Engineering Mathematics II
15MAT21
QUESTION PAPER SOLUTION MODULE - 1 DIFFERENTIAL DIFFERENTIAL EQUATIONS – EQUATIONS – I I 1) Solve 4
d4y dx
d3y
4
4
dx
23
3
d2y dx
2
12
dy
36
dx
0 (July 2015)
Sol Sol : The The given given equa equati tion on is writ writte ten n as 4 D 4
4D3
23D 2 12D 36
A.E is 4m 4
4m 3
23m 2 12m 36
solving we get m yc
d3y
2) Solve
dx
6
3
d2y dx
2
0
2, 2, 2, 3 / 2, 3 / 2
c2 x e 2 x
c1
0
11
dy dx
c3
c4 x e 3/ 2
6y
e x 1
July 2015
Sol : Th The given equation is written as as D3 A.E is m3 m3
m2
6m 2 11m 6 5m2
m2 m 1 m2 m
1,
yc
c1e
y p
5m 6 m 6
0
2, x
3 c2 e
2x
c3 e
3x
ex
6 D 2 11 11D 6
e x
1
24
6
GS is y
0
0, m 1 0
e x 1 D3
0
0
5m( m 1) 6( m 1)
5m 6
6 D2 11D 6 y
c1e
x
c2 e
2x
D3
c3 e
Department of Mathematics, Mathematics, SJBIT
e0 x .1
6 D2 11 11D 6
3x
e x
1
24
6
D3
6 D 2 11 D 6
Page 1
Engineering Mathematics II 3) Solve 2
d2y dx
2
2
dy dx
15MAT21
e x sin x
July 2015
Sol : Th The given equation is writtenas nas 2D 2 2D y e x sin x A.E is m 2 2m 0 m(m 2) 0 m 0,2 0, 2 yc
c1 c2e 2 x
let PI as y e x ( A cos x b sin x ) Differentiating Differentiating w.r .t x y ' e x ( A sin x b cos x ) e x (A cos x b sin x ) y"
e x
A B cos x ( A B ) sin x )} e x {( A B ) sin x (A B ) cos x
Sustituti Sustituting ng these these values valuesin inthe the given equati equations ons and solving solving we get A 0, B
G S is y y c 4) Solve : y
D 2
AE : m2 yc y p
12 y
e2 x
c2 e
0
x
x
e
e2 x m
D2
3 sin sin 2 x
2,
6
4 D 12
4 D 4
e2 x
3 D 4 sin 2 x
8
4 D 2 16
yc
Jan 2016
3sin2 x
3 2 co cos x 4 ssiin 2 x
8 General solution y
sin x
3sin2 x
D 4
xe2 x
2
6x
4 D 12 2 x
1
3 sin 2 x
e2 x D 2
sin x
4 D 12 y
4m 12
c1e2 x
2
c1 c 2e 2 x e x
yp
4y
So Soll : Wehave Wehave
1
y p e x
1/ 2
yp
80
y
c1e
2 x
Department of Mathematics, Mathematics, SJBIT
c2e
6x
xe2 x
3 2 cco os x 4 ssiin 2 x
8
80 Page 2
Engineering Mathematics II 3) Solve 2
d2y dx
2
2
dy dx
15MAT21
e x sin x
July 2015
Sol : Th The given equation is writtenas nas 2D 2 2D y e x sin x A.E is m 2 2m 0 m(m 2) 0 m 0,2 0, 2 yc
c1 c2e 2 x
let PI as y e x ( A cos x b sin x ) Differentiating Differentiating w.r .t x y ' e x ( A sin x b cos x ) e x (A cos x b sin x ) y"
e x
A B cos x ( A B ) sin x )} e x {( A B ) sin x (A B ) cos x
Sustituti Sustituting ng these these values valuesin inthe the given equati equations ons and solving solving we get A 0, B
G S is y y c 4) Solve : y
D 2
AE : m2 yc y p
12 y
e2 x
c2 e
0
x
x
e
e2 x m
D2
3 sin sin 2 x
2,
6
4 D 12
4 D 4
e2 x
3 D 4 sin 2 x
8
4 D 2 16
yc
Jan 2016
3sin2 x
3 2 co cos x 4 ssiin 2 x
8 General solution y
sin x
3sin2 x
D 4
xe2 x
2
6x
4 D 12 2 x
1
3 sin 2 x
e2 x D 2
sin x
4 D 12 y
4m 12
c1e2 x
2
c1 c 2e 2 x e x
yp
4y
So Soll : Wehave Wehave
1
y p e x
1/ 2
yp
80
y
c1e
2 x
Department of Mathematics, Mathematics, SJBIT
c2e
6x
xe2 x
3 2 cco os x 4 ssiin 2 x
8
80 Page 2
Engineering Mathematics II
15MAT21
5) By the method of undetermined coefficients solve y
Sol:
we have D 2 1 y
2 cos x
(Jan 2016)
2 cos x
AE : m 2 1 0 yc
y
m
i
c1 cos x c2 sin sin x
Assume Assume y p y p
x
y p x
x A cos x B sin sin x A sin sin x B cos x
A cos x B sin sin x
A cos x B sin sin x 2 A sin sin x 2 B cos x
substituting in given equation and comparing coefficients we get B 1 and A y p
0
x sin sin x
G.S is given by y
c1 cos x c2 sin sin x x sin sin x
6) By the method of variation of parameters solve
Sol:
D 2
4 y
AE : m 2 yc
4 y
tan 2 x
(Jan 2016)
tan 2 x
4
c1 cos 2 x
y
y
0
m
2i
c2 si sin n 2 x
A cos 2 x B si sin n 2 x W A B
cos 2 x 2 si sin n 2 x
4
si sin n 2 x
dx
2 cos 2 x tan 2 x
si sin n 2 x
2
2 cos 2 x
si sin n 2 x tan 2 x
2 G.S is given by
y
sin 2 x
4 cos 2 x
dx
4
log sec 2 x
tan 2 x
4
Department of Mathematics, Mathematics, SJBIT
lo log g sec 2 x
tan 2 x
k 1
4 k 2
k 1 cos 2 x
cos 2 x 4
k 2 si sin n 2 x
Page 3
Engineering Mathematics II 7) Solve ( D 4
m4 ) y
0
2
solution : A.E D 2
( Jan 2016) 2
m2
0
( D 2
m2 )2
2 D 2m 2
D 2
m2
2Dm D 2
2a
D
2ai 2
a
a
0 m2
y
2
e
c1 cos
8) Solve ( D 4
2
2ai
a
x c2 sin
m4 ) y
Sol : A.E D 2
2
a
2
2
m2
a
m2
2a
, D
y=e
2
c1cos
9) Solve
( D 2
a 2
a
7 D 12) y
Sol: We have ( D
AE : m2 yc
c1e
2
3 x
c2e
a
2Dm
a
2
2
a 2
x
0
i
ax
x +e
2
c1 cos
cosh x
7 D 12) y
7m 12
2
x c 2 sin
2
i
x+c 2sin
a
2ai
2 2 2 Thegeneral solution is given by ax
c1 cos
0
2Dm D 2
2ai
a
2
e
0
D 2
2
x
( Jan 2016)
m 2 ) 2 2 D 2m 2
2a
ax
a
( D 2
D
i
2
0
m2
0
2
i
a
2Dm
2a
, D
2 2 2 The general solutionis givenby ax
15MAT21
0
a 2
x c 2 sin
a 2
x
(Jan 2016)
e x
e
x
2
m
3,
4
4x
Department of Mathematics, SJBIT
Page 4
Engineering Mathematics II y p
1
e x
2 D 2
7 D 12
1 e x
e
2 20
15MAT21 x
e D2
7 D 12
x
6
General solution y
yc
y
yp 3 x
c1e
c2 e
1 e x
4x
e
2 20
x
6
10) By the method of variation of parameters solve Sol : D 2 1 y
x sin x
AE : m2 1 0
m
yc
c1 cos x c2 sin x
y
A cos x W A B
y
y x sin x
(Jan 2016)
i
B sin x
cos x
sin x
sin x
cos x
x sin 2 x 2
1 1 x2
dx
2
x sin x
cos 2 x
2
4
x cos 2 x
sin 2 x
4
8
2
x sin x cos xdx
k 1
k 2
G.S is givenby y
11)
1 x 2 2
x sin x
cos 2 x
2
4
2
Solve (4 D4
8 D3
k 1 cos x
7 D2 11D 6) y
4 3 Solution: The AE is 4m 8m
x cos 2 x
sin 2 x
4
8
0
7m 2 11m 6
k2 sin x
(June 2015)
0
By inspection method m=-1 and m=2 are two roots then we get 4m2 y
c1e
x
c2 e 2 x
4m 3 c3e
Department of Mathematics, SJBIT
1 x 2
0 c1e
m
1 3 , 2 2
3 x 2
Page 5
Engineering Mathematics II 2 12) Solve ( D
x2
4) y
2 Solution: The AE is m
yc
4
1 D 1 4 1
1
m
2i
1 D
2
1
D 2
x
4
4
4
4
x
2
......... x
.2 0
e x
1
5
4 1
x
4
5 1
2
2
x
e
2
x
5
1
e x
2
5
2m
2
0
(June 2015)
m 1 i
e x c1 cos x c2 sin x
yc
e x cos x u u
e x cos x
v x
v
e
vf ( x) w
e x sin x
v
sin x
uf ( x)
e
x
e x sin x.e x tan x
dx
dx
e x sin x
cos x e 2 x
sec x cos x dx B
x
e
2
e x tan x using method of var iation of parameters.
2 D 2) y
2 Solution: The AE is m
x
5
D2
1
e
2
D 2
2
4
e
c1 cos 2x c2 sin 2x
2 13) Solve ( D
A
(June 2015)
0
x2
4
1
x
4 y
2
1
4
w
x
c1 cos 2x c2 sin 2x
y p
u
e
15MAT21
2 x
w e Hence general slolution is e x A cos x
y
e x c1 cos x c2 sin x
dx
cos x sin 2 x
dx
cos x
sin x log sec x
e x cos x.e x tan x
y
sin x
e 2 x dx
tan x
sin xdx
k 1 cos x k 2
B sin x
Department of Mathematics, SJBIT
e x cos x log sec x
tan x
Page 6
Engineering Mathematics II 14) Solve ( D
3
2e x
D) y 3
Solution: Given ( D
m3 m( m 3
4cos x 2e x
D) y
A.E is ( D3 m 0
0, m 2
m
0, 1,1
c1eox
C.F
c1
1
x
c2 e
m
0 x
c2 e
2e x
P.I
0
0
m C.F
(Jan 2015)
4cos x
D) y
1)
2e x
D
2e x
4 cos x 3
D
4 cos x D3
D
2e x
4 cos x
1 1
D
2 xe x
4 cos x 2D
2 xe x
2 cos x
3.1 1
D
2 xe x
D
D
3 D 2 1
2sin x
2 xe x
c3 e x
4 cos x
D3
I1
c3 e x
f ( D)
P.I
15MAT21
2sin x
The solution is y = C.F. +P.F y = c1 15) Solve: ( D
2
Solution: L( D)
2 3 x 2 )y= x e
D2
1 L( D)
x
c3e x
xe x
2sin x
e x cos2 x
( Jan 2015)
2, which leads to
C.F . A cos 2 x Also,
P.I .
c2 e
x2 e3 x
B sin 2 x.
e x cos 2 x
Department of Mathematics, SJBIT
1 L( D)
e3 x x2
1 L( D)
ex cos 2 x
Page 7
Engineering Mathematics II e3 x e3 x
1 L( D
x2
3)
1 ( D
3)
2
2
D 3 x 1
ex
x2
ex
2
6 D 11 1 e 1 D2 11 11 1 11 1
e3 x 1
11 1 11 1
e3 x 1
e3 x x 2
e 3 x x 2
1
6D
D 2
6D
11
1 12
( D 1) 2 D
2
2D
1
x2
ex
1
cos2 x
1 112
cos2 x 1 2
2
2 D 2
6D
D4
12 D 3
1
ex
72
2
11 1
3
D2
112
50
x
2
1
2 12 x
11
cos2 x
1
6D
D2
11 1
L( D 1)
x2
1
e3 x
1
e3 x
15MAT21
3
cos 2 x
...... x 2
ex
36 D 2
x2
2 D 1 4( 22 ) 1
1 2 D 1
cos 2 x
2 D 1
e x
4 D 4
1
cos 2 x
cos 2 x
e x ( 4 sin 2 x cos 2 x)
11 11 121 17 Therefore, the general solution of the given equation is 1 3 x 2 12 50 Y=C.F.+P.I.=Acos 2 x B sin 2 x e x x 11 11 121
1 17
e x (4 sin 2x
16) Solve the simultaneous equation (D+5) x-2y=t and (D+1) y+2x=0 Solution: The given equations are (D+5) x-2y=t (D+1) y+2x=0
cos 2x ).
( Jan2015)
From equation (ii), we get (D+5) (D+1)y +2(D+5)x=0. Using equation (i), this becomes (D+5) (D+1)y +2(t+2y)=0. 2
For this equation, the A.E. is (m+3) = 0 whose toots are -3, -3. Therefore -3t C.F. = (c1+ c2t) e P.I .
1 ( D 3) 2 9
1
2
( 2t ) D 3
1
1
9 (1 D / 3) 2
2
(t )
2 9
1
2 3
( 2t ) D ...... t
2 9
t
2 3
Therefore, the general solution of equation (iii) is 4 -3t 2 Y = C.F. + P.I. = (c1+ c2t)e - t 9 27 Using this in equation (ii), we get
Department of Mathematics, SJBIT
Page 8
Engineering Mathematics II 2x =- (D+1)y=-(D+1)
c 2 e -3t -
c1
( 3) c1
c 2 e -3t
c2e
( 2) c1
c 2 e -3t
c2e
2c1 -c 2 x
15MAT21
1
c1-
2
2ct e c2
2
3t
9
c2t e
2 9
9
4
t
27
2
3t
c1 c 2t e -3t
9 2
3t
9
t
2 9
t
4 27
2 27
2
t
3t
2
27
t
1 27
Expressions (iv) and (v) constitute the general solution of the given system. It is given that x=0 and y=0 when t = 0. Using this condition in expressions (iv) and (v), we get.
0
4
c1
27
, 0
1
c1
2
1
c2
27
From these, we get 4
c1
27
, c2
1
2 c1
2
27
3
2
27
9
Substituting these values of c 1 and c2 into expressions (V) and (iV), we get
1
x
27 1
y
27
1
3t
(1 6t )e (2 3t )e
27 1
3t
(3t 1)
27
(3t 2)
These constitute the required solution of the given system.
2) 2 y
17) Solve ( D
8 e 2 x
A.E is m 2 Solution:
m
P.I
c1
sin 2 x
0
xe
2
c2 x e2 x
8
f ( D)
8
(June 2014)
2,2,
C.F
8 e2 x
2
sin 2 x
D
e2 x
sin 2 x
D 2
2 x
e2 x
8
2
D 2
sin 2 x 2
4D
sin 2 x 2
D2
4D 4 2
8
x e
2
2x
cos 2 x 8
4 x 2e 2 x cos 2 x The solution is y = C.F. +P.F y = c1
c2 x e2 x
4 x 2 e2 x
Department of Mathematics, SJBIT
cos 2 x
Page 9
Engineering Mathematics II 18) Solve: y
2y
y
15MAT21
x cos x
(June 2014)
Solution:
A.E is m2
2m 1 0
m 1, 1 C.F
P.I
c1
c2 x e x
x cos x D
2
x
2D 2
x
2D 1
D
2 D 1
cos x
2
2 D
D 1
2
2D 1 D
sin
x
2 x sin
D
1
cos
x
2 The solution is y = C.F. +P.F
y = c1 dx
2 x
y
sin x
D 1
2
x sin x
(D
2 sin
1
x
2
D 5 y
2
x sin x
x
1 cos
x
sin
x
x sin x cos x sin x
2
dy
0,
1) sin D
2x 5 y 0 dt dt Sol : The given equations can be written as D 7 x y 0.............(1) 19) Solve
7 x
2D 1
2
1
c2 x e2 x
2
x
x sin x
x
cos x 2
(June 2014)
0..............(2)
Operating (2) by (D-7) and multiplying (1) by 2 we have
2 D 7 x 2 y 2 D 7 x
D 5 D 7 y
Adding we get , or
D 5 D 7
D 2 12D 37 y
A.E is y
m2 12m 37
1 d 2 dt
0 2 y
0
0 0
m
6 i
e6t c1 cos t c 2 sin t ............(3)
By considering
x
0
dy dt
e 6t c1 cos t
2 x 5 y
c2 sin t
Department of Mathematics, SJBIT
0 we get, x
5e 6 t c1 cos t
1 dy 2 dt
5y
c 2 sin t
Page 10
Engineering Mathematics II 1
e6 t
2
c1 sin t
1
x
2
6e6t c1 cos t
c2 cos t
c2 e6t cos t
c1
15MAT21
c2
5e6t c1 cos t
c2 sin t
c2 sin t
c1 e6t sin t .................(4)
(3) and (4) represents the general solution of given system of equations
dx
2 y
dy
2 x sin 2t , given that x dt dt Sol: The given equations can be written as Dx 2 y cos 2t .............(1) 20) Solve
cos 2t ,
1, y
0 at t
0
(Dec2013)
2 x Dy sin 2t ..............( 2) Operating (1) by D and multiplying (2) by 2 we have
D 2 x 2 Dy
2sin 2t
4 x 2 Dy
2sin 2t
Adding we get, D2 m2
A.E is x
c1 cos2t dx
By considering
2 dt 1 y
0
cos 2t we get, y
c1 cos 2t
2c1 sin 2t
2
4
0
c2 sin 2t ............(3)
2 y
dt
1 d
y
4 x
c1 sin 2t
c2 sin 2t
1 dx 2 dt
cos2t
cos 2t
2c2 cos 2t
cos 2t
c2 1/ 2 cos2 t .............(4)
(3) and (4) represents the general solution of given system of equations It is given that x=1 and y=0 when t = 0. Using this condition in expressions (3) and (4), we get.
from (3)1 c1 from (4) 0
0
0
c1
1
c2 1/ 2
c2
1/ 2
Substituting these values of c 1 and c2 into expressions (3) and (4), we get
x
cos 2t
sin2t 2
; y
Department of Mathematics, SJBIT
sin 2t
Page 11
Engineering Mathematics II
15MAT21
21) Using the method of variation of variation of parameters solve y
6y
9y
e3 x x 2
.
( June 2014, Dec 2013)
We have D
Solution: ;
m2
A.E. yc
2
e3 x
6D 9 y
6m 9 0
x 2 3,3
m
c2 x e3 x
c1
A Bx e3 x be the complete solution of the given equation where A(x),B(x) are to
y
be found We have y 1=e3x and y2=xe3x
y2
A '
x
,
W 3 x
3x
xe .e / x
A '
e 1
A '
x
y1 ( x)
B'
W
2
3x
,
6 x
B'
,
3x
e .e / x e
B'
2
6x
1 x
2
Integrating we get
A
log x k1
B
1 x
k 2
A Bx e3 x we have,
Using the expression of A and B in in y
2
y
k1
''
xy'
22) Solve: x y
1
log x k1 e 3 x
y
k 2 x e 3 x
x
k 2 xe3 x
e 3 x log x
2cos2 (log x) .
y
( Dec 2013)
t
Solution: Put logx=t or e = x
xy
Dy, x2 y
D( D 1) y
Dy
D( D 1) D 1 D 2
2D 1 y
D( D 1) y y
where D
d dt
2
2cos t 2cos 2 t
2cos 2 t
Department of Mathematics, SJBIT
Page 12
Engineering Mathematics II m2
15MAT21
2m 1 0
m 1 yc yp
c2t )e t
(c1
2cos2 t D 2 2 D 1 1 cos 2 t D 2
2D 1 1
D2 P 1
P 2
D2
2D 1
2
2D 1
eox
1
D cos 2t D2
cos 2t
2D 1 D cos 2t
2
2D 1 4 2D 1 (2 D 3) cos 2t
P1
P 2
eox
2 D 1 0 2(0) 1 cos 2t
1 1
1
2D 3
(2 D 3)(2 D 3) 2 D(cos 2t ) 3cos 2t (4 D 2
9)
4 sin 2t 3 cos 2 t 25 4 sin 2 t
3 cos 2 t
25
Department of Mathematics, SJBIT
Page 13
Engineering Mathematics II
15MAT21 MODULE-2
DIFFERENTIAL EQUATIONS - II dx 1) Solve the simultaneous equations dt y = 1 when t = 0
Sol: Given Dx 2 y
sin t ,
2 x Dy
4
yc y p y
Sol:
yc y p
(Jan 2016)
4 y
3 sin t
2i
3 sin t
sin t 2 D 4 c1 cos 2t c2 sin 2t sin t dy
x y et
2 y x sin log x
(Jan 2016)
t log x
et sin t
then we have D( D 1) D 2 y AE m 2
2 x cos t 0 given that x = 0,
in the given equation we get dt c1 sin 2t c2 cos 2t cos t
x
put x
m
dt
c1 cos 2t c2 sin 2t
Substituting
2 x 2) Solve y
0
dy
cos t
Solving the equations we get D 2 m2
2 y sin t 0,
2m 2
0
m 1 i
et c1 cos t c2 sin t et sin t
............. D D 1 2 D 2 sin t sin t et 2 e t 2 ( Dr 0) D 1 1 1 D 2
sin t e t 2 D t
GSis y
e t t cos t 2
t
e c1 cos t c2 sin t
e t t cos t 2
y x c1 cos(log x) c2 sin(log x)
Department of Mathematics, SJBIT
x log x cos log x 2
Page 14
Engineering Mathematics II dy
dx
x
y
3) Solve dx
dy
y
x
15MAT21 (Jan 2016, June 2015)
Sol: The given equation can be written as
dy
we note that p p
1
x y
p
y x
p 2 p
x
dx
y
1 0
y x
Solving for p we get p
x y
now
y
or p
x
dy x dx
xdx
y
then x 2 y 2 Similarly
c
ydy
k
0
dy
y
dx
dy
dx
x
x
y
then log x log y then xy c G S is x
k
log c
0
2
y
2
c xy c
2
x a y
0
4 x a y
6 y x
(July 2015)
4) Solve
Solution:
et
put x
put x
t log x et
a
t
log x
a
then we have D( D 1) 4 D 6 y AE m yc y p
2
5m 6
c1e2t
m
a
2,3
c2 e3t
et D 2
0
et
a 5D 6
et
a
2
6
Department of Mathematics, SJBIT
Page 15
Engineering Mathematics II
GS is y
c1e
y
2t
c2e 2
c1 x a
5) Solve p
3t
et
a
2
6
c2 x a
3
15MAT21
x a
a
2
6
p
tan x
(Jan 2016)
1 p 2
Sol: The given equation can be written as
x
1
tan
p
p
1 p 2
.......(1)
Differentiating both the sides dx dy
1
1 p 2
dp
1 p 2 dy
p
1
dp
dy
2
1 p 2
dp
1 p2
1
dy 1 p 2
dy
1 p 2
dp dy
2
dp dy 2
2
2 1 p 2
2 p
dy
1 p
2
2
2
dp......... put 1
p2
t
dt / t 2
d y y
p 2
1 p dy
dp
dy
p 2p
1 p 2
dp 1
dp
1 t
c
y
1 1 p2
c......(2)
(1) and (2) arethe general solution Department of Mathematics, SJBIT
Page 16
Engineering Mathematics II
15MAT21
6) Find the general and singular solution of the equation y
px p 3
(Jan 2016)
Solution: The given equation is Clairaut’s equation
the general solution is obtained by replacing p by c y
cx c
3
diff w.r.t c to get the singular solution 0
x 3c y
x 3
2
x
c
3
x x
ix
3
7) Solve ( px
3
i
i
y )( py
Solution: Put u=x 2
du
Put v=y 2
p
a 2 pby pby reducing toCla oClairaut ' sequ sequation.
x)
2 xdx
dv
du
dx
2 ydy
(June 2015)
2 x dv
dy
2 y
dy
dv / 2 y
x dv
x
dx
du / 2 x
y du
y
P
where P
dv du
substituting this in the given equation we get x y
P x
2
x P
x
y
y
y
v
P x
x
y 2 )( P 1) Pa
a
2
x y
Pa 2 x
x y
P
P
uP v ( P 1)
Pa 2
2
......this this is Claira Clairaut ut ' s form P 1 TheGeneral TheGeneral Solutionof Solutionof this this equationis equationis v
uP
a2
x
2
y x( x 2 P
Py
uc
ca
2
c 1
or y
2
Department of Mathematics, Mathematics, SJBIT
cx
2
ca
2
c 1
Page 17
Engineering Mathematics II 2
8) Solve (1 x) y
(1 x) y e z
Put 1 x
Solution:
y z
15MAT21
2 sin log x
(June 2015)
log(1 x)
givenequationbecomes D( D 1) y
Dy
y
The AE is m2 1 0 yc
D2 1 y
2 sin z m
i
c1 cos z c2 sin z 2 sin z
y p
2 sin z
sin z
z cos z log(1 x) cos log(1 x) D 2 1 2D y c1 cos llo og(1 x) c2 sin llo og(1 x) lo log(1 x) cco os llo og(1 x)
9) Solve y
Solution:
y 2 p3 by solving fo for x.
2 px
(June 2015)
y 2 p3
y
x
2z
c1 cos llo og(1 x) c2 sin llo og(1 x)
2 p
differentiating w.r.t y we get dx
1
dy
or
p 1 y 2 .3 p 2
dp dy
2
1 p
p 3.2 y p
1 2p
2
p 3y2 p3
dp dy
y
y2p3
dp dy
2
2 p4y
y
dp dy
y2p3
dp dy
oncross on cross multiplicatio multiplication n p 2 y 2 p3
2 p
or p 1 2 yp
p
y
dp dy
3
dp
2 p4 y
dy y
dp dy
1 2 yp3
on int egra egrati tio on p
y
dp dy
1 2 yp
3
0 or
0 d dy
( py)
0
c y
Department of Mathematics, Mathematics, SJBIT
Page 18
Engineering Mathematics II
15MAT21
substituting substituting in givenequationwe givenequation we get y
2.
c
2
x
y
y .
c3
y
y3
2.
c
c3
x
y
y
or y 2
2cx c3 whichisth ichisthee req required red solu luti tio on
d2y
dy
10) Solve x
2
Solution:
Put x
dx
4x
2
ex
2y
dx
e z
z
(June 2015)
log x
givenequationbecomes given equationbecomes D( D 1) y
4 Dy 2 y
The AE is m2 yc
c1e
3m 2
z
2 z
c2e
D
0
1, 2
m
c1
c2
x
x2
z
y p
ee
1
1
D 1 D 2
D 2
D 1
1
y
2
2 sin z
z e z
e e
D
2
c1
c2
e
x
x 2
x2
11) Solve
Solution:
dx dt
e
2z
ez
z
e e
3D 2 y
e
1
z
ee
e
D 2
2z
e dz
e
e z
z
z
e e e z dz
e x
2 z ez
e
x
2
x
7x
Put D Dx 7 x
y
d dt y
or ( D 7) x
0,
dy dt
2x 5 y
0
(June 2015)
the given given equationsbecom equationsbecomes es 0 ; Dy 2 x 5 y y
0;
0
2 x ( D 5) y
0
solving for y we get D2 12 D 37 y
The AE is m2 12m 37 37
Department of Mathematics, Mathematics, SJBIT
0
m
0
6 i
Page 19
Engineering Mathematics II e6t c1 cos t
y
c2 sin t dy
substituting y in x
1 dy e6t
c1
2
12) Solve p
2x 5 y
dt 1
5y
2 dt
e6t
2
c2 cos t
2
15MAT21
0
c1 sin t c2
4 x5 p 12 x 4 y
6e 6t c1 cos t
c2 cos t
c2 sin t
5e 6t c1 cos t
c 2 sin t
c1 sin t
0 , obtain the singular solution also. .
(Jan 2015)
Sol: The given equation is solvable for y only.
p 2
4 x 5 p 12x 4 y p 2
y
4x5 p
0 ...........(1)_
f ( x, p )
12 x 4
Differentiating (1) w.r .t.x, 2 p dp dx
dp
dp
4 x5
dx
20 x 4 p 12 x 4 p 48x 3 y
dx 5
3
(2 p 4 x ) 8 x ( xp
( p 2 x5 )
dp
2p
dx
x
dp
2p
dx
x
p2
4x5 p 2x 4
)
0
( p 2x5)
0
Integrating log p
log x
k
p
c2 x2
equation (1) becomes
c4
4c 2 x3
12 y
Setting c 2
k the general solutionbecomes
k2
12 y
4kx3
0
Differentiating w.r.t k partially we get
Department of Mathematics, SJBIT
Page 20
Engineering Mathematics II 2k
4 x3
0 2 x 3 in general solution we get
Using k x 6
3y
13) Solve
15MAT21
0 as the singular solution px
y
py
x
2 p , by reducing into Clairaut's form,taking the 2
substitution X
x2
Solution: Let X
x2
Y
x ,Y
dX
dy
X
p
2y dy dY dX
dx
Y
P X
( PX Y
y ) ( py
x) X
Y
Y )( P 1) PX
and let P
dY dx
P
Y Consider ( px X
(Jan 2015)
y .
dY dX dx 1 x . p.2 x or p P 2 y y
P
.
2x
dx dX
dy
Now, p
2
Y
2p
P Y
X
2
X Y
P
2 P
2 P P 1
Is in the Clairaut’s form and hence the associated general solution is
Y
cX
2c c 1
Thus the required general solution of the given equation is y
14) Solve p
3
4 xyp 8 y 2
0 by solving for x. .
2
cx 2
2c c 1 (Jan 2015)
Solution: The given equation is solvable for x only.
p3 x
4 xyp 8 y 2 p3
8 y2
4 yp
0 f ( y, p)
Differentiating (1) w.r.t. y, dp dp 1 3 p 2 4 xy 4 yp. 4 px 16 y dy dy p
Department of Mathematics, SJBIT
0
Page 21
Engineering Mathematics II dp dy
(3 p 2
dp
3 p
dy
4 xy) p3
2
4 px 12 y
8 y2
p3
p
dp 2 p3
p3
p dy 2 dp
( p
12 y
4 y2
p
2 dp
8 y2 y
8 y2
dy
y 3
p 3
2
4y )
4 y2 y
1
p dy
y
2 log p
log y
U sin g P cy cy
log c
cy in (1) we have, 8 y 2
4 xy cy
0
Dividing throughout by y y y c c
4 x c
c (c
8 y
4 x)
15) Solve p( p
2
we have,
8 y
y)
4 x) 2
64 y
x (x y)
Sol: The given equation is, p
y2
y
3
0
Thus the general solution isc(c
p
15MAT21
(June 2014)
2
py x( x y )
4 x( x
y)
4 xy
y2
0
2 4x2
y
p
y
2
ie., p
2( y x ) 2
x or p
(y
(2 x 2
y)
x)
We have, dy dx
x
Also,
dy dx
x2
y
k
2 y
x
Department of Mathematics, SJBIT
Page 22
Engineering Mathematics II dy
ie.,
y
dx
x, is alinear d.e ( similar tothe previous problem)
P 1, Q
x; e
Hence ye x x
ie., ye
15MAT21
P dx
e x
xex dx c ( xe
x
x
e ) c, int egrating by parts.
Thus the general solution is given by (2 y
x
2
x
c) e ( y
x 1) c
0
16) Obtain the general solution and singular solution of the equation
y
p2 y .
2 px
(June 2014)
Solution: The given equation is solvable for x and it can be written as
2 x
y
py........(1)
p
Differentiating w.r.t y we get
2 p
1
y dp
p 1 p
p dy
p
Ignoring
1 1
y
y dp
dp dy
0
p dy
p whichdoesnot contain
p
dy
dp
p dy y Integrating we get
p
1
yp
y dp
p
2
0 or
dp dy
, this gives
0
c........(2)
substituting for p from 2 in (1) y 2
2cx c 2
17) Obtain the general solution and singular solution of the Clairaut’s equation xp
3
yp 2 1 0 . (Dec 2013)
Solution: The given equation can be written as
y
xp3 1 p
2
y
px
1 p2
whosegeneral solutionis y Thus general solutionis y
i s in theClairaut ' s form y cx cx
px
f ( p)
f (c) 1
c2 Differnetiating partially w.r.t. c we get
Department of Mathematics, SJBIT
Page 23
Engineering Mathematics II 2
0 x
2
c
c3
15MAT21
1/3
x
Thus general solutionbecomes 2
y
1/3
x
x
or 4 y 3
2/3
x
22/3 y
2
3 x2/3
27 x 2
18) Solve p 2
2 py cot x y 2 . 2 Solution: Dividing throughout by p , the equation can be written as
y 2 p
2
y 2 p
2y p 2y
2
p y
or
p y p
cot x 1 adding cot 2 x tob.s cot x cot 2 x 1 cot 2 x 2
dy / dx
cos ec2 x
cot x
cot x
y
(Dec 2013)
cos ecx
cot x cos ecx
dy
sin x
y
cos x 1
dx and
dy
sin x
y
cos x 1
Integrating these two equations we get y (cos x 1) c1 and y(cos x 1) c2
general solutionis y (cos x 1) c y(cos x 1) c
Department of Mathematics, SJBIT
0
Page 24
Engineering Mathematics II
15MAT21
MODULE 3 PARTIAL DIFFERENTIAL EQUATION 1) Form the partial differential equation of Z = y f (x) + xg(y) where f and g are arbitrary functions. (Jan 2016)
Sol :
z
g ( y)
x z
yf '( x);
xg '( y)
y
g'( y ) and f '( x)
Substituting 2
xy
z
x y
x
2
xy
z
x y
f ( x)
x
z x
z x
y
y
z y
z y
[ xg ( y)
yf ( x)]
z
This is the required pde. 2. Derive one dimensional heat equat i on as
u t
2
c
2
u
x2
.
(Jan 2016, July 2015)
Sol:
Consider a heat conducting homogeneous rod of length L placed along x-axis. One end of the rod at x=0(Origin) and the other end of the rod at x=L.
Department of Mathematics, SJBIT
Page 25
Engineering Mathematics II
15MAT21
Assume that the rod as constant density
and uniform cross section A. Also assume that the rod
is insulated laterally and therefore heat flows only in the x direction. The rod is sufficiently thin so that the temperature is same at all points of any cross sectional area of the rod. Let u(x, t) be the temperature of the cross section at the point x at any time t. The amount of heat crossing any section of the rod per second depends on the area A of the cross section, the thermal conductivity k of the mate rial of rod and the temperature gradient
u x
i.e., the rate of change of temperature with respect to distance normal to the area. Therefore q1 the quantity of heat flowing into the cross section at a distance x in unit time is q1
kA
u x
persecond x
Negative sign appears because heat flows in the direction of decreasing temperature (as x increases u decreases ) q2 the quantity of heat flowing out of the cross section at a distance x
(i.e, the rate of heat flow at cross section x q2
kA
u x
x
x )
persecond x
x
The rate of change of heat content in segment of the rod between x and x+ x must be equal to net heat flow into this segment of the rod is q1
q2
kA
u
u
x
x
x
x
per second.............(1) x
But the rate of increase of heat in the rod
s A x Where S is the specific heat,
u t
...............(2)
the density of material.
From (1) & (2)
Department of Mathematics, SJBIT
Page 26
Engineering Mathematics II u
s A x
u
kA
t
u
x
x
x
x
u u
or s
x
k
t
s or
k
t u t
u
or
x2
x
u x
x
x
x
x
Taking limit as x 2
0 , we have
u
k
t
s
2
u
x2
2
c
2
15MAT21
u ...........................(3) where c 2 2 x
k s
Is known as diffusivity constant. Equation (3) is the one dimensional heat equation which is second order homogenous and parabolic type. 2
2
3. From the function f(x + y , z-xy) = 0 form the partial differential equation . 2
(July 2015)
2
Sol: Let u = x + y and v= z-xy so that the given relation is f(u,v) = 0 Differentiating this partially w.r.t x and y, we get f u f u
(2 x) (2 y )
f
z
v
x
f
z
v
y f
Eliminating 2 x 2 y or x or x
z x z y z y z y
u
y
0
x
0 f
and
v
fromthese equations, we get
y 0 x x y
y z x
z x x2
y
0 y2
This is the required partial differential equation. Department of Mathematics, SJBIT
Page 27
Engineering Mathematics II
15MAT21 2
4) Derive one dimensional wave equation as
t
u 2
2
c
2
u
x2 .
(July 2015)
Sol:
Consider a tightly stretched elastic string of length l stretched between two points O and A and displaced slightly from its equilibrium position OA. Taking O as origin and OA as x axis and a perpendicular line through O as Y- axis. We shall find the displacement y a function of the distance x and the time t. We shall obtain the equation of motion of string under the following assumptions. The string is perfectly flexible and offers no resistance to bendin g Points on the string move only in the vertical direction, there is no motion in the horizontal direction. The motion takes place entirely in the X Y plane . Gravitational forces on the string are neglected. Let m be the mass per unit length of the string. Consider the motion of an element PQ of length s . Since the string does not offer resistance to ben ding, the tensions T1 At P and T2 at Q are tangential to the curve. Since the is no motion in the horizontal direction, some of the forces in the horizontal direction must be zero. i.e., - T1cos + T2cosβ=0 or T1cos = T2cosβ=T=constant…..(1)
Department of Mathematics, SJBIT
Page 28
Engineering Mathematics II
15MAT21
Since gravitational force on the string is neglected , the only two forces acting on the string are the vertical components of tension - T1sin at P and T2sinβ at Q with up[ward direction takes as positive. Mass of an element PQ is m s . By Newton’s second law of motion , the equation of motion in the vertical direction is Resultant of forces = mass *acceleration 2
T2sinβ - T1sin =m s
2 T sin gives 2 1 T2 cos or tan 2
T
t2
m s
2
T 1 sin
y
t2
m s
x
T
2
m
y .
y x
x
x
x
x
arethe slopes of the curveof the string at x and x
y x
x
x
x
x
Taking Limit as x 2
y
T
2
m x
t
x to a first approximation and tan , tan
y
2
t
t
y t
y 2
tan
T
2
2
2
T
tan
T 2
m s
y
( s
m s
T1 cos
tan
y
y ..............(2) . t 2
y 2
0 2
or
t
y 2
2
c
2
y
x
...................(3) wherec 2 2
T m
Which is the partial differential equation giving the transverse vibrations of the string . Equation (3) is the one dimensional wave equation which is second order homogenous and parabolic type. 5) Solve z xy
sin x sin y for which z y
when y is an odd multiple of
2sin y when x
.
2
0 and z 0 (Jan 2015)
Solution: Here we first find z by integration and apply the given conditions to determine the arbitrary functions occurring as constants of integration.
The given PDF can be written as
z x
Department of Mathematics, SJBIT
y
sin x sin y Page 29
x)
Engineering Mathematics II
15MAT21
Integrating w.r.t x treating y as constant,
z y
sin y sin x dx
f ( y)
sin y cos x
f ( y)
Integrating w.r.t y treating x as constant
z
cos x sin y dy
z
f ( y) dy g ( x)
cos x ( cos y)
where F ( y ) Thus z
F ( y)
g ( x),
f ( y) dy.
cos x cos y
Alsoby data,
F ( y)
z
2sin y when x
y
2 sin y ( sin y).1 Hence F ( y )
g ( x) 0. U sin g this in (1)
f ( y) (cos 0 1)
f ( y) dy
sin y dy cos y
With this, (2) becomes z
cos x cos y cos y
U sin g the condition that z 0
cos x cos(2n 1)
2
0 if y
g ( x)
(2n 1)
cos x c(2 n 1)
2
But cos (2n 1)
0. and hence 0 0 0 2 Thus the solution of the PDE is given by
2
in (3) we have
g ( x) g( x)
z=cos x cosy + cosy 6) Solve: x
2
y2
z2 P
2 xyq
2 xz
(Jan 2015)
Solution: The given equation is of the form Pp +Qq = R. The auxiliary equations are
dx 2
2
dy
dz
2
2 xy 2 xz x y z Taking the second and third terms we have, dy dz dy dz or 2 xy 2 xz y z Integrating we get, log y = log z + log C 1 Log (y/z) = logc1 Using multipliers x, y, z each ratio in (1) is equal to
Department of Mathematics, SJBIT
Page 30
Engineering Mathematics II xdx x 3
y dy
xy 2
xz 2
15MAT21
z dz
x dx
2 xy 2
2 xz 2
x3
y dy xy 2
z dz xz 2
x dx
y dy
x x2
y2
z dz z2
Let us consider dy
xdx
y dy
x( x 2
2 xy
z dz
y2
z 2 ) log ( x 2
Integrating we get , log y
y2
z 2 ) log c2
Thus a general solution of the PDE is given by
( y / z , y / x 2
y 2
z 2 )
7) Solve by the method of variables 3u x Solution: Given 3
u
0 2u y
0, giventhat u( x, 0)
4e
x
(Jan 2015)
u
2
0 x x Assume solution of (1) as U=XY whereX=X(x); Y
3
u x
3Y Let
u
( xy) 2 dX
x
2 X
dx
3 dX
3log X
e3
X 2 dY
Kdy 2
3 dX
0
dy
2 dY
X dx
3dX
Y dy
kdx
X
Kx
log X
3
c1
c1
dY
k
Y dy
log Y
dY
kx c1 kx
Let
( xy) 0
K
X dx
Y ( y)
Kdy
Y
2 ky
c2
Y
e
2
c2
Substituting (2)& (3) in (1) K
U
e
x y 3 2
Also u ( x1o)
c1 c2
4e
x
Department of Mathematics, SJBIT
Page 31
Engineering Mathematics II i.e., 4e
k
x
15MAT21
2 x
kx
6
Ae
4e
Comparing we get A 3
U
x
Ae 3
4 & K
3
y
3 2
4e
x
is required solution.
8) Form the partial differential equation by eliminating the arbitrary functions from z = f(y-2x)+g(2y-x) (June 2014) Solution: By data, z = f(y-2x)+g(2y-x) z p 2 f ( y 2 x) g (2 y x) x z q f ( y 2 x) 2 g (2 y x) y 2
r
z
4 f (y
x 2
2x)
g (2 y
x)...............(1)
2
s
z
2 f (y
x y
2 x) 2 g (2 y
x).........(2)
2
t
z
f (y
y 2
2 x) 4 g (2 y
(1) 2 (2)
2r
(2) 2
2 s t
(3)
s
6 f (y 3 f (y
x)................(3) 2 x )..............(4) 2 x )............(5)
Nowdividing (4) by (5) we get 2r
s
2 or 2r
2 s t
5s
2
Thus
9) Solve: x 2
yz P
y2
2
z
x 2 zx q
2
5
z
x y
z2
2t 0 2
2
z
y2
0 is the required PDE
xy
(June 2014, Dec 2013)
Solution: The given equation is of the form Pp +Qq = R. The auxiliary equations are dx dy dz
x 2 yz y 2 zx z 2 xy Equivalently we can write in the form, dx dy dy dz
x 2
y2
z x
y
y2
z2
Department of Mathematics, SJBIT
x y
z
dz dx z2 x2
y z
x
Page 32
Engineering Mathematics II dx dy
i.e.,
x
y
x
dx dy
or
x
dy dz
y
z
y
dy dz
y
15MAT21
y
z
z
dz
dx
z
x
x
dz
y
z
z
x
dx x
y
z
...........(1)
Fromthe first and sec ond terms of (1) we have, d x
y
x
d y
y
z
y
log
z
Similarly we get
y
x
y
y
z
log c1
x
y
y
z
c1
z
c2 z x Thus the general solutionis x
y y z , z z x
y
0
11) Solve by the method of variables 4
u
u
dx
y
3u, given that u (0, y)
2e 5 y
(Jan 2015, June2014) u
Solution: Given 4
u
3u x y Assume solution of (1) as
u
XY where X
4
x
( XY )
4Y
y
dX
X
dx
Y ( y)
( XY ) 3 XY
dY
3 XY
dy
4 dX
Let
X ( x); Y
1 dY
X dx
Y dy
3
1 dY
k X dx Y dy Separating var iablesand int egrating we get kx
log X
c1 ,
4 kx
e4
X
k,
4 dX
c1
3
log Y
and Y
e kx
Hence u put x
XY
0 and u
e
c1 c2
e
3 k y c2
4
3 k y c2 kx
3 k y
Ae 4
3 k y
where A
e
c1 c2
2e5 y
The general solutionbecomes 2e5 y
Ae
3 k y
A
2 and k
2
Particular solutionis x
u
2e
2
5 y
Department of Mathematics, SJBIT
Page 33
Engineering Mathematics II
15MAT21
2
11)
z
12) Solve
x
2
z
0 giventhat when x
0, z
e y and
z x
1
(Dec 2013)
Solution: Let us suppose that z is a function of x only, the PDE assumes the form of ODE
d 2z dx
2
z
A.E. m
0 2
D
2
1 0
1 z m
0 whereD
d dx
i
Thesolutionof the ODE z
c1 cos x c2 sin x
z
f ( y) cos x
Now put x e y
g ( y) sin x....(1)by replacing c1 & c2 by functions of y
0 and z
e y in (1)
f ( y) cos 0 g ( y )sin 0
Again put x
0 and
ey
f ( y)
z
1 x Differentiating (1) partially w.r.t x z
f ( y) sin x x g ( y ) 1
g ( y) cos x
1
f ( y) sin 0
g ( y) cos 0
equation (1) becomes z
e y cos x sin x
Department of Mathematics, SJBIT
Page 34
Engineering Mathematics II
15MAT21
MODULE-4
INTEGRAL CALCULUS 1 2 x
xy dy dx changing the order of integration.
1. Evaluate 0 x
(July 2015)
2
1 2 x
Sol : Let I
xy dy dx 0 x
2
Toidentify theregion of integration R let usfind thepoint of intersectionof thecurves y
x 2 and y
x
1, x
2 x
2 and y
1, y
4
On changing theorder wemust haveconstant limits for y and variable limits for to cover thesame region y
1
2 2 y
I
x y dx dy
x y dx dy
y 0 x 0 1
y y 0 1
y 2
y 0
2
6
y
x 2 2
0
2
dy y 1
2y 2 y
2
dy
2
x 0
2
1
1
2 y
x2
y
x 0
dy
1
y 3
I
y 1x 0
y (2 y) 2 dy 1
2
4 y 3
y4
3
4
2
1
3 8 c b
a
x2
2. Evaluate c b
y2
z 2 dxdydz
(Jan 2016)
a
c b 2
Solution : I
2
x z
y z
z 3
c b
2ax
2ay
dydx
3
c b 2
a
2
c b
a
2a3 3
c
2ax 2 y 2a y 3 / 3 c
Department of Mathematics, SJBIT
dydx 2a 3 3
b
y
dx b
Page 35
Engineering Mathematics II c
4abx
2
15MAT21
4a3b
3
4a b / 3
3
c 3
4a b / 3 x
8abc 3
8ab3c
8a 3bc
3 8abc
3
3
a2
3
b2
z
x y
Sinx sin y, for which
when y is anodd multipleof
Sol :
z x
3
c
x c
c2
2
3) Solve
4a 3b
3
4ab x / 3
dx
z y
2sin y when x
0 and z 0 Jan 2016, July2015
2
Sinx sin y ,
y
Integrating w.r.t x treating y as constant z y
sin y sin xdx
z
sin y cos x
y
f ( y) f ( y).........(1)
integrating w.r.t y treating x as constant z
- cos x( cos y) z
cos x cos y
F ( y)
g ( x) whereF ( y)
f ( y) dy
F ( y ) g ( x)........(2)
by given data (1) becomes -2siny
-siny.1 f(y)
hence F(y) (2)
z
-sinydy
f(y) cosy
cos x cos y cos y
again by data 0
-siny
g ( x)........(3)
cos x cos(2n 1)
but cos(2n 1)
0, hence g( x ) 2 solution of given PDE is z
cos(2n 1)
2
2
g ( x)
0
cos y (cos x 1)
Department of Mathematics, SJBIT
Page 36
Engineering Mathematics II
15MAT21
1 z x z
x
4. Evaluate
y
z dy dx dz .
y
z dy dx dz
(July 2015)
1 0 x z 1 z x z
Sol : Let I
x 1 0 x z
1 z
I
y 2
xy
2
10
x z
zy
dx dz x z
1 z
4 xz
2 z 2 dx dz
10 1
z (2 x 2 ) 2 z 2 (x)
z 0
dz
1 1
4 z 3 dz 1
z 4
1 1
0 xy dx dy 5. Evaluate
R
, where R is the region bounded by x- axis, the ordinate x=2a and the
parabola x2=4ay.
(Jan 2016)
2
Sol: x =4ay is a parabola symmetrical about the y-axis. The point of intersection of this curve with x=2a is to be found. Hence 4a2=4ay or y=a The point of intersection is (2a,a )
Department of Mathematics, SJBIT
Page 37
Engineering Mathematics II I
15MAT21
xy dx dy R x 2 2a
4a
xy dy dx x
y
0
0
x 2 2a
y 2
x x
2a
2 x
0
x5 16a 2
0
1
x 6
32a 2
6
I
dx
2
0
1
4a
a
dx
2a
0
4
3
e
6. Evaluate
(x 2 y2 )
dx dy by changing into polar coordinates.
(Jan 2016)
0 0
Department of Mathematics, SJBIT
Page 38
Engineering Mathematics II
15MAT21
2
2
7) Find the area between the parabolas y =4ax and x = 4ay
(July 2015)
2
Sol: We have y =4ax ………………… (1) and x2 = 4ay…………………(2). Solving (1) and (2) we get the point of intersections (0,0) and (4a,4a) . The shaded portion in the figure is the required area divide the arc into horizontal strips of width y y 2
x varies from p,
4a
to Q 4ay and then y varies from O, y=0 to A, y=4a .
Therefore the required area is 4 ay
4a
dy 0
dx y
2
4a
4 ay
4a
dy x 0
y
2
4a
Department of Mathematics, SJBIT
Page 39
Engineering Mathematics II
15MAT21 4a
3 4a
4ay 0
y 2
dy
4a
1 y3
y2 4a . 3
4a 3
2 4 a 3 32 3
a2
3
4a 16 3
1
2
16
a2
3
4a
12a
0
a2
3
8) DefineGamma function and Beta function. Prove that 1/ 2 Sol :
(Jan 2016)
e x xn 1dx, (n 0) is called Gamma function.
n 0
1
x m 1(1 x) n 1 dx, ( m, n 0) is called Beta function.
(m, n) 0
We have (m, n)
m
n
m n
, put m n 1/ 2
1/ 2
(1/ 2,1 / 2)
1/ 2 1
(1/ 2,1/ 2)
but 1
1/ 2
1
2
/2
nowconsider (m, n) 2 sin 2m 1 cos 2n
1
d
0 /2
/2
(1/ 2,1/ 2) 2 sin
0
0
1/ 2
2
Department of Mathematics, SJBIT
cos
0
d
2 1d 0
1/ 2
Page 40
Engineering Mathematics II m
9) Prove that (m, n )
15MAT21
n
(Jan 2016)
m n
Solution :We haveby the definitionof Beta and the Gamma function /2
( m, n)
sin 2 m
2
1
cos 2n
1
d .........(1)
0
( n)
x 2
2 e
x 2 n 1dx...............(2)
0
( m)
y 2
2 e
y 2 m 1 dx.............(3)
0
( m n)
2 e
r 2
r
2 m n
1
dx........( 4)
0
( m). ( n)
4
e
x 2 y 2
x 2 n 1 y 2m 1dxdy.........(5)
0 0
evaluating RHS by changing into polars put x x 2
y2
r 2 , dxdy /2
(5)
( m). ( n)
4
rdrd , r :0
r 2
e
r co s
2n 1
r cos , y
r sin
and :0
/2
2m 1
r sin
rdrd
0 0 /2
4
r2
e
r 2m
2n 1
cos 2 n
1
sin
2m 1
drd
0 0
2 e
r2
/2
r
2 m n 1
dr
sin 2 m
2
0
1
cos 2 n
1
d
0
( m n). ( m, n) by using (1) and (4) m
Thus ( m, n)
n
m n
10) Show that the area between the parabolas y 2 Sol: Solving y 2 y :0
4 ax and x 2
required area
16 3
a2
(July 2015, Jan 2016)
4ay
4 ay
dy 0
4 ay is
4ay we get
4a and x : y 2 / 4a 4a
4 ax, and x 2
4a
dx y 2 /4 a
dy x
4 ay y 2 /4 a
0
4a
4ay
y 2 / 4a dy
0
Department of Mathematics, SJBIT
Page 41
Engineering Mathematics II 4a
y 3/ 2
1 y3
3
4a 3
4 a 3 32 3 16
a2
4a 16 3
15MAT21
4a
0
1
3/2
12a
4a
3
a2
a2
3 2 2 2 2 2 2 11) Find the volume common to the cylinders x +y =a and x +z =a Sol: In the given region z varies from a2
to
a2
x 2 to
a2
x 2 and y varies from
(Jan 2016) a2
x 2
x 2 .for z=0, y=0 x varies from – a to a
Therefore, required volume is a2 x 2
a
a2 x2
v
dzdydx x
a y
a2 x2 z a2 x2
a 2 x2
a
a2 x2
z a
a2 x 2
dydx a2 x 2
a 2 x2
a
2 a2 a
x 2 dydx
a2 x 2
a2 x 2
a2
a
2
x 2 dy dx
a 2 x2 a
a 2 x2
a
a2
2
x2 y
dx a2 x2
a
a
a2
2
x2 2 a2
x 2 dx
a
Department of Mathematics, SJBIT
Page 42
Engineering Mathematics II
15MAT21
a
a
4
2
x
2
dx
x
2
a x
4
3
a
a
a
3
4 2a
3
4
0
Solution: /2
2a
3
dx 1 x 4
Put x 2
2 sin 1
1 2 2
a
a
3
3 3
4
using beta and gamma function
sin
1 sin 2
a
3
16a
3
cos
I 0
a
3
1
12) Evaluate
3
3
x
sin /2
d 0
1 2
sin
dx 1 2
1 2 sin
(June 2015)
cos d
d
4 3 4
Department of Mathematics, SJBIT
Page 43
Engineering Mathematics II
15MAT21
MODULE-5 LAPLACE TRANSFORMS 2t
1. Find L e
sin 3t et t cos t .
Sol. L sin 3t 2t
L e
(Jan 2016)
3 s 2
9 3
sin 3t
s 2
4 s 12
s
L cos t
s 2 1
L t cos t
1
d
s
ds s 2 1
s 2 1
s (2s )
s 2 1 1 s 2
s2 1
s 2 1
s2 1
s 1
t
L e tcos t
s 1 s 2 s
L e
2t
2
2
2 2
1 1
2s 2s
2
sin 3t
e t cos t
s 2
3
t
s 2
4 s 12
2. Find the inverse Laplace transform of
s2
2s 2s
2
4 s 5 s 1
2
s 2
.
(Jan 2016)
4 s 5
Sol .L 1
s 1
2
4 s 5 s 1
2
s
2
s
2 A
B
s 1
s 1
Department of Mathematics, SJBIT
C 2
s 2
Page 44
Engineering Mathematics II 4 s 5 A s 1 s 2 on solving , A L1
3, B
1, C
4 s 5 s 1
2
B s 2
t
3e 3e
3 L1
s 1
s 2
t
6 L y '(t)
s 2 L y (t)
s
3e
3L 1
1 s 2
2t
2 t
1
3t
9 L y(t)
sy(0)
2
2
by Laplace transform method with y(0) = 0 =y (0). (Jan 2016)
Sol. y ''(t) 6 y '(t) 9 y(t) 12t 2 e L y ''(t)
1
e t .t 3e 3t
1 s 1
e tL1
2 3. Solve y '' 6 y ' 9 y 12t e
2
C s 1
1
3L 1
15MAT21
y '(0)
12 L e 3t t 2
6 sL y(t)
y(0)
9 L y(t)
12 2! s 3
3
using given initial conditions we obtain, s 2
6s
L y (t)
y (t)
y (t)
24
9 L y (t)
s 3
3
24 s 3
L1
24 e
5
24 s 3 3t
L
5
1
1
s y (t)
e
5
24 e
3t
t 4 4!
3t 4
t
cost ,0 t 4. Exp ress f (t)
1
,
t 2
in terms of unit step function and hence find its Laplace
sin t , t 2 transform .
Department of Mathematics, SJBIT
(Jan 2016)
Page 45
Engineering Mathematics II sol : f (t ) cos t
(1 cos t)u(t
L{ f (t )} L {cos}t Let F (t
)
F (t ) 1 cos t 1
F ( s)
s
s
s
L[ F (t
)u(t
L[G (t
2 )u ( t
; G(t
; G( t )
; G( t )
2
)]
s
e
2 )]
L[(sint 1)u (t
)} L{(sin t 1)u( t 2 )}.....(1)
2 )
sin t 1
sin( t 2 ) 1
sin t 1 1
; G ( s)
1
s
2
1 1
s
F ( s) 2 s
e
)]
e
s
2 )]
e
2 s
L[(1 cos t ) u(t
) (sin t 1) u( t 2 )
L{(1 cos t )u(t
) 1 cos t
F (t ) 1 cos(t
15MAT21
G( s) 1
s
s
s2 1 1
s
2
1 1
s
equ(1) becomes L{ f (t )}
s s
2
1
1
s
e
s
s
2
s
5. Solve y '' 6 y ' 9 y 12t e
3t
2
1
6L y '(t)
s 2 L y (t)
s
2
1 1 s
1
3t
9L y (t)
sy(0)
1
2 s
by Laplace transform method with y(0) = 0 =y (0). (Jan 2016)
Sol. y ''(t) 6 y '(t) 9 y(t) 12t 2 e L y ''(t)
e
y '(0)
12L e 3t t 2
6 sL y(t)
y(0)
9 L y(t)
12 2! s 3
3
using given initialconditions weobtain, s 2
24
6s 9 L y(t)
s 3 L y (t)
y (t)
y (t)
3
24 s 3
L1
24 e y (t)
5
24 s 3 3t
5
1
L
1
e
3t 4
s
5
24 e
3t
t 4 4!
t
Department of Mathematics, SJBIT
Page 46
Engineering Mathematics II 6. Find L
cos at cos bt
(Jan 2016)
t cosat cosbt
Sol : Let f (t )
s
f ( s)
2
s
s a
2
s
2
b
2
f (t )
hence L L
15MAT21
s
t
s
s
cosat cosbt
1
t
2
2
s a
2
2
s
b
ds
2
log( s 2 a 2 ) log( s 2 b 2 )
s
1 log( s 2 a 2 ) 2 log( s 2 b 2 ) 1 2
2 1 a
lim log
L
cosat cosbt
log
( s
2
3
(s 2 a 2 )
s 4 t
Hence L(te
thus L(te
4 t
2
9
L(te
L{ f (t )}
L(eat )
( s 2 b 2 )
2
. 4 t
eat sin3t and
s2
e
at
(July 2015)
t
3
sin 3t ) 3
ds
3 2
9
s
2
8s 25
3(2s 8)
8s 25
s 2
8s 25
2
6( s 4)
sin 3t ) eat
4 t
d
sin 3t )
( s 2 b 2 )
2
s 4
s (ii ) Let f (t )
log
a )
7. Find the Laplace transform of te
Sol :(i ) L(sin 3t )
s
( s 2 a 2 )
( s 2 b 2 )
( s 2 b 2 )
log
t
s 2
( s 2 a 2 )
log1 log
2
2
2 1 b
s
1
1
s
e
2
8s 25
2
at
L(e
1
at
)
s a
Department of Mathematics, SJBIT
1 s
a
F ( s)
Page 47
Engineering Mathematics II L
eat
at
e
1
t
1
s a
s
s
a
ds
s a
lim log
s a s a
a
1
s a
lim log
s
log(s a) log(s a) s
log
s a
s
15MAT21
1
log
s a s a
s
s a
log
s a
8. Express f(t) in terms of unit step function and find its Laplace transform given that
f (t )
t 2 ,0
t 2
t 4t , 2
t 4
8 ,t
Sol : f (t )
4
f1 (t) t2
f 2 (t)
2)
L{t 2} L
Let F1 (t 2)
4(t 2) t 2
L{F1 (t )}
f 3(t)
f 2 (t) u (t b)
8 4t u (t 4)
4t t 2 u (t
2)
L 8 4t u (t 4) .....(1)
4t t 2 2
t 2
4t 8 t 2
L
f 1(t) u (t a )
4t t 2 u (t
L{ f (t )}
F1 (t)
(July 2015)
4t 4
4
4
2
s
s3
4t t 2 u (t
2)
e
2 s
4
2
s
3
s
........(2)
Let F2 (t 4) L 8 4t u (t
4)
e 4e
4
8
s 2
s
4 s
4 s
1
2
2
s
s
F2 (t)
8 4t 8 4 t
L{F2 (t )} ......(3)
4
4t 8
4
8
s 2
s
substituting (3) and(2)in (1) 2!
4
2
Department SJBIT L{ f (t )} of Mathematics, e 2 s 4e 3 3
s
s
s
4s
1
2
2
s
s
Page 48
Engineering Mathematics II
1
9. Find L 1
s 1 s 2 1
Sol : Let F ( s) 1
f (t )
L
g (t )
L 1 G( s)
L
1
using convolution theorem.
9
F ( s)
L
s 1 s
2
9
1
e
s 1 1
9
9
1
L
t
1
2
s
1 2
s
1
L1
3
sin 3t
F ( s ) G( s)
f (t ) * g( t )
t
t
f (t u )g (u ) du
e
0
1 3
(July 2015)
1
and G ( s)
s 1
15MAT21
t u
3
0 t
e
t
1
u
e sin3udu
3
0
1 15 1 15
t
e
e
1
t
.sin 3udu
e u (sin 3u 3 cos 3u) 5
t
0
et (sin3t 3cos3t ) 3
sin 3t 3 cos 3t 3e
t
10. A periodic function f(t) with period 2 is defined by f (t )
t
,0
t 1
2 t ,1 t 2
find L{f(t)} (July 2015)
Sol :We haveT L f (t )
2 and L f (t )
1 e
e
sT
st
f (t ) dt
0
2
1 1 e
T
1
e
sT
f (t ) dt
0 1
1 1 e
st
2 st
2 t e st dt
te dt
2 s 0
Department of Mathematics, SJBIT
1
Page 49
Engineering Mathematics II 1
t.
2 s
1 e
2
s
L f (t )
s 2 1 e
11. Find L
Sol : L
3 s
3 s
L 1
2
5 s 2
5 s
5 s 2 3 s
2
4s 8
4s
s
8
1
as /2
tanh(as / 2)
s2 1
s2 1
2
3 s
16
3
3
3
s2
2
(July 2015)
L
5s 2
1
3s
2
L
4s 8
1
log
1 s
2
1
...(1)
,
2
1
( 1)
st
2s
e 1
1
1
log
4s 8 2
s 2 e as /2
log
4s 8
5 s 2 3 s 2
s
s .2cosh( as / 2)
4s 8
consider 3s
L 1
2
e
as / 2
e
2
5 s 2
1
eas /2
5 s 2
1
st
s
0
1 e
2sinh(as / 2)
L f (t )
2 t .
s2
e
s
1 e
s
1 e
1
st
e
2
s
1 e s 1 e
s
2e
2 s
s 1 e
(1)
s
1 2
st
e
15MAT21
3
2
5 s
8
s
2
3
3
s
2
2
20
3
9
16
3
L1 s
4
2
3 2
20
3
9
2 t 3
e
s
5 L 1
3
16 20
s 2
2
3
1
L1 s2
3
20
2
3
2 t
e
3
3
5cos
Department of Mathematics, SJBIT
20 3
t
16 20
sin
20 3
t .... .(2)
Page 50
Engineering Mathematics II f ( s) f '( s) L
1
1 s 2
log 2
s s
s
s2 1
f '( s)
2L
1
s
s
s2 1
1
2(1 cosh t )
f (t )
log(1 s 2 ) 2log s
2
1
t
15MAT21
........(3) 2 t
L 1
(1)
5 s 2 3 s
2
log
4s 8
1 s
2
3
e
1
5cos
3
12. Solve using Laplace transform method
d2y dt
2
2
dy dt
20 3
y
t
16 20
sin
20 3
t
2(1 cosht ) t
te t with y(0) 1, y '(0)
2
(July 2015)
Sol : y ''(t) 2 y '(t) y (t) te L y ''(t) s 2 L y (t)
2L y '(t) sy (0)
t
L e t t
L y (t) y '(0)
2 sL y (t)
y (0)
L y (t)
1 s 1
2
using given initialconditions weobtain, s 2
2s 1 L y (t)
s
1
2 2
s 1 2
s 1 L y (t)
L y (t) y (t)
1
s
s 1
2
s s 1
1 2
s 1
s 1 s 1
2
4
1 2
s 1
Department of Mathematics, SJBIT
1 2
s 1
4
Page 51
Engineering Mathematics II e
t
1
L1
y (t)
s
e
t
1
L1
2
s
t 3 6
1 4 1
cos3 t )
(Jan 2015)
3
(3cos t cos 3t) 1
2
s
2s
1)
s2 9
f ( s) we have,
ds 2 s
4 ( s
3 s
4 s2 1
9 d
3
cos t )
4
3
4 s 2 1 s 2
3
L t (sin t
1
(3sin t sin 3t )
U sin g the property : L tf (t ) 3
4
cos 3 t )
sin3 t cos3 t
L(sin3 t
s
1 t
13) Find L t (sin 3 t Solution:
1
L1
15MAT21
2
(s
2
1
9)
2
4
3.
(s
2
1) 2 s
(s
2
1)
2
2
(s
2
(s
9) 2 s 2
9)
2
2
Thus 3
3 s
3
L t (sin t cos t )
1
2 ( s 2 1) 2
1
1
( s 2 9) 2
4
3.
(1 s 2 )
9 s2
( s 2 1) 2
( s 2 9) 2
14) Express f(t) in terms of unit step function and hence find the Laplace transform
t2 given that f (t )
0 t 2
4t 2 t 4 8
Solution:
t 4 t2
f (t ) L f (t )
L t 2 s 2 3
s 2 s
(Jan 2015)
3
(4t t 2 )u(t 2) (8 4t) (t 4)
2
2
e
3
e e
2 s
2 s
2 s
L (4t t )u(t
2)
L 4(t
2) 2
L
t2
4
2
s
3
s
Department of Mathematics, SJBIT
2)
(t
4
e e
4s
4s
L (8 4t ) (t 4) e
4s
L
4t 8
4
8
2
s
s
L 8 4 (t 4)
Page 52
Engineering Mathematics II 3
15) Find the value of t e
t
15MAT21
sin tdt using Laplace transforms
(Jan 2015)
0
Solution: We have st 3
e t sin tdt
3
L(t sin t )
1
d3
3
ds
0
d 3
L sin t
3
ds
2
d2
2s
ds 2
s 2 1
2 d s 1 .2 2
ds
d
2s
s2 1
ds
s 2 1
4 s3
8s 2
s 2 1 (4s 2)
2s 3
1
4
2s 2 s2 1
8s
s
2
2s.2 s 2 1
3
2
1
3
2s .3 s 2 1
2
6
2
4
s 2 1
Putting s = 1 in this result, we get t 3
e t sin tdt 0
1 8
This is the result as required. 16) Find Laplace transform of a periodic function f (t )
Solution:
Here T = 2
=
1 e
(Jan 2015)
t 2
2
e
2 s
st
f (t )dt
0
st
2 s
te dt
(
t )e st dt
0
1 1 e
t,
t
2
1 1 e
0
. Therefore
1
L f (t )
t,
2 s
t
st
e
s
Department of Mathematics, SJBIT
e
st
s 2
t 0
e
st
s
1
e
st
2
s2
Page 53
Engineering Mathematics II 1 2 s
1 e
2 s
1 e (t
s
s2
e
2 s
e
as
a)
s
e
s
1
17) Prove that L
s
e
15MAT21 e
s2
s e
(t
a)
s
e
s2
s2
2 s
s2 (June 2014)
(t
Solution: We shall first find the Laplace transform of
L
2 s
e
s
1 2e
s
e
s
1
e
st
e
st
(t
a)
a)dt
0 a
a
(t
a)dt
0
(t
(t
a)dt
a a
i.e., L
st
e
a)
st
e
1
e
st
(t
a)dt
a
dt, by u sin g definition
a
1 e
a
st
s L
(t
But L
a) (t
e
a)
t,
s ( a
)
e
as
s
s L lim
(t
0
e
18) If f (t )
e
s
a
1 e
as
1
0
t
a
2a t , a
t
2a
as
lim
a)
e
s
, where f (t
(t
0
a)
s
1 e
0
lim L
2a)
as
( By applying L ' Hospital ' s rule)
f (t ), showthat L f (t )
E s
tanh
as 2
(June 2014)
Solution:
L f (t )
Here T = 2 a. Therefore
=
a
1 1 e
1 e
e
2 as
st
f (t )dt
0
2a
te
2 as
2a
1
st
dt
0
Department of Mathematics, SJBIT
(2a t )e
st
dt
a
Page 54
Engineering Mathematics II
1
1
s 2 1 e 2
s 1 e
1 2e
2 as
a s
(1 e
s
(eas /2
as
2
L f (t ) 1,
e
1
2
s
as
) s
as
as /2
2 as
e
2
s
2
)
e
s
s
L f (t )
1
s 2 .2cosh(as / 2)
s2
1
t2,
t 2
t
as
2
)
1 e
as
tanh(as / 2)
L (t 2
L (t 1)u(t 1)
e s L t e
t)u(t 2)
s
1 s
2
e e
2s
L t2
2s
3t 2
2
3
2
3
2
s
s
t )u(t 2)
s
1
2
20) Find the inverse Laplace transform of tan (2 / s ) Solution:
as
(June 2014)
L 1
s
a
2 int erms of unit step function and hence find its
1 (t 1)u(t 1) ( t 2
s 1
s2
t 1
t,
1
e
2a
as 2
Laplacetransform Solution: f (t )
s
1
st
( multiplied Nr.& Dr bye as /2 )
as /2
e
as
(1 e
2sinh(as / 2) 0
ae
s2 1 e as / 2
e
st
0
2 as
e
e
2a t
s 2
e
s
1
19) Express f (t)
as
a
st
e
s
ae 2 as
1 e
e
t
2 as
1 e
st
15MAT21
(June 2014)
Let f ( s) tan 1 (2 / s 2 )
f ( s) Hence L 1
1
.
4
4 s
tan 1 (4 / s 4 ) s3 f (s ) tf (t )
Department of Mathematics, SJBIT
L1 L1
s4
4
4 s s 4
4 4 s
s 4
4
Page 55
Engineering Mathematics II Now s 4
(s2
4
Also 4 s ( s 2
4s 2
(s 2
2 2s ) (s 2
4 s
Hence 4 s
2) 2
4
15MAT21 2 2s ) ( s 2
2 2s )
(s 2
2 2s ) ( s 2
2 2s )
2
2
2 2s )
(s
2 2s ) ( s
1 s 2 L 1
4 s s 4
1 s2
2 2s
L1
4
2 2s )
2 2s
1 s2
L1
2s 2
1 s2
2s 2
U sin g (1) in L.H .S we have, 1
L1
tf (t )
( s 1) 2 1 1
tf (t ) et L 1
s
2
tf (t ) et sin t
1
L1
( s 1) 2 1 1
e t L 1
1
e t sin t
s
2
1
sin t (e t
e t )
tf (t ) sont.2sinh t 2sin t sinh t
f (t ) 21) Find L
t
S
1
( s 1)(s 2
using convolut ion theorem
4)
(June 2014)
Solution:
L 1
S ( s 1)( s 2
F(s) =
1
s
c1 (s)
s-1
f (t ) L 1 F ( s) L 1
using convolution theorem.
4)
e
( s t
( s 1)( s
4)
then
and g (t ) L 1 G (s )
S 2
2
L 1 F (s ) G(s)
4) t
cos 2t
f (t )* g (t ) t
et
f (t u) g (u )du 0
u
cos 2u du
0 t
t
e u cos2u du
e
0
Department of Mathematics, SJBIT
Page 56
Engineering Mathematics II e
12
et 5 1 5
t
u
e
t
15MAT21
( 1) cos 2u
22
2 sin 2 u 0
e t (2 sin 2 t cos 2t ) 1
(2sin 2t
e t )
cos 2t
22) Solve the following initial value problem by using Laplace transforms: d2y dy 4 4 y e t ; y(0) 0, y (0) 0 2 dt dt Solution: The given equation is y (t )
4 y (t ) 4 y (t )
(Dec 2013)
t
e
Taking Laplace transform on both sides we have,
L y (t )
4 L y (t )
2
s L y (t )
sy (0)
4 L y( t ) y (0)
1
L e
4 sL y(t )
y(0))
4 L y(t )
1 s 1
using the given initial conditions we obtain, L y (t ) s 2 y (t ) L 1 Let
4s 4
1 s 1
1 ( s 1)( s 2) 2 1
A
( s 1)( s
2)
2
B
s 1
s
C 2
s
2
2
2
Multiplying with ( s 1) ( s 2) weobtain 1 A ( s 2) 2
B ( s 1) ( s 2) C( s 1)
Putting s
1 we get A 1
Putting s
2 we get C
Putting s Hence
1
0 we have 1 1(4) 1
1
( s 1)( s 2) 2
s 1
Department of Mathematics, SJBIT
B(2) 1(1) 1
s 2
B
1
1 s 2
2
Page 57
Engineering Mathematics II L 1
1
t
y (t ) e
Thus y (t )
5 s s
5 s
2 s
e t L 1
s
2 s
=
5
A
( Bs
1
5 s
3 5 e
5
t
e t
1 s 1
s 2
t
put s 1 and s
L 1
1
3 L 1
5
0 we get A 1, C 2, B
1
4
3 2
L 1
4
3
s 1
s
1
2
4
s s 2
4
. sin 2t cos 2t
24) U sin g convolution theorem evaluateL
L 1 f ( s)
2 s
1)
2
L 1
e
s 2
C
C )( s 1)
s
2 s
Bs
s 1
s 1
Solution : f (t )
2
(Dec 2013)
5
2 s
2t
(1 t ) e
5
5
3
1 s 2
t
e
( s
s 2 4
s 2
s 2
3
e t
s 2
1
1
1
3 s
5 s
2 s
2 s
s 1 s 2
L1
3
2
2
1
L1
s 1
e 2t t
3
3
2
Thus L 1
L 1
1 s
3 A s 2
5 s
e t
1 s
L1
2t
e
5 s s
Solution:
2 t
e t
e
1
23) Find L L
s 1 s
2t
e
1
L1
( s 1)( s 2) 2
15MAT21
cos at; g (t)
s 2
1
s 2
L 1 g ( s)
a 2 s 2
a2
(Dec 2013)
cos bt
Now applying convolution theorem we have,
L
t
s 2
1
s 2
a2
s2
a2
cos au.cos(bt bu) du u 0
Department of Mathematics, SJBIT
Page 58
Engineering Mathematics II 1
f (t ) g (t )
t
t
cos( au bt bu)
2u
sin( au bt
a b 1
2 a b 2
sin at.
sin at 2a a2
25) Solve y
'''
L
2 y''
b2
1
sin bt
2b b2
1 a2
u 0
sin at sin bt
a b
sin bt. 2 a
b2
s 2
1
y'
t
bu)
a b
Thus L f (t ) g (t )
Thus
bu) du
u 0
2
1
cos( au bt
0
1 sin(au bt bu)
1
15MAT21
a.
a s 2
a2
b.
s 2
b s 2
b2
s 2
a 2 s 2
a2
f ( s). g ( s)
a sin at b sin bt
s 2
a2
s2
2y
0 given y(0)
a2
b2 y' (0)
b2 0 and y'' (0)
6 by usin g
Laplace transform method (Dec 2013)
Taking laplace on both sides L [ y ''' (t )] L[2 y '' (t )] L[ y ' (t)] 2 L[ y( t)]
Solution:
s 3 l [ y(t ) s 2 y(0) sy ' (0) y '' (0)
2 s 2 Ly(t ) sy(0) y' (0)
sL[ y(t ) y(0
l (0)
2 L[ y(t )] 0
U sin g initial conditions we have Ly(t )
y (t )
6 s 2 s 1 s 1 L1
6 s 2 s 1 s 1
6 s
2 s 1 s 1
A
B
C
s 2
s 1
s 1
Department of Mathematics, SJBIT
Page 59
