9

Laplace Transforms and Its Applications

SOLVED PROBLEMS

1. Let f(t) be a function of t defined for all t 0. What is the Laplace transform of f(t) ? Write the
Laplace transforms of 1, tn, eat, sin at, cos at, sinh at and cosh at.
Sol. Laplace transform of f(t) denoted by
st
L {f(t)}, is defined as L{f(t)} = e f ( t) dt
0
provided that the integral exists, ‘s’ is a parameter which may be real or complex. L{f(t)} is a
function of s and is briefly written as f (s). Some authors use the letter ‘p’ for the parameter
instead of ‘s’.
Laplace transforms of 1, tn, eat, sin at, cos at, sinh at and cosh at :
st
st e 1
(i) L(1) = e . 1 . dt , if s 0
0 s s
0

st
(ii) L(tn) = e tn . dt
0
Put st = x, dt = dx/s
n
x x dx 1 (n 1)
L(tn) = e 1
xne x
dx = , provided s > 0 and n + 1 > 0
0 s s sn 0 sn 1

If n is a positive integer, (n 1) = n !
n!
Hence, L(tn) = 1 .
sn
st ( s a )t 1
(iii) L(eat) = e . e at . dt e . dt , if s > a
0 0 s a
st
e a
(iv) L(sin at) = e–st sin at dt = 2
( s sin at a cos at )
0 s a2 0
s2 a2

st
e s
(v) L(cos at) = e–st . cos at dt = 2 2
( s cos at a sin at ) = 2
0 s a 0
s a2

1 at at 1 1 1 1 1 1 a
(vi) L(sinh at) = L (e e ) = L(eat) – L(e–at) = =
2 2 2 2 s a 2 s a s2 a2

1 at at s
(vii) L(cosh at) = L (e e ) = .
2 s2 a2

642
LAPLACE TRANSFORMS AND ITS APPLICATIONS 643
at n at
2. State first shifting theorem. Applying this theorem, write the Laplace transforms of e .t ,e
at at at
sin bt, e cos bt, e sinh at and e cosh at.

Sol. First shifting theorem states that if L {f(t)} = f (s), then L {eat f(t)} = f (s – a)
Similarly L {e–at
f(t)} = f (s + a)
Applying this theorem, we get the useful results :

n! n!
L(eat . tn) = L(tn ) 1
(s a )n 1 sn
b s a
L(eat sin bt) = 2 2 ; L(eat cos bt) = 2
(s a) b (s a) b2

b s a
L(eat sinh bt) = 2 2 ; L(eat cosh bt) = 2
.
(s a) b (s a) b2
3. What is change of scale property ? Prove it.
1 s
Sol. If L {f(t)} = f ( s) , then L {f(at)} = . f
a a

L {f(t)} = e–st f(t) dt = f (s)

0

s
u du
L {f(at)} = e–st f(at) dt = e a f (u ) . | Put at = u ; a dt = du
0 0 a
1 1 s
= L {f(u)} = . f .
a a a
4. Find the Laplace transforms of
(i) sin 2t cos 3t (ii) e–3t (cos 4t + 3 sin 4t).
1 1
Sol. (i) Since sin 2t cos 3t = (2 cos 3t sin 2t) = (sin 5t – sin t)
2 2
1 1
L {sin 2t cos 3t} = L (sin 5t sin t ) = [L (sin 5t) – L (sin t)]
2 2

1 5 1 2 ( s2 5)
= = .
2 s2 52 s2 12 (s 2
25 )( s2 1)
(ii) L {e–3t (cos 4t + 3 sin 4t)}
= L {e–3t cos 4t} + 3L {e–3t sin 4t}
s 3 4 s 15
= 2 2
3. 2 2 2
.
(s 3) 4 (s 3) 4 s 6s 25
5. Find the Laplace transform of t2 sin at.
2! 2
Sol. Since L(t2) = 3
s s3
By first shifting theorem, we have
2 2( s ia )3
L (t2 ei at) = 3 3
(s ia ) (s ia ) ( s ia )3
644 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

2 [( s3 3a 2s ) i( 3as2 a 3 )]
or L {t2 (cos at + i sin at)} = 2 2 3
(s a )
Equating the imaginary parts on both sides, we get
2a( 3s 2 a2 )
L {t2 sin at} = .
( s2 a 2 )3
6. Find the Laplace transform of te–4t sin 3t.
1
Sol. Since L(t) =
t2
By first shifting theorem, we have

1 (s 3i )2
L {te3it} =
(s 3i )2 {( s 3i )( s 3i )}2

( s2 9) 6is
or L{t( cos 3t + i sin 3t)} =
( s2 9) 2
Equating the imaginary parts on both sides, we get
6s
L {t sin 3t} = 2
(s 9 )2
Again applying the first shifting theorm, we have
6( s 4) 6( s 4)
L {e–4t . t sin 3t} = 2 2
= 2
.
[( s 4) 9] (s 8s 25 )2

t
, when 0 t T
7. Find the Laplace transform of f(t) defined as f(t) = T
1 , when t T.

T t
st st st
Sol. L {f(t)} = e f(t) dt = e . dt e . 1 . dt
0 0 T T

T
st T st st
1 e e e
= t. 1. . dt
T s 0 s s
0 T

T
sT
1 T sT 1 e st e e sT
e sT
1 e sT
1 e sT
= e =– .
T s s s s s Ts 2 s Ts 2
0
2t –2t 3
8. Find the Laplace transform of 7e + 9e + 5 cos t + 7t + 5 sin 3t + 2.
Sol. L (7e2t + 9e–2t + 5 cos t + 7t3 + 5 sin 3t + 2)
= 7L(e2t) + 9.L(e–2t) + 5L(cos t) + 7.L(t3) + 5L (sin 3t) + 2L(1)
1 1 s 3! 3 1
=7. 9. 5. 7. 5. 2.
s 2 s 2 s2 1 s4 s2 9 s

7 9 5s 42 15 2
= 2 4 2 .
s 2 s 2 s 1 s s 9 s
LAPLACE TRANSFORMS AND ITS APPLICATIONS 645

9. Find the Laplace transforms of the following :

x sin x
(i) t7/2 e5t (ii) dx (M.D.U. Dec., 2009)
0 x
1
Sol. (i) L (e5t) =
s 5
d3 1 ( 1)7 / 2 . 5 ! 120
L (t7/2 e5t ) = (– 1)7/2 = = .
3
ds s 5 (s 5)4 (s 5)4
x sin x
(ii) f(x) = . dx
0 x
1 sin x 1 1 1
L(sin x) = , L = . ds = [tan s]s = tan s
s2 1 x s s2 1 2
x
sin x 1
L . dx = tan 1 s .
x0 s 2
10. Find the Laplace transform of
(i) sin3 2t (ii) cosh3 2t (iii) (1 + te–t)3.
Sol. (i) Since sin 6t = 3 sin 2t – 4 sin3 2t
3 1
sin3 2t = sin 2t sin 6t
4 4
3 1
L(sin3 2t) = L(sin 2t ) L (sin 6t )
4 4
3 2 1 6 48
= . . = 2 .
4 s2 4 4 s2 36 (s 4 )( s2 36)
(ii) Since cosh 6t = 4 cosh3 2t – 3 cosh 2t
3 1
cosh3 2t =
cosh 2t cosh 6 t
4 4
3 1
L(cosh3 2t) = L (cosh 2t) L (cosh 6 t)
4 4
3 s 1 s s( s2 28 )
= . . 2 = 2 .
4 s2 4 4 s 36 (s 4 )( s2 36 )
(iii) (1 + te–t)3 = 1 + t3e–3t + 3te–t (1 + te–t) = 1 + t3e–3t + 3te–t + 3t2e–2t
L {(1 + te–t)3} = L(1) + L(t3 e–3t) + 3L(te–t) + 3L (t2e–2t) ...(1)
Determination of L(t3 e–3t)
3!
L(t3) =
s4
3! 6
L(t3 e–3t) = (Using first shifting property)
(s 3 )4 (s 3 )4
Determination of L(te–t)
1
L(t) =
s2
1
L(te–t) = (Using first shifting property)
(s 1) 2
646 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

Determination of L(t2e–2t)
2!
L(t2) =
s3
2! 2
L(t2 e–2t) =
(s 2)3 (s 2)3
1
Also, L(1) =
s
Substituting these values in equation (1), we get
1 6 3 6
L {(1 + te–t)3} = .
s (s 3 )4 (s 1) 2 (s 2)3
11. Find the Laplace transform of
3
1
(i) et t–1/2 (ii) t (iii) sinh2 2t.
t
1
2
Sol. (i) Since, L(t–1/2) =
s1/ 2 s

1/ 2
L(e t . t )= (Using first shifting property)
s 1
3 3
1 1 1
(ii) t ( t )3 3 t (Using (a + b)3 expansion formula)
t t t
3
1
or t = t3/2 + t–3/2 + 3t1/2 + 3t–1/2
t
3
1
L t = L(t3/2) + L(t–3/2) + 3.L(t1/2) + 3.L(t–1/2)
t

5 1 3 1
3.
2 2 2 2
= 3.
s3 / 2 s 1/ 2
s3 / 2 s1/ 2
3 1 1
.
2
= 2 52/ 2 3 . 2 3/ 2 +3.
s s 1/ 2
s s1/ 2
1 1 3 1 5 3
2 , , and
2 2 2 2 2 4
3
1 3 3 3
or L t 2 s .
t 4 s5 / 2 2s3/ 2 s
2
e 2t e 2t 1 4t 4t
(iii) sinh2 (2t) = = [e e 2]
2 4
1
L(sinh2 2t) = [L(e4t) + L(e–4t) + L(– 2)]
4
1 1 1 2 1 s 1 8
= = 2 2
.
4 s 4 s 4 s 2 s 16 s s( s 16)
LAPLACE TRANSFORMS AND ITS APPLICATIONS 647

12. Find the Laplace transform of

sin t
(i) t sin3 t (ii) e–t . (M.D.U. May, 2007)
t
Sol. (i)L{ f (t)} = L {t sin3 t}
3 sin t sin 3 t
Let F(t) = sin3 t =
4
3 1 3 1 1 3
L{ F(t)} = L sin t L sin 3 t = . .
4 4 4 s2 1 4 s2 9

d 3 1 1 3 3 2s 1 2s
L{t F(t)} = (– 1) . . = . .
ds 4 s2 1 4 s2 9 4 (s2 1) 2 4 (s2 9)2

3s s s 3 ( s2 9)2 ( s2 1)2
= 2 2 2 2 =
2(s 1) 2(s 9) 2 ( s2 2
1) ( s 2
9)2

s 3 ( s4 81 18 s2 ) ( s4 1 2 s2 ) s 2 s4 242 52 s2
= 2 2 2 2 =
2 (s 1) (s 9) 2 (s2 1)2 (s2 9)2

s4 26s2 121
=s 2 2 2 .
(s 1) (s 9) 2

1
(ii) L(e–t sin t) =
(s 1)2 1
sin t 1
L e t
= ds = tan 1 ( s 1) = – tan–1 (s + 1) = cot–1 (s + 1).
t (s 1)2 1
s s 2
13. Find the Laplace transform of (i) te–t sin 2t (ii) t2 et sin 4t.
1
Sol. (i) L(t) =
s2
By first shifting property,

1 (s 2i )2 s2 4 4 is
L(t.e2it) = or L(t cos 2t + it sin 2t) =
(s 2i ) 2 ( s2 4 )2 ( s2 4 )2
Equating imaginary parts, we get
4s
L(t sin 2t) =
( s2 4 )2
Again applying first shifting property, we get
4( s 1) 4s 4
L (e–t . t sin 2t) = 2 2 2 .
{( s 1) 4} (s 2s 5 )2
2
(ii) L(t2) =
s3
By first shifting property
2 2( s 4 i )3
L(t2 e4it) =
(s 4 i )3 ( s2 16 )3
648 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

2 [ s3 64i 12is( s 4i )] 2[( s3 48s ) i(12s 2 64 )]

or L(t2 cos 4t + it2 sin 4t) = 2 3
= 2
(s 16) (s 16 )3
Equating imaginary parts on both sides,
8(3s2 16 )
L(t2 sin 4t) =
( s2 16 )3
Again applying first shifting property, we get

8[ 3( s 1)2 16 ] 8(3s2 6s 13 )
L(et . t2 sin 4t) = = .
[( s 1) 2
16] 3 ( s2 2s 17)3
14. Find the Laplace transform of
1, 0 t 1
cos t , 0 t t, 1 t 2 .
(i) f(t) = (ii) f(t) =
0 , t
t2 , 2 t

st
Sol. (i)L {f(t)} = e–st . f(t) dt = e–st . cos t . dt + e . 0 dt
0 0

st s s
e e 1 s(1 e )
= 2
( s cos t sin t ) = .s ( s) = 2
s 1 2 2 s 1
0
s 1 s 1

1 2
(ii) L {f(t)} = e–st f(t) dt = e st
. dt t.e st
. dt t2 e st
. dt
0 0 1 2
1 2
st st st st st
e e e 2 e e
= t + t 2t . dt
s
0
s s2 1
s
2
2 s

s 2s s s 4 2
1 e 2 2s e e e 2s st
or L {f(t)} = e + e te . dt
s s 2 s 2 s s
s s 2

s 2s
1 2 2s e e 2 2 2s 1 e st
= e 2 2
e
s s s s s s s s
2

s 2s
1 2 2s e e 4 2s 2 2s
= e 2 2 2
e (0 e )
s s s s s s3
s
1 2 2s e 3 2s 2
= e e 2s . 2 2
e
s s s s s3
3t 4
15. Find the Laplace transform of e + 5t – 2 cos t + 3 sin t (M.D.U. Dec., 2007)
3 t 4
Sol. L(e ) + L(5t ) – L(2 cos t) + 3L(sin t)
1 4! s 1 1 120 2s 3
= 5. 5
2. 2
3. 2 = 5 2 2 .
s s 3s 1 s 1 s 3 s s 1 s 1
16. Find the Laplace transform of the function f(t), where

sin t , t
3 3 (t 1)2 , t 1
(i) f(t) = (ii) f(t) = . (U.P.T.U., 2006)
0 , 0 t 1
0 , t
3
LAPLACE TRANSFORMS AND ITS APPLICATIONS 649

s s 1
3 . L (sin t ) 3 .
Sol. (i) L{f(t)} = e e 2
s 1
2
(ii) L{f(t)} = e–s . L (t2) = e–s . .
s3
t, 0 t c
17. Find the Laplace transform of function f(t) given by f(t) = .
2c t, c t 2c
(M.D.U. May, 2007)
t, 0 t c
Sol. L f(t) = 2c t, c t 2c
Since f(t) is periodic with period 2 c
2c
st
e f (t) dt c 2c
0 st st
L[ f (t)] = 2 cs = e f (t) dt e f (t) dt
1 e 0 c
c 2c
st st
= e . t dt e (2c t) dt
0 c

st c st 2c
te c e e st 2c e st
= . dt (2 c t) ( 1) dt
s 0 s ( s) c ( s)
0 c

cs st c cs st 2 c
ce e ce e
= 0 0
s s2 0
s s2 c
cs cs cs 2 cs cs cs 2 cs
ce e 1 ce e e 2e 1 e
= 2 2 2 2 2 2
s s s s s s s s s2
1 cs 2
2 cs cs 1 (1 e )
= 2
e 2e 1 =
s 2 2 cs
s (1 e )
cs cs
cs 2 cs 1 e2 e 2 1 cs
1 (1 e ) 1 1 e
= = = = tanh .
s2 (1 e cs
) (1 e cs
) s2 1 e cs s2 cs cs
s2 2
e2 e 2

sin t 1 1 sin at
18. Given that L tan , find L . (M.D.U. Dec., 2009; U.P. May, 2005)
t s t
Sol. By change of scale property

sin at 1 1 1
L tan
at a s/a

1 sin atq 1 1 a sin at 1 a

L tan L tan .
a t a s t s
19. Find the Laplace transform of sin kt . sinh kt.
e kt e kt
Sol. As sinh kt =
2

( e kt e kt
) sin kt
L [sin kt . sinh kt] = L
2
650 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

( e kt e kt
) e kt . sin kt e kt
. sin kt
R.H.S. = L sin kt = L L
2 2 2

k k k (s k )2 k2 (s k )2 k2
= = 2 2 2
2[( s k) 2
k ] 2
2[( s k) 2 2
k ] 2 {( s k) k } {( s k) k2 }

k 4sk 2k 2 . s
= = 4
.
2 s4 4 k 4 s 4k 4
20. Find the Laplace transform of

t2 , 0 t 2
(i) e–t sin2 t (ii) f(t) = t 1, 2 t 3. (U.P., 2007, 2009)
7 , t 0

1 cos 2t
Sol. (i) sin2 t =
2
t
e (1 cos 2t ) e t e t
. cos 2t
L L L
2 2 2

1 1 ( s 1) 1 ( s 1)2 4 ( s 1)2
= 2
=
2( s 1) 2 ( s 1) 4 2 ( s 1)( s2 2s 5)

2 2
= 2 3 2
(s 1)( s 2s 5) s 3s 7s 5
2
t , 0 t 2
(ii) f(t) = t 1, 2 t 3
7 , t 0

2 3
L {f(t)} = e–st f(t) dt = t2 e st
dt (t 1)e st
. dt 7.e st
dt
0 0 2 0

2 3
2s st
e 2 te 2 1 . e st (t 1)e st 3 e st e st
=4. dt – 1. dt 7
s s s 0 ( s) s 2 ( s) s
0 2 3

2 3
4 2s 2 2 . e 2s 1 e st e 3s e 2s 1 e st e 3s
=– e 2. + +7.
s s s s s ( s) ( s) s s s
0 2

4 2s 4 2s 2 2s 2 e 3s
e 2s 1 e 3s e 2s e 3s
= e e e –2. + s ( s) +7. .
s s 2
s 3
s 3 s s ( s) s

2s 3s 3s 2s 2s 3s
2 e e e e e e
= 3 3
(2 4s 4s2 ) 2. 2 2
7.
s s s s s s s

2s 3s
2 e e
= 3 3
(2 3s 3s2 ) 2
( 5s 1) .
s s s
LAPLACE TRANSFORMS AND ITS APPLICATIONS 651

21. Find Laplace transform of each of the following:

(i) e4t sin 2t cos t (ii) sin h(t) cos2 t (iii) e–t cos2 t.
(M.D.U. May, 2008; U.P.T.U., 2006)
1
Sol. (i) sin 2t cos t = [sin 3t + sin t]
2
1 3 1
L(sin 2t cos t) =
2 s2 9 s2 1
1 3 1 1 3 1
Now, L(e4t sin 2t cos t) = = .
2 (s 4)2 9 (s 4)2 1 2 s2 8s 25 s2 8s 17

1 1 1 s
(ii) L(cos2 t) = L(1 + cos 2t) = 2
2 2 s s 4
t t
e e 1 1
L(sin h t cos2 t) = L . cos2 t = L{et cos2 t} – L {e–t cos2 t}
2 2 2

1 1 s 1 1 1 s 1
= 2 2
4 s 1 (s 1) 4 4 s 1 (s 1) 4

1 1 1 1 s 1 s 1
=
4 s 1 s 1 4 (s 1)2 4 (s 1) 2
4

1 1 1 1 s 1 s 1
= 2 2
4 s 1 s 1 4 s 2s 5 s 2s 5

1 1 1 s
(iii) L(cos2 t) = L(1 + cos 2 t) = 2
2 2 s s 4

1 1 s 1 1 1 s 1
L(e–t cos2 t) = 2
= 2 .
2 s 1 (s 1) 4 2 s 1 s 2s 5
22. Find the Laplace transform of the function f(t) defined as
f(t) = | t – 1 | + | t + 1 | + | t + 2 | + | t – 2 |, f 0. (M.D.U., 2005)
Sol. For 0 t 1, | t – 1 | = – t + 1, |t+1|=t+1
| t + 2 | = t + 2, |t–2|=–t+2
f(t) = 6
For t > 1, | t – 1 | = t – 1, | t + 1 | = t + 1, | t + 2 | = t + 2,
| t – 2 | = t – 2, f(t) = 4t.
1
1 st st st
st st e te e
L {f(t)} = e . 6 . dt e 4t dt = 6 4 . dt
0 1 s s s
0 1

6 s s s s
e e 6 6 s 4e e
L {f(t)} = (1 – e–s) + 4 2 = e 4.
s s s s s s s 2

s s s
6 2e 4e 2 s e
= = 3 e 2 .
s s 2 s s
s
652 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

23. Find the Laplace transform of the following functions :

0, 0 t 5 at bt
e e
(i) g(t) = t 3, t 5 (ii) . (M.D.U. Dec., 2008)
t
5
st st st
Sol. (i) L[ g (t)] = e . g(t) dt = 0.e . dt (t 3) e dt
0 0 5

st st 5s 5s
e e 2e e
= (t – 3) (1) =0–0–
s s2 s s2
5

2e 5 s e 5s
2 1 5s
= 2 = e .
s s s s2
1 1
(ii) L[e–at – e–bt] =
s a s b
at bt
e e 1 1
L = ds = [log (s a) log (s b)]s
t s s a s b

a a
s a 1 1 s b
s s s a
= log = log = log 1 – log = – log = log .
s b b b s b s a
s 1 1
s s
s
24. Write the inverse Laplace transforms of
1 1 1 1
(i) (ii) (iii) (iv)
s s a s n (s a)n
1 s 1 s
(v) (vi) 2 2 (vii) 2 2 (viii) 2
s 2
a 2 s a s a s a2
1 s a s 1
(ix) (x) (xi) (xii) .
(s a) 2 b2 (s a) 2 b2 (s2 a 2 )2 (s2 a 2 )2

Sol. If L {f(t)} = f (s) , then f(t) is called the inverse Laplace transform of f ( s ) and is denoted by
L–1 {f(s)} = f(t)
Here L–1 denotes the inverse Laplace transform.
1 1 1
For example, Since L (e5t) = ,L = e5t
s 5 s 5
From solution of Question 1, it follows that inverse Laplace transforms of various functions are as
given below :
1 1
(i) L–1 =1 (ii) L–1 = eat
s s a

1 tn 1 tn 1
(iii) L–1 n
= , if n is a positive integer and = , otherwise.
s ( n 1) ! n

1 tn 1 1 1
(iv) L–1 n
e at . (v) L–1 sin at
(s a) ( n 1) ! s 2
a 2 a

s 1 1
(vi) L–1 = cos at (vii) L–1 = sinh at
2 2 2 2 a
s a s a
LAPLACE TRANSFORMS AND ITS APPLICATIONS 653

s 1 1 at
(viii) L–1 = cosh at (ix) L–1 e sin bt
s2 a2 (s a )2 b2 b

s a s 1
(x) L–1 = eat. cos bt (xi) L–1 2 2 2 t sin at
(s a) 2
b 2 (s a ) 2a

1 1
(xii) L–1 2 2 2 (sin at – at cos at).
(s a ) 2a3
25. Find the inverse Laplace transforms of the following functions :

3(s2 1) 2 3s 5 2
(i) (ii)
2s 5 s2 8
4s 15 s 1
(iii) (iv) .
16s2 25 s2 s 1

3( s2 1)2 3s4 6s 2 3 3 1 1 3 1
Sol. (i) = . 3. 3
.
2s 5
2s 5 2 s s 2 s5

3( s2 1)2 3 1 1 1 1 3 1 1 3 t2 3 t4
L–1 5
L 3L 3
L 5 = (1) 3
2s 2 s s 2 s 2 2! 2 4!

3( s 2 1)2 3 3 2 1 4
or L–1 5
t + t .
2s 2 2 16

3s 5 2 s 1
(ii) Since 3. 5 2.
s2 8 s2 ( 2 2 )2 s2 ( 2 2 )2

3s 5 2 1 s 1 1
L–1 3L 5 2 .L
s2 8 s2 ( 2 2 )2 s2 ( 2 2 )2

1
= 3 cos (2 2 t) + 5 2 . sin (2 2 t)
2 2
5
= 3 cos (2 2 t) + sin (2 2 t).
2
4s 15 4s 15 1 s 15 1
(iii) Since . .
16s2 25 16 s 2 25 4 5
2 16 5
2

16 s2 s2
4 4

4s 15 1 1 s 15 1 1
L–1 2
L .L 2
4 2 16
16s 25 5 5
s2 s2
4 4

1 5 15 1 5 1 5 3 5
cosh t . . sinh t cosh t sinh t .
4 4 16 5 4 4 4 4 4
4
654 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

1 1 1
s s
s 1 2 2 2 1 1
(iv) Since .
s2 s 1 1
2
3 1
2
3
2 2
1
2
3
2
s s s
2 4 2 2 2 2

1
s
s 1 1 2 1 1 1
L–1 L L
s2 s 1 1
2
3
2 2
1
2
3
2
s s
2 2 2 2

1 1 1
t 3 1 1 t 3 1 t 3 3
= e 2 . cos t . .e 2 . sin t = .e 2 3 cos t sin t .
2 2 3 2 3 2 2
2

1
26. Find the inverse Laplace transform of 2 2
. (M.D.U. Dec., 2007)
s (s a2 )

1
Sol. L–1 2 2
s (s a2 )

1 – 1 1
Compare with f (s) , f (s) = , f(t) = sin at
s2 s2 a2 a
t t t t
1 1 1 ( cos at)
L–1 2
f (s) = sin at dt = . dt
s 0 0 a 0 a a 0
t 1 1 t
= 2
. (1 cos at) dt = 2
(1 cos at) dt
0 a a 0
t
1 sin at 1 sin at
= t = 2
t .
2 a a a
a 0
27. Find the inverse Laplace transforms of

2s2 4 2s 2 1 21s 33
(i) (ii) (iii) .
(s 1)(s 2)(s 3) (s 2
1)(s 2
4) (s 1)(s 2)3
Sol. (i) Here the denominator has non-repeated linear factors.

2s2 4 A B C
Let
(s 1)( s 2)( s 3) s 1 s 2 s 3
Multiplying both sides by (s + 1) (s – 2) (s – 3), we have
2s2 – 4 = A(s – 2)(s – 3) + B(s + 1) (s – 3) + C(s + 1) (s – 2)
1
Putting s = – 1, we get – 2 = A (– 3)(– 4) A=–
6

4
Putting s = 2, we get 4 = B(3)(– 1) B=–
3
LAPLACE TRANSFORMS AND ITS APPLICATIONS 655

7
Putting s = 3, we get 14 = C(4)(1) C=
2

2s2 4 1 4 7
Thus
(s 1)( s 2)( s 3) 6( s 1) 3( s 2) 2( s 3)

2s2 4 1 1 1 4 1 1 7 1 1
L–1 L L L
(s 1)( s 2)( s 3) 6 s 1 3 s 2 2 s 3

1 t 4 2t 7 3t
=– e e e .
6 3 2

2 2p 1
(ii) Put s = p. The given fraction becomes, which has non-repeated linear factors
(p 1)( p 4 )
in the denominator.
Resolving it into partial fractions ;
2p 1 A B
Let or 2p – 1 = A(p + 4) + B(p + 1)
(p 1)( p 4 ) p 1 p 4
Putting p = – 1, – 4, we get A = – 1, B = 3
Substituting the values of A and B and replacing p by s2, we get
2s2 1 1 3
=–
(s 2
1)( s 2
4) s2 1 s2 4

2s2 1 1 1 1 1 3
L–1 2
( 1) L 3L = – sin t + sin 2t.
(s 1)( s2 4) s2 12 s2 22 2
(iii) Here the denominator has a linear factor repeated thrice.
21s 33 A B C D
Let
(s 1)( s 2 )3 s 1 s 2 (s 2 )2 (s 2)3
Multiplying both sides by (s + 1) (s – 2)3, we get
21s – 33 = A(s – 2)3 + B(s + 1) (s – 2)2 + C(s + 1) (s – 2) + D(s + 1)
Putting s = – 1, we get – 54 = A(– 27) A=2
Putting s = 2, we get 9 = D(3) D=3
Equating co-efficients of s3, 0 = A + B B=–A=–2
Putting s = 0, we get – 33 = – 8A + 4B – 2C + D
or – 33 = – 16 – 8 – 2C + 3 or C = 6
21s 33 2 2 6 3
Thus 3 2
(s 1)( s 2) s 1 s 2 (s 2) (s 2 )2

21s 33 1 1 1 1 1 1 1
L–1 2L 2L 6L + 3L–1
(s 1)( s 2) 3 s 1 s 2 (s 2) 2 (s 2)3

t t2
= 2e–t – 2e2t + 6.e2t . 3 . e 2t .
1! 2!

1 1 tn 1
Since L e at .
(s a )n ( n 1) !

3 2 2t
= 2e–t – 2e2t + 6te2t + t e .
2
656 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

28. Find the inverse Laplace transforms of

5s 3 s
(i) (ii)
(s 1)(s2 2s 5) s4 s2 1
s s3
(iii) 4 4
. (iv) 4
s 4a s a4
Sol. (i) Here the denominator has a non-factorisable quadratic factor s2 + 2s + 5.
5s 3 A Bs C
Let
(s 1)( s2 2s 5) s 1 s2 2s 5
Multiplying both sides by (s – 1) + 2s + 5), we get (s2
5s + 3 = A(s2 + 2s + 5) + (Bs + C)(s – 1)
Putting s = 1, we get 8 = 8A A=1
Equating the coefficients of s2, 0 = A + B B=–A=–1
Putting s = 0, we get 3 = 5A – C C = 5A – 3 = 2
5s 3 1 s 2
Thus
(s 1)( s 2 2s 5) s 1 s2 2s 5
1 (s 1) 3 1 s 1 1
= = 3.
s 1 (s 1)2 4 s 1 (s 1)2 22 (s 1) 2 22

5s 3 1 1 1 s 1 1 1
L –1 L L 3L
(s 1)( s2 2s 5) s 1 (s 1)2 22 (s 1)2 22

3 –t
= et – e–t cos 2t +
e sin 2t
2
(ii) Since s4 + s2 + 1 = (s4 + 2s2 + 1) – (s2) = (s2 + 1)2 – s2 = (s2 + 1 + s)(s2 + 1 – s)
s s
s4 s2 1 ( s2 1 s )( s 2 1 s)
s As B Cs D
Let 2 2 2 2
(s 1 s )( s 1 s) s s 1 s s 1
Multiplying both sides by (s2 + s + 1) (s2 – s + 1), we get
s = (As + B)(s2 – s + 1) + (Cs + D)(s2 + s + 1)
3
Equating coefficients of s , 0 = A + C or C = – A ...(1)
Equating coefficients of s2, 0=–A+B+C+D ...(2)
Equating coefficients of s, 1=A–B+C+D ...(3)
Putting s = 0, 0 = B + D or D = – B ...(4)
From (2), 0=–A+B–A–B
A = 0, From (1), C=0
1
From (3), 1=A–B–A–B B=–
2
1
From (4), D=
2

1 1
s 2 2 1 1 1
Thus 4 2 2 2
= 2 2
s s 1 s s 1 s s 1 2 1 3 1 3
s s
2 4 2 4
LAPLACE TRANSFORMS AND ITS APPLICATIONS 657

s 1 1 1 1 1
L–1 L L
s4 s2 1 2
1
2
3
2
1
2
3
2
s s
2 2 2 2

1 1
1 1 t 3 1 t 3
= e 2 sin t e 2 . sin t
2 3/2 2 3 /2 2

1 1
t t
2 3 e2 e 2 2 3 t
= sin t = sin t . sinh .
3 2 2 3 2 2

(iii) Since s4 + 4a4 = (s2 + 2a2)2 – (2as)2 = (s2 + 2as + 2a2) (s2 – 2as + 2a2)
s s
4 4 2
s 4a (s 2as 2a )( s 2
2
2as 2a 2 )
1 1 1
=
4a s2 2as 2a 2
s 2
2as 2a 2

1 1 1
= 2 2
4a ( s a) a (s a )2 a2

s 1 1 1 1 1
L–1 L L
s4 4a 4 4a (s a )2 a2 (s a )2 a2

1 1 at 1 at 1 e at e at
= e sin at e sin at = 2
sin at
4a a a 2a 2

1
= sin at sinh at.
2a 2

s3 s2 1 1 1 1 s s
(iv) s s 2 =
s 4
a 4
(s 2 2
a )( s 2
a ) 2 2 s a2 s 2
a 2 2 s2 a 2 s2 a2

s3 1 1 s 1 s 1
L–1 4
L L = (cosh at + cos at).
s a4 2 s2 a2 s2 a2 2

s2
29. Find the inverse Laplace transform of . (M.D.U. May, 2007)
(s 1)3
Sol. Put s + 1 = u or s = u – 1 or s2 = (u – 1)2 = u2 – 2u + 1
s2 (u 1)2 u2 2u 1 1 2 1 1 2 1
= = =
(s 1) 3 3
u u3 u u2 u3 s 1 (s 1)2 (s 1)3

s2 1 1 2 1 1 t2 3t
L–1 = L–1 L 2
L 3 = e– 2t – 2te– 2t + e .
3 s 1 (s 1) (s 1) 2
(s 1)
658 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

30. Find the inverse Laplace transforms of

6 3 4s 8 6s 1 4 2s 18
(i) (ii)
2s 3 9s 2
16 16s 2
9 s3/2 s 2 s2 9
2s 9 12 1
(iii) 2
.
s 9 4 3s s

6 3 4s 8 6s
Sol. (i) L–1 2 2
2s 3 9s 16 16s 9

1 1 1 1 4 1 s
= 3L–1 L 2
L 2
s 3/2 3 4 9 4
s2 s2
3 3

1 1 1 3 1 s
+ L 2
L 2
2 3 8 3
s2 s2
4 4

3
t 1 3 4 4 4 1 4 3 3 3
= 3 e2 . sinh t . cosh t . sin t cos t
3 4 3 9 3 2 3 4 8 4
3
t 1 4 4 4 2 3t 3 3
= 3e2 sinh t cosh t sin cos t .
4 3 9 3 3 4 8 4

1 4 2s 18 1 1 1 1 s 1 1
(ii) L–1 3/ 2 2 = L–1 3/ 2
4L 2L 2
18L 2
s s 2 s 9 s s 2 s 9 s 9

t1/ 2
= + 4e2t + 2 cosh 3t – 6 sinh 3t
( 3/2)

t 3
=2 + 4e2t + 2 cosh 3t – 6 sinh 3t
2 2

2s 9 12 1 2s 9 1 12 1 1
(iii) L–1 2 = L–1 2
L L
s 9 4 3s s s 9 3s 4 s

s 1 3 1 1 1 1
= 2L–1 3L 4L L
s 2
9 s 2
9 4 s
s
3
4
t 1
= 2 cos 3t + 3 sin 3t – 4 e 3 .
t
31. Find the inverse Laplace transforms of
2
2s 1 s 1 1
(i) (ii) (iii) .
s(s 1) s 2s 3
LAPLACE TRANSFORMS AND ITS APPLICATIONS 659

2s 1 s ( s 1) 1 1 1 t
Sol. (i) L–1 = L–1 L e 1
s( s 1) s( s 1) s 1 s
2
s 1 s 1 2 s 1 1 1 2
(ii) L–1 = L–1 L
s s2 s s2 s3/ 2

1 1 1 1 1 t1/ 2 t
= L–1 L 2
2L 3/ 2 =1+t–2. =1+t–4 .
s s s 3
2

1 1 1 1 1 1
(iii) L–1 L .L
2s 3 2. s 3/2 2 3
s
2
3 3
1 t 1 1 t
2 1 2
= e .L .e .
2 s 2 t
32. Find the inverse Laplace transforms of
1 s 15 s 8
(i) (ii) (iii) (iv) .
s a 2s 2 8 s2 4s 13 s2 4s 5

1/ 2 at
1 at 1 1 at t e
Sol. (i) L–1 e .L e . .
s a s 1 t
2

s 1 1 s 1
(ii) L–1 2
L 2 cos 2t
2s 8 2 s 4 2

15 1 15 15
(iii) L–1 2
L 2 = e–2t . L–1 2
s 4s 13 (s 2) 9 s 9
[By first shifting property]
1
= e–2t . 15 . . sin 3t = 5 . e–2t . sin 3t
3

s 8 1 (s 2) 6 s 2 1 1
(iv) L–1 L = L–1 6.L
s2 4s 5 (s 2)2 1 (s 2 )2 1 (s 2 )2 1

s 1
= e–2t . L–1 + 6.e–2t . L–1
s2 1 s2 1
= cos t + e–2t 6e–2t . sin t = e–2t . (cos t + 6 sin t).
33. Find the inverse Laplace transforms of
s 2 2s 3 1 2s
(i) (ii) 2
s(s 3)(s 2) (s 2) (s 1) 2
1 s
(iii) (iv) .
(s )(s ) (s2 1) 2
660 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

s 2 2s 3 A B C
Sol. (i) Let,
s( s 3)( s 2) s s 3 s 2
Multiplying both sides by s(s – 3)(s + 2), we have
s2 + 2s – 3 = A(s – 3)(s + 2) + B(s + 2)s + Cs(s – 3)
Comparing the coefficients, we get
A + B + C = 1, – A + 2B – 3C = 2, – 6A = – 3
1 4 3
Solving these, we get A= ,B= ,C=–
2 5 10
s2 2s 3 1 4 3
s( s 3)( s 2) 2s 5( s 3) 10( s 2 )
2
s 2s 3 1 1 1 4 1 1 3 1 1 1 4 3t 3 2t
L–1 L L L = e .e .
s( s 3)( s 2) 2 s 5 s 3 10 s 2 2 5 10

1 2s A B C D
(ii)
(s 2 )2 ( s 1)2 s 2 (s 2 )2 (s 1) (s 1)2
1 + 2s = A(s + 2)(s – 1)2 + B(s – 1)2 + C(s + 2)2(s – 1) + D(s + 2)2
= A (s3 – 3s + 2) + B(s2 – 2s + 1) + C(s3 + 3s2 – 4) + D(s2 + 4s + 4)
Comparing the coefficients of s3, s2, s and constants on both sides, we have
A + C = 0, B + 3C + D = 0, – 3A – 2B + 4D = 2, 2A + B – 4C + 4D = 1
1 1
On solving these, we getA = 0, C = 0, B = – ,D=
3 3
1 2s 1 1
2 2 2
(s 2) ( s 1) 3( s 2) 3( s 1) 2

1 2s 1 1 1 1 1 1
L–1 L L
(s 2)2 ( s 1)2 3 (s 2 )2 3 (s 1)2

1 2t 1 t t t 2t
=– .e .t e . t = (e e )
3 3 3

1 1 1 1
(iii) We have
(s )( s ) s s

1 1 1 1 1 t
L–1 .L 1
= . (e e t)
(s )( s ) s s

1 1 1 1 1
(iv) 2 2 2 2
(s 1) (s 1) ( s 1) 4s ( s 1)2 (s 1)2

s 1 1 1
2 2 4 ( s 1)2
(s 1) (s 1)2

s 1 1 1 1 1 1 t
L–1 2 2
L 2
L = . ( et . t e t
. t) . sinh t.
(s 1) 4 (s 1) (s 1)2 4 2
LAPLACE TRANSFORMS AND ITS APPLICATIONS 661

2s2 6s 5
34. Find the inverse Laplace transform of the function f(s) = 3 2
.
s 6s 11s 6
Sol. We observe that denominator,
s3 – 6s2 + 11s – 6 = (s – 1) (s – 2) (s – 3)

2s2 6s 5 A B C
Let 3
s 6s2 11s 6 s 1 s 2 s 3
2s2 – 6s + 5 = A(s – 2)(s – 3) + B(s – 1)(s – 3) + C(s – 1)(s – 2)
1
Putting s = 1, we get 2A = 1 or A =
2
Putting s = 2, we get – B = 1 or B = – 1
5
Putting s = 3, we get 2C = 5 or C =
2

2s2 6s 5 1 1 5
Hence 3
s 6s 2 11s 6 2( s 1) s 2 2( s 3)

2s2 6s 5 1 1 5 1 1 t 5 3t
L–1 L 1
L 1
L 1
= e e 2t e .
3 2 2( s 1) s 2 2 s 3 2 2
s 6s 11s 6

–1 s 1 1 32s
35. If L t sin t, find L .
(s 2 1) 2 2 (16s2 1) 2

as 1 t t
Sol. Replacing s by as, we get L–1 . sin
( a 2s2 1)2 2 a a

Now, writing a = 4
4s 1 t 1 –1 32s 1 t
L–1 t sin or L t sin
(16s2 1)2 8 4 8 (16s2 1)2 8 4

32s t
L–1 t sin .
(16s2 1)2 4

5s 3
36. Find the inverse Laplace transform of 2
. (U.P.T.U., 2005)
(s 1)(s 2s 5)

5s 3 5(1) 3 As B
Sol. Let
(s 1)( s2 2s 5) (s 1)(12 2 5) s2 2s 5
Multiplying both sides by (s – 1) + 2s + 5), (s2
5s + 3 = 1 . (s2 + 2s + 5) + (As + B) (s – 1)
Equating the coefficients of s2 from both sides,
0=1+A A=–1
Similarly comparing coefficient of s from both sides,
5 = 2 – A + B or B = 3 + A = 2

5s 3 1 1 1 s 2 1 1 (s 1) 3
L–1 L L = L–1 L
(s 1)( s 2
2s 5) s 1 s 2
2s 5 s 1 (s 1)2 4
662 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

1 1 s 1 1 1
= L–1 L 3L
s 1 (s 1)2 4 (s 1)2 4

3 –t
= et – et cos 2t + e sin 2t.
2
37. Find the inverse Laplace transforms of
s 2 4s 5 s 3 s2
(i) (ii) (iii) (iv) .
s2 4s 13 (s 1) 2 (s 2) s2 4s 13 (s 2)3

s 2 1 s 2 1 s 2 4
Sol. (i) L–1 L L
s2 4s 13 (s 2)2 9 (s 2 )2 9

s 2 1 1 4 2t
= L–1 4L = e2t cos 3t + e sin 3t.
( s 2) 2
9 ( s 2) 2 2
3 3

4s 5 A B 4( 2 ) 5
(ii) 2 2 2
(s 1) ( s 2) s 1 (s 1) ( 2 1) ( s 2)
Multiplying both sides by (s – 1)2 (s + 2),
1
4s + 5 = A(s – 1)(s + 2) + B(s + 2) – (s – 1)2
3
Putting s = 1, we get 9 = 3B B=3
1 1
Equating the coefficients of s2 from both sides, 0 = A – , A=
3 3

4s 5 1 1 1 1 1 1 1 1 1 t 1
L–1 2
L 3L 2
L = e + 3tet – e–2t.
(s 1) ( s 2) 3 s 1 (s 1) 3 s 2 3 3
s 3 s 2 5
(iii) 2 2 2
s 4s 13 (s 2) 3 (s 2 )2 32

s 3 1 s 2 5 1 3
L–1 L L (Using shifting property)
s2 4s 13 (s 2 )2 32 3 (s 2)2 32

5 2t
= e2t cos 3t + e . sin 3t
3
s2 (s 2 )2 4( s 2) 4 1 4 4
(iv) 3 3 2
(s 2) (s 2) s 2 (s 2) (s 2 )3

s2 1 1 1 1 1 1
L–1 L 4L 4L
(s 2)3 s 2 (s 2)2 (s 2 )3

= e 2t 4 e2t . t 2. e 2t . t 2 (Using shifting property)

38. State the theorems regarding Laplace transforms of integrals and derivatives.
Sol. (a) Theorem on Laplace transform of derivatives is :

If f (t) be continuous and L {f(t)} = f ( s) ,

st
then L {f (t)} = s f ( s) – f(0), provided Lt { e . f ( t)} 0
t
LAPLACE TRANSFORMS AND ITS APPLICATIONS 663

(b) Theorem on Laplace transforms of integrals states :

t 1 t
1 f ( s) f (u ) du .
If L {f(t)} = f ( s ) , then L f ( u ) du f ( s) or L–1
0 s s 0

2ws
39. If L {t sin wt} = , evaluate
(s2 w2 ) 2
(i) L {wt cos wt + sin wt} (ii) L {2 cos wt – wt sin wt}. (U.P.T.U., 2005, 2006)
Sol. Let f(t) = t sin wt
Then f (t) = wt cos wt + sin wt
f (t) = 2w cos wt – w2t sin wt
Also f(0) = 0, f (0) = 0, f (0) = 2w
2ws
f ( s) = 2
(s w 2 )2
Since L {f n(t)} = sn f ( s) – sn–1 . f(0) – sn–2 . f (0) – sn–3 . f (0) .....
(i) L {f (t)} = s f ( s ) – f(0)
2ws 2ws 2
L {wt cos wt + sin wt} = s . 0
( s2 w2 )2 ( s2 w2 ) 2
(ii) L {f (t)} = s2 f ( s) – s.f(0) – f (0)
2ws 2ws3
L {2w cos wt – w2t sin wt} = s2 . s.0 0
( s2 w2 )2 ( s2 w2 )2
2s3
L {2 cos wt – wt sin wt} = 2
.
(s w 2 )2
40. Find the inverse Laplace transforms of
1 1
(i) (ii) .
s3 (s2 a2 ) s(s 1)3

1 1
Sol. (i) Since, L–1 sin at
s2 a2 a
therefore by ‘‘Laplace transform of integral’’ Theorem, we have
t t
1 1 1 1
L–1 sin au du = – 2 cos au = (1 cos at )
s (s 2
a ) 2 0 a a 0
a2
t t
1 1 1 sin au 1 sin at
L–1 (1 cos au ) du u = t
s 2 ( s2 a2 ) 0 a2 a2 a 0 a2 a
t
1 t 1 sin au 1 u 2
cos au 1 t2 cos at 1
L–1 u du = 2 2
3
s (s 2
a ) 2 0 a 2 a a 2 2 a 2
0
a 2 a a2

1 e t
. t2 1 2 t
(ii) Since, L–1 3
t .e
(s 1) 2! 2

1 1 t 1 2 t
L–1 3
t2 e t
. dt = t ( e t) 2t ( e t ) 2( e t )
s (s 1) 2 0 2 0

1 t2
= [( t 2 2t 2 )e t
2 ] = 1 – e–t 1 t .
2 2
664 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

2 3 –3t
41. Find the Laplace transforms of (i) t sin 3t, (ii) t . e . (M.D.U. 2009, U.P.T.U., 2005, 2007, 2008)
1 cos 6t 1 1 s 18
Sol. (i) Since L (sin2 3t) = L = =
2 2 s s2 36 s( s 2 36)

d 18
L{t sin2 3t} = (– 1)
ds s( s 2 36 )

54( s2 12 )
= – 18(– 1) (s3 + 36s)–2 (3s2 + 36) = 2 2
.
s (s 36 )2
1
(ii) Since L(e–3t) =
s 3

d3 1 ( 1)3 . 3 ! 6
L {t3 . e–3t} = (– 1)3 . =– .
ds 3 s 3 (s 3 )4 (s 3 )4
s 1
42. Find the inverse Laplace transform of log .
s 1

s 1
Sol. Let L–1 log = f(t)
s 1
s 1
Then L {f(t)} = f ( s) = log .
s 1
d s 1 d
L {t . f(t)} = (– 1) log =– [log (s + 1) – log (s – 1)]
ds s 1 ds
1 1 1 1
=– = L {et – e–t} = L{2 sinh t}
s 1 s 1 s 1 s 1
2 sinh t s 1 2 sinh t
t . f(t) = 2 sinh t or f(t) = . Hence L–1 log = .
t s 1 t
Note. In questions 41 and 42, we have used the result :
dn
If L {f(t)} = f ( s) then L {tn . f(t)} = (– 1)n [ f ( s )] , where n = 1, 2, 3, ....
dsn
(M.D.U., U.P.T.U., 2009)
Thus this is useful to find the transform and inverse Laplace transform of the function f(t) when
1
multiplied by tn. Also when there is a division of f(t) by t, then L f (t ) f ( s ) ds , provided
t s
the integral exists.
43. Find the Laplace transforms of
at bt
e e cos at cos bt
(i) (ii)
t t
t
e . sin t cos 2t 1
(iii) (iv) .
t t
(U.P.T.U., 2006, A.U.U.P., 2009 Uttarakhand June, 2007)
1 1
Sol. (i) Since L {e–at – e–bt} =
s a s b
at bt
e e 1 1
L ds [log (s a) log (s b)] s
t s s a s b
LAPLACE TRANSFORMS AND ITS APPLICATIONS 665

a
a 1
s a 1 s s a
= log log s = log 1 – log b = 0 – log s
s b b 1 b.
s 1 s
s s
s b
= log
s a
s s
(ii) Since L (cos at – cos bt) = 2 2 2
s a s b2
cos at cos bt s s
L ds
t s s2 a2 s2 b2

1 1 1 s2 a2
= log ( s2 a2 ) log ( s2 b2 ) = 2 log 2
2 2 s
s b2 s

a2
1 1 s2 a2 1 s2 b2
1 s2
= log = 2 log 1 log 2 2 = 2 log 2 .
2 b2 s b s a2
1
s2 s
1
(iii) L (e–t. sin t) =
(s 1)2 1

t
e . sin t 1 1
L 2
. ds = tan (s 1)
t s (s 1) 1 s

1 1
= tan (s 1) cot (s 1)
2
1 s
(iv) L (1 – cos 2t) = 2
s s 22

1 cos 2t 1 s 1
L 2 ds = log s log ( s2 4)
t s s s 4 2 s

1 s2 1 s2 1 s2 4
= log 2 = log 1 log = log .
2 s 4 s
2 s2 4 2 s2
s (s 1)
44. Find inverse Laplace transform of the function log . (M.D.U. May, 2009)
s2 4

s (s 1) s (s 1)
Sol. L–1 log = f (t) L{ f(t)} = log = f (s)
s2 4 s2 4
d 1 1 2s
L[t f (t)] = [log s + log (s + 1) – log (s2 + 4)] = – 2
ds s s 1 s 4

1 1 2s
t f (t) = – L–1
s s 1 s2 4
t . f (t) = – [1 + e–t – 2 cos 2t] = – 1 – e–t + 2 cos 2t
t
1 e 2 cos 2t
f (t) = .
t
666 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

45. Find the Laplace transforms of

1 et
(i) t2 sin at (ii) .
t
a
Sol. (i) Since, L (sin at) = 2
s a2

d2 a d 2as 2a(3s 2 a2 )
L (t2 sin at) = (– 1)2 = 2 2 3
ds2 s2 a2 ds ( s 2 a 2 )2 (s a )
1 1
(ii) Since, L(1 – et) = L(1) – L(et) =
s s 1

1 et 1 1
L ds = log s log ( s 1)
t s s s 1 s

s 1 s 1
= log = – log 1 = log .
s 1 1 s
s
s
46. Evaluate :

2t sin mt t et sin t
(i) te sin t dt (ii) dt (U.P. May, 2005) (iii) L dt .
0 0 t 0 t
(A.U.U.P., 2008)

Sol. (i) te–2t . sin t dt = e–st (t sin t) dt, where s = 2

0 0
= L(t sin t) by definition
d 1 2s 2 2 4
= (– 1) = 2 2
ds s2 1 (s 2
1) 2
(2 1) 25
m
(ii) Since, L (sin mt) = = f(s), say
s2 m2
sin mt m . ds 1 s
L = f(s) ds = 2 2
tan
t s s s m m s

sin mt s
or e–st .
dt = tan 1
0 t 2 m
s
Now Lt tan 1 = 0 if m > 0 or if m < 0
s 0 m
Thus taking limits as s 0, we get
sin mt
dt = if m > 0 or – if m < 0.
0 t 2 2
sin t ds
(iii) Since, L = 2
[tan 1 s ]s = – tan–1 s = cot–1 s
t s s 1 2
sin t
L et = cot–1 (s – 1), by shifting property
t
t e t sin t 1
Thus, L . dt cot–1 (s – 1).
0 t s
LAPLACE TRANSFORMS AND ITS APPLICATIONS 667

2t
e sin 2 t
47. Evaluate : (i) t3 e–t sin t dt (ii) dt.
0 0 t

Sol. (i) t3 e–t sin t dt = e–st (t3 sin t) dt, where s = 1

0 0
= L(t3 sin t) by definition

d3 1 d2 2s d ( s2 1)2 2 2s 2( s2 1) 2s
= (– 1)3 3 2 = 2 2 2 =
ds s 1 ds (s 1) ds (s 2
1) 4

d 2( s2 1) 8s2 d 2(1 3s2 ) ( s2 1)3 ( 6s ) (1 3s 2 ) 3( s2 1)2 2s

= = ds =2.
ds ( s2 1)3 ( s2 1)3 (s 2
1) 6

6s( s2 1) 6s(1 3s2 ) 24s( s2 1)

=2. = = 0 as s 1.
(s 2
1) 4 ( s2 1)4

e 2t
sin 2 t st sin 2 t
(ii) dt = e dt, where s = 2.
0 t 0 t

sin 2 t
=L by definition
t

1 cos 2t 1 s2 4
=L = log as in Q. 35 part (iv).
t 2 s2
1
= log 2 as s 2.
2
t 3t
e e
48. Evaluate : . dt (M.D.U. Dec., 2007)
0 t
t 3t
e e 1 1
Sol. dt = L L
0 t t t s 1 t s 3

1 1 3 1
= ds ds = ds = (log s)13 = log 3 – log 1 = log 3.
1 s 3 s 1 s

sin 2 t
49. Evaluate : (i) e t
. dt (ii) e–st t3 cos t dt (iii) e–3t t sin t dt.
0 t 0 0

1 1 1 s
Sol. (i) L (sin2 t) = L (1 – cos 2t) =
2 2 s s2 4

sin 2 t 1 1 s 1 s2 4
L ds = log
t 2 s s s2 4 4 s2
By definition

sin 2 t 1 s2 4
e–st dt = log
0 t 4 s2

sin 2 t 1
Put s = 1 . e–st dt = log 5.
0 t 4
668 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

s
(ii) L (cos t) =
s2 1
d3 s d2 1 s2 d 2s3 6s
L (t3 cos t) = (– 1)3 3 2 =– 2 2 2 =–
ds s 1 ds (s 1) ds ( s2 1)3

(1 s2 )3 . ( 6s2 6) ( 2s3 6s ) . 3 (1 s2 )2 . 2s 6( s4 6s 2 1)
=– 2 6 =
(s 1) (1 s2 )4
6( s4 6s 2 1)
e–st . t3 cos t dt = 2 4
(By definition)
0 (1 s )
1
(iii) L (sin t) = 2
s 1
d 1 2s
L (t sin t) = –
ds s2 1 ( s2 1)2
2s
e–st . t sin t dt = (By definition)
0 ( s2 1)2
6 3
Putting s = 3, we obtain e–3t . t sin t dt = or .
0 100 50
50. Find the Laplace transform of the following functions :
t
(i) t2 e–2t cos t (ii) e–t cos t dt (iii) sin t
0

cos t t/2 2x
1 e
(iv) (v) . dx.
t 0 x
s 2 s 2
Sol. (i) L(e–2t cos t) =
(s 2 )2 1 s2 4s 5
2
d s 2
L(t2 e–2t cos t) = (– 1)2 2 2
ds s 4s 5

d ( s2 4s 5) (s 2)( 2s 4) d s2 4s 3
= = ds 2
ds (s 2
4s 5) 2 (s 4s 5 )2

( s2 4s 5 )2 ( 2s 4) ( s2 4s 3 ) . 2( s2 4s 5 )( 2s 4)
=– 2 4
(s 4s 5)
3 2
2( s 6s 9s 2)
=
( s2 4s 5)3
s
(ii) L (cos t) =
s2 1

s 1
L (e–t cos t) = = f(s), say
(s 1)2 1

t 1 s 1
t
L e cos t dt f (s)
0 s s( s 2 2s 2)
LAPLACE TRANSFORMS AND ITS APPLICATIONS 669

( t )3 ( t )5 t3/ 2 t5/ 2
(iii) sin t t – ...... = t1/2 – – ......
3! 5! 3! 5!
3 5 7
1/ 2 t3/ 2 t5/ 2 2 2 2
L(sin t) = L t ...... = ......
3! 5! s3/ 2 3 ! s5 / 2 5 ! s7/ 2
2
1 1 1
= 1 ......
2 s3 / 2 22 . s 2 ! 22 . s

1
4s
L (sin t)= .e
2 s3 / 2
d
(iv) Now, L (sin t ) s [ L (sin t )] 0 [See Q. 38]
dt
1
cos t 4s
L .e [See part (iii) above]
2 t 2 s
1
cos t 4s
L .e
t s
t/ 2 e 2x
1
(v) . dx
0 x
Put 2x = u, 2dx = du
t u
1 e
Integral becomes du
0 u
1 1
Now, L (1 – e–u) = –
s s 1
u
1 e 1 1 s s s 1
L ds = log = 0 – log log
u s s s 1 s 1 s 1 s
s
t u 1 s 1 1
1 e 1
L du = log = log 1 .
0 u s s s s
Note. (For Part iii) Q. No. 50.
3 5 7 1/2
The values for , and have been used to calculate Laplace transform of t ,
2 2 2
3/2 5/2
t and t . Here,
(n 1)
L (tn) = n 1 ; if n > – 1 and s > 0.
s
3
2
L(t1/2) =
s3/ 2
3 5 7
To calculate , , , we may use an important result of Gamma Function
2 2 2

1 ( 2n 1) .
(Reduction Formula) which is n , (n 1) = n !
2 22n . (n 1)
670 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

3 (3 ) . 2! 1
Put n = 1 2 =
22 . ( 2) 22 . 1 ! 2

5 (5 ) . 4! 12 3
Put n = 2 4 4 4
=
2 2 . ( 3) 2 .2! 2 22
7 ( 7) . 6! 120
Put n = 3
2 26 . ( 4 ) 26 . 3 ! 26
On using these values and simplification gives the final results.
51. Find the inverse Laplace transforms of
1 s 2 s 2 s
(i) (ii) (iii) (iv)
s(s 2 a2 ) s2 (s 1)(s 2) (s2 4s 5) 2 (s2 a 2 )2

s2 1 s2 1
(v) (vi) 2 2 2
(vii) log .
2 2 2 (s a ) s(s 1)
(s a )
1 1
Sol. (i) Since L–1 2 2
sin at
s a a

1 t 1 1 t 1 cos at
L–1 sin at dt = 2 [ cos at ]0
s( s 2 2
a ) 0 a a a2

s 2 A B
(ii) Let
(s 1)( s 2) s 1 s 2
s + 2 = A(s – 2) + B(s + 1)
Putting s = 2, – 1, we get A = – 1/3, B = 4/3
s 2 1 4
(s 1)( s 2) 3( s 1) 3( s 2)
s 2 4 1 1 1 1 1 4 2t 1 t
L–1 L L = e e
( s 1)( s 2) 3 s 2 3 s 1 3 3

s 2 t s 2
1
Now, L–1 L dt
s( s 1)( s 2) 0 (s 1)( s 2)
t 4 2t 1 2 2t 1
t t
= e e dt = e e 1
0 3 3 3 3

s 2 t s 2
1
Again, L–1 2
L dt
s (s 1)( s 2) 0 s( s 1)( s 2)

t 2 2t 1 1 2t
t
= e e 1 dt = (e – e–t – t)
0 3 3 3
1 1 1 2t
(iii) L–1 2
L e . sin t
s 4s 5 (s 2)2 1

d 1
Now,L–1 = (– 1)1 . t . e–2t . sin t
ds s2 4s 5

(2 s 4) s 2 1 2t
i.e., L–1 2 = – t . e–2t sin t or L–1 t.e . sin t
(s 4 s 5) 2 (s 2
4s 5) 2 2
LAPLACE TRANSFORMS AND ITS APPLICATIONS 671

s
(iv) If f(t) = L–1 2 , then
(s a 2 )2

f ( t) s 1 2s 1 1 1 1
L ds . ds = – .
t s ( s2 a 2 )2 2 s ( s2 a 2 )2 2 s2 a 2
s
2 s2 a 2

f (t ) 1 1 1 sin at 1
L , Hence f(t) = t sin at.
t s2 a 2 2 2a 2a
(v) In part (iv) we have proved that
s 1
L–1 t sin at = f(t), say
( s2 a 2 )2 2a
Since, f(0) = 0
s2 s d d 1 1
L–1 L 1 s. 2 { f (t )} = t sin at = (sin at at cos at )
(s 2
a 2 )2 (s a 2 )2 dt dt 2a 2a
(vi) In part (iv) we have shown that
s 1
L–1 2 2 2 t sin at = f(t), say
(s a ) 2a

1 1 s t
1
L–1 L . f ( t) dt
( s2 a 2 )2 s ( s 2 a 2 )2 0

t t
t 1 1 cos at cos at
= t sin at dt = t 1. dt
0 2a 2a a 0 0 a

1 t cos at sin at 1
= = {sin at – at cos at}
2a a a2 2a3
s2 1
(vii) If f(t) = L–1 log , then
s( s 1)

d s2 1
t f(t) = L–1 log
ds s( s 1)

d d d
= – L–1 log ( s2 1) L 1
log s + L–1 log ( s 1)
ds ds ds
2s 1 1 1
= – L–1 + L–1 L = – 2 cos t + 1 + e–t
s2 1 s s 1
1
Thus, (1 + e–t – 2 cos t).
f(t) =
t
52. Apply convolution theorem to evaluate :
s s2
(i) L–1 2 2 2 (ii) L–1 2
.
(s a ) (s a )(s2
2
b2 )
Sol. Convolution theorem states that
If L–1 { f ( s) } = f(t) and L–1 { g ( s ) } = g(t)
t
Then L–1 { f ( s ) . g ( s ) } = f(u) g(t – u) du = F G
0
672 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

F * G is called the convolution of F and G.

s 1 1
(i) Since L–1 = cos at and L–1 sin at
s2 a2 s2 a2 a
By convolution theorem, we get
t 1 t
s 1 sin au
L–1 . cos a( t u) . du = { 2 cos a( t u ) sin au } du
2 2 2 2 a 2a
s a s a 0 0

Using formula ;
2 sin A cos B = [sin (A B) sin (A B)]
1 t
R.H.S. = [sin at sin ( 2au at)] du
2a 0
t
1 1
= u sin at cos (2au at)
2a 2a 0

s 1
Hence L–1 2 2 2 = t sin at
(s a ) 2a

s s
(ii) Since L–1 = cos at, L–1 = cos bt
s2 a2 s2 b2

s2 1 s s
By convolution theorem, we have L–1 2
L . .
(s a )( s2
2
b2 ) s2 a 2 s2 b2
53. Find the inverse Laplace transform of
1 s 3
(i) log 1 (ii) cot–1 (M.D.U. May, 2009)
s 2 2
2 s s a
(iii) tan–1 (M.D.U. May, 2008) (iv) cot–1 (v) log .
s2 2 s b
1
Sol. (i) Let f(s) = log 1 = L{F(t)}, then
s2
1 2 2 1 s
f (s) = 2
1
1 s3 s( s2 1) s s2 1
s2
1 s
L–1 {f (s)} = – 2L–1 2
= – 2(1 – cos t)
s s 1
But L–1 {f (s)} = – tF(t)
2 (1 cos t )
– t F(t) = – 2 (1 – cos t) F(t) =
t
s 3
(ii) Let f(s) = cot–1 = L{F(t)},
2
1 1 2
then f (s) = 2
.
s 3 2 (s 3 )2 4
1
2
LAPLACE TRANSFORMS AND ITS APPLICATIONS 673

1 1
L–1 {f (s)} = – 2 L–1 = – 2e–3t L–1 = – e–3t . sin 2t.
(s 3)2 4 s2 4)
But L–1 {f (s)} = – t F(t)
3t
e sin 2t
– tF(t) = – e–3t . sin 2t F(t) =
t

2 1 2
(iii)Let, f(s) = tan–1 2
tan
s 1 s2 1

2 1 1 1
= tan–1 = tan–1 tan
1 (s 1)( s 1) s 1 s 1

1 1 1 1 1
L–1 {f(s)} = L–1 tan L tan
s 1 s 1

1 1 t 1 1 1
= et L–1 tan e .L tan
s s
1 1 sin t
= 2 sinh t L–1 tan = 2 sinh t .
s t
2
F(t) = sin t sinh t
t
s 1 1 2
(iv) Let, f(s) = cot–1 = L {F(t)}, then f (s) = 2
. 2
2 s 2 s 4
1
4
1
L–1 {f (s)} = – 2 L–1 = – sin 2t
s2 4
But, L–1 {f (s)} = – t F(t)
sin 2t
F(t) =
t
s a
(v) Let, f(s) = log = log (s + a) – log (s + b) = L{F(t)}, then
s b
1 1
f (s) =
s a s b
1 1 1
Now, L–1 {f (s)} = L–1 L = e–at – e–bt
s a s b
But, L–1 {f (s)} = – t F(t)
bt at
e e
– t F(t) = e–at – e–bt F(t) = .
t
54. Find inverse Laplace transforms of the following functions :
s 1
(i) (ii)
(s2 1)(s2 4) s(s 2 a2 )
s s2
(iii) 2 2 (iv) .
(s 4) (s2 a 2 )(s2 b2 )
(M.D.U., Dec., 2008, 2009)
674 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

1 s
Sol. (i) Let f(s) = 2 g(s) = 2
s 4 s 1
1
F(t) = sin 2t, G(t) = cos t
2
1
F(u) = sin 2u, G(t – u) = cos (t – u)
2
s t 1 1 t
L–1 2 2 sin 2u . cos (t – u) du = [sin (u + t) + sin (3u – t)] du
(s 1)( s 4) 0 2 4 0

t
1 cos (3u t)
= cos (u t)
4 3 0

1 cos 2t cos t
= cos 2t cos t
4 3 3

1 4 4 1
= cos 2t cos t = [cos t – cos 2t]
4 3 3 3
1 1
(ii) Let f(s) = 2 2
and g(s) =
s a s
1
F(t) = sinh at, G(t) = 1
a
1
F(u) = sinh au, G(t – u) = 1
a
t t
1 1 1 cosh au
L–1 sinh au . 1 . du =
s( s2 a2 ) 0 a a a 0

1 1
or L–1 (cosh at – 1)
s( s2 a2 ) a2
s 1 s
(iii) 2 2 2
. 2
(s 4) s 4 s 4
1 s
Let f(s) = , g(s) =
s2 4 s2 4
1 1
F(t) = L–1 {f(s)} = L–1 sin 2t
s2 4 2

s
and G(t) = L–1 {g(s)} = L–1 2 = cos 2t
s 4
1
Now, F(u) =
sin 2u, G(t – u) = cos 2(t – u)
2
By convolution theorem, we have
t 1 t
s 1
L–1 sin 2u . cos 2(t – u) du = [sin 2t + sin (4u – 2t)] du
( s2 4 )2 0 2 4 0

t
1 cos ( 4u 2t) 1
= u sin 2t = t sin 2t.
4 4 0
4
LAPLACE TRANSFORMS AND ITS APPLICATIONS 675

s2 s s
(iv) 2
.
(s a )( s2
2
b2 ) s2 a2 s2 b2
s s
Let f(s) = . g(s) =
s2 a2 s2 b2

s
F(t) = L–1 {f(s)} = L–1 = cos at
s2 a2

s
and G(t) = L–1 {g(s)} = L–1 2 = cos bt
s b2
Now F(u) = cos au, G(t – u) = cos b(t – u)
By convolution theorem, we have

s2 t
L–1 2 2 2 2
cos au . cos b(t – u)du
(s a )( s b ) 0

1 t
= [cos {(a – b)u + bt} + cos {(a + b)u – bt}] du
2 0
t
1 sin {( a b )u bt } sin {( a b)u bt } a sin at b sin bt
= = .
2 a b a b 0 a2 b2
t
55. Using the convolution theorem, evaluate sin u cos (t – u) du.
0
Sol. By convolution theorem, we have
t
sin u cos (t – u) du = sin t * cos t ...(1)
0
Here, F(t) = sin t, G(t) = cos t.
1
f(s) = L {F(t)} = L (sin t) = 2
s 1
s
and g(s) = L {G(t)} = L (cos t) =
s2 1

s
sin t * cos t = L–1 {f(s), g(s)} = L–1 2 ...(2)
(s 1)2

s
Now, L–1 = cos at
s2 a2
Differentiating this w.r.t. a, we get
2as
L–1 2
= – t sin at
(s a 2 )2

s t
If a = 1, L–1 sin t
( s2 1)2 2

t 1
From (1) and (2), sin u . cos (t – u) du = t sin t.
0 2
676 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

1 1
56. Prove that : L–1 2 3 [(3 – t2) sin t – 3t cos t].
(s 1) 8

1 1
Sol. We have L–1 sin at
s2 a2 a
Differentiating w.r.t. a, we get

d 1 d sin at 2a 1 t
L–1 2 2 L–1 2 2 2 =– 2
sin at cos at
da s a da a (s a ) a a

1 1 t
L–1 sin at cos at
( s2 a 2 )2 2a 3 2a2
Differentiating again w.r.t. a, we get

4a 3 t t t2
L–1 sin at + cos at cos at sin at .
( s2 a 2 )3 2a4 2a3 a3 2a 2
Putting a = 1, this yields
1 1 1
L–1 2 3 [3(sin t – t cos t) – t2 sin t] = [(3 – t2) sin t – 3t cos t].
(s 1) 8 8

–1 s2
57. Apply convolution theorem to evaluate : L 2 .
(s 4) 2

s
Sol. Since L–1 = cos 2t
s2 4
By convolution theorem
s s t
L–1 2
. 2 cos 2(t – u) cos 2u du
s 4 s 4 0

t
= cos 2u . [cos 2t . cos 2u + sin 2t sin 2u] du
0
t t
= cos 2t cos2 2u du + sin 2t cos 2u . sin 2u du
0 0

1 t 1 t
= cos 2t (1 + cos 4u) du + sin 2t sin 4u . du
2 0 2 0
t t
cos 2t sin 4u sin 2t cos 4u
= u
2 4 0 2 4 0

cos 2t sin 4t sin 2t

= t [1 cos 4t ]
2 4 8
t cos 2t sin 2t 1
= [sin 4t cos 2t – sin 2t cos 4t]
2 8 8
1 1 1 [ 2t cos 2t sin 2t ]
= t cos 2t + sin 2t + sin 2t = .
2 8 8 4
LAPLACE TRANSFORMS AND ITS APPLICATIONS 677

cos 2t cos 3t et 1
58. Find the Laplace transform of the function . (M.D.U., 2005)
t
Sol. Let f(t) = cos 2t – cos 3t – et + 1
s s 1 1
L{f(t)} = L [cos 2t – cos 3t – et + 1] =
s2 4 s2 9 s 1 s

1 s s 1 1
L . f ( t) ds
t s s2 4 s2 9 s 1 s

1 1
= log ( s2 4) log ( s 2 9) log ( s 1) log s
2 2 s

4
1
1 s2 4 s 1 1 s2 1
= log 2 log = log log 1
2 s 9 s 2 9 s
s s 1 s
s2 s

4
1
1
log 1 log s2 log 1 log 1
1
=
2 9 s
1
s2

1 s2 4 s 1 1 s2 9 s 1
=– log 2 log or log 2 log .
2 s 9 s 2 s 4 s

e at cos bt
59. If F(t) = , find the Laplace transform of F(t).
t
1 s
Sol. L (eat) = , L (cos bt) =
s a s2 b2
1 s
L (eat – cos bt) =
s a s2 b2

e at cos bt 1 s 1
Now, L ds = log ( s a) log ( s2 b2 )
t s s a s 2
b 2 2 s

2
a
2 1
1 1 (s a) 1 s
= log ( s a )2 log ( s2 b2 ) = log 2 = log
2 2 s b2 s
2 b2
s 1
s2
s
2
1 ( s a) 1 s2 b2
=– log 2 = log .
2 s b2 2 ( s a )2

sin at cos at
60. Find the Laplace transform of . Does the Laplace transform of exist ?
t t
(U.P.T.U., 2005)
sin at sin at
Sol. Since lim = a, the Laplace transform of exists.
t 0 t t
678 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

a
Now, L (sin at) =
s2 a2
sin at a 1 s 1 s s a
L 2 2
ds = tan tan = cot–1 = tan–1 .
t s s a a 2 a a s
s
cos at
The function is discontinuous at t = 0, so that the Laplace transform does not exist.
t
61. Apply the method of Laplace transform to solve differential equation :
d2x
dx
+2 + 5x = e–t sin t, x (0) = 0, x (0) = 1.
dt2 dt
Sol. The given equation is x + 2x + 5x = e–t . sin t
Taking the Laplace transform of both sides, we get
1
[s2 x – s.x(0) – x (0)] + 2[s x – x(0)] + 5 x =
(s 1)2 1
1 2 1
Since, L {f n(t)} = sn . f ( s ) sn . f ( 0) sn . f ( 0 ) ...... f n ( 0)
Using the given conditions x(0) = 0, x (0) = 1, it reduces to
1
(s2 + 2s + 5) x – 1 =
s2 2s 2
1 1
or x =
( s2 2s 2)( s 2 2s 5) s2 2s 5
1 1 1 1
=
3 s2 2s 2 s 2
2s 5 s 2
2s 5

1 1 2 1 1 2
= = 3
3 s2 2s 2 s2 2s 5 (s 1)2 1 (s 1)2 4
Taking the inverse Laplace transform of both sides, we get

1 1 1 1
x= L 2.
3 (s 1)2 1 (s 1)2 22

1 1 1 t
. e t sin t 2 . . e t . sin 2 t
= or x = e (sin t sin 2 t)
3 2 3
which is the required solution.
3t
62. Solve the equation : y – 3y + 2y = 4t + e , where y(0) = 1, y (0) = – 1.
Sol. Taking the Laplace transform of both sides, we have
4 1
[ s2 y sy( 0 ) y ( 0 )] 3 [ sy y( 0 )] 2y 2
s s 3
Using the given conditions, it reduces to
4 1 4 1
( s2 3s 2) y s 1 3 or ( s2 3s 2) y s 4
s2 s 3 s2 s 3

s4 7s3 13s2 4s 12
y 2
s (s 1)( s 2 )( s 3)
LAPLACE TRANSFORMS AND ITS APPLICATIONS 679

3 2 1 2 1
or y 2 (Resolving into partial fractions)
s s 2( s 1) s 2 2( s 3)
Taking inverse Laplace transform of both sides, we get
1 1 1 1 1 1 1 1 1 1 1
y = 3L–1 2L 2
L 2L L
s s 2 s 1 s 2 2 s 3
1 t 1 3t
e 2e 2t
y = 3 + 2t – e
2 2
1
or y = 3 + 2t + (e3t – et) – 2e2t, which is the required solution.
2
63. Solve the D.E. by the method of Laplace transformation :
y – 8y + 15y = 9 te2t, y(0) = 5, y (0) = 10. (M.D.U. Dec., 2009)
Sol. Taking L.T. of both sides of the D.E., we get
9
[s2 y – s.y(0) – y (0)] – 8[s y – y(0)] + 15 y =
s2 ( s 2)
Under the given conditions, it reduces to
9
[s2 y – 5s – 10] – 8 (s y – 5) + 15 y =
s2 ( s 2)
9
(s2 – 8s + 15) y = + 5s – 30
s2 ( s 2)

9 9 s2 (5 s 30)(s 2)
(s – 5) (s – 3) y = + 5s – 30 =
2
s (s 2) s2 ( s 2)

9 s 2 [ 5s 2 10s 30s 60] 9 s 2 [ 5s 2 40s 60]

= 2 =
s (s 2) s2 ( s 2)

9 5 s4 40 s3 60s2 5s4 40 s3 60 s2 9
y = 2 = 2
s (s 2)(s 3)(s 5) s (s 2)(s 3)(s 5)
3 2 3 7 2.44
=
s s2 4(s 2) s 3 s 5
(Resolving by Partial Fractions)
Taking the inverse L.T. of both sides, we get
1 1 3 1 1 1 1
y = 3L–1 + 2.L–1 + L + 7L–1 – 2.44 L–1
s s2 4 s 2 s 3 s 5
3 2t
or e + 7e3t – 2.44 e5t
y = 3 + 2t +
4
which is the required solution.
64. Solve the equation :

d3 y d2 y dy dy d2 y
2. 2y 0 , where y = 1, 2, 2 at t 0.
dt3 dt 2 dt dt dt 2
Sol. The given equation is y + 2y – y – 2y = 0
Taking the Laplace transform of both sides, we get
[ s3 y s2 y ( 0 ) sy ( 0 ) y ( 0 )] 2[ s2 y sy( 0 ) y ( 0 )] – [ sy y( 0 )] 2y 0 ...(1)
680 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

Using the given conditions y(0) = 1, y (0) = 2, y (0) = 2, equation (1) reduces to
( s3 2s 2 s 2) y s2 2s 2 2s 4 1 0 or (s3 + 2s2 – s – 2) y = s2 + 4s + 5
s2
s2 4 s 5 4s 5 5 1 1
y 3 2 =
s 2s s 2 ( s 1)( s 1)( s 2 ) 3( s 1) s 1 3( s 2 )
(After resolving into partial fractions)
Taking the inverse Laplace transforms of both sides, we get
5 1 1 1 1 1 1 1 5 t 1 –2t
y= L L L = e – e–t + e
3 s 1 s 1 3 s 2 3 3

1
or (5et + e–2t) – e–t which is the required solution.
y=
3
65. Using Laplace transformation method, solve the differential equation :
d2 x
+ 9x = cos 2t, if x(0) = 1, x = – 1.
dt2 2
Sol. The given equation is x + 9x = cos 2t
Taking the Laplace transform of both sides, we get
L(x ) + 9L(x) = L (cos 2t)
s s
[ s2 x s . x ( 0) x ( 0 )] 9x (s2 + 9) x s A | where A = x (0)
s2 4 s2 4
s s A
x
( s2 4 )( s2 9) s2 9 s2 9

1 s s s A
=
5 s2 4 s2 9 s2 9 s2 9
Taking the inverse Laplace transform of both sides, we get
1 A
x= (cos 2t – cos 3t) + cos 3t + sin 3t
5 3

But x =–1 (given)

2
1 A
x=–1=
(– 1 – 0) + 0 + (– 1)
5 3
1 A 12
–1=– – or A =
5 3 5
1 4
x(t) = (cos 2t – cos 3t) + cos 3t + sin 3t
5 5
1
or x= (cos 2t + 4 cos 3t + 4 sin 3t)
5
This is the required solution.
66. Solve the following D.E. by using Laplace transformation y + y = t cos 2t, y(0) = 0, y (0) = 0.
(M.D.U. Dec., 2008)
Sol. y + y = t cos 2t,
Taking L.T. on both sides,
L(y ) + L(y) = L(t cos 2t)
d s s2 4 s2 4
s2 – s (y(0) – y (0) + y = or (s2 + 1) y = or y = 2
ds s2 4 (s 2
4) 2
(s 1) (s2 4) 2
LAPLACE TRANSFORMS AND ITS APPLICATIONS 681

s2 4 A B C
Now, 2 2 = 2
(s 1) (s 4) 2 s 1 s2 4 ( s2 4) 2
Let, s2 = P
P 4 A B C
2
= ...(1)
(P 1)(P 4) P (P 4) 2
1 P 4
2
P – 4 = (P + 4) A + (P + 1) (P + 4) B + (P + 1) C
5
Putting P = – 1, A=
9
8
Putting P = – 4, C=
3
Putting P = 0, – 4 = 16 A + 4B + C
80 8 80
8
–4= 4B or – 4 + = 4B
9 3 39
80 36 24 20 5
4B = = or B =
9 9 9
Put these values of A, B and C in (1)
P 4 5 5 8
2 =
(P 1)(P 4) 9 (P 1) 9 (P 4) 3 (P 4) 2
5 5 8
Putting P = s2, y =
9 (s2 1) 9 (s2 4) 3 ( s2 4) 2
Taking L.T. on both sides,
5 5 1 8 1
y= sin t . sin 2t (sin 2t 2 cos 2t)
9 9 2 3 2 8
1
which on simplification gives y=
[4 sin 2t – 5 sin t – 3 t cos 2t].
9
67. Using Laplace transformation, solve the differential equation :
(D2 + n2)x = a sin (nt + ) ; x = Dx = 0 at t = 0.
Sol. The given equation is (D2 + n2)x = a sin (nt + )
Taking Laplace transform of both sides, we get
L(x ) + n2 L(x) = L{a sin (nt + )}
n s
[ s2 . x s . x ( 0) x ( 0] n2x a cos . 2 2
a sin . 2
s n s n2
an cos as sin
(s2 + n2) x 2 2
s n s2 n2
an cos as sin
x ...(1)
( s2 n 2 )2 ( s2 n 2 )2
Taking the Inverse Laplace transform of both sides, we get

n 1 s
x = (a cos ) L–1 ( a sin ) L ...(2)
( s2 n 2 )2 ( s2 n 2 )2

1 1
We know that L–1 2 2 sin nt ...(3)
s n n
682 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

d 1 nt cos nt sin nt 2n nt cos nt sin nt

L–1 2 2 2
or L–1 2 2 2 2
dn s n n (s n ) n

n 1
L–1 (sin nt – nt cos nt)
( s2 n 2 )2 2n 2

d 1 1
Again, from (3), L–1 =–t. sin nt
ds s2 n 2 n

2s t s t
L–1 2 2 2 sin nt L–1 2 2 2 sin nt
(s n ) n (s n ) 2n

1 t
From (2), x = (a cos ) . (sin nt – nt cos nt) + (a sin ) . . sin nt
2n2 2n

a
or x=
[cos sin nt – nt cos ( + nt)]
2n2
This is the required solution.
68. Solve the differential equation by Laplace transform

y – 2y + 5y = 0 ; y = 0, y = 1 at t = 0 and y = 1 at t = .
8
Sol. Taking Laplace transform on both sides, we get
L(y ) – 2L(y ) + 5 . L(y ) = L(0)

[ s3 y s 2 y( 0 ) s . y ( 0) y ( 0 )] 2 . [ s2. y s . y( 0) y ( 0 )] + 5 [ sy y(0)] 0
(s3 – 2s2 + 5s) y s A 2 0 |y (0) = A, say
A 2 s A 2 1 s 2 1
y 2
=
5 s 2 2
s (s 2s 5) s 2s 5 s 2s 5

A 2 1 A 2 (s 1) 1 1
= 2
5 s 5 (s 1) 4 (s 1) 2 4

A 2 1 A 2 s 1 A 3 2
y
5 s 5 (s 1)2 4 10 (s 1)2 4
Taking inverse Laplace transform on both sides, we get
A 2 A 2 A 3 t
y= [ e t cos 2t ] e sin 2t
5 5 10
Since y = 1 when t = /8
A 2 A 2 /8 1 A 3 /8 1
1= e . e .
2 5 5 10 2
A (On simplification)
The required solution therefore is y = 1 + et (sin 2t – cos 2t).
69. Solve by the method of Laplace transforms, the differential equation
y + 2y – y – 2y = 0 given y(0) = y (0) = 0 and y (0) = 6.
Sol. Taking the Laplace transform of both sides, we get
[ s3 y s2 y ( 0 ) sy ( 0 ) y ( 0 )] 2 [ s2 y sy ( 0 ) y ( 0 )] [ sy y ( 0)] 2y 0
LAPLACE TRANSFORMS AND ITS APPLICATIONS 683

Using the given conditions, it reduces to (s3 2s2 s 2) y 6

6 6 6 6
y
( s 1)( s 1)( s 2 ) ( s 1)6 ( 2 )( s 1) 3( s 2)
On taking inverse Laplace transforms, we get
1 1 1 1 1
y = L–1 3.L 2L
s 1 s 1 s 2
or y= – + et 3e–t 2e–2t
which is the required solution of the given equation.

d2 x dx dx
70. Solve : 2
2 + x = et with x = 2, = – 1 at t = 0.
dt dt dt
Using Laplace transform method.
Sol. Taking the Laplace transform of both sides, we get
1
[ s2 x s. x(0) x (0)] 2[ sx x(0)] x
s 1
Using the given conditions, it reduces to
1 2 s2 7s 6
( s2 2s 1) x 2s 5
s 1 s 1
2
2s 7s 6
x
(s 1)3
2 3 1
x on resolving into partial fractions.
s 1 (s 1)2 (s 1)3
On taking inverse Laplace transforms of both sides, we get
1 1 1 1 1
x = 2L –1 3L L
s 1 (s 1)2 (s 1)3
t t2 1 2 t
et .
x = 2et – 3et .
or x = 2et – 3tet + t .e.
1! 2! 2
which is the desired solution of the given equation.
71. Solve (D3 – 3D2 + 3D – 1)y = t2et given that y(0) = 1, y (0) = 0, y (0) = – 2.
Sol. Taking Laplace transforms of both sides, we get
2
[ s3 y s2 y(0) sy (0) y (0)] 3 [ s2 y sy(0) y (0)] 3[ sy y(0)] y
(s 1)3
n!
Since, L(eat . tn) =
.
( s a) n 1
Using the given conditions, it reduces to
2
( s3 y s2 2) 3(s2 y s) 3(sy 1) y
(s 1)3
2
(s3 3 s2 3s 1) y s2 2 3s 3
(s 1)3
2
( s3 3 s2 3s 1) y s2 3s 1
(s 1)3
684 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

2
(s 1)3 y ( s2 3s 1)
(s 1)3

s2 3s 1 2 (s 1)2 (s 1) 1 2
y =
(s 1) 3
(s 1) 6 (s 1)3 (s 1)6
1 1 1 2
or y
s 1 (s 1)2 (s 1)3 (s 1)6
Taking Laplace transform inversion, we obtain
3
1 1 1 1 1 1 1
y = L–1 L L 2L
s 1 (s 1)2 s 1 (s 1)6
1 2 1 5
or y = et 1t t t
2 60
which is the required solution of the given equation
72. Solve the differential equation ty + 2y + ty = cos t given that y(0) = 1.
Sol. Taking Laplace transform of both sides of the equation and noting that
d
L{tf(t)} = – [L{ f (t)}] , we get
ds
d 2 d s
– [s y sy(0) y (0)] 2[ sy y(0)] ( y) 2
ds ds s 1
dy d s
or – s2 2s y y(0) 0 2sy 2 y(0) ( y) 2
ds ds s 1
dy s dy 1 s
or (s2 + 1) 1 or
ds s2 1 ds s2 1 ( s2 1) 2
On taking inverse Laplace transforms and noting that L–1 { f ( s)} = – tf(t), we get
1 1 2
– ty = – sin t – t sin t or y = 1 sin t which is the solution.
2 2 t

d2x
73. Solve + 9x = cos 2t, given that x(0) = – 1 and x (0) = 2 using Laplace transform.
dt 2

d2 x
Sol. + 9x = cos 2t
dt2
Taking Laplace transform of both sides of the given equation, we have
L(x ) + 9.L(x) = L(cos 2t)
s
s2 L(x) – sx(0) – x (0) + 9.L(x) =
s2 4
s
(s2 + 9) L(x) – s(– 1) – 2 =
s2 4
s
or (s2 + 9) L(x) = –s+2
s2 4
s s 2
or L(x) =
( s2 4)(s2 9) ( s2 9) ( s2 9)
s s s 2
=
5(s2 4) 5(s2 9) ( s2 9) ( s2 9)
LAPLACE TRANSFORMS AND ITS APPLICATIONS 685

s 6s 2
=
5(s2 4) 5(s2 9) s2 9
1 –1 s 6 1 s 1 1
x= L L 2.L
5 s2
4 5 2
s 9 2
s 9
1 6 2
or x=
cos 2t – cos 3t + sin 3t
5 5 3
which is the desired solution of the given equation.
74. Solve : y + y – 2y = t, given that y(0) = 1 and y (0) = 0, using Laplace transform.
dy d2 y
Sol. Given equation is – 2y = t
dt2 dt
Taking Laplace transform of both sides
L(y ) + L(y ) – 2L(y) = L(t)
1
s2L(y) – sy(0) – y (0) + sL(y) – y(0) – 2L(y) = 2
s
1 1
s2L(y) – s – 0 + sL(y) – 1 – 2L(y) = 2 or (s2 + s – 2) L(y) = 2 + s + 1
s s
1 s 1
L(y) = 2 2
s (s s 2) s2 s 2
Taking inverse Laplace transform, we have
1 1 s 1
y = L–1 2 2
L 2
s (s s 2) s s 2
1 1 1 1
or y = L–1
s 1 4(s 2) 2 s2 4s
(After resolving into partial fractions)
1 2t t 1
or ey = et +
4 2 4
is the desired solution of the given equation.
75. Solve : y + 4y + 3y = e–t, y(0) = y (0) = 1 using Laplace transform.
Sol. Taking Laplace transform on both sides, we get
L(y ) + L(4y ) + 3.L(y) = L(e–t)
1
[ s2 y sy(0) y (0)] 4(sy y(0)) 3y
s 1
1
( s2 4s 3) y s 1 4
s 1
1
( s2 4s 3) y s 5
s 1

s2 6s 6 s2 6 s 6 s2 6 s 6
y =
(s 1)(s2 4s 3) (s 1)(s 1)( s 3) (s 1)2 ( s 3)
2
s 6s 6 A B C
Let
(s 1)2 ( s 3) s 1 (s 1)2 s 3
s2 + 6s + 6 = A(s + 1)(s + 3) + B(s + 3) + C(s + 1)2
686 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

1
Put s = – 1, 2B = 1 B=
2
3
Put s = – 3, 4C = – 3 C=
4
Comparing constant terms on both sides, we have
3 3 7
3A + 3B + C = 6 or 3A + =6 A=
2 4 4
7 1 1 1 3 1
y .
4 s 1 2 ( s 1)2 4 s 3
Taking inverse Laplace transform, we have
7 t 1 3 3t
e y= .t e t e
4 2 4
which is the desired solution of the given equation.
76. A voltage Ee–at is applied at t = 0, to a circuit of inductance L and resistance R. Show that the
E at Rt / L
current at time t is (e e ).
R aL
Sol. Let I be the current in the circuit at any time t, then by Kirchhoff ’s law, we have
dI
+ R.I. = E.e–at, where I(0) = 0
L
dt
Taking Laplace transform of both sides, we have
E
L . [s I I(0)] + RI
s a
Using the given condition, it reduces to
E E E 1 L
(Ls + R) I or I
s a (s a)(Ls R) R aL s a Ls R
Taking the inverse Laplace transform of both sides, we get

E E 1 1
I= . L–1 . [ e at e Rt /L ] .
R aLR R a L s a
s
L
77. Solve the simultaneous equations (D2 – 3)x – 4y = 0, x + (D2 + 1)y = 0 for t > 0, given that x = y
dy dx
= = 0 and = 2 at t = 0.
dt dt
Sol. Taking the Laplace transforms of the given equations, we get
s2 . x sx(0) x (0) 3x 4y 0 or (s2 3) x 4y 2 ...(1)
and x s2 y sy(0) y (0) y 0
or x (s 2
1) y 0 ...(2)
Solving (1) and (2) for x and y , we get
2( s2 1) 1 1
x =
( s2 1)2 (s 1)2 (s 1)2
2 1 1 1 1 1
and y=– 2 2 2
(s 1) 2 s 1 s 1 (s 1) (s 1) 2
LAPLACE TRANSFORMS AND ITS APPLICATIONS 687

Taking the inverse Laplace transforms of both sides, we get

1 1 et e t
x = L–1 2 2
= tet + te–t = 2t or x = 2t cosh t
(s 1) (s 1) 2

1 1 1 1 1 1 1
and y=– L =– . (e–t – et – te–t + tet)
2 s 1 s 1 (s 1)2 (s 1)2 2

et e t
= (1 – t) = (1 – t) sinh t
2
Hence the solution is given by x = 2t cosh t and y = (1 – t) sinh t.
dx dy
78. Solve the simultaneous equations : – y = et ; + x = sin t, given that x(0) = 1, y(0) = 0.
dt dt
Sol. Taking the Laplace transforms of the given equations, we get
1 1
{ sx x(0)} y i.e., sx 1 y
s 1 s 1
1 s
sx y 1 ...(1)
s 1 s 1
1 1
and { sy y(0)} x 2 or x sy 2 ...(2)
s 1 s 1
Solving (1) and (2) for x and y , we get

s2 1 1 1 s 1 1
x =
(s 1)(s2 1) ( s2 1)2 2 s 1 s2 1 s2 1 ( s2 1)2

s s s 1 1 s 1
and y
( s2 1)2 (s 1)(s2 1) (s2 1)2 2 s 1 s2 1 s2 1
Taking the inverse Laplace transforms of both sides, we get
1 1 1 s 1 1 1
x= L L
2 s 1 s2 1 s2 1 ( s2 1)2
1 1
= . {et + cos t + sin t} + (sin t – t cos t)
2 2
1 1
L–1 (sin at – at cos at)
( s2 a2 ) 2 2a 2
1 t
x= [e + cos t + 2 sin t – t cos t]
2
s 1 1 1 s 1
y = L–1 2 2
L 2 2
(s 1) 2 s 1 s 1 s 1

1 1 t 1 s 1
= t sin t – {e – cos t + sin t} L 2 2 2
t sin at
2 2 (s a ) 2a
1
= {t sin t – et + cos t – sin t}
2
688 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

1 t 1
Hence x=
(e + cos t + 2 sin t – t cos t) ; y = (t sin t – et + cos t – sin t)
2 2
is the required solution of the given equation.

d2 x dy dx d2 y
79. Solve the simultaneous equations : 5 x t, 2 4y 2, given that when t = 0,
dt 2 dt dt dt 2
dx dy
x = 0, y = 0, 0, 0.
dt dt
Sol. Let L{x(t)} = x (s), L{ y(t)} y (s)
Taking Laplace transform of the given equations, we get
1
{ s2 x x(0) x (0)} 5{ sy y(0)} x
s2
2
and 2 { sx x(0)} { s2 y y(0) y (0)} 4y
s
Using the given initial conditions, these equations reduce to
1 2
(s2 – 1) x 5sy ...(1) and 2 sx (s2 4) y ...(2)
s2 s
Eliminating y between (1) and (2), we have

s2 4
{(s2 – 1)(s2 – 4) + 10s2} x + 10
s2

11s2 4 1 5 4
x 2
s (s 2
1)( s2 4) s2 s2 1 s2 4
Taking inverse Laplace transform, we get
x = – t + 5 sin t – 2 sin 2t ...(3)
Again eliminating x between (1) and (2), we have
2 2(s2 1)
{10s2 + (s2 – 1) (s2 – 4)} y
s s
4 2 s2 1 2s s
y or y 2 2
2 s 1 s 2 s 4
s(s 1)( s 4)
Taking inverse Laplace transform, we get
y = 1 – 2 cos t + cos 2t
Hence the desired solution of the given simultaneous equations is
x t 5 sin t 2 sin 2t
y 1 2 cos t cos 2t. .

dx dy
80. Use Laplace transform to solve the simultaneous equations y sin t, x cos t, given that
dt dt
x = 2, y = 0 at t = 0.
Sol. Taking Laplace transform of the given equations, we get
1 1
sx x(0) y sx y +2 ...(1)
s2 1 s2 1
s s
and sy y(0) x 2
x sy 2 ...(2)
s 1 s 1
LAPLACE TRANSFORMS AND ITS APPLICATIONS 689

Solving (1) and (2) for x and y , we get

2s 1 2
x ,y
s2 1 1 s2 1 s2
1 1 1 1 1
x ,y
s 1 1 s2 s 1 s 1
s 1
Taking Laplace transform on both sides, we get
x = e–t + et, y = sin t + e–t – et which is the desired solution.
81. The co-ordinates (x, y) of a particle moving along a plane curve at any time t are given by
dy dx
+ 2x = sin 2t, – 2y = cos 2t ; (t > 0).
dt dt
It is given that at t = 0, x = 1 and y = 0.
Show using Laplace transforms that the particle moves along the curve 4x2 + 4xy + 5y2 = 4.
Sol. The given equations are
dy dx
+ 2x = sin 2t ...(1) – 2y = cos 2t ...(2)
dt dt
Above equations may be rewritten as
d
2x + Dy = sin 2t or Dx – 2y = cos 2t, where D =
dt
Taking Laplace transform of equation (1) on both sides, we get
2
2x sy y(0) , where x L( x), y L( y)
s2 4
2
2x sy 2 ...(3)
s 4
Again taking Laplace transform of equation (2) on both sides, we get
s s
sx x(0) 2y 1 sx 2y ...(4)
s2 4 s2 4
Multiplying equation (3) by 2 and equation (4) by s and then adding, we get
4 s2 1 s 1 s
4x s2 x + s = 1 + s or x
s2 4 s2 4 4 s2 4 s2 4 s2
Taking inverse Laplace transform, we get
1
x= sin 2t + cos 2t ...(5)
2
Similarly, we solve equations (3) and (4) to obtain y as follows :
Multiplying equation (3) by s and equation (4) by 2 and then subtracting equation (4) from
equation (3), we get
2s 2s 2
s2 y 4y 2 or y
s2 4 s2 4 s2 4
Taking inverse Laplace transform, we get
y = – sin 2t ...(6)
1
Now, 4x2 = 4 sin 2 2t cos2 2t sin 2t cos 2t
4
5y2 = 5 sin2 2t
690 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

1
4xy = 4 sin 2t cos 2t ( sin 2t)
2
or 4xy = – (2 sin2 2t + 4 sin 2t cos 2t)
4x2+ 5y2
+ 4xy = 4 sin2 2t + 4 cos2 2t or 4x2 + 5y2 + 4xy = 4.
This shows the particle moves along this curve since the co-ordinates (x, y) given by equations
(5) and (6) satisfy the curve.
82. Solve y + ty – y(t) = 0 if y(0) = 0 and y (0) = 1.
Sol. Taking the Laplace transform of both sides of the given equation, we get
L(y ) + L(ty ) – L(y) = L(0)
d d
or { s2 y sy(0) y (0)} L( y ) y 0 s2 y 1 { sy y(0)} y 0
ds ds
d dy 2
s2 y 1 (sy) y 0 –s + (s 2) y 1
ds ds
dy 2 1
+ s y
ds s s
dy
This is a linear differential equation of the type + Py = Q, where P and Q are functions of x
dx
(or s) alone.
2
s ds
s2 / 2
I.F. = e s s2 . e
Solution of the differential equation is
s2 / 2 1 2 s2 / 2 s2 / 2
y s2 e s e C = se ds C
s

=C+ e s2 / 2 , where C is a constant.

C must vanish if y is a transform since y 0 as s .

1 1
y = or y = L–1 =t
s2 s2
Hence y = t is the desired solution.
dx dy
83. Solve the simultaneous equations + 5x – 2y = t, + 2x + y = 0, given that x = y = 0, when t = 0,
dt dt
by the Laplace transform method.
Sol. Taking the Laplace transform of the given equations, we get
1 1
[ sx x(0)] 5x 2y i.e., (s + 5) x 2y ...(1)
s2 s2
and sy y(0) 2x 2y 0 i.e., 2x (s 1) y 0 ...(2)
Solving (1) and (2) for x ,
s 1 1 1 1 2
x = =
s2 ( s 3) 2 27 s 9 s2 27(s 3) 9(s 3) 2
Substituting the value of x in (2), we get
2 4 2 4 2
y=– or y
s 2 (s 3)2 27 s 9 s2 27( s 3) 9(s 3) 2
LAPLACE TRANSFORMS AND ITS APPLICATIONS 691

On taking inverse Laplace transform, we get

1 t 1 3t 2 3t 4 2t 4 3t 2 3t
x= e te , y= e te
27 9 27 9 27 9 27 9
which is the desired solution of the given simultaneous differential equations.
dx dy
84. Solve the equations: – 6x + 3y = 8et, – 2x – y = 4et given that x(0) = – 1, y(0) = 0, by Laplace
dt dt
transformation. (M.D.U. May, 2009)
dx
Sol. – 6x + 3y = 8et ...(1)
dt
dy
– 2x – y = 4et ...(2)
dt
x(0) = – 1, y(0) = 0.
(D – 6)x + 3y = 8et
(D – 1)y – 2x = 4et ...(3)
Eliminating y in (3),
(D – 6) (D – 1)x + 3(D – 1)y = 8(D – 1)et or 3(D – 1)y – 6x = 12 et
(D – 6)(D – 1)x + 6x = 8(D – 1)et – 12et or [(D – 6)(D – 1) + 6]x = 8(D – 1)et – 12et
(D2 – 7D + 12)x = – 12et
Its A.E. is D2 – 7D + 12 = 0 D = 4, 3
C.F. = C1e–4t + C2e–3t
12 et 12et
P.I. = = = – 2et
D2 7D 12 1 7 12
x = C1e–4t + C2e–3t – 2et
dx
= – 4 C1e–4t – 3C2e–3t – 2et
dt
dx
Putting the values of x and in equation (1)
dt
(– 4C1e–4t – 3C2e–3t – 2et) – 6 (C1e–4t + C2e–3t – 2et) + 3y = 8et
or – 10 C1e–4t – 9C2e–3t + 10et – 8et = – 3y
10 2 t
or y= C1e–4t + 3C2e–3t – e.
3 3
2 2 2
85. Solve by transform method, the simultaneous differential equations D x + 3x – 2y = 0, D x+D y
– 3x + 5y = 0 Given x = 0, y = 0, Dx = 3, Dy = 2 when t = 0.
Sol. Taking Laplace transform of both the equations, we get

[ s2 x sx(0) x (0)] 3x 2y 0
i.e., (s2 + 3) x 2y 3 ...(1)
and [ s x2 2
s x(0) x (0)] [s y sy(0) y (0)] 3x 5y 0
2
i.e., (s2 – 3) x (s 5) y 5 ...(2)
Solving (1) and (2) for x and y , we get
3 s2 25 11 1 1 1 2 s2 24 11 1 3 1
x = 2 2
. 2 . 2 and y = 2 . 2 . 2
(s 1)(s 9) 4 s 1 4 s 9 (s 1)(s2 9) 4 s 1 4 s 9
692 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

On taking inverse Laplace transform, we get

11 1 11 1
x= sin t sin 3t and y= sin t sin 3t
4 12 4 4
which is the desired solution.

d2y dy dy
86. Solve the equation : t 4ty 0 given that y = 3 and = 0 when t = 0.
dt 2 dt dt
Sol. Taking Laplace transform on both sides of the given equation, we have
L(ty ) + L(y ) + 4L(ty) = L(0)
d d
–. L(y ) + L(y ) – 4 . L(y) = 0
ds ds
d dy
– { s2 y sy(0) y (0)} { sy y(0)} 4 0
ds ds
dy
(s2 + 4) sy 0
ds
dy sds
Separating the variables, we have 2 = 0
y s 4
1 C
On integration, we get log y + log (s2 + 4) = log C y
2 s 2
4
Taking inverse Laplace transform
1 1
y = C.L–1 2
or L–1 = J0 (t),
s 4 s2 1
called Bessel function of order zero.
t2 t4 t6
J0(t) = 1 – 2 2 2 + ......
2 2 .4 2 . 42 . 62
2

y = C J0(2t)
Since y(0) = 3, from above y(0) = C. J0 (0) = C, C=3
Hence the required solution is y = 3 . J0 (2t).
87. Using Laplace transform, solve the differential equation y + 2t y – y = t, when y(0) = 0 and y (0)
= 1.
Sol. Taking Laplace transform on both sides, we get
L(y ) + 2L(ty ) – L(y) = L(t) ...(1)
d 1
s2 y sy(0) y (0) 2 L( y ) y
ds s2
d 1
s2 y 1 2. { sy y(0)} y
ds s2
dy 1
s2 y 1 2. s y y
ds s2
dy 1
– 2s y ( s2 3) 1
ds s2
dy 3 s2 1 1
y 3
...(2)
ds 2s 2s 2s
LAPLACE TRANSFORMS AND ITS APPLICATIONS 693

This is a linear differential equation of order 1 in y and s.

1 3 1 s2
s ds 3 log s
2 s 2 2 ( s 2 /4 )
I.F. = e = e e . s3 / 2
Solution to equation (2) is
s2 / 4 1 1 1 3/ 2 s2 /4 1 1 s 2 /4
y .e . s3 / 2 s .e ds = – s e ds
2 s 3 s 2 s3 / 2
dt
Put s2 = 4t s = 2 t , ds =
t
1 1 s 2 /4
R.H.S. integral s 3/2
e . ds
2 s
1 1 dt 1 1
= 2 t 1/ 4 t 3/4
e t
. = t 1/ 4
t 5/ 4
e t
. dt
2 2 2 t 2 4
t
1 1/4 e 1 5/4 t 1 5/ 4 t
=– t . .t .e dt t .e dt
2 ( 1) 4 4
1
1 t 1/4 1 s2 / 4 s2 4 1 s 2 / 4)
= e .t = e e(
2 2 4 s

s2 / 4 1 ( s 2 / 4) 1
y.e . s3 / 2 e y C, where C is a constant.
s s2
C must vanish if y is a transform since y 0 as s .

1
y
s2

1
Taking inverse Laplace transform, we get y = L–1 = t, which is the required solution.
s2

dx dy d2x dy
88. Solve the simultaneous equations – 2x + 2y = 1 – 2t, 2 x 0 . Given that
dt dt 2 dt
dt
dx
x = 0, y = 0, 0 when t = 0, by using Laplace transforms. (M.D.U., 2005)
dt
dx dy
Sol. – 2x + 2y = 1 – 2t ...(1)
dt dt
dy d2 x
2 +x= 0 ...(2)
dt 2 dt
Taking Laplace transform on both sides of equation (1), we get
1 2
(sx x(0)) [ sy y(0)] 2x 2y
s s2
s 2
sx sy 2x 2y 2
s
s 2 1 1
(s 2) x y (s 2) 2
x y or y x
s s 2
s2
694 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

Taking Laplace transform of both sides of equation (2), we get

[ s2 x sx(0) x (0)] 2[ sy y(0)] x 0
2
s x 2sy x 0 or (s2 + 1) x + 2s y = 0
1
Putting the value of y , y = x , we get
s2
1 2 2
(s2 + 1) x 2s x = 0 or (s2 + 2s + 1) x or x
s2 s s( s 1)2
Taking inverse Laplace transform
2
x = L–1
s(s 1)2

2 2 2
x = L–1 or x = 2 – 2.e–t – 2t . e–t
s s 1 (s 1)2
x = 2(1 – e–t – te–t)
1 2 1
Now y x or y
s2 s( s 1) 2 s2
Taking inverse Laplace transform of this equation

2 1 1
y = L–1 L = 2(1 – e–t – te–t) – t
s(s 1)2 s2
or y = 2(1 – e–t) – t(1 + 2e–t)
Hence the solution of the given simultaneous equations is
x = 2(1 – e–t – te–t)
y = 2(1 – e–t) – t(1 + 2e–t).
d2 x d2 y
89. By using Laplace transform, solve the equation y 5 cos 2t , x 5 cos 2t , where
dt 2 dt 2
x(0) = x (0) = y (0) = 1, y(0) = – 1. (M.D.U., 2005)
2
d y
Sol. Taking Laplace transform of both sides of the equation + x = 5 cos 2t, we have
dt 2
s
[ s2 y sy(0) y (0)] x 5. 2
s 4
5s
s2 y s 1 x ...(1)
s2 4

d2 x
Now, + y = – 5 cos 2t
dt 2
Taking Laplace transform on both sides, we get
5s
[ s2 x sx (0) x (0)] y
s2 4
s
or s2 x – s – 1 + y = – 5 . . ...(2)
s2 4
LAPLACE TRANSFORMS AND ITS APPLICATIONS 695

To eliminate y from (1) and (2), we multiply equation (2) by s2 and subtract from equation (1) and
we have
5s 5s3
s2 y + s – 1 + x – s4 x + s2(s + 1) – s2 y =
s2 4
5s 5s3
(1 – s4) x + s – 1 + s2 + s3 = 2
s 4
5s 5s3
(1 – s4) x = 2 – s + 1 –s2 – s3
s 4
5s(1 s2 )
or (1 – s4) x = 2 + (1 – s2) – s(1 + s2)
(s 4)
5s(1 s2 ) 1 s2 s(1 s2 ) 5s 1 s
x = 4 2 4 =
(1 s )(s 4) 1 s 1 s4 (1 s2 )(s2 4) 1 s2 1 s2

5s s 1 5s s( s2 4) 1
= =
(1 2
s )(s 2
4) 1 s2
1 2
s (1 s )(s2
2
4) 1 s2
s 1
x =
s2 4 1 s2
Taking inverse Laplace transform, we get
s 1 1
x = L–1 L or x = cos 2t + sin t
s2 4 1 s2
To eliminate x from equations (1) and (2), we multiply equation (1) with s2 and subtract equa-
tion (2) from the same. Thus, we get
5s(1 s2 )
(s4 – 1) y = 2 + s2 – 1 – s(1 + s2)
s 4
1 s 5s s 1
or y= y =– 2
( s2 1)( s2 4) s2 1 s2 1 s 4 s2 1
Taking inverse Laplace transform, we have
y = – cos 2t + sin t.
Hence the solution of the given simultaneous differential equations is given by
x sin t cos 2t
y sin t cos 2t .
90. Discuss and describe some other useful functions and how to find their Laplace transforms.
(M.D.U. May, 2009)
Sol. 1. Unit Step Function or Heaviside’s Unit Function
Definition : The unit step function u(t – a) is defined as
0 , for t a
u(t – a) = 1 , for t a , where a 0

0 , for t 0
As a particular case, u(t) = 1 , for t 0
0 , for t 0
The product, f(t) . u(t – a) = f (t) , for t 0
1 as
L {u(t – a)} = e
s
696 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

1
In particular, L{u(t)} =
s
If L {f(t)} = f ( s), then L {f(t – a) . u (t – a)} = e–as. f ( s)
If a = 0, L {f(t) . u(t)} = f ( s) = L{f(t)}.
2
Example. Find the Laplace transform of (t – 1) . u(t – 1).
Sol. Comparing with f(t – a) u(t – a), we have a = 1 and f(t) = t2
2
f ( s) = L{f(t)} = L(t2) = 3
s
s
2e
L {(t – 1)2 u(t – 1)} = e–s . f (s) =
.
s3
2. Unit Impulse Function or Dirac-delta Function
0 , for t a
1
(t – a) = , for a t a
0 , for t a
As 0, the function (t – a) tends to be infinite at x = a and zero elsewhere, with the charac-
teristic property that its integral across t = a is unity. If (t – a) represents a force acting for a
short duration at time t = a, then the integral.
a
Lt (t a) dt = 1, represents unit impulse at t = a. Hence the limiting form of (t – a) as
0 a
0 is expressed as unit impulse function denoted by (t – a).
Thus, the unit impulse function (t – a) is defined as follows :
, for t a
(t – a) =
0 , for t a

Such that, (t a) dt 1.
0
Laplace transform of unit Impulse function
st sa
L { (t – a)} = e (t a) dt e .
0
L ( (t)) = 1.
3. Periodic Functions
If f(t) is a periodic function with period
1 T
st
T, i.e., f(t + T) = f(t), then L {f(t)} = sT
e f (t) dt .
1 e 0
4. The Error Function
2 t 2
The error function denoted by erf t is defined as erf t = e d
0
Note. Lim erf x = 0 and lim erf x = 1.
x 0 x
t t 4
2 2 2 2
L(erf t ) = L e d =L 1 ... d
0 0 2!

3 5
2 t3 / 2 1 2 2 2
=L t . t5 / 2 ... = ...
3 5.2 ! s3 / 2 3 s5 / 2
LAPLACE TRANSFORMS AND ITS APPLICATIONS 697

1 1 1 1.3 1
= . . ...
s3 / 2 2 s5 / 2 2.4 s7 / 2
1/2
1 1 1 13
. 1 1 1 1
= 1 . . ... = 1
s3/ 2 2 s 2.4 s2 s3/ 2 s s s 1
1 1 2
Also, L(erf 2 t ) = .
4 s s s s 4
1
4 4
d 2 3s 8
and L(t erf 2 t ) = – 2
ds s s 4 s (s 4)3 / 2
5. The Bessel Function (M.D.U., 2005; U.P., June, 2006, 2009; A.U.U.P., 2008)
Bessel function of order n is given by
n 2r
xn x2 x4 ( 1)r x
Jn(x) = n
1 =
2 (n 1) 2 . (2n 2) 2.4(2n 2)(2n 4) r! (n r 1) 2
r 0
Bessel function of order zero is given by
x2 x4 x6
J0(x) = 1 – ...
22 22 . 4 2 22 . 4 2 . 6 2
t2 t4 t6
J0(t) = 1 – 2 2 2
...
2 2 .4 2 . 4 2 . 62
2

1 1 2! 1 4!
L {J0(t)} = . . ...
s 22 s3 22 . 4 2 s5
1/ 2
1 1 1 1. 3 1 1. 3 . 5 1 1 1 1
= 1 . ... = 1
s 2 s2 2.4 s4 2.4.6 s6 s s2 s 2
1
By the recurrence relation, J0 (t) = – J1(t), where J1(t) is the Bessel function of order one.
6. The Exponential Integral Function
The exponential integral is defined as

e u 1
Ei(t) = . du or L {Ei (t)} = log (s + 1)
t u s
7. Laguerre Polynomial

e x dn
This polynomial is defined as Ln(x) = . (e–x . xn)
n ! dxn

(s 1)n
Laplace transform of Ln (t) i.e., L {Ln(t)} = n 1 ; s > 1 and n is a positive integer.
s
8. The Sine Integral
t sin y
Si(t) = dy
0 y
9. The Cosine Integral
cos u 1
Ci(t) = du, L {Ci(t)} = log (s2 + 1)
0 u 2s
698 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

10. The Complementary Error Function

2 u2
This is defined as erfc (t) = 1 – erf t = e . du
t

1 1
L(erfc t ) = L [1 – erf t ] = L(1) – L {erf t } = .
s s s 1
91. Express the following function in terms of unit step function and find its Laplace transform.
(M.D.U., May, 2009; U.P.T.U, 2007)
0, 0 t 1
Sol. Here, f(t) = t 1, 1 t 2
1, t 2
or f(t) = (t – 1) {u(t – 1) – u(t – 2)} + u(t – 2) = (t – 1) u(t – 1) – (t – 2) u(t – 2)
If L {f(t)} = f (s) then L {f(t – a) u(t – a)} = e–as f (s)
1
Here, L {f(t)} = L(t) =
s2
1 1
L {(t – 1) u(t – 1)} = e–s . and L {(t – 2) u (t – 2)} = e–2s .
s2 s2
s 2s
e e
Hence, L {(t – 1) u(t – 1) – (t – 2) u(t – 2)} = .
s2
–t
92. Find the Laplace transform of the function e {1 – u(t – 2)}.
Sol. Expressing e–t as a function of t – 2, we have
e–t = e–{t – 2 + 2} = e–2 . e–(t – 2)
e . u (t – 2) = e–2 . e–(t – 2) . u(t – 2)
–t

Comparing e – 2) u(t – 2) with F(t – a)u(t – a), we get a = 2 and F(t) = e–t
–(t

1
f(s) = = L {F(t)}
s 1
1
L [e–t . u(t – 2)] = e–2 . e–2s .
s 1
2(1 s)
1 1 1 e
L[e–t {1 – u(t – 2)}] = – e–2 . e–2s . = .
s 1 s 1 (s 1)

t 1, 1 t 2
93. Find Laplace transform of the function F(t) = 3 t, 2 t 3. (U.P.T.U., 2009)

Sol. F(t) = (t – 1) {u(t – 1) – u(t – 2)} + (3 – t) {u(t – 2) – u(t – 3)}

= (t – 3) u(t – 3) – 2(t – 2) u(t – 2) + (t – 1) u(t – 1)
Hence, L {F(t)} = L {(t – 3) u(t – 3)} – 2L {(t – 2) u(t – 2)} + L {(t – 1) u(t – 1)}
e 3s
2e 2s
e s
e s
(1 e s )2
= = .
s2 s2 s2 s2

t2 , 0 t 2
94. Express the function f(t) = in terms of unit step function and obtain the Laplace
4t, t 2
transform.
Sol. In terms of unit-step function,
f(t) = t2 {u(t) – u(t – 2)} + 4t . u(t – 2) = t2 u(t) + (4t – t2) u(t – 2)
LAPLACE TRANSFORMS AND ITS APPLICATIONS 699

L {f(t)} = L {t2 u(t)} + L [{4 – (t – 2)2} u(t – 2)]

2s
2 2s 2 2s 4 2 2(1 e ) 4. e 2 s
=– e L(4 t2 ) = e . = .
s3 s3 s s3 s3 s
95. Find the Laplace transform of the following periodic functions :
t t
(i) f(t) = , for 0 < t < T (saw-tooth wave of period T) (ii) f(t) = sin for 0 < t < a (Rectified
T a
sine wave of period a).
1 T
st
Sol. (i) Here, L {f(t)} = sT
e f (t) dt
1 e 0

T
1 T t st T st
st 1 te e
= sT
e . dt = 1. . dt
1 e T sT s s
0 T(1 e ) 0
0

sT sT sT
1 e 1 e 1 e
= sT 2 = 2 sT .
1 e s s .T s T s(1 e )
1 a t
st
(ii) L {f(t)} = as
e sin dt
1 e 0 a
a

st
a
st t e t t
Let I= e . sin dt = 2
s sin cos
0 a a a a
s2
a2 0

as as
e 1 (1 e )a
= 2 2 =
2 a 2 a a s 2 2 2
s 2
s 2
a a
(1 e as
)a ( eas / 2 e as / 2
) . (a )
L {f(t)} = = as / 2 as / 2
(1 e as
)(a s 2 2 2
) (e e )( a2 s2 2
)
a as
= coth .
a2 s2 2
2
96. Find the Laplace transform of the triangular wave function of period 2c given by
t , 0 t c
f(t) =
2c t, c t 2c
(M.D.U. May 2009, Uttarakhand, June 2007)
1 2c
st 1 c
st
2c
st
Sol. L {f(t)} = 2 cs
e f (t) dt = e . t dt e . (2c t) dt
1 e 0 1 e 2 cs 0 c

2 cs
1 ce cs e cs
1 e ce cs e cs
= 2 cs 2 2 2 2
1 e s s s s s s
cs 2 cs cs 2 cs
1 1 2e e 1 (1 e ) 1 1 e
= 2 cs 2 = 2 cs cs 2
. cs
1 e s s (1 e )(1 e ) s 1 e
1 cs
= 2
tanh .
s 2
700 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

97. Find the Laplace transform of the rectified semi-wave function defined by

sin wt, 0 t
w
f(t) = 2
0 , t .
w w
2
Sol. Here, f(t) is a periodic function with period
w
1 2 /w 1 /w 2 /w
st st st
L {f(t)} = 2 s
e . f (t) dt = 2 s
e sin wt dt e . 0 . dt
0 0 /w
1 e w 1 e w

st /w
1 e ( s sin wt w cos wt)
= 2 s
s2 w2 0
1 e w

s
we w w w
= s s
= s
.
w w 2 2 w 2 2
1 e 1 e (s w ) 1 e (s w )

98. Find the Laplace transform of the square-wave function of period a defined as :
a
1, 0 t
f(t) = 2.
a
1, t a
2
1 a 1 a/ 2 a
st st st
Sol. L {f(t)} = as
e . f (t) dt = as
e 1 . dt e ( 1) dt
1 e 0 1 e 0 a/ 2

st a / 2 st a
1 e e 1 as / 2 as as / 2
= as = as
[1 e e e ]
1 e s s s (1 e )
0 a/ 2

as / 2
1 as / 2 2 1 1 e
= as
[1 e ] = as / 2
s (1 e ) s 1 e

1 eas / 4 e as / 4
1 as
= . tanh .
s eas / 4 e as / 4 s 4
99. A condenser of capacity C is charged to potential E and discharged at t = 0 through an inductive
CE t
resistance L, R. Show that the charge q at time t is given by q = e ( sin nt + n cos nt), where
n

R 2
1 R2
= and n = > 0.
2L LC 4L2
Sol. Let q be the charge and i be the current in the circuit at time t then the potential difference
di q
across the inductance L, resistance R and capacitance C are respectively L , Ri and .
dt C
LAPLACE TRANSFORMS AND ITS APPLICATIONS 701

When discharging, the applied potential is zero,

di q d2 q dq q dq
L + Ri + =0 L R =0 ...(1) i
dt C dt 2 dt C dt
The initial conditions are q = EC and i = 0 when t = 0.
Hence Laplace transform of (1) is
R
EC s
2 1 L EC (s 2 )
Ls Rs q = (Ls + R) EC q =
C 2
s
R
s
1 s2 2 s 2
n2
L LC
Now, taking inverse Laplace transform, we get

s
q = EC
(s )2 n2 (s )2 n2
t
e
or q = EC . (n cos nt + sin nt).
n
100. In an electrical circuit with e.m.f. E(t), resistance R and inductance L, the current i builds up at
di
the rate given by L + Ri = E(t). If the switch is connected at t = 0 and disconnected at t = a, find
dt
the current i at any instant. [U.P.T.U., (C.O.) 2005]
E, for 0 t a
Sol. Here, i = 0 at t = 0 and E(t) = 0, for t a
Taking Laplace transform of both sides of the given equation, we have

st
L [s i – i(0)] + R i e E(t) dt
0

a
st E E E . e as
or (Ls + R)i e E dt = (1 – e–as) i
0 s s(Ls R) s(Ls R)
Taking inverse Laplace transform of both sides, we have

E 1 E . e as
i = L–1 L ...(1)
s(Ls R) s(Ls R)

E 1 1 1 L
Now L–1 L E.
s(Ls R) R s Ls R

E 1 1 1 1 E
= L L = (1 – e–Rt/L)
R s s R/L R

E . e as E
and L–1 s(Ls R) = [1 – e–R(t–a)/L] u(t – a)
R
E E
From (1), we have i= (1 – e–Rt/L) – [1 – e–R(t–a)/L] u(t – a)
R R
E E –Rt/L
Hence, for 0 < t < a, i= (1 – e–Rt/L) and for t > a, i= e . (eRa/L – 1).
R R
702 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

101. An impulsive voltage E (t) is applied to a circuit consisting of L, R, C in series with zero initial
conditions. If I be the current at any subsequent time t, find the limit of I as t 0 and justify your
result.
dI q dq
Sol. The equation for the circuit is L RI = E. (t) ; = I, with zero initial conditions.
dt C dt
q
Laplace transform of these equations are (Ls + R) I + = E, qs = I
C
1 Rs 1 E
Multiplying by s, Ls 2 Rs I Es or s2 I .s
C L LC L
E R 1
(s2 + 2ms + m2 + n2) I = s, where = 2m, = m2 + n2
L L LC
E (s m) m E m mt
This gives, I so that, I = . cos nt sin nt e
L (s m)2 n2 L n

E
We see that I , where t 0.
L
Even though the initial current is zero, a large current will develop instantaneously in the circuit
due to the impulsive voltage applied at t = 0. It will be the limit of this current which we have
found above.
102. A particle of mass m can perform small oscillations about a position of equilibrium under a restor-
ing force mn2 times the displacement. It is started from rest by a constant force F which acts for a
2F nT
time t and then ceases. Show that the amplitude of subsequent oscillation is sin .
mn 2 2
Sol. The force F acting for 0 < t < T, and ceasing afterwards can be represented by F {1 – u(t – T)}
Hence the equation of motion of mass m is
d2 x d2 x F
m = – mn2x + F{1 – u(t – T)} or + n2x = {1 – u(t – T)}
dt 2 dt 2 m
F
Its Laplace transform is (s2 + n2) x (1 e sT )
ms
The initial displacement and velocity are zero.
sT
F(1 e ) F sT 1 1 s
x 2 2 x (1 e ).
ms( s n ) m n2 s s2 n2
Taking inverse Laplace transform, we get
F
x= [(1 – cos nt) – {1 – cos n(t – T)} u(t – T)]
mn2

F
= (1 – cos nt), for 0 < t < T
mn2

F
and for t > T, x= [(1 – cos nt) – {1 – cos n(t – T)}]
mn2

F F nT T
= 2 [cos n(t – T) – cos nt] = 2 . 2 sin . sin n t
mn mn 2 2

2F nT
Amplitude of oscillation is 2
sin .
mn 2
LAPLACE TRANSFORMS AND ITS APPLICATIONS 703

103. A body falls from rest in a liquid whose density is one fourth that of the body. If the liquid offers
resistance proportional to the velocity and the velocity approaches a limiting value of 9 metres/sec.,
find the distance fallen in 5 seconds.
Sol. The equation of motion of the body is
dv 1 dv 3
m = mg – mg – mkv + kv = g ...(1)
dt 4 dt 4
The initial conditions are v = 0 when t = 0
Taking Laplace transform of equation (1) on both sides, we get
3 g 3g 1 1
sv kv. or v =
4 s 4k s s k
Taking inverse Laplace transform,
3g
v= (1 – e–kt) ...(2)
4k
3g
When s ,v
4k
But limiting velocity is given to be = 9 m/sec
3g g
= 9 or k =
4k 12
Equation (2) becomes
dx
v or = 9(1 – e–kt) ...(3) and the initial condition is x = 0 when t = 0
dt
Taking Laplace transform of (3), we get
1 1
sx 9
s s k

1 1 1 1 1 1
i.e., x 9 2
x 9 2
s s (s k) s k s s k
Taking inverse Laplace transform, we get
kt
1 e 12(1 e gt / 12
)
x 9 t x=9 t
k g
Putting t = 5 sec , g = 9.8 m/sec2, the distance travelled in 5 sec is given by

12(1 e 49 / 12 )
x= 9 5 x = 34.17 metres.
9.8
104. Discuss application of Laplace transforms to deflection of beams. Assuming that the deflection of
d4 y
a beam loaded by a weight w(x) per unit length is given by EI = w(x), obtain the deflection of
dx 4
a beam, clamped horizontally at one end and free at the other, carrying a uniform load P per unit
length.
Sol. Suppose a beam is kept along x-axis and its ends are x = 0, x = l. Suppose the beam suffers a
transverse deflection y(x) which is produced by applying a vertical load w(x) per unit length. The
d4 y
w( x)
deflection is given by 4 , 0 < x < l, where E is Young’s modulus of elasticity for the beam
dx EI
and I is the moment of inertia of a cross-section of the beam about x-axis.
704 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

Following are the boundary conditions

(i) If beam is hinged or simply supported ends, then y = y = 0
(ii) If beam is clamped or built in or fixed end, then y=y =0
(iii) If the beam has free end, then y =y =0
d4 y P
In the given problem, the equation is 4
...(1)
dx EI
Taking Laplace transform, we get
P 1
s4 y – s3y(0) – s2y (0) – sy (0) – y (0) = .
EI s
a
or s4 y – s3y(0) – s2y (0) – sy (0) – y (0) = ...(2)
s
P
where, a = (say).
EI
Boundary conditions are y(0) = 0 = y (0) ...(3)
y (l) = 0 = y (l) ...(4)
From equation (2), in view of (3),
a a
s4 y – sy (0) – y (0) = s4 y = bs + c + , where y (0) = c(say), y (0) = b(say)
s s
b c a
y
s3 s4 s5
Taking inverse Laplace transform, we get
bx 2 cx3 ax 4
y= ...(5)
2 6 24

ax2
y = + b + cx ...(6) and y = ax + c ...(7)
2
Subjecting this to (4), we obtain
al 2
y (l) = + b + cl = 0
2
y (l) = 0 = al + c
al 2
On solving, we getc = – al and b =
2
ax 4 al2 2 al 3 x4 x2 l 2 lx3 P
y= x x =a = (x4 – 4lx3 + 6x2l2)
24 4 6 24 4 6 24EI
P
y= . x2 (x2 – 4lx + 6l2)
24EI
This gives the deflection of the beam.
105. A beam which is hinged at its ends x = 0 and x = l carries a uniform load w0 per unit length. Show
w0 x 2 2
that the deflection at any point is y(x) = (l – x)(l + lx – x ).
24EI
d4 y w0
Sol. The deflection y(x) is given by 4
= a(say) ...(1)
dx EI
Boundary conditions are, y(0) = 0 = y (0) ...(2)
y(l) = 0 = y (l) ...(3)
LAPLACE TRANSFORMS AND ITS APPLICATIONS 705

Taking Laplace transform on both sides of equation (1)

a a
s4 y – s3y(0) – s2y (0) – sy (0) – y (0) = s4 y – s2y (0) – y (0) = ...(4)
s s
Let y (0) = b and y (0) = c, then from (4),
a b c a
s4 y = bs2 + c + y =
s s2 s4 s5
Taking inverse Laplace transform,
cx3
ax4 cx3 ax 4
y = bx + = bx + ...(5)
3!
4! 6 24
cl2 al 4
Subjecting to (3), y(l) = 0 = bl + ...(6)
6 24
cl2
y (l) = 0 = cl + ...(7)
2
al al3
Solving equations (6) and (7), we get c = – ,b
2 24
Then equation (5) becomes
al3 x alx3 ax 4 ax 3 ax
y= y= (l – 2lx2 + x3) y= (l – x)(l2 + lx – x2)
24 12 24 24 24
w0 x
or y= (l – x)(l2 + lx – x2). Hence shown.
24 EI
106. A beam is simply supported at its end x = 0 and is clamped at the other end x = l. It carries a load
l
w at x = . Find the resulting deflection of the beam at any point.
4
d4 y w l
Sol. The equation for the deflection is 4
= . x
dx EI 4
w –ls/4
Taking Laplace transform, s4 y – s3y(0) – s2y (0) – sy (0) – y (0) = e
EI
Using the conditions y(0) = 0, y (0) = 0 and taking y (0) = c1 and y (0) = c2, we get

c1 c2 w e ls / 4
y 2 4
.
s s EI s4
On taking inverse Laplace transform, it gives

x3 w ( x l/4)3
y = c1x + c2 . . . u (x – l/4)
3! EI 3!

1 1 w l
i.e., y = c1x + c x3, for 0 < x < l/4 = c1x + c x3 + (x – l/4)3, for < x < l.
6 2 6 2 6EI 4
Using the conditions y(l) = 0, y (l) = 0, we get

1 9wl3 1 9wl2
0 = c1l + c2l3 + and 0 = c1 + c2l2 +
6 128 EI 2 32EI
c1 = 9wl2/256 EI or c2 = – 81w/128 EI
Substituting the values of c1 and c2, we get
3
45wl2 27wx 3 w l l
y=q .x x u x .
128EI 256EI 6EI 4 4
706 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS

13
107. A weight less beam of length l is freely supported at ends when a concentrated load W acts at
692

d4 y
x = a. Obtain the deflection of the beam. Equation for deflection is EI = W. (x – a).
dx 4
Sol. Taking Laplace transform of the equation, we get
W
s4 y – s3y(0) – s2y (0) – sy (0) – y (0) = . e–as
EI
Using the conditions y(0) = y (0) = 0 and taking y (0) = c1, y (0) = c2, we get

W c1 c2 W e as
s4 y – c1s2 – c2 = . e–as or y 2 4
.
EI s s EI s4
Taking inverse Laplace transform, we have
x3W ( x a)3
y = c1x + c2 . . u (x – a) ...(1)
3!EI 3!
1 W
y = c1 + . c2x2 + (x – a)2 u(x – a)
2 2EI
W
y = c2x + (x – a) u(x – a)
EI
Using the conditions y(l) = 0, y (l) = 0, we get
l3 W b3
0 = c1l + c2 .
6 EI 6
W
0 = c2l + .b [Since u(x – a) = 1, for x a and l – a = b]
EI
W b 1 l3 W b3
c2 = – . , c1 = – c2 .
EI l l 6 EI 6

1 W bl2 W b3 Wab
=– . . = (l + b)
l EI 6 EI 6 6EIl
From (1), the solution is
Wab W b 3 W ( x a)3
y= (l b) x . x . . u (x a)
6EIl 6EI l EI 3!
When 0 < x < a, u(x – a) = 0
W ab(l b) x b 3
y= x ...(2)
6EI l l
When a < x < l, u(x – a) = 1
W ab(l b) b 3
y= x x (x a) 3 ...(3)
6EI l l
From (2) and (3), we have
W W a2 b W a 2b
y(a) = [a2b(l + b) – ba 3] = [l + b – a] = (2b)
6EIl 6EIl 6EIl

1 W a 2 b2
= . . .
3 EI l