<

1.5

=

y(1.5)

2.053338827.



0(h 4 ) we expect

£b.l/£fl.0S

«

18-14.

+ If

we


2

2"

0.5,

-,

(c+1) is

(x+l)


for

is

h2

1

(b) Since

^

1

=-

Using y"

With

Eo.i/Eom =

actually have

-3

2.053222989.

3/(1.5) f»


=

>.

0.0000855.

Using the fourth-order Runge-Kutta method with h

With h

16.

(5)

=

is

—

(

c+

l)2

*

< ~

,„

(0

+

=

1

for

<

c

<

0.5

1)2

and

<

y"(c)

(c)

Using the Euler method with k y(0.5)

(d) Since £0.05

w

(a)

Using

=

0.0069.

£b.i/£fl.05

17.

0.1

we obtain

0.4198.

y(0.5)

With h

=

0.05

we obtain

0.4124.

j/{0.5)

=

=

(1)^^ = 0.005.

=

0.4055, the error for

With

h

=

0.1

£

is

=

.i

0.0143, while the error for h

we expect

global truncation error 0(h)

i?o.i/£b.05

=

0.05

is

^ 2- We actually have

2.06.

y"'

(ar+l) 3

we


is

h3

1

'

6 *

(b) Since (l

+

is

a decreasing function for

(c

+

l)

3

0.5,

3

-,

—

(c

l)

+

1)

and ft

m i)

3

y"'(c)~<

(1)

3

=

0.000333.

3

< -

,„

(0

*

+

.

,

1)

„ 3

=

1 for

<

c

<

0.5

— Exercises 9.6

(c)

Using the improved Euler method with

we obtain (d) Since

E

05

=

j/(0.5)

=

«

y(0.5)

With

-

have Eo.i/Eom

-tj

-.

+

(x

—+—^ 3

(x

is

is

y(0.5) as 0.405281.

With

=

ft

0.05

=

Boa

2 )


we expect Eo.i/Eom

=

ft

We

4.

0.05

is

actually

we


is

I)

1 = _J_ !t

>»

-,

=

ft

global truncation error 0(ft

v

(b) Since

0.1

we obtain

3.98.

2

—

18. (a) Using y'"

0.1

0.405419.


0.000046.

=

ft

ic)


<

for

x <

0.5,

-.

—^—^ <

(c+

1)

1)

J

(0

V,,J =

+

<

1 for

c

<

0.5

1)

and y"'{c)

(c)

t/(0.5)

^0.05

=

=


19.

(a)

Using j/ 5 '

(b) Since ^

'

-,

With

0.000195.

have £*. t /£t».o5

=

—+

(x

1)

^

5

0.000333.

A

ft

=

0.1

we obtain

yf0.5)

«

0.404643.

With

ft

=

h

=

0.05

0.05

& 0.405270.

j/(0.5)

Since

=

(1)

Using the three-term Taylor method with

we obtain (d)

^6 <

=

=

0.1

is


Eb.i

0(h

=


2

we expect £o.l/£b.05

)

We

4.

is

actually

4.22.

+\) 5

is

ft

We

See

t

^ at

^ e '° Ca

V

W 120


'

truncat ^ on error ' s

'

(c+1) 5

<

for

x

<

5

—+

^—tt

0.5, r (c

5

1)

<

—-ttt =

t: (1

+

5

1

for

<

c

<

0.5

1)

and y (c)

i5)

(c)

—<

(1)

~^ =

5

ft

=

0.1 is

B

Using the fourth-order Runge-Kutta method with

With (d) Since ft

=

16.

ft

=

0.05

j/(0.5)

0.05

We

is

=

we obtain

y(0.5)

«

£0.05

0.000000003.

actually have Eo.\/Eqx>5

0.1

we obtain

y{0.5)

=

0.405465168.

0.405465111.


=

0.000002.

5

With

=

=

.i

=


global truncation error 0(ft 4 )

1T.64.

392

we expect

£b.i/£b.05

«

7 Exercises

9.

Exercises 9.7, 1.

The

substitution y'

—

u leads to the iteration formulas

=Vn+ hu„, The

initial

The we 2.

conditions are yo

— —2

and uq

=

yi

=

w + 0.1«o =

ui

=

ii

+

V2

=

tfi

+ O.ltii -

-2

+ 0.1(1) =

-4y

0.1(4u

find c\

and 1

The

substitution y

The

initial

c2 =

—

5.

-1.9

u leads

conditions are

j/o

)

=

+

1

-1.9

y

is

—

+ 8) =

0.1(4

+ 0.1(2.2) -

= -2e 2x + 5xe 2x

Thus y

).

Then

1.

general solution of the differential equation

= -2

=u n + h(4u n - 4yn

u n+1

c\e 2x

and

2.2

-1.68.

+ C2xe 2x from .

the initial conditions

y(0.2) ss 1.4918.

to the iteration formulas

=

=

4 and uo

Then

9.

2/1

=i/D

+ O.I1/0 -4 + 0.1(9) =

U1

=

+ 0.1 (Ju - ^w) =

uo

9

4.9

+ 0.1(2(9) -

2(4)]

=

10

m = yi + O.Imi = 4.9 + 0.1(10) = 5.9. The general conditions 3.

The

solution of the Cauchy-Euler differential equation

we

find

a — —1

substitution y'

—

u

and

ci

=

Using formulas

(5)

ml

0.2000

m2 4400

y,

y

=

c\x

and y(l.2)

=

+ C2X 2 FVom .

the

6.

and

—

(6) in the text

w,

u'

=

4u

— 4y.

with x corresponding to

f,

y corresponding to x, and

we obtain

m3 5280

9072

k2

kl

n\4

2

4000

3

.2800

k3 3

k4

.5360

4

8064

X

y

u

0.00 -2 0000 1 0000 0.20 -1 .4928 4 4731

Ruiwe -KuttB method wi th h= 0.1 ml

0.1000 0.2443

m2 1600 3298

initial

leads to the system y'

corresponding to

Thus y

5.

= — x + 5x 2

is

ml 1710 3444

m4 2452 4487

k2

kl

1

1

2000 7099

1 2

4200 0031

k3

*4

1

4520

1

2

0446

2

7124 3900

X

y

u

0.00 -2 0000 1 0000 0.10 -1 S321 2 4427 0.20 -1 4919 4 4753

u

7 Exercises

4.

The

9.

substitution y'

=

u

leads to the system ,

Using formulas

(5)

corresponding to

y,

and

2

2

x

xl

the text with x corresponding to

(6) in

(,

y corresponding to

x,

and u

we obtain Runqe-Ku^ta method with h=0.2

m2

m3

2.0000

2.0017

ml 1.8000

kl

k2

2.0000

2.0165

m4

2.1973

k3

k4

1.9950

1.9865

1.00 4.0000 1.20 6.0001

9.0000 11.0002

Runge-Kutta method with h-0.1

0.9000 1.0000 5.

The

m3

m2

ml

0.9500 1.0500

9501 0501

0. 1.

substitution y'

m4

=

u

0.9998 1. 0998

(5)

and

1

.0000

1.0000

1.0023 1.0019

X

k4

k3

0.9979 0.9983

9996 9997

y

u

1.00 4 0000 1.10 4 9500 1.20 6 0000

0000 10 .0000 11 .0000 9.

leads to the system y'

Using formulas

k2

kl

(6) in

=

u,

u'

= 2u —

2y

+

e'

cost.

u corresponding to

the text with y corresponding to % and

y,

obtain

ml

m2

.4000

4600

ml

m2

m3 .4660

m4 5320

kl

6000

k2

k3

6599

k4

6599

y

t

0.7170

u

0.00 1.0000 0.20 1.4640

2 2

0000 6594

Runqe-Kutta method with h= 0.1 !tl3

m4

.2000 .2315

2150 2480

2157 .2487

2315 2659

ml

m2

m3

!t>4

kl

3000 3299

k2

k3

k4

3150 3446

3150 3444

k2

k3

y

t

0.3298 0.3587

u

0.00 1.0000 0.10 1.2155 0.20 1.4640

2 2 2

0000 3150 6594

Runge-Kutta method with h^Q.l

^-

10 0000 8

.7500 -2

10 1563 -4 13 2617 -6 17 9712 -8

0000 5000 3750 3672 S867

kl

5000 -20 0000 13 .4375 -28 7500

0000 -5 0000

4

17. 0703 -40 0000

-8 7500

5

22. 9443 -55 1758 -12 7344

6

31.3507 -75 9326 -17 7734

8

12

.

5

0000 -5 0000 22 3750 -10 6250 29 07 81 -16 0156 10 6309 -22 5488 55 9856 -31 2024 75

394

k4

5000 6875 .3516 3076 9B21

00 10

20 30 40 50

i2

il

t

0000 7500 2 5 7813 2 0703 7 4023 6104 9 1919 -1 .5619 11 4877 2

0000 SOOO 8125

3

we

Exercises

Runoe-Kutta method with h=0.2

2

.0000

m3

m2

ml 2

2800

2

3160

mi

kZ

kl

2 6408

1

2000

1

.4000

k3 1

k4

.4280

1

6632

x

t

0.00 6.0000 0.20 8.3055

y 2.0000 3.4199

Runae -Kutta method with h=0.1 ml 1 1

.0000 .1494

1 1

0700 2289

m4

m3

ra2

1 1

0745 .2340

1 1

1496 3193

k3

k2

kl

k4

6000 7073

.6500 .7648

.6535 .7688

kl

k2

k3

7075 8307

x

t

0.00 6.0000 0.10 7.0731 0.20 8.3055

y 2.0000 2.6524 3.4199

Runae -Kutta method with h=0.2

m2

ml

9400

.6000

m3 1

1060

m4 1

7788

1

4000

2

.0600

2

k4

3940

3

7212

t

x

y

0.00 0.20

1.0000 2.0785

1.0000 3.3382

Runae -Kutta method with h=0.1

m2

171]

3000 5193

m4

m3

3850 6582

4058 6925

m2

m3

k2

kl

5219 8828

1

7000 1291

1

8650 4024

k3

1

k4

9068 4711

1 1

1343 8474

x

t

0.00 1.0000 0.10 1.4006 0.20 2.0845

y 1.0000 1.8963 3.3502

Runge-Kutta method with h-0.2 ml

m4

k2

kl

k3

k4

X

t

y

0.00 -3 0000 5 0000 -1.0000 -0.9200 -0. 9080 -0.8176 -0.6000 -0.7200 -0.7120 -0.8216 0.20 -3 9123 4 2857

Runqe- Kutta method with h= 0.1 m2

ml

m4

ia3

k2

kl

k3

k4

X

t

0.00 -3 0000 .10 -3 4790 -0.5000 -0.4800 -0.4785 -0.4571 -0.3000 -0.3300 -0.3290 -0.3579

-0.4571 -0.4342 -0.4328 -0.4086 -0.3579 -0.3858 -0.3846 -0.4112 0.20 -3 9123

y 5

0000

4 6707 4 2857

Runoe-Kutta method with h=0.2 m2

ml

6400

1

27 60

in4

ra3

1

7028

3

.3558

k2

kl

-

1

3200

1

7720

k3 2

k4

1620

3

5794

X

t

O.00 5000 0.20 2 1589

y 2

2000 3279

Runge -Kutta me thod with h= 0.1

m2

ml .3200 .7736

1

4790 0862

m3 5324 1

1929

m4 .7816 1 .6862

1

6600 0117

k3

k2

kl

1

.7730 .2682

1

8218 3692

k4 1 1

0195 7996

X

t

0.00 0.10 0.20

1 2

5000 0207 1904

y 2000 0115 2 3592 1

9.

Exercises 9.8

Exercises 9.8, 1.

We

identify

equation

=

P{x)

0,

=

Q{x)

9,

=

f(x)

0.0

We

equation

1.0

0.5

identify

X

P(x)

=

0,

Q(x)

=

-1, f(x)

3.

identify

equation

0.25

We

x 2 and h ,

We

=

P(x)

2,

Q(x)

=

1

0-25.

Then the

finite difference

2

0.0625a:,

.

f{x)

1,

00

.

=

and

5x,

-

1.96?/;

ft

=

(1

-

0)/5

=

0.2.

Then

the finite difference

Then


+ 0.8yi_i - 0.0400-

0.4

0.2

0.6

1.0

0.8

-0.2259 -0.3356 -0.3308 -0.2167

identify

P(x)

=

-10, Q(x)

=

25, f(x)

=

0.0000

and h

1,

=

-

(1

=

0)/5

0.2.

is

+

2Vi-l

=

0-04.

solution of the corresponding linear system gives

0.0

0.2

1.0000

1.9600

identify

P(x)

=

0.4 3.SB00

-4, Q(x)

0.6

o.a

7.7200

15.4000

=

4,

1.0 0.0000

=

(1

+

1.8889yi

+

1.3333yt _i

f{x)

xje 2", and

h

=

(1

-

-

=

0.2778(1


0.0000 3.0000

0)/6

=

0.1667.

is

0.6667yi+i

X y

0)/4

0.0000

0.0000

difference equation

The

-

(1

-

2.0625yi

-Vi

5.

=


equation

The

=

+ ^1 =0.

is

0.0

4.

=

0.75

0. 50

1.2w + i

The

finite difference


0.00

We

Then the

is

0.0000 -0.0172 -0.0316 -0. 0324

Y

0.5.

2.0 1.0000

1.5 6.3226

-5.6774 -2.5807

W+i -

The

=

0)/4


4.0000 2.

-

(2

is

yi+l +0.25^

The

=

and h

0,

0.1667 3.3751

0.3333 3.6306

0.5000 3.6448

0.6667 3.2355

0.8333 2.1411

396

1.0000 0.0000

+ Xi)e 2x

'.

Then the

finite

Exercises 9.8

6.

We

identify

=

P(x)

difference equation

=

Q{x)

5,

-

=

l)/6

0.1667.

Then

the finite

- 2 W + 0.5833yi_! = 0.2778(4^).

1.1667 1.3333 1.5000 1.6667 1.8333 2.0000 -0.S918 -1.1626 -1.3070 -1.2704 -1.1541 -1.0000

We

identify

P(x)

= is

+

The X

—

3/x 2

0.1875\ J

-2-

Vi+i -

=

f{x)

,

and

0,

= (2-

ft

—

0.1875\

0.0469

+

Vi

x%

\

1 (

tt-i

=

o.

J


y

1.000 5.0000

We

identify

1.125 3.8842

P(x)

=

difference equation

1.250 .9640

2

=

1.625 1.0681

1.500 1.5826

1.750 0.6430

x~ 2 f(x) = lnx/x 2 and ,

,

ft

=

1.875 0.2913

-

(2

=

l)/8

2.000 0.0000

Then the

0.125.

finite

is

0.0625\

/

1.375 2.2064

-1/x, Q(x)

—) K+l +

^1

The

=

Q(x)

3/a:,

Xii

/ (

0.0156\

"2„ +

/,

W+

0.0625N

— ^-J W~l

+

(l

=

0.01561a

a*.


Y

1.375 1 500 1 625 1.750 1.875 2.000 1. 000 1.125 1 .250 0.0000 -0.1988 -0.4168 -0.6510 -0.8992 -1.1594 -1.4304 -1.7109 -2.0000

We

identify

X

.

equation

P(x)

The

0.0

identify

equation

=

x,

+ 0.05(1 - Xi)]yi+ + i

0.1 0.2660

P{x)

=

.

x,

0.2 5097

Q(x)

f(x)

=

x,

and

=

ft

(1

- 0)/10 =

0.1.

Then

[-2

+ O.OliilK +

[1

-

- Xj)l»-i =

0.05(1

=

0.3 .73 57

1,

f{x)

.

=

0.4 9471

a:,

1


0-Olxi.

and

0.6 1.3353

0.5 146 5

.

0.7

0.8

0.9

1.0

1.5149

1.6855

1.8474

2.0000

ft

=

(1

-

0)/10

=

0.1.

Then the

is

(1

The

Q{x)


0.0000

We

= l-x,

.

is

[1

10.

(2

1.0000 1.0000

1

9.

=

ft

Y

difference equation

8.

and

,


X

7.

Ay/x

is

1.4167jfc+i

The

=

f(x)

0,

+

0.05xi)yi+i

-

l.Wtf,

+

(1

-


0.05a:i)!K-i

=

O.Olarj.

finite difference

Exercises 9.8

11.

We

0.0

0.1

0.2

0.3

1.0000

0.8929

0.77 89

0.6615

identify

equation

The X

12.

0.7 .2225

0.

and h

=

0)/8

tt + i-2.0625»j

+

=

-4, f(x)

equation

0.250 0.7202

0.125 0.3492

identify P(r)

The

=

Q(x)

0,

0.6

0.3216

0,

0.8

-

(1

1347

.

-

0.9 0601

1.0

0.0000

Then the

0.125.

finite difference

J

«_ l

=

0.


0.000 0000

We

=

0.5 0.42 96

is

0.

Y

P(x)

0.4 0.5440

=

0.375 1.1363

0.500 1.6233

2/r, Q(r)

=

0,

('l

+

Û

f(r)

=

and h

0,

0.750 .9386

0.625 2 .2118

=

0.875 3.8490

2

-

(4

=

l)/6

1.000 5.0000

Then the

0.5.

finite difference

is

+1

-2u,+

fl-Û_

=0.

1


1.0

2.0

1.5

3.0

2.5

4.0

3.5

50.0000 72.2222 83.3333 90.0000 94.4444 97.6190 100.0000 13.

(a)

The

difference equation

1

is

+

+

yt+ i

(-2

+

^

h 2 Qi) yi

«_! = h 2 fi

The equations

the same as the one derived on page 530 in the text.

are the

same because

the derivation was based only on the differential equation, not the boundary conditions. allow jfo-i-

i

to range from

Since yn

(b) Identifying

j/o

n—

1

we obtain n equations

one of the given boundary conditions,

is

=

to

y(0),

y~i

—

y(0

—

ft),

1

=

~ V-i]

[yi

and

yi

=

if'(0)

= 1

y(0

in

the

it is

+ h) we

or

j/i

n+

1

unknowns y-i,

not an unknown.

have from

- y-i =

(5) in

2ft.

2ft

The

difference equation corresponding to

1

becomes, with y-i (l

or

+ ~Po)

=

yi

+ ^Pbj

+

yi

(-2

i

=

0,

+ h 2 Qo)ya +

(l

-

\p

V-i

\

=

2 ft

/o

— 2ft, yi

+

2»i

(-2

+

+ h2 Q

("2

)

yo

+

+ h?Qo)yo = 398

h a /o

Jp )

(in

+ 2h~

ya,

-

Pq.

2ft)

=

h?fo

the text

If

j/i,

.

we . .


Alternatively,

n

obtaining (c)

Using n

=

5

+

we may simply add 1

equations in the

the equation 3/1— y-i

n

+

1

unknowns y_i

,

=

2h to the

yo>

we obtain 0.2

0.0

0.4

O.B

0.6

1.0

-2.2755 -2.0755 -1.8589 -1.6126 -1.3275 -1.0000


h- 0.1

X (n) 1 .00 1 .10 1 .20 1 .30 1 1

.40 .50

EULER 0000 1386 3097 5136 7504 0201

2 2 2 2 2 3

h=0.05 x(n> 1.00 1

.05

1.10 1

.15

1.20 1.25 1.30 1.35 1.40 1.45 1.50

EULER 2 0000 2 2 2 2 2 2 2 2 2 3

0693 1469 2328 3272 4299 5409 6604 7883 9245 0690

IMPROVED 3 -TERM EULER TAYLOR 0000 1549 2.3439 2 .5672 2 .8246 3 .1157

0000 1556 2 3446 2 5680 2 8255 3 1167

2

.

2

2

.

2

IMPROVED 3 -TERM EULER TAYLOR .0000 .0735 2 .1554 2 .2459 2

2

2

2

2.3450 2

.4527

2 2 2 2

2.5689 2.6937 2.8269 2.9686

2 2

1187

3

3

.

2 2

0000 0735 1555 2460 3451 4528 5690 6938 8271 9688 1188

RUNGE KUTTA 2 2

0000 .1556 .

2.3454 2.5695 2 8278 3.1197 .

RUNGE KUTTA 0000 0736 2 1556 2.2462 2

.

2

.

.

2.3454 2

.4532

2.5695 2 6944 2.8278 2 9696 .

.

3

.1197

2/i>

-

-

-

,

list

of n difference equations

I/n-l-


4.

O— u J. x(n) n uu nn u .

EULER 0000 1000 2010 3049 4135 5279

.

1ft

ft

0.20 0.30 0.40 50

.

h=0. 05 x(nl 0.00

EULER 0000 0500 1001 1505 2017 2537 3067 3610 4167 4739 5327

.05

0.10 0.15 0.20 0.25 .30

0.35 .40 .45

0.50

.

X fW

EULER

.50 60 .70 80 .90 1 00

5000 6000 7095 8283 9559 0921

.

.

.

h=0 .05 x(ni 0. 50

55 0. 60 65 .70 .75 .80 .

1

EULER 5000 5500

.

0.85

1

.90 .95 00 .

1 1

6024 6573 7144 7739 8356 8996 9657 0340 1044

FTTT CTJ ft

ft ft ft

3 -TERM TAYLOR

n

0000 1000 2025 3087 4202 5377

0.2030 0.3092 0.4207 0.5382

IMPROVED 3 -TERM EULER TAYLOR 0000 0501 1004 .1512 .2027 .2552 0.3088 0.3638 0.4202 0.4782 5378

.0000 .0500 .1003

0.

.

.

.

0.1511 0.2027 0.2551

Iff ft

iqupa

n

ftft

1001

0.2026 0.3087 0.4201 0.5376

RUNGE KUTTA 0. 0000 0.0500 0. 1003 0. 1511

.2026

0.2551

.3088

.3087

0.3638

0.3637 0.4201 0.4781 0. 5376

.4202 4782 0. 5377 0.

IMPROVED 3 -TERM TAYLOR EULER

RUNGE KUTTA

5000 6050 7194 8429 9754 1 1166

.5000 .6049 .7194

.5000 .6048 .7191

0.8427 0.9752 1.1163

IMPROVED 3 -TERM EULER TAYLOR 0.5000 5000 0.5512 5513 0.6049 6049 0.6609 6610 7194 0.7193 7801 0.7800 8430 0.9082 0.9755 1. 0451 1 .1168 0.

1 1

8430 9082 9756 0451 1168

0.8431 0.9757 1.1169

RUNGE KUTTA 5000 0.5512 0.6049 0.6610 0.7194 0.7801 0. 8431 0. 9083 0. 9757 1.0452 1. 1169 0.

400


6.

hj£

TMPROVRD 3 -TERM EULER TAYLOR

.

EULER

In) ILL-

1 00 1 10 1 .20 1 .30

1

.

1

0000 2000 4760 8710 4643 4165

.

1

1 2

1 .40 1 .50

3

1 1

3 5

1. 1.

1.05 1.10

1

1.20 1.25 1.30 1.35 1.40 1.45 1

7.

0000 1000

.2183

1.3595 1.5300 1.7389 1.9988

1. 15

2

.3284

2.7567 3.3296 4.1253

.50

2

3 5

0000 1.1091 1.2405 1.4010 1.6001 1

.8523

1

2.1799 2. 6197 3.2360 4 1528

.2745 .8338

1

.8515

.

.

.2340 .1497 5.6350

4

.6404

.0000

1.1095 1.2415 1.4029 1 .6036 1 8586 2.1911 2.6401 3.2755 4.2363 5. 8446

3

.

5

RUNGE KUTTA

2.1789 2 6182

Using

=

+ hu n

Vn

= we obtain (when h =

=

0.2) yi

y(0.2)

Vl=IfO

xlnl 0.00 0.10 0.20 0.30 0.40

9.

5

1

Vn+\

8.

.

.

3

.0000 1.1088 1.2401 1.4004 1.5994

1.

L-<

!

0000 1.2415 1. 6036 2 1909 1

IMPROVED 3 -TERM EULER TAYLOR

EULER

1 .00

.1524 .1458 .2510

1

KUTTA

0000 2350 5866 1453 1329 2208

1

1 1

1.5910 2

h=0.05 x(n)

0000 .2380 .

RTTWHR XViyi VJ

Using xq

+ h(2x„ +

=

l)y„,

+ hu =

y

+ 0.1uo =

3

ui

= u + 0.1(2xo +

VZ

=

pi

,

+ O.lwi -

+

+

(0.1)l

l)yo

3.1

3

+

(0.2)1

=

yo

=

3

u

=

1

=

3.2.

3.1

= 1 + 0.1(1)3 =

0.1(1.3)

3.23.

Yin,

2.0000 2.4734 3 17E1 4.3925 6.7689 7.0783 .

—

1,

yo


=

2,

and h

=

0.1

we have

xx

=

xa

+

10

-

yo

+ h(x -

h{x Q

+ yQ = 1+0.1(1+2) )

y

)

=

2

+

0.1(1

When

-2) =

1.3

1.9

1.3

h

=

0.1

we have


and '

Thus, i(0.2) 10.

We

identify

«

1.62

=

P{x)

difference equation

x2

=

xi

+

h(xi

+ yi) =

1.3

+ 0.1(1.3 +

V2

=

Vi

+

/i{x!

- yi -

1.9

+

and 0,

3/(0.2)

=

Q(x)

«

)

1.9)

=

1-62

0.1(1.3-1.9)

=

1.84.

-

0)/10

1.84.

6.55(1

+

x),

f(x)

=

1,

and h

=

(1

=

0.1.

Then

is

K+i + [-2 + 0.0655(1

+ Xi)]vi + yi-i =

0.001

or y i+ i

The

+

(0.0655xj

-

1.9345)^

+ yi_i =

0.001.


0.0 0000

.

0.1

0.2

4.19B7

8.1049

0.3 0.5 0.4 11.3640 13.7038 14.7770

14.4083

0.7 12.5396

0.8

0.9

9.2847

4.9450

402

0.6

1.0 0.0000

t.

Differential-Equations-by-Zill-3rd-Edition-Solutions-Manual(ciitbooks.blogspot.com)_text.pdf

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