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 + c^e
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!
= + 2^OX
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
differential equation
— = — —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
differential equation
is
A'(t)
=
kA(t) where k
11.
The
differential equation
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)
differential equation
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
k^A.
—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 + Ac^e 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
+ 23e 2x
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
equation will have a unique
+ 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
equation will have a unique
differential
equation will have a unique
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
differential equation
has a unique solution at (1,4).
18.
The
differential equation
is
not guaranteed to have a unique solution at (5,3).
19.
The
differential equation
is
not guaranteed to have a unique solution at
20.
The
differential equation
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 + ^e~ 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
solution of the initial-value problem
In
solution of the initial-value problem
46. From
tan
or
2
-1 ' 2
solution of the initial-value problem
_!
*j
x +
J/
c
+ c or
dx we obtain
solution of the initial-value problem
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
^
56. Separating variables
.
Setting z
we obtain
^
=
y -
solution
=
1)^+ and y
^—
Then
1)
we obtain c — —100. The
55. Separating variables
— 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
solution of the initial-value problem
-~ + -^—) dy = dx we y-3)
|
—
I
-4.
\y +
1
=
-
y y is
—+
or -
solution of the initial-value problem
~
dx we obtain
iy
Ifi,(0)
From
The
.
+c
1|
we obtain
~)
~~
f ~~2
(a)
(c)
50.
-
In |i
= ~—y~ ^ =
(it,
1
49.
=
l\
solution of the initial-value problem
— 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:|
— ^u 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
x^if + 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
+ ^u 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
solution of the initial-value problem
+ 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
solution of the initial-value problem
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~y^is
=
u4
~ x^e 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
solution of the initial-value problem
(uV + 2ua;
Using y(l)
4
=
c
=
r=
\x\
H
t
—
c
solution of the initial-value problem
is a;
we have (x
— ux —
v?t 2 xj dx
+
(i
+
v^a:) (udx
dx
+ xdu) =
+ x (l + V") du =
^+ In
x
(l
+
du
=
+ u + ^u3 ^ 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
solution of the initial-value problem
(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
solution of the initial-value problem
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.
solution of the initial-value problem
\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
solution of the initial-value problem
= 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
.
solution of the differential equation
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 /
differential equation is
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.
solution of the differential equation
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
an integrating 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
^e 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,%
an integrating factor
—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.
an integrating factor
+ 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.
x an integrating factor
— tt/2 < x <
2 cos
00.
00.
an integrating factor
sec
=
and y
an integrating factor
-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.
x an integrating factor
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
-
an integrating factor
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
an integrating factor
— + - -s H e>
1
4 y2
ce _B
y
s2
<
for
w
an integrating factor
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
an integrating factor
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.
an integrating factor
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#} -
an integrating factor
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)
an integrating factor
and y
=
|(»
+
2)" 1
an integrating factor
+
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
an integrating factor
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
an integrating factor
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
an integrating factor
c
is
= e
150 and
/(l+2/*)
T=
£
x
c
1
—^e
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
an integrating factor
-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)
-
an integrating factor
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;)
an integrating factor
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
an integrating factor
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+ -
x an integrating factor
=
= -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'
{
an integrating factor
+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
=
^u;
=
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 = -^e 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
differential equation
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
= ^e1 + 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.
The equation becomes
dx
+w+
1
=
uV T
— + u = uV*. dx
or
j
w = u _1
to obtain
dx
w—
e
-3*.
An.
_I so e ,
^ = 2™'.
The equation becomes
~ +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
The equation becomes
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
The equation becomes
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
The equation becomes
= - => 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
The equation becomes
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.
The equation becomes
y
du
c.
x efy. The equation becomes du
—— = y-—
Separating variables
= — 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)\
|
The equation becomes
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
—^i— = 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)
.
Separating variables we obtain
j/
= — In
I
cj.
Eliminating the
x\ 5 /2
24 /
==> u
xa
+ C2.
u — C\X
is
_
^/C j
_
The equation becomes
_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
The equation becomes
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'
The equation becomes
y".
,,,,,, In u = In +c
—U = -
1
=
xu'
,
u=
a;
Separating variables we obtain
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
The equation becomes
y".
—(tana;) dx
it
=
25. Let u
=
y' so that u'
=
u
Separating variables
0.
—
du
The equation becomes w—
.
=*
cicosi
I-
=
2yu
y
=
c\
ainx
we
obtain
+ C2-
Separating variables
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
_
Separating variables we obtain
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".
The equation becomes
/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
,
Separating variables
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
+
differential equation
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
Chapter 2 Review Exercises
10.
The
^ + 6x +
differential equation
^
„
.
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-le^u]
If 3,(0)
so
1,
u 3 we obtain
also exact.)
= xdy + y dx. The differential
that du
du -
An
1
w =
Using
Separating variables we obtain
e"du If 3/(0}
+
Bernoulli.
-^r + r^-T
=> ™ =
-9* a
du — =x+ dy
that
« 6x
is
is
1
differential equation
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
14. Write the equation in the form
y
y
c
is
is
(ar
\i
+ 41
,
so
= \ + c (x 2 + 4)
e~ 2y so ,
dy
dy 15.
The
[ *
e
-2
"x1
=
ye"
differential equation
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) '
.
Chapter 2 Review Exercises
16. Writing the differential equation in the form y
=
with /(f)
(t
+
l)
A
2 .
family of solutions
solution of the initial-value problem
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
Separating variables
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'
The equation becomes
?/'.
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
so that the differential equation of the orthogonal family
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
so that the differential equation of the orthogonal family
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.
=
so that the differential equation of the orthogonal family
x
Then ydy 2.
3.1
—
=
o^x.
so that the differential equation of the orthogonal family
is
+ ciso that the differential equation of the orthogonal family 2
and 2y \ay
— y 2 = —2x 2 + ci.
so that the differential equation of the orthogonal family
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.
so that the differential equation of the orthogonal family
— — 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?.
so that the differential equation of the orthogonal family
— 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'
so that the differential equation of the orthogonal family
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 .
so that the differential equation of the orthogonal family
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.
so that the differential equation of the orthogonal family
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
.
so that the differential equation of the orthogonal
'
2- "
j/
^=
x5 / 3
+ c2
.
—
x.
An
x e y
=
2e
1
+ 3e~ x
x2
=
'a
+ c2
.
so that the differential equation of the orthogonal
+ y-1
-xe 1 =^
3cix we obtain y'
2 In cosx\
orthogonal family
so that the differential equation of the 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
.
.
= —_
so that the differential equation of the orthogonal
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
.
so that the differential equation of the orthogonal
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
so that the differential equation of the orthogonal
Ifien
-z
l
+
cos0
dr 1
-
r— dr
20.
.
.
.
,
=
tan20 so that the
differential
equation of the orthogonal
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#"
so that the differential equation of the orthogonal family
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
so that the differential equation of the orthogonal family
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 — =>
l^u
±
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/3±u?/V3 — —=
,
+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
20. Separating variables
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 = e^e - ™
-
" 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= e^e - ™"".
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
^ot
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
^
^i
— 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
Chapter 3 Review Exercises
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
~ cos^o).
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
.
Chapter 3 Review Exercises 1.
+ Cij =
From y(x 3 family
— —x 2 y 2
y'
so that the differential equation of the orthogonal
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
equation of the orthogonal
+
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)
so that the differential equation of the orthogonal
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
.
Chapter 3 Review Exercises
,
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^/'
Separating variables we obtain
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.
Separating variables we obtain
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
+ c^e -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)
= ^e 1 — ^e
=
+ 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\
the functions are linearly dependent.
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.
the functions are linearly dependent.
+ 3) =
+x
0.
The
Thus we require
+ c 2 sin 5A =
/
1.
2.
0.
y
is
=
the functions are linearly dependent.
— x
^
the family of solutions
the functions are linearly dependent.
+
(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-
the functions are linearly dependent.
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
functions satisfy the differential equation
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
functions satisfy the differential equation
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
functions satisfy the differential equation and are linearly independent since
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
functions satisfy the differential equation and are linearly independent since
-co < x <
The
+ C2 sinh 2x.
cosh 2x
cos 2x, e
y 37.
2
functions satisfy the differential equation and are linearly independent since
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
functions satisfy the differential equation
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.
which has the integrating factor
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 + Je^u 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
12. Identifying P(x)
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'
which has the integrating factor
-/(2/,)
13. Identifying P(x)
+
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
16. Identifying P(x)
is
y%
=
= -2x/
+ x 4-
x2
2.
- x 2 ) we
(l
have
B = / c -/-^Mi--*}^ = | e -Ki-A
second solution
17. Identifying P(x)
is j/2
=
j
= — 1/x
— 1
.
=
A
second solution
18. Identifying P(x)
e
=
= — 3/x
/ j i"1
^
A
second solution
19. Identifying P{x)
=
4x/(l
y2
A
second solution
20. Identifying P(x)
V2
=
=
is
^
+
+
A
second solution
21. Identifying P{x)
is
J yi
/
J
- / ~3dx/x
x^o^x)
J
x a x^snrfmx) .
—
x)
dX
=
x3
^^iiWH 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
= xsai^ax)
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
22. Identifying P{x)
=
is j/2
=
xln|x].
we have B -J0
A
second solution
23. Identifying P(x)
=
^2
is
x5
= — 5/x
.
we have
-5 — /(lnx) —
fx— 5
(ir/l
x
A
second solution
24. Identifying P(x)
?/2
=
=
cos(ln x)
6
= x3
y%
is
dx
r^-
st;
2
=
In
second solution
25. Identifying P(x)
/ -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
27. Identifying P(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
29. Identifying P(x)
=
is 1/2
x
second solution
=
is j/2
=
-(2
have
= j e' i =
30. Identifying P(x)
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.
33. Identifying P(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
34. Identifying P(x)
+ 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= c^e*
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 v^a:/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 =
= -^e 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^+^e
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'"
the differential equation
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
differential equation is
+ c2 e _4l sinx + c^e1*.
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"
Substituting into the differential equation
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
Substituting into the differential equation
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
Substituting into the differential equation
.
+ 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
Substituting into the differential equation
.
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
differential equation
+
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?
differential equation
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
Substituting into the
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
—
,
differential equation
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\
differential equation
Prom
we obtain
Thus
- l2a;V + ^e 5 *.
.
Substituting into the
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
+
^e
\
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
differential equation
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
=
to the differential equation
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
to the differential equation
=
+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
to the differential equation
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
to the differential equation
+ 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.
and the general solution
,
x
D=
4
x
3
2
+ C2xe + x + 6x +
is
22x
+ 32.
we obtain
to the differential equation
-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
to the differential equation
+ c^e~ 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:
Substituting yp into the differential equation yields
A—
1/10.
The y
11. Applying
=
1)
general solution
—
e~
x
(c\
cos a;
to the differential equation
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$e x + c4 _
96
=
5c
61 .
Equating
Exercises 4.6
and
j/p
= Ae x +
Equating
B. Substituting yv into the differential equation yields
coefficients gives
A = —1
=
y
D 2 {D + 2)
12. Applying
B=
and
c\e
3x
The
3.
general solution
— 4Ae x — 3B =
1
4c
—
9.
is
+ c2 e~ x -e x + 3. we obtain
to the differential equation
D 2 (D + 2)(D 2 + 6D + &)y = D 2 (D + 2) 2 (D + 4)y =
0.
Then y
=
cie~
+ c2 e~ ix + c$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
to the differential equation
(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
to the differential equation
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
into the differential equation yields
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
to the differential equation
(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
to the differential equation
(D -
2
1}
{D 2
+ 3D -
is
- —xe 4* +
4x .
we obtain
= D 2 (D -
10)?,
^e
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
to the differential equation
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
+ c^e~ x + -x 3 e x - -x 2 e x + 6
to the differential equation
(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
to the differential equation
{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
-28/225, and the general solution
=
cie~
x
is
^ + C2xe~*ft - -^e 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-
into the differential equation yields
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
to the differential equation
(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$x cos x + c%x sin x
a:
a;
a:
Vc
and
i/p
=
(B Equating
+ Bs'mx + Cxcos x + Dxs'mx.
Substituting yp into the differential equation yields
+ C + 2D)cosx + Dxcosx + {-A -2C + D)s'mx - Cxsinx =
xsinx.
coefficients gives
B + C + 2D = O= -A - 2C + D =
-C= Then
j4
=
2,
5=
1,
C=
-1, and
ci
24. Writing cos 2 x
=
h(1
+ cos2:r)
D = 0,
— v/3
cos
2:
+
1.
and the general solution
V3 —
C2 sin
and applying
2:
D(D 2 + 4)
+
2 cos
x
is
+ sin x —
x cos x.
to the differential equation
D{D 2 + 4)(D S + 4) = D(D 2 +
2
4)
=
we obtain
0.
Then y
and
j/p
=
-4i cos 2x
-4Aain2x and
C=
=
ci
C32 cos 2x
+
C4X sin 2x
1/8.
+ C5
+ Bx sin 2x + C. Substituting yp into the differential equation yields 5 + 5 cos2x. Equating coefficients gives A = 0, B = 1/8,
+ 4£?cos2i + 4C = The general
solution
j/
25. Applying
+ C2 sin 2z +
cos 2z
=
ci
is
cos2i
+ C2 sm2x + ^1 sin 2a: +
D 3 to the differential equation we
8
obtain
D S (D A + 8D = D b (D + 8) = 0. 2
)
Then y
~
c\
-\-
cix
+
C3e
81
+ c4 x 2 + c 5 x 3 + c$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=
-1/16, and the general solution
we obtain
to the differential equation
(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$e~ x + <%xe x + 31
j/
2 x
c?x e
Vc
and yp
= A + Be~ x + Cxe x + Dx 2 e x
.
Substituting yp into the differential equatioi
Wxe x + {2C + AD)e x - 4Be~ x -A = xe*~ e~ x + 7. Equating
coefficients gives
4£>
=
1
2C + 4D =
-4B = -1
Then
A=
-7,
B=
1/4,
C=
-1/2, and
=
cie*
+ C2Cosi + cgsinx -
y
27. Applying
D 2 (D — 1)
£>
=
1/4,
to the differential equation
D 2 (D -
1)(Z>
3
and the general solution 7
+ ^e" 1 - ^xe x +
^
1)
= D 2 (D -
l)
4
=
Then y
=
cie
x
2 I e
we obtain
- 3D 2 + 3D -
+ C2xe x + c$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/
into the differential equation yields
Equating
coefficients gives
-A + 3B =
16
~B =
-1
6C = Then 4 = -13,
B—
C=
and
1,
y
28. Writing (e*
=
and the general solution
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)
to the differential equation
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$xe 2x +
cee'
2)
{D
+
1)(2D
-
1}
=
0.
Then
=
y
x
cit~
2*
Vc
21
= A + Bxe + Ce~ 2A + 9Be 2x - 2QCe~ 2x = 2 + and yp
and
C=
2x
e
Substituting yp into the differential equation yields 2x _2x Equating coefficients gives = = e
+
=
D(D —
1) to
cie~
x
A
.
—1/20, and the general solution
y 29. Applying
.
+ c2 e 2x + cze x/2 + 1 +
4 1)(D -
B
1/9,
is
the differential equation
D(D -
1,
^xe
2x
-
^
e
~ 2 "-
we obtain
2D 3 + D 2 = D 3 (D - if =
0.
)
Then y
and yp
= Ax 2 + Bx 2 e x
Equating
.
coefficients gives
D 3 (D — 2)
ci
,
+ c3£ x + c^e 1 +
+
2
C5X
+ C6i 2 e I
Substituting yp into the differential equation yields
A= y
30. Applying
=
=
1/2 and
ci
+
c2 x
B—
1/2.
The
general solution
+ C3e x + e^are* + ^x 2 4- iUV
is
1
to the differential equation
we obtain
D 3 {D - 2){D A - AD = D h (D - 2} 2
)
.
2
(D
+ 2) =
0.
2A + 2Be x =
1
+
3
e
.
.
Exercises 4.6
Then
=
y
+ c2 x + ae 2* + f^e"^ + c 5 x 2 + cex 3 +
ci
c7 x
4
+ cgxe 2*
Vc
4
21
= Ax + Bx + Cx + Die Substituting yp 2 (-SA + 24C) 24Bx 4SCx + IQDe 2* = 5z 2 - e 2x 2
and yp
3
into the differential equation yields
.
'.
Equating coefficients gives
-8A + 24C = -24B = -48C -
160= Then
A=
-5/16,
B = 0, C = =
S 31. Applying (2D
-
1)
-5/48, and
O=
to the differential equation
(2D - 1)(16D 4 -
1)
-1.
-1/16, and the general solution
+ c2 x + c$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
.
Substituting yp into the differential equation yields
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%* + c 4 e~ 2x -
^ 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
Substituting yp into the differential equation
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
c^e~ 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 + ^e 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\ + c^e 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
+
c^e
1 .
Using
D 2 {D -
1) to annihilate
Substituting yp into the differential equation
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 - ^e~ 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
^e 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 -
^e
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| — ^e 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\
= -^e 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
+ c^e
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
29. Write the equation in the form
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
30. Write the equation in the form
+ 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 +
31. Write the equation in the form
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.
Substituting into the
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
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
12. Identifying P(x)
—
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
=
Tj^e
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
to the differential equation
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
Substituting yp into the differential equation yields
.
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)
to the differential equation
we obtain (D -
1)
3
Then y
=
cie
1
c^xe?
4-
+
2 I
c3 x e
+
3
c4 x e
x
+ cs^e 1
yc
and
j/
p
= Ax
2
e
x
+ Bx
3 x
+ Cx
e
A x
e
.
Substituting yp into the differential equation yields
\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—
+ c^e 1 +
to the differential equation
D{D 2 +
l)(D 3
1/12.
iV.
The general
^^e 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.
Chapter 4 Review Exercises
and yp
= Ax + B cosx + C sinx. 6,4
+ (5B + 5C) cos x + (-5B + 5C) sinx = A=
Equating coefficients gives
y 22. Applying
Substituting yp into the differential equation yields
=
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/
into the differential equation
=
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
solution of the initial-value problem
-x -
and
W=
„ 3.
- -=
c2
= — | The
C2
e
2
0,
so yc
=
= -e 2x cos x
.
cje1 sin x
+ cze? cosx
Chapter 4 Review Exercises
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
ca^e
-
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 the general solution
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--^e- 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
~ -^e" 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^^A !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
—
>
=
^e" 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, - -^e~'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-
^e _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.
6.
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
differential equation is
=
100 and 7
=
60.
^g"
is
z aV
z*
V 11.
and
Substituting into the differential equation
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
(
differential equation
-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
Chapter 5 Review Exercises
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 2k t
+
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 .
Chapter 5 Review Exercises
+
15. Writing gi"
§£
=
+ sin 7*
cos yt
w = 64/3. From Example when7 = \/6473 = 8/\/3. 2
xp = A/u 2
16. Clearly
17.
—
j^e
',
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
differential equation
(c)
conditions x(0)
initial
To compute the maximum
=
we obtain x c
^
sin
=
2v^t)
e~ 41
(c\ cos
+ ^e~
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
into the differential equation
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
into the differential equation
+ 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
into the differential equation
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»
differential equation
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~^intj
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
-K^Ini = 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
=
—x^e?
+
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 '
^e*
- \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
+ c^e 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.
i^i-lwe obtain dy
=
dx
dy dt
and 2
_
d y dx 2
d dy dx
Substituting into the differential equation
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\
Substituting into the differential equation
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
converges by the alternating series
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
converges by the alternating series
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
converges by the alternating series
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
^I= ^ 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 +
22. Separating variables
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 +
23. Separating variables
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
24. Separating variables
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
25. Separating variables
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 +
26. Separating variables
=
co [l
+ x + z2 +
i
= E^o
+
•
•]
=
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
— £n^o
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
and so on. Thus, two solutions are
=
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
and so on. Thus, two solutions are
=
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
into the differential equation
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.
Thus, two solutions are
= l+ 3x 2 +x 4 -~x b +
yi
and
y2
5 11. Substituting y
=
2
+ x*.
00
+ 2)y" + 3xy' -y=-£n(n-
lj^i" +
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
into the differential equation
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
differential equation
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"
= £J^o
+ (smx)y =
dii" into the
1,
3
£ n(n - l)^' 2c2
+ 6c3 i +
4
-x 6 + --).
so 1
1
5
7
differential equation
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*
Thus, two solutions are
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
^^(T^A^*'
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
into the differential equation
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.
Regular singular point: x
=
0.
5.
Regular singular points: £
=
0,
±2i.
6.
Irregular singular point:
x
=
5.
Regular singular point: x
=
0.
7.
Regular singular points: x
=
—3,
8.
Regular singular points: x
=
0,
±i.
9.
Irregular singular point:
x
=
0.
Regular singular points: x
10. Irregular singular point:
x
=
—1. Regular singular points: x
3. Irregular singular point:
11. Substituting y
2xy" -y'
= X^o
+ 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
into the differential equation
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
into the differential equation
(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
=
the recurrence relation
+
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
the recurrence relation
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
indicial roots are r
=
1/2 and r Cfc
and For r
c2
=
1
=
the recurrence relation
=
=
1,
2
(2r
-
+ r) ci = ~
2r
so ci
3)
=
+
0.
-^h)> c3
-~co,
=
-
l)(r
=
1/2 the recurrence relation
=
=
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)^o
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.
the recurrence relation
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
=
1/3 the recurrence relation
and
r
+ r)(3fc + 3r -
and
and For r
-i I
0,
which implies
indicial roots are r
360
co^"*
£ P(* +
+
The
I
into the differential equation
(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
indicial roots are r
=
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,
1/3 the recurrence relation
=
£*_i
2/3 the recurrence relation
ft
,
+k
3k 2
~
Cfc
l60 general solution
on
{0,
oo)
is
1/3
17. Substituting y
= E^Lo
+r
Cn%"
into the differential equation
and
collecting terms,
[2(ft
+ r)(ft + r -
we obtain
00
2xy" -
(3
+
2x)y'
+ y=
(2r
2
-
=
and
The
(ft
indicial roots are r
=
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.
the recurrence relation
3) Cfc _!
-
5)
o
fc(2fc
and
r(2r
-
= ^o,
5/2 the recurrence relation
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+
+
equation and collecting terms, we obtain
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
indicial roots are r
=
—2/3 and
r
~~
3k(3k
and
C2
For r
=
-jCo,
2/3 the recurrence relation
and
The
=
C2
general solution on
19- Substituting y 9a:
V
(0,
oo)
a
2/3, so
- 4)
-- 0.
*
,
=
For r
= —2/3
the recurrence relation
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
equation and collecting terms, we obtain
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.
1/3 the recurrence relation
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
into the differential equation
=
-
2
4-
(3*-2)^
"
2/3 the recurrence relation
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
indicial roots are r
= -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 the recurrence relation
1/2.
For r
l)(r
+
Ije*
+ 2c
= -1
<=
2
fc
_i
=
the recurrence relation
= -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=
into the differential equation
-
2
2r
and
\{k
indicial roots are r
-
= —1/2
2
-
-
r
and r
the recurrence relation
=
=
=
1
1)
(2r
-
~, 2k Ck-l
c2
=
'
c^o,
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
into the differential equation
+ 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,
the recurrence relation
and
13.
ci
general solution on
Substituting y
=
xy"
(0,
oo)
Y,%Lo cn x"
+
2y'
+r
-xy=
=0.
l)cfc+1
3/2 the recurrence relation
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
the recurrence relation
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
into the differential equation
=
(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
1/2 the recurrence relation
is
Exercises 6.4
and
c2
=
-jjjco
C3
=
c5
C4
=
=
=
c7
•
•
•
=
C0 ^!
(-1)"
C2n= For T2
= — l/2
the recurrence relation
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
equation and collecting terms, we obtain
-(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
the recurrence relation
- 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.
the recurrence relation
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
+
indicial roots are ri
solution
=
f"2
=
+C2 y2{x).
C\y\{x)
differential
equation and collecting terms, we obtain
+ 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
into the differential equation
+ 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
+
...).
equation and collecting terms, we obtain
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
the recurrence relation
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.
the recurrence relation
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).
into the differential equation
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
equation and collecting terms, we obtain
+ 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
=
into the differential equation
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
the recurrence relation
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
^
into the differential equation
and
collecting terms,
we obtain
00
It follows
x 3 y"
+y=
=
and
that co
cox
r
+
Ck
The only
solution
= En^o
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>
into the differential equation
= ^[{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
into the differential equation
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^^ix).
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
=
1/4 the general solution
y
=
ci ^1^3(63;)
6. Since
i/
7. Since
i/
8. Since 9.
If
j/
=
f
1/16 the general solution
y
is
is
—
ciJi/ 4 (x)
+
+ c^J-i^ix).
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
\
a solution of Bessel's equation)
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)]
a solution of Bessel's equation)
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)
Chapter 6 Review Exercises
=
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.
Chapter 6 Review Exercises
and so on. For
co
=
and
=
ci
1
we obtain c3
=0
Ca
=
1
12
=
C5
C7=
=
fk
and so on. Thus, two solutions are
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
=
.
.
.
.
=
Chapter 6 Review Exercises
and
so on. Thus, two solutions are
and 2
I » = C ( I+
15. Substituting
=
?/
,
2
+
X
»
4
*
+
X
YH
3
into the differential equation
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
so on. Thus, two solutions are
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
=
=
Chapter 6 Review Exercises
obtain C3
=
ce
=
Cg
=
C4
=
c7
=
Cio
C5
=
eg
= cn =
-
•
=
= •
=
..
and so on. Thus, two solutions are
=
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
indicial roots are r
=
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.
the recurrence relation
=
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*
Chapter 6 Review Exercises
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
indicial roots are r
=
=
and r
For r
Ct
=
=
1/2 the recurrence relation
Two
=
=1
>
2 3 >
is
-" 1
= -—c
c3
^co,
b
.
90
is
= ~*(2fc
Cl
*
1
=
c2
-co,
Cfe
the recurrence relation
= ~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
=
Chapter 6 Review Exercises
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
..-).
into the differential equation
+ 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
Chapter 6 Review Exercises
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
indicial roots are ri
-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 + ^e-"' )=-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
<
= J^ismt^+^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}
\
- ^in(} =
{|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
^U
.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 +
^e 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|^e +
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- ^e- 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/'sinr^U^if/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 e^e^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 = ^e" 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
The Laplace transform
s
—
1
= -1 + e'.
of the differential equation
is
3
Solving for
X {y} we obtain „.
1-s 2 2 s {s + 2)
,
^_J +
Thus, 3.
The Laplace transform
we obtain t£{y} =
1
-75
-.
(s
Thus, 4.
+ 4) 2 y
The Laplace transform
11
3
As
2s
As + 2'
=
2
(
is
1
+ 4i?{s} = s
+4
2
+
+ 4'
u+
of the differential equation
2e
_4t
:
.
is
s
Solving for f£{y}
s
2
+
tete~
1
_3 e -a.
l
of the differential equation
*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
of the differential equation
*{y} -
*s(0)
-
2
(s-3) 2
2
+ 22
= --e 3t sin2(.
T/
The Laplace transform
Solving for ^£{y)
39 + 4'
- 6 \s2{y) - 2,(0)] +
y'(0)
Thus,
2
11
1
is
3
s
0.
1 -4t - -e
of the differential equation
s2
7.
+ 42{y] =
y(0)}
3s+l
4 -t
=
j/
%{y) we
is
obtain
The Laplace transform
Solving for
+ sin().
- y'(0) + 5 [sX{y} -
sy(0)
Thus,
s
^(cosf
of the differential equation
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 +
of the differential equation
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.
The Laplace transform s
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
of the differential equation
sy(0)
-
y'(0)
Thus, y
.
of the differential equation
X{y} -
-
_
= j^e 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.
The Laplace transform
^^
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.
The Laplace transform
— ^ 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
The Laplace transform s
2
=
J_
1
16 a
115 00344 +
16
16
of the differential equation
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.
The Laplace transform s
z
^
-
r 4-
^e* cos
of the differential equation
X{y} -
-
53/(0)
-
j/(0)
2
^e' 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
of the differential equation
- 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)
of the differential equation
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.
The Laplace transform
of the differential equation
s*X{y} Solving for 18.
%{y} we
obtain
The Laplace transform s
Solving for
4
%{y) we
- *V(0) -
3
s y(0)
is
g
%{y) =
of the differential equation
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.
The Laplace transform
%{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.
The Laplace transform
= 5%t-l)-5e-^°U{t-l).
of the differential equation
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.
The Laplace transform
= l- e -'-2[l-e-('-
of the differential equation
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.
The Laplace transform
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.
=
The Laplace transform
~ cos 2t ^
sin It
- r i-icos2(i-l) «(t-l).
of the differential equation
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.
The Laplace transform s
Solving for
2
%{y} we
=
coe 2f
+
\
Bin(f
-
2jt)
-
of the differential equation
^{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.
The Laplace transform
2
-
6
1)
+
-
2(
e
is
g-jre 2
31
3
of the differential equation
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
The Laplace transform s
2
Solving for
2{y) -
-
-
[1
of the differential equation
-
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
The Laplace transform
+ 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{/} =
The Laplace transform
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)
The Laplace transform
2
+
|*
l
.
e*
of the given equation
5%/5s 2
sin \/5
t.
4
1
te *
|
1
(s
-
l)
2
+
(S
Thus,
11 l)
3
8 a
is
+
1)
=
/(*)
The Laplace transform
3
5
4e~'
of the given equation
+
(s
1
%{f} we
obtain
The Laplace transform
#{/} =
.
S
+2 {sinf},
l)
2+
%
(
is
=
Thus, /(()
if{l}.
=
e"'.
1
of the given equation
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)
of the given equation
,2
^
2
=
/(()
The Laplace transform
Solving for
s
is
if{cosi}
+if{e-
we obtain
254
t
}i!?{/}.
(s
+
1)
3
'
+
1
Exercises 7.5
Thus, 35.
=
f{t)
The Laplace transform
+ sint.
cost
of the given equation is
= #{1} +%{t} + Solving for
X {/} we obtain s^s +
Thus,
l)
-
16
/(f)
=
s
4
8s-2
The Laplace transform
Solving for if {/}
a
2
^
f1
~
f{t)
The Laplace transform
of the given equation
+
\ 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
The Laplace transform
=
^
2
fl*
sin*
of the given equation
—
^
,
+ (
l
1
2s
2 (s 2
+
Solving for if{/}
we obtain %{y)
=
—+
7
(s
Prom equation
(3) in
0.005
The Laplace transform
—+
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
of the given equation
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
*»-^iU-.-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
differential equation
is
10^ +
IO5
=
-
30e'
30e'^/(f
-
1.5).
OX
The Laplace transform
of this equation
is
3e
3
Solving for if {q}
ls
_ liBa
we obtain
^-(-'i)-^T + i^"(^^)--Thus,
43.
The
differential equation
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
differential equation
=
—
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)]
differential equation is
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
The Laplace transform
(<-|) + *. (« -
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,
^i = 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)
- ^e" 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
^A 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
differential equation
The Laplace transform
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
The Laplace transform
=
sinf
-
sm(i
of the differential equation
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
of the differential equation
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
The Laplace transform
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£
^
The Laplace transform
0.
of the differential equation
\
-
+ 24 _ 2 + 9)2-3
4s
~
{s 2
2(3)a
2(3)
12
+
3
(77^5
27
Thus,
55.
The
=
2
-ts'mZt
differential equation
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
^^i
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
differential equation
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
The Laplace transform
c\
= wqL s /24EI
and
of the differential equation
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.
The Laplace transform
of the differential equation
~
\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)
The Laplace transform
.
of the differential equation
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.
The Laplace transform
of the differential equation
is
of the differential equation
is
so that
2.
The Laplace transform
p~ s
2
so that
S 3.
The Laplace transform
=
2e-
1
+
e-<
i_1
of the differential equation
^(t-l).
is
" ?TT( + e
*<»> =
1
a
")
so that
=
i/
4.
The Laplace transform
sinf
+
sini^((-27r).
of the differential equation is
*<»>so that 3/
5.
The Laplace transform
=
^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
The Laplace transform
(
S
i
of the differential equation
is
so that
y
=
cost
+ sint^fi. 268
2?r)
+*U(t
-
4w)].
(
Exercises 7.6
7.
The Laplace transform
of the differential equation
2 i
is
+ 2s
11
11
2 s
2 s
+
(l
2}
+
O
so that
_
1 8.
The Laplace transform s
2
i e -2£*-i) /{(-
of the differential equation
-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.
The Laplace transform
O
Of
J.
-I]*(t-2).
of the differential equation
is
„-2tts
so that
=
!/
10.
The Laplace transform
e
-2(t-2,)
sini 9/(f-27r).
of the differential equation
%{y} =
is
+
{s
1)=
so that
y 11.
The Laplace transform
=
(t-l)e- {t
of the differential equation
+ 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.
The Laplace transform
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.
The Laplace transform
l
e
6(t-4)
m{t
-
4).
5
of the differential equation
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
^iv"(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.
The Laplace transform s
2
of the differential equation
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(
-
The Laplace transform
e~
(t
~ 3w)
sin((
- 3ir)%t -
3jt)
of the differential equation
=
(
e" cost
is
* <»> = i so that
y
Note that 19.
y'{0)
=
— — eiiiwt. w
1.
The Laplace transform
of the differential equation
is
so that
Li
Note that 20.
i{0)
=
1/L
^
The Laplace transform
0.
of the differential equation
s
+
is
s
5
so that
V
=
S(t)
-
5e" 5t
.
+5
+
e
- (t - 3,,)
sin t°lt(t
-
3tt).
Chapter 7 Review Exercises
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=-^-^e- 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 /
Chapter 7 Review Exercises
18
= 20
.
cos7r(t
-
-
+sin7r{(
1)
-
-
1)
= H-^(t-^1 = ^^^t ^{ht^I 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)
Chapter 7 Review Exercises
-
(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
Chapter 7 Review Exercises
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
^o
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 -
^e 1
ZJ 2 y
,
8.
.
= jP 4 a:
\e l
4)x
=
,
and
Then
.
ci cos 2t 4- c 2 ain 2t
+ c^e 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
(
+
^ic
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
—sinv^A + 6
/
^-cosv^f + ^sin v^ij +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
Taking the Laplace transform of the system gives
{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
Taking the Laplace transform of the system gives
+ \)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.
Taking the Laplace transform of the system gives
+ 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
Taking the Laplace transform of the system 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.
Taking the Laplace transform of the system gives
-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 + ^i 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 = | - ^e- 100*cosh50\/2t -
law we write
Kirchoff's second law
=
ii
12
+ ^3.
Since
E{t) E(t)
=
=
ii
and summing the voltage drops
^e-
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.
bOe^e-^^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
into the system yields
from which we obtain ai
—
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
Corresponding eigenvectors are
|_M
=
K,
4.
K 2 =(|
and
Thus
Substituting
into the system yields
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
=
Corresponding eigenvectors are
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
into the system yields
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.
Corresponding eigenvectors are
A
+
21
-
(A
-
3)(A
7)
=
A
3 and A2
K '=L
Ka=
and 3)
(g
Thus
xe =c 1(
e
_;)
-+
ri^.
c2
Substituting
into the system yields
+ ^1 =
3«i 9ai
from which we obtain ai
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.
into the system yields
— 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
=
Corresponding eigenvectors are
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/
into the system yields
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)
=
Corresponding eigenvectors are
Thus
Xc =
-
e'
+ C2
339
„2i
8.
7 Exercises
8.
Substituting
into the system yields
-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^_ ^e + 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}
Corresponding eigenvectors are
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.
Corresponding eigenvectors are
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
=
Corresponding eigenvectors are
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*
)•
Chapter 8 Review Exercises 1.
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.
Chapter 8 Review Exercises
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.
Chapter 8 Review Exercises
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
^e 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
.
'
Chapter 8 Review Exercises
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
^WJ^U
-(\-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=
^
Chapter 8 Review Exercises
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
Chapter 8 Review Exercises
'
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
differential equation
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
differential equation
.
— 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)
of the differential equation
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 . U^UU 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
11. Write the equation in the form
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
an integrating factor
=
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
initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector
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
initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector
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
an increasing function,
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
= COOl^e 2 ". 6 e '
2' 2
,1
=
'
=
e
'
2
<
for
c
<
0.1.
Thus an upper bound
0.001628.
1.221403, the actual error
is
y(0.1)
-
yl
=
0.001403 which
is less
0.001628.
(d)
h
the case.
is
see that the local truncation error
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
halved we expect the error for h
Comparing 0.000378 with 0.001403 we 6.
=
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
Comparing 0.000378 with 0.001403 we 7.
(b)
=
Using j/ '(x)
32e
2ir
we
32e
)
is
an increasing function,
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
see that the local truncation error
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
=
halved we expect the error for h
is
With
0.000378.
<
for
2c .
<
c
0.1.
Thus an upper bound
0.000003257.
1.221402758, the actual error
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
=
0.8234, the actual error
(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.
see that the local truncation error
10e
0.05 to be one-sixteenth the
0.8.
Using the improved Euler method we obtain y(0.1)
(b) Using y'"
=
global truncation error
0.025.
halved we expect the error
is
=
yi
-acM! =
a decreasing function, e
=
~
see that the local truncation error
Comparing 0.0109 with 0.0234 we 9.
j/(0.1)
With
0.000000187.
Comparing 0.000000187 with 0.000002758 we
the local truncation error
step size
we obtain
halved we expect the error for h
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~
the local truncation error
is
2c
0.001667(1)
<
=
e°
=
1
for
0.001667.
388
<
c
<
0.1.
Thus an upper bound
for
Exercises 9.6
0.823413, the actual error
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.
halved we expect the error for h
Comparing 0.000305 with 0.001587 we
y'"
Using
= — 10e
we
0.001587, which
—
With
y2
is less
=
than 0.001667.
0.823781.
global truncation error
0(h 2 ),
0.05 to be one-fourth the error
=; y\
see that the local truncation error
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 <
the local truncation error
0.001667(1)
0.823413, the actual error
=
Since y(0.1)
(d)
Using the three-term Taylor method with h
(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
Comparing 0.000305 with 0.001587 we
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;
see that the local truncation error
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
0.823413441, the actual error
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.
halved we expect the error for h
—
With
global truncation error
0.05 to be one-sixteenth the
Comparing 0.000000185 with 0.000003225 we see that
_3 ^ I_1 '
we "
y
see that the local truncation error
{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
see that the local truncation error
h3
(b) Since
0.1
0.19.
1.9424.
(d) Since y(l.5) #0.05
=
=
-3 * 1-1
'
is
= IHe" 3 ^"
a decreasing function
^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
—
=
2.053216, the error for h
With
0.005950.
=
have Uo.l/^o.05
0.1
=
0.1 is
global truncation error
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
see that the local truncation error
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)
Using the three-term Taylor method with h
we obtain
y(\.$)
(d) Since y(1.5) £fl,05
=
=
w
0.005950.
=
0.1
0.019.
we obtain
y(1.5)
w 2.080108.
With h
=
0.05
2.059166.
2.053216, the error for h
have .Eb.i/£b.05
=
=
With
=
0.1 is
global truncation error
4 -52.
390
Eq A
0(h
=
0.026892, while the error for
2
we expect Eq.\/Eq.q$
)
fa 4.
h
We
=
0.05
is
actually
Exercises 9.6
15. (a) Using j/
5
3<- x
= -1026e
'
we
see that the local truncation error
w 120
y
3 (b) Since e~ <
1_1
a decreasing function
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.
0.000122595, while the error for
global truncation error
0(h 4 ) we expect
£b.l/£fl.0S
«
18-14.
+ If
we
see that the local truncation error
2
2"
0.5,
-,
(c+1) is
(x+l)
a decreasing function
for
is
h2
1
(b) Since
^
1
=-
Using y"
With
Eo.i/Eom =
actually have
-3
2.053222989.
3/(1.5) f»
2.053216232, the error for h
=
>.
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
see that the local truncation error
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 )
0.000184, while the error for
we expect Eo.i/Eom
=
ft
We
4.
0.05
is
actually
we
see that the local truncation error
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.
0.405465, the error for
0.000046.
=
ft
ic)
a decreasing function
<
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
=
=
0.405465, the error for
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
global truncation error
Eb.i
0(h
=
0.000823, while the error for
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
a decreasing function
'
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.
0.405465108, the error for h
=
0.000002.
5
With
=
=
.i
=
0.000000060, while the error for
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
the finite difference
+ 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
solution of the corresponding linear system gives
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.
=
solution of the corresponding linear system gives
equation
The
=
+ ^1 =0.
is
0.0
4.
=
0.75
0. 50
1.2w + i
The
finite difference
solution of the corresponding linear system gives
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
solution of the corresponding linear system gives
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
solution of the corresponding linear system gives
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*.
solution of the corresponding linear system gives
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
the finite difference
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)
solution of the corresponding linear system gives
0.0000
We
= l-x,
.
is
[1
10.
(2
1.0000 1.0000
1
9.
=
ft
Y
difference equation
8.
and
,
solution of the corresponding linear system gives
X
7.
Ay/x
is
1.4167jfc+i
The
=
f(x)
0,
+
0.05xi)yi+i
-
l.Wtf,
+
(1
-
solution of the corresponding linear system gives
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.
solution of the corresponding linear system gives
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
+
^U
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-^U_
=0.
1
solution of the corresponding linear system gives
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 . .
Chapter 9 Review Exercises
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
Chapter 9 Review Exercises
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-
Chapter 9 Review Exercises
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
Chapter 9 Review Exercises
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
initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector
=
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
Chapter 9 Review Exercises
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.
solution of the corresponding linear system gives
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.