Laplace Transformations

Definition of Laplace transformation: Let f(t) be a function of t

defined for all positive values of t. Then the Laplace
transformation of f(t), denoted by L[f(t)], is defined by

L[f(t)] =  e  st f (t )dt ; provided the integral exists.
0

Example (1): Find the Laplace transform of f(t) = 1.

Solution: By the definition of Laplace transformation, we have
 
e  st 1 1
L[1] =  e .1 dt =  st
= – (0 – 1) = [Ans]
0
s 0 s s
Example (2): Find the Laplace transform of tn.
Solution: By the definition of Laplace transformation, we have

L[t ] =  e  st t n dt
n Let st = p
p dp
0
or, t =  dt =
 n s s
 p  dp
= e   p

0 s s
 
1
 e p dp [ (n) =  e  x x n 1 dx ]
p n
=
s n 1 0 0

1 n!
= n 1 . (n  1) = n 1 [Answer]
s s

Example (3): Find the values of (i) L[ e at ], (ii) L[sin at],

(iii) L[cos at], (iv) L[sinhat], (v) L[coshat].
Solution: By the definition of Laplace transformation, we have
  
e  ( s a )t 1
(i) L[ e ] =  e e dt =  e
at  st at ( s  a ) t
dt = = ; s>a
0 0
 ( s  a) 0 sa

1
 
(ii) L[sin at] =  e  st sin at dt
0

e ax
[  e ax sin bxdx = [a sin bx  b cos bx] ]
a2  b2

e  st
= 2 ( s sin at  a cos at )
s  a2 0
1 a
= 2 [0  a ] = 2 [Answer]
s a 2
s  a2

(iii) L[cos at] =  e  st cos at dt
0

e ax
[  e ax cos bxdx = [a cos bx  b sin bx] ]
a2  b2

e  st
= 2 (a sin at  s cos at )
s  a2 0
1 s
= 2 [0  s ] = 2 [Answer]
s  a2 s  a2
 
 st  e  e  at 
at
(iv) L[sinhat] =  e  st
sinh at dt = e    dt
0 0  2 
 
1
2 0
= [ e ( s  a )t dt –  e ( s  a )t dt ]
0
 
1 e  ( s a )t e  ( s  a )t
= [ – ]
2  ( s  a) 0  ( s  a) 0
1 1 1 a
= [ – ]= 2 ; s>a [Answer]
2 sa sa s  a2
 
 e at  e  at 
(v) L[coshat] =  e  st cosh at dt =  e  st   dt
0 0  2 
2
  
1
Or, L[coshat] = [  e ( s  a )t dt +  e ( s  a )t dt ]
2 0 0
 
1 e  ( s a )t e  ( s  a )t
= [ + ]
2  ( s  a) 0  ( s  a) 0
1 1 1 s
= [ + ]= 2 ; s>a [Answer]
2 sa sa s  a2

Example (4): Find the Laplace transform of

e4t + 4t3 – 2sin 3t + 3 cos 5t + 1 + 0
Solution: Since integration is a linear transformation, Laplace
transform is also a linear transformation and hence we have,
L[e4t + 4t3 – 2sin 3t + 3 cos 5t + 1 + 0]
= L[e4t] + 4L[t3] – 2L[sin 3t] + 3L[cos 5t] + L[1] + L[0]
1 3! 3 s 1
= + 4. 4 – 2. 2 + 3. 2 + +0
s4 s s 3 2
s 5 2
s
1 24 6 3s 1
= + 4 – 2 + 2 + [Answer]
s4 s s 9 s  25 s


e
 st
Inverse Laplace transform: If L{f(t)] = f (t )dt ; L[f(t)]
0

being clear a function of s, is briefly written as L[f(t] = f (s)

 f(t) = L1[ f ( s)] . So, f(t) is called the inverse Laplace
transformation of f (s) and L1 is called the inverse Laplace
transformation operator.
Theorem (1): (Shifting theorem) If L[f(t)] = f (s), then
L[eat f(t)] = f (s – a).
Proof: By the definition of Laplace transform, we have

L[e f(t)] =  e  st e at f (t ) dt
at

3
 
Or, L[e f(t)] =  e ( s a )t f (t ) dt
at
[Put (s – a) = r]
0

=  e  rt f (t ) dt = f (r) = f (s – a) [Proved]
0
Note: Using theorem (1) in examples (2) and (3), we have
n!
(a) L[eat tn] = ; s>a
( s  a) n 1
b
(b) L[eat sin bt] = ; s>a
( s  a) 2  b 2
s
(c) L[eat cos bt] = ; s>a
( s  a) 2  b 2
b
(d) L[eat sinh bt] = ; s>a
( s  a) 2  b 2
s
(e) L[eat cosh bt] = ; s>a
( s  a) 2  b 2

Example (5): If f(t) is defined by

t : 0  t  4
f(t) = 
5 : t  4
Find the Laplace transformation of f(t).
Solution: From definition of Laplace transform, we have

L[f(t)] =  e  st f (t )dt
0
4 
= e  st
f (t )dt +  e  st f (t )dt
0 4
4 
=  e  st t dt +  e  st 5 dt
0 4

4
 4 
te  st e  st e  st
4
Or, L[f(t)] =
s 0
– 0  s dt + 5
s 4
4 s 4 s 4 s
e e 1 e
=–4 2
–
+ 2 +5
s s s s
1 1 1
= e 4 s ( – 2 ) + 2 [Answer]
s s s

Example (6): Find the Laplace transformation of f(t), where

sin t : 0  t  
f(t) =  .
0 : t  
Solution: From definition of Laplace transform, we have

L[f(t)] =  e  st f (t )dt
0
 
=  e  st f (t )dt +  e  st f (t )dt
0 
 
= e  st
sin t dt +  e  st .0 dt
0 

=  e  st sin t dt .............. (1)
0

Now let, I =  e  st sin t dt = – e  st cos t – s  e  st cos t dt

= – e  st cos t – s e  st sin t – s 2  e  st sin t dt

= – e  st (cos t  s sin t ) – s 2 I
 e  st
 I= (cos t  s sin t )
1 s2

5
 
 e  st e s  1
Hence, L[f(t)] = (cos t  s sin t ) = [Answer]
1 s2 0
1 s2

Theorem (2): If L[f(t)] = f (s), then

(i) L[f/(t)] = s f (s) – f(0);
(ii) L[f//(t)] = s2 f (s) – s f(0) – f/(0); and hence in general
(iii) L[fn(t)] = sn f (s) – sn – 1 f(0) – sn – 2 f/(0) – ... – f n – 1(0).
Proof: (i) By the definition of Laplace transform, we have
 

L[f/(t)] =  e  st f / (t )dt = e  st f (t ) –  ( s)e  st f (t )dt
0
0 0

= 0 – f(0) + s  e  st f (t )dt = – f(0) + s f (s)
0
/
Therefore, L[f (t)] = s f (s) – f(0) [Proved]
(ii) We have from the definition of Laplace transform,
 

//
L[f (t)] = e  st //
f (t )dt = e  st /
f (t ) + s  e  st f / (t )dt
0
0 0
= 0 – f/(0) + s L[f/(t)]
= – f/(0) + s[s f (s) – f(0)] [Using (i)]
// 2 /
Therefore, L[f (t)] = s f (s) – s f(0) – f (0) [Proved]
(iii) Generalizing (i) and (ii), we expand them as follows:
L[fn(t)] = sn f (s) – sn – 1 f(0) – sn – 2 f/(0) – sn – 3 f//(0) – ... – fn – 1(0).

Example (7): If f(t) = sin t then f/(t) = cos t and then find L[cos t].
Solution: We have,
if L[f(t)] = f (s), then L[f/(t)] = s f (s) – f(0).
So, L[cos t] = s L[sin t] – sin(0)

6
 1 a 1
Or, L[cos t] = s – 0 [ L[sin at] = 2  L[sint]= 2 ]
s 1 2
s a 2
s 1
s
= 2 [Answer]
s 1
d
Theorem (3): If L[f(t)] = f (s), then (i) L[t f(t)] = (– 1) [ f (s)];
ds
2 n
2 2 d n n d
(ii) L[t f(t)] =(– 1) [ f (s)]; (iii) L[t f(t)] = (– 1) [ f (s)]
ds 2 ds n
Proof: (i) Since L[f(t)] = f (s)

Or,  e  st f (t )dt = f (s) [By definition of Laplace transform]
0
Differentiating with respect to s, we have

 d
[  e  st f (t )dt ] = [ f (s)]
s 0 ds

 d
 s (e
 st
Or, ) f (t )dt = [ f (s)]
0
ds

d
 (t )e
 st
Or, f (t )dt = [ f (s)]
0
ds

d
e
 st
Or, [tf (t )]dt = (– 1) [ f (s)] ................... (a)
0
ds
d
 L[t f(t)] = (– 1) [ f (s)] [Proved]
ds
(ii) Differentiating equation (a) with respect to s, we get

 d2
[  e  st [tf (t )]dt ] = (– 1) 2 [ f (s)]
s 0 ds

d2
 (t )e [tf (t )]dt = (– 1)
 st
Or, [ f (s)]
0
ds 2
7
 
d2
0 e [t f (t )]dt = (– 1) ds 2 [ f (s)]
 st 2 2
Or,

d2
 L[t2f(t)] =(– 1)2 [ f (s)] [Proved]
ds 2
(iii) Parts (i) and (ii) show that 3rd part of the theorem is true for
n = 1 and 2. Let it is true for n = m, i.e.,
dm
L[tm f(t)] = (– 1)m m [ f (s)]
ds
 m
m d
0
 st m
Or, e [t f (t )]dt = (– 1) [ f (s)]
ds m
Again differentiating with respect to s, we get

 d m 1
[  e  st [t m f (t )]dt ] = (– 1)m m 1 [ f (s)]
s 0 ds

d m 1
 (t )e [t f (t )]dt = (– 1)
 st m m
Or, [ f (s)]
0 ds m 1

d m 1
 e [t f (t )]dt = (– 1)
 st m 1 m+1
Or, [ f (s)]
0 ds m 1
m+1 m+1 d m 1
Or, L[t f(t)] = (– 1) [ f (s)]
ds m 1
That is, the theorem is true for n = m + 1. The mathematical
induction method concludes that the theorem is true for all positive
dn
integral values of n. Hence, L[tn f(t)] = (– 1)n n [ f (s)] [Proved]
ds
1  cos at 1
Example (8): If L[ ] = , then show that
a 2
s(s  a 2 )
2

t (1  cos at ) 3s 2  a 2
L[ ]= 2 2 .
a2 s (s  a 2 ) 2

8
Proof: We know that
d
if L[f(t)] = f (s), then L[t f(t)] = (– 1) [ f (s)].
ds
1  cos at 1
Since given that L[ ]= , hence,
a 2
s(s  a 2 )
2

t (1  cos at ) d 1
L[ ] = (– 1) [ ]
a 2
ds s( s  a 2 )
2

3s 2  a 2
= [Proved]
s 2 (s 2  a 2 ) 2
Example (9): Find the Laplace transform of t2 cos at.
Solution: We know that if L[f(t)] = f (s), then
2 2d2
L[t f(t)] = (– 1) [ f (s)].
ds 2
s
We also know that L[cos at] = .
s  a22

2 d
2
s d ( s 2  a 2 ).1  s(2s)
 L[t cos at ] = (– 1)
2
[ ] = [ ]
ds 2 s 2  a 2 ds (s 2  a 2 ) 2
d a2  s2
= [ ]
ds ( s 2  a 2 ) 2
( s 2  a 2 ) 2 (2s)  (a 2  s 2 )2( s 2  a 2 )(2s)
=
(s 2  a 2 ) 4
2s( s 2  3a 2 )
= [Answer]
(s 2  a 2 ) 3
t 2 8(3s 2  6s  13)
Example (10): Show that L[e t sin 4t] = .
( s 2  2s  17) 3
4 4
Proof: We know that, L[sin 4t] = 2 = 2
s 4 2
s  16
By theorem (1), we have
9
 4 4
L[et sin 4t] = = 2
( s  1)  16
2
s  2s  17
By theorem (3(ii)), we have
d2 4 d 8  8s
L[et t2 sin 4t] = (– 1)2 2 [ 2 ]= [ 2 ]
ds s  2s  17 ds ( s  2s  17) 2
( s 2  2s  17) 2 (1)  (1  s)2( s 2  2s  17)(2s  2)
= 8[ ]
( s 2  2s  17) 4
8(3s 2  6s  13)
= [Answer]
( s 2  2s  17) 3

1 
Theorem (4): If L[f(t)] = f (s), then L  f (t ) =
t 
 f (s) ds
s

Proof: Since L[f(t)] = f (s)



e
 st
Or, f (t )dt = f (s)
0
Integrating with respect to s over the interval s to  , we have
  

 [ e  f (s)ds
 st
f (t )dt ]ds =
s 0 s
  

 [ e  f (s)ds
 st
Or, ds] f (t )dt =
0 s s
  st  
e
Or, [
0
t
] f (t )dt =  f (s)ds
s
s
 
1
e  f (s)ds
 st
Or, f (t )dt =
0
t s

1
 L[ f(t)] =
t  f (s)ds
s
[Proved]

10
 sin at
Example (11): Find the Laplace transform of . And show
t
cos at
that the Laplace transform of does not exist.
t
a
Solution: First part: We know that L[sin at] = 2 .
s  a2
By theorem (4), we have
 
sin at a 1 s 1 s
s s 2  a 2 ds = tan a s = tan   tan a
1
L[ ]=
t
 s
=  tan 1
2 a
s
= cot 1 [Answer]
a
s
Second part: Again we know L[cos at] = 2 . And hence,
s  a2
 
cos at s 1
L[ ]=  2 ds = ln( s 2
 a 2
) does not exist, because
t s s  a 2
2 s

lt ln( s 2  a 2 ) is infinity.
s 

cos at
Hence, L[ ] does not exist. [Proved]
t
t 1
Example (12): Given that L[2 ]= , then show that
 3
2
s
1 1
L[ ]=
.
t s
Proof: By the definition of Laplace transformation, we have
 1

1  st 1
L[ ] = e t 2 dt
t 0 

11
 
1 1
2 
1 1 t t2
Or, L[ ]= [ e  st   ( s)e  st dt ]
t  1 0
1
2 0 2

1 1
Or, L[ ]= [0  s  e  st 2 t dt ]
t  0

1 t
Or, L[ ] = s  e  st 2 dt
t 0

1 t
Or, L[ ] = s L[2 ] [By definition of Laplace transform]
t 
1 1 t 1
Or, L[ ] = s. [ L[2 ]= ]
t
3
2
 3
2
s s
1 1
 L[ ]= [Proved]
t s
Example (13): Find the Laplace transform of e2t(sin32t).
3 1
Solution: We know that sin32t = sin 2t – sin 6t
4 4
3 1 3 1
 L[sin32t] = L[ sin 2t – sin 6t] = L[sin 2t] – L[sin 6t]
4 4 4 4
3 2 1 6 48
= . 2 – . 2 = 2
4 s  22 4 s  62 ( s  4)( s 2  36)
48
= 4 = f (s) (say)
s  40s 2  144
Applying the shifting theorem, we have
L[e2t(sin32t)] = f (s – 2)
48
= [Answer]
( s  2)  40( s  2) 2  144
4

12
Theorem (5): (Laplace transform of integrals) If L[f(t)] = f (s)
t
1
then L[  f (t )dt ] = f (s)
0
s
t
Proof: Let F(t) =  f (t )dt
0
 F/(t) = f(t) and F(0) = 0.

So, L[F/(t)] = L[f(t)]

Or, s F (s) – F(0) = f (s)
Or, s F (s) = f (s) [ F(0) = 0]
1
Or, F (s) = f (s)
s
1
Or, L[F(t)] = f (s)
s
t
1
Therefore, L[  f (t )dt ] = f (s)
0
s
t
sin t
Example (14): Find the Laplace transform of 
0
t
dt .

1
Solution: We know that L[sin t] =
s 1
2

By theorem (4), we have


sin t 1 
L[ ] =  2 ds = tan 1 s
s s 1
t s

= tan 1   tan 1 s
 
=  tan 1 s [ tan 1 s + cot 1 s = ]
2 2
= cot 1 s
t
1
We also know that L[  f (t )dt ] = f (s).
0
s
13
 t
sin t 1 1 1
 L[  dt ] = cot 1 s = tan 1 [Answer]
0
t s s s

Theorem (6): (Laplace transform of periodic functions) If f(t)is a

periodic function with period T, i.e., f(t + T) = f(t) then
T
1
 sT 
L[f(t)] = e  st f (t )dt
1 e 0

Proof: We have L[f(t)] =  e  st f (t )dt
0
T 2T 3T
Or, L[f(t)] =  e  st f (t )dt +  e f (t )dt +
 st
e
 st
f (t )dt + ...
0 T 2T
Putting t = u + T in the 2nd integral, t = u + 2T in the 3rd integral
and so on , we have
T T
L[f(t)] =  e  st
f (t )dt +  e  s (u T ) f (u  T )du
0 0
T
+  e  s (u  2T ) f (u  2T )du + ...
0
[Since f(t) is a periodic function with period T ]
T T T
Or, L[f(t)] =  e  st f (t )dt + e  sT  e  su f (u )du + e 2 sT  e  su f (u )du + ...
0 0 0
b b
[Since  f (t )dt =  f (u)du ]
a a
T
Or, L[f(t)] = [1 + e  sT + e 2 sT + ...]  e  st f (t )dt
0
T
1
e
 st
Therefore, L[f(t)] = f (t )dt [Proved]
1  e  sT 0

14
Example (15): Find the Laplace transformation of f(t) where f(t) is
 
sin wt when 0  t  w
defined by f(t) =  and also it is a
0  2
when  t 
 w w
2
periodic function with period .
w
2
Solution: Since f(t) is a periodic function with period , so
w
2 / w
1
1  e 2 s / w 0
L[f(t)] = e  st f (t )dt

 /w 2 / w
1
 e sin wt dt + 
 st  st
=  2 s / w
[ e .0 dt ]
1 e 0  /w
 /w
1
e
 st
= sin wt dt .................. (1)
1  e  2 s / w 0

Now let I =  e  st sin wt dt

 cos wt  cos wt
= e  st ( ) –  ( s)e  st ( ) dt
w w
cos wt s
=  e  st –  e  st cos wt dt
w w
cos wt s e  st sin wt s2
2 
=  e  st – 2
– e  st sin wt dt
w w w
 st
cos wt s e sin wt s2
=  e  st – – I
w w2 w2
e  st ( s sin wt  w cos wt )
I=
s 2  w2
Then from (1), we have

15
  /w
1 e  st ( s sin wt  w cos wt )
L[f(t)] =
1  e  2 s / w s 2  w2 0

1 we s / w  w
= .
1  e  2 s / w s 2  w2
w
= 2 [Answer]
( s  w )(1  e  s / w )
2

Example (16): Find the Laplace transformation of the periodic

sin t , 0  t  
function f(t) =  with period 2  .
0 ,   t  2
Solution: Since f(t) is a periodic function with period 2  , so
2
1
 2s 
L[f(t)] = e  st f (t )dt
1 e 0
 2
1
1  e  2s 0 e
= [ e  st . sin t dt +  st
.0 dt ]


1 e  st ( s sin t  cos t )
= [ + 0]
1  e  2s s2 1 0

1 e s  1
= .
1  e  2s s 2  1
1
= 2 [Answer]
( s  1)(1  e s )
1 s
Theorem (7): If L[f(t)] = f (s) then (i) L[f(at)] = f ( ) and
a a
t
(ii) L[f( )] = a f (a s).
a
Proof: (i) By definition of Laplace transform, we have

L[f(at)] =  e  st f (at )dt
0

16
  u
s 1 u 1
Or, L[f(at)] =  e a
f (u ). du Let at = u  t = & dt = du
a a a
0
 s
1  t
=  e a f (t ).dt [Putting u = t]
a 0
1 s
= f( ) [Proved]
a a
(ii) By definition of Laplace transform, we have

t t
L[f( )] =  e  st f ( at )dt Let = u  t = au & dt = a du
a 0 a

=  e  sau f (u ).a du
0

= a  e  ( as)t f (t ).dt [Putting u = t]
0

= a f (a s) [Proved]
1
Example (17): If L[sin t] = 2 then find the value of L[sin 8t].
s 1
1
Solution: Given that L[sin t] = 2 .
s 1
1 s
We know that if L[f(t)] = f (s) then L[f(at)] = f( )
a a
1 1 8
So, L[sin 8t] = . s 2 = 2 [Answer]
8 (8) 1 s  82
1  cos t
Example (18): Calculate the value of L[ ].
t2
1 s
Solution: We know that L[1 – cos t] = – 2
s s 1
From theorem (4), we have

17
 
1 
if L[f(t)] = f (s), then L  f (t ) =
t 
 f (s) ds
s
  
1  cos t 1 s 1
s s [ s  s 2  1]ds ds = s ln s  2 ln( s  1) s ds
2
So, L[ ]=
t2

 

 
1  s  2
1  1 
=  ln  2  ds =  ln   ds
2 s  s 1 s 2 s 1 1 2 
 s s
 
1  s2  1
=– s ln s 2  1 ds = 2 s [ln( s  1)  ln s ].1ds
2 2

2

1  2s 2s
= [ {ln( s 2  1)  ln s 2 }s   ( 2  2 ) s ds ]
s s 1
2 s s
 
1 1 1
= [ s ln(1  2 ) ] +  2 ds
s s 1
2 s s
1 1 
= [ .0  s ln(1  2 ) ] + tan 1 s
2 s s

1 1 
= – s ln(1  2 ) +  tan 1 s
2 s 2
s 1
= cot 1 s – ln(1  2 ) [Answer]
2 s
Example (19): Using Laplace transform find the value of

 e  at  e bt 
0  t  dt
1 1
Solution: Since L[ e  at  e bt ] = –
sa sb
 at  bt
e e 
1 1
 L[ ] = (  ) ds
t s
sa sb

18
 
 at
e  bt 1 a 
= ln  s
e 
Or, L[ ] = ln( s  a)  ln( s  b) s
t  1  b 
 s s
sa
= ln(1) – ln( )
sb
sb
= ln( )
sa
  at
 st e  e bt sb
 e ( )dt = ln( ) [Using definition]
0
t sa
Taking limit as s  0, we have
  at
 st e  e bt sb
s 0 
lt e ( )dt = lt [ ln( )]
0
t s  0 sa

 e  at  e bt  b
Therefore, 0  t  dt = ln  a  [Answer]
Example (20): Find inverse Laplace transforms of
1 1 s 1 1
(i) , (ii) ln( ), (iii) , (iv) ln(1+ 2 ),
s ( s  2) s s(s  a )
2 2
s
1 1
(v) 3 2 , (vi)
s ( s  1) s( s  1) 3
Solution: (i) Inverse Laplace transform
1 1 1 1
L1 [ ] = L1 [ (  )]
s ( s  2) 2 s s2
1 1 1
= [ L1 ( ) – L1 ( )]
2 s s2
1
= [1 – e 2t ] [Answer]
2
1 s
(ii) Let L[f(t)] = f (s) = ln( ) = ln(1 + s) – ln s ....... (1)
s
19
 d
 L[t f(t)] = (–1) [ln(1 + s) – ln s]
ds
1 1
Or, L[t f(t)] = (–1) [ – ]
1 s s
1 1
Or, t f(t) = L1 ( ) – L1 ( ) = 1 – e t
s 1 s
t
1 e
Or, f(t) =
t
1 s
Equation (1) implies f(t) = L1 [ ln( )]
s
1 1 s 1  e t
 L [ ln( )] = [Answer]
s t
a 1 1
(iii) We know that L[sin at] = 2 or, L[ sin at] = 2
s a 2
a s  a2
1 1
 L1 [ 2 ] = sin at
s a 2
a
t
1
We know that L[  f (t )dt ] = f (s)
0
s
t
1 1 1
Or, L[  sin at dt ] = . 2
0
a s s  a2
t
1 1 1 1
1
] =  sin at dt = 2  cos at 0
t
Or, L [ . 2
s s a 2
0
a a
1 1
Or, L1 [ ] = 2 (1  cos at ) [Answer]
s(s  a )
2 2
a
1
(iv) Let L[f(t)] = f (s) = ln(1+ 2 ) = ln(s2 + 1) – ln(s2)
s
d 1 s
 L[t f(t)] = (–1) [ ln(s2 + 1) – ln(s2)] = 2[ – 2 ]
ds s s 1

20
Or, L[t f(t)] = 2 L[1 – cos t] = L[2 – 2 cos t]
Or, t f(t) = 2 – 2 cos t
2(1  cos t)
Or, f(t) =
t
1 2(1  cos t)
Therefore, L1 [ ln(1+ 2 )] = [Answer]
s t
1
(v) We know that L1 [ 2 ] = sin t
s 1
t
1
 L [1
] =  sin t dt [By theorem (5)]
s( s 2  1) 0

=  cos t 0 = 1 – cos t
t

t
1
1
] =  (1  cos t ) dt = t  sin t 0 = t – sin t
t
Again, L [ 2 2
s ( s  1) 0
t t
1 t2
1
And L [ 3 2 ] =  (t  sin t ) dt =  cos t
s ( s  1) 0
2 0

t2
=  cos t – 1 [Answer]
2
2!
(vi) We know that L1 [ 3 ] = t2
s
2
 L1 [ ] = e – t t2 [By theorem (1)]
( s  1) 3

1 1
Or, L1 [ ] = e – t t2
( s  1) 3
2
t
1 1
And L1 [  2t e t dt
2
]= [By theorem (5)]
s( s  1) 3 0

1 2 t
= t (e t )  2t e t  2e t
2 0

21
 1 1
Therefore, L1 [ ] = [2 – t2e– t – 2te– t – 2e– t] [Answer]
s( s  1) 3
2

Theorem (8): (Convolution theory) If L1 [ f (s)] = f(t) and

t
L1 [ g (s)] = g(t) then L1 [ f (s). g (s)] =  f (u) g (t  u)du
0
t
Proof: Let  (t) =  f (u) g (t  u)du
0
So, by definition of Laplace transform, we have
 t
u
L[  (t)] =  e [  f (u ) g (t  u )du ]dt
 st
u=t
0 0
t 
 t
=   e  st f (u ) g (t  u ) du dt
0 0 0 u=0 t
The above integration is in the area which is made by u = 0 and
u = t straight lines. By changing the order of the integrations, we

get L[  (t)] = e
 st
f (u ) g (t  u ) dt du
0 u
 
=  e  su f (u )[ e  s (t u ) g (t  u ) dt ] du
0 u
 
=  e  su f (u )[  e  sv g (v) dv] du [Putting t – u = v]
0 0
 
=  e  su f (u ) g ( s) du = g (s).  e  su f (u ) du = g (s). f (s)
0 0

Or,  (t) = L [ f (s). g (s)]

1

t
Therefore, L1 [ f (s). g (s)] =  f (u) g (t  u)du
0
[Proved]

22
Example (21): By using convolution theory find inverse Laplace
s2
transforms of 2 .
( s  a 2 )(s 2  b 2 )
s s
Solution: Since L1 [ 2 ] = cos at and L1 [ 2 ] = cos bt,
s a 2
s  b2
we get by convolution theory
t
s s
L1 [ 2 .
s  a2 s2  b2
] = 0 cos au cos b(t  u) du
t
1
=  [cos(au  bt  bu )  cos(au  bt  bu ) ]du
2 0
1 sin(au  bu  bt ) sin(au  bu  bt )
t

= 
2 a b ab 0

1 sin at  sin bt sin at  sin bt

= [  ]
2 a b ab
a sin at  b sin bt
= [Answer]
a2  b2
s
Example (22): Evaluate L1 [ 2 ] using convolution theory
(s  a 2 ) 2
s 1 sin at
Solution: Since L1 [ 2 ] = cos at and L1 [ 2 ]= .
s a 2
s a 2
a
Therefore, by using the convolution theorem, we have
sin a(t  u )
t
s 1
1
L [ 2 . 2 ] =  cos au du
s a s a
2 2
0
a
t
1
a 0
= [cos au (sin at cos au  cos at sin au ) ]du

t t
1 1
= sin at  cos 2 au du – cos at  cos au sin au du
a 0
a 0

23
 t t
1 1
= sin at  (1  cos 2au ) du – cos at  sin 2au du
2a 0
2a 0

1 sin 2at 1
= sin at (t  )– cos at (cos 2at  1)
2a 2a 4a 2
t sin at sin at 2 sin at cos at cos at
= + . – .2sin2at
2a 2a 2a 4a 2
t sin at sin 2 at cos at sin 2 at cos at
= + –
2a 2a 2 2a 2
s s t sin at
Hence, L1 [ 2 . 2 ]= [Answer]
s a s b
2 2
2a

Example (23): Find the inverse Laplace transforms of

s3 2s 2  6s  5
(i) 4 and (ii)
s  a4 s 3  6s 2  11s  6
Solution: (i) The inverse Laplace transform
s3 s2
L1 [ 4 ] = L 1
[s{ }]
s  a4 ( s 2  a 2 )( s 2  a 2 )
s 1 1
= L1 [ { 2  2 }]
2 s a 2
s  a2
1 s s
= L1 [ 2  2 ]
2 s a 2
s  a2
1
= (cos at + cosh at) [Answer]
2
(ii) We have s3 – 6s2 + 11s – 6 = (s – 1)(s – 2)(s – 3)
2s 2  6s  5 A B C
Let = + + .............. (1)
( s  1)( s  2)( s  3) s 1 s  2 s3
Multiplying (1) by (s – 1)(s – 2)(s – 3), we have
2s2 – 6s + 5 = A(s – 2)(s – 3) + B(s – 1)(s – 3) + C(s – 1)(s – 2)
..................... (2)
Putting s = 1, 2 and 3 in (2) we get
24
 1 5
A= , B = – 1 and C =
2 2
From (1), we get
2s 2  6s  5 1 1 5
= – +
( s  1)( s  2)( s  3) 2( s  1) s  2 2( s  3)
Hence the inverse Laplace transform
1 2s 2  6s  5 1 1 1 5 1
L [ 3 ] = L1 [ ]– L1 [ ]+ L1 [ ]
s  6s  11s  6
2
2 s 1 s2 2 s3
1 5
= et – e2t + e3t [Answer]
2 2

Theorem (9): (Majumdar-Shahidul's partial fraction method) If

q(s) contains a repeated quadratic factor of the form (s2 + a2)3, then
corresponding to this factor, the terms in the partial fractions
p(s)
decomposition of are
q(s)
p ( s ) As  B C ( s 2  a 2 )  Ds E (3s 2  a 2 )  Fs ( s 2  3a 2 )
=  
q(s) s 2  a 2 (s 2  a 2 ) 2 (s 2  a 2 ) 3
Discussion: From the theory of the Laplace transforms, we know
2as s2  a2
that L{t sin at}  2 , L{t cos at}  ,
(s  a 2 ) 2 (s 2  a 2 ) 2
2a(3s 2  a 2 ) 2s( s 2  3a 2 )
L{t 2 sin at}  , L{t 2
cos at} 
(s 2  a 2 ) 3 (s 2  a 2 ) 3
Taking these formulas into consideration, Majumdar and Shahidul
p(s)
give this new method of decomposing the rational function
q(s)
into partial fractions that would be fruitful in finding the inverse
p(s)
Laplace transform of . They prove that if q(s) contains a
q(s)

25
repeated quadratic form (s2 + a2)2 or, (s2 + a2)3 then the
p(s)
decomposition of will be
q(s)
p(s) As  B C ( s 2  a 2 )  Ds
= 2  or,
q(s) s  a2 (s 2  a 2 ) 2
p ( s ) As  B C ( s 2  a 2 )  Ds E (3s 2  a 2 )  Fs ( s 2  3a 2 )
=  
q(s) s 2  a 2 (s 2  a 2 ) 2 (s 2  a 2 ) 3
and it makes easy to find out the inverse Laplace transform of that
type of rational functions.
s2
Example (24): Evaluate L1 [ 2 ] using Majumdar-Shahidul
( s  4) 2
method.
Solution: Since the denominator of the given rational function
contains (s2 + 4)2, considering the Majumdar-Shahidul method let
s2 As  B C ( s 2  4)  Ds
=  ......... (1)
( s 2  4) 2 s 2  4 ( s 2  4) 2
To determine the unknown constant, A, B, C and D, we multiply
throughout by (s2 + 4)2 to get
s2 = (As + B)(s2 + 4) + {C(s2 – 4) + Ds} .......... (2)
and then equate from both sides of the identity (2) the coefficients
of s3, s2, s and s0. Thus, we get the following system of four
equations in four unknowns:
0 A   A  0,
 B  1 ,
1 BC   2
 ==> 
0  4A  D  C  2 ,
1

0  4 B  4C   D  0.
With these values, we now have from (1),
s2 1 ( s 2  4)
= 
( s 2  4) 2 2( s 2  4) 2( s 2  4) 2
26
 1 s2  1 1  1  1 1  s 2  4 
So, we have L  2 2 
 L  2  L  2 2 
 ( s  4)  2  ( s  4)  2  ( s  4) 
1 1
= sin 2t + t cos 2t [Answer]
4 2
8
Example (25): Evaluate L1 [ 2 ] using Majumdar-Shahidul
( s  1) 3
method.
Solution: Since the denominator of the given rational function
contains (s2 + 1)3, considering the Majumdar-Shahidul method let
8 As  B C ( s 2  1)  Ds E (3s 2  1)  Fs ( s 2  3)
=  
( s 2  1) 3 s2 1 ( s 2  1) 2 ( s 2  1) 3
........... (1)
To determine the unknown constant, A, B, C, D, E and F, we
multiply throughout by (s2 + 1)3 to get
8 = (As + B)(s2 + 1)2 + {C(s2 – 1) + Ds}(s2 + 1)
+ E(3s2 – 1) + Fs(s2 – 3) ........... (2)
and then equate from both sides of the identity (2) the coefficients
of s5, s4, s3, s2, s and s0. Thus, we get the following system of six
equations in six unknowns:
0 A   A  0,
0  BC   B  3,
 
0  2 A  D  F  C  3,
 ==> 
0  2 B  3E   D  0,
0  A  D  3F   E  2,
 
8  B  C  E   F  0.
With these values, we get from (1),
8 3 3( s 2  1) 2(3s 2  1)
=  
( s 2  1) 3 s 2  1 ( s 2  1) 2 ( s 2  1) 3
And the inverse Laplace transform is given by

27
  8 
L1  2 3
= 3 sin t  3t cos t  t 2 sin t. [Answer]
 ( s  1) 
Example (26): Using Majumdar-Shahidul method find the inverse
400s 2
Laplace transform of 2 .
( s  1)[(s  1) 2  4]2
400s 2
Solution: We assume that =
( s 2  1)[(s  1) 2  4]2
As  B C ( s  1)  D E[(s  1) 2  4]  F ( s  1)
  ........... (1)
s2 1 ( s  1) 2  4 [(s  1) 2  4]2
which gives, after multiplying throughout by (s2 + 1)[(s + 1)2 + 4],
the following identity:
400s2 =(As + B)[(s + 1)2 + 4]2 + [C(s + 1) + D][(s + 1)2 + 4](s2 + 1)
+ [E{s + 1)2 – 4} + F(s + 1)](s2 + 1) ............. (2)
Equating from both sides of (2) the coefficients of s5, s4, s3, s2, s
and s0 successively, we get
0  AC   A  16,
0  4A  B  C  D  E   B  12,
 
0  14 A  4 B  8C  2 D  2 E  F  C  16,
 ==> 
400  20 A  14 B  8C  6 D  2 E  F   D  51,
0  25 A  20 B  7C  2 D  2 E  F   E  55,
 
0  25B  5(C  D)  3E  F   F  40.
Therefore, from (1), we get
400s 2 s 1 s 1
= 16 2 – 12 2 – 16
( s  1)[(s  1)  4]
2 2 2
s 1 s 1 ( s  1) 2  4
1 ( s  1) 2  4 s 1
+ 51 – 55 – 40
( s  1)  4
2
[( s  1)  4]
2 2
[( s  1) 2  4] 2
and the inverse Laplace transform is given by

28
 
1 400s 2 
L  2 2 
= 16 cos t  12 sin t  16e t cos 2t
 ( s  1)[(s  1)  4] 
2

51
 e t sin 2t  55te t cos 2t  10te t sin 2t. [Answer]
2
Example (27): By the inverse Laplace transformation solve the
t
equations (i) y(t) = t2 +  y(u) sin(t  u)du
0
t
y (u )du
(ii) 
t u
0
= 1 + t + t2

Solution: (i) Taking Laplace transformation, we have

t
L[y(t)] = L[t ] + L[  y (u ) sin(t  u )du ]
2

2 1
Or, y (s) = + y (s). 2
s 3
s 1
2( s  1)
2
2 2
Or, y (s) = 5
= 3 + 5
s s s
Taking inverse Laplace transform on both sides, we get
1 4
y(t) = t2 + t [Answer]
12
t
y (u )du
(ii) Given that  = 1 + t + t2
0 t u
t
1
Or,  y(u) (t  u)
0
2
du = 1 + t + t2

Taking Laplace transformation, we have

( 1 )! 1 1 2
y (s). 12 = + 2 + 3
s 2 s s s

29
 1 1 1 2
Or, y (s) = [ 1 + 3 + 5 ]
( 2 ) s 2
1
s 2 s 2
Taking inverse Laplace transformation on both sides, we get
1 t 1 / 2 t 1 / 2 2t 3 / 2
y(t) = [  1  3 ]
( 12 ) ( 12 ) 2! 2!

1 1 t 2t 2
= [   ]
  t 1
2  t 3
4  t
1
= [3 + 6t + 8t2]
[Answer]
3 t
Note: The main purpose of the Laplace transformation is to solve
various types of differential equations.

Example (28): Using Laplace transformation solve the equation

d 2x dx
2
–2 + x = et; given that x(0) = 2, x/(0) = – 1.
dt dt
Solution: Taking Laplace transformation on both sides of the
given equation, we get
d 2x dx
L[ 2 ] – 2 L[ } + L[x} = L[et]
dt dt
1
Or, [s2 x – s x(0) – x/(0)] – 2[s x – x(0)] + x =
s 1
1
Or, s2 x - 2s + 1 – 2s x + 4 + x =
s 1
1
Or, x (s2 – 2s + 1) = + 2s – 5
s 1
2s 2  7 s  6
Or, x =
( s  1) 3
2 3 1
Or, x = – +
( s  1) ( s  1) 2
( s  1) 3
30
Taking inverse Laplace transformation, we have
1
x = 2et – 3tet + t2et [Answer]
2
Example (29): By using the method of Laplace transformation
solve the equation y// + y/ – 2y = t; y(0) = 1, y/(0) = 0.
Solution: Taking Laplace transformation on both sides of the
given equation, we have
1
[s2 y – s y(0) – y/(0)] + [s y – y(0)] – 2 y = 2
s
1
Or, y (s2 + s – 2) = 2 + s + 1
s
1 s  s
2 3
Or, y = 2 2
s ( s  s  2)
1  s2  s3
Or, y =
s 2 ( s  2)( s  1)
1 1 1 1
Or, y = 2   
2s 4s 4( s  2) s  1
Taking inverse Laplace transformation on both sides, we get
1 1 1
y =  t   e 2t  e t [Answer]
2 4 4
Example (30): Using Laplace transformation solve the equation
d2y dy
2
+2 + 5y = e–tsint, given that y(0) = 0, y/(0) = 1.
dt dt
Solution: Taking Laplace transformation on both sides of the
given equation, we have
1
[s2 y – s y(0) – y/(0)] + 2[s y – y(0)] + 5 y =
( s  1) 2  1
1
Or, y (s2 + 2s + 5) = 2 +1
s  2s  2

31
 s 2  2s  3
Or, y =
( s 2  2s  2)(s 2  2s  5)
1 1 2
Or, y = [ + ]
3 ( s  1)  1 ( s  1) 2  2 2
2

Taking inverse Laplace transformation on both sides, we get

1
y = [e– t sint + e– t sin 2t] [Answer]
3
Example (31): By using the method of Laplace transform solve
the equation y/// + 2y// – y/ – 2y = 0; y(0) = 1, y/(0) = 2, y//(0) = 2.
Solution: Taking Laplace transformation on both sides of the
given equation, we have
[s3 y – s2 y(0) – s y/(0) – y//(0)] + 2 [s2 y – s y(0) – y/(0)]
– [s y – y(0)] – 2 y = 0
Or, y (s + 2s2 – s – 2) = s2 + 4s + 5
3

Or, y (s + 2)(s + 1) (s – 1) = s2 + 4s + 5
s 2  4s  5
Or, y =
( s  2)( s  1)( s  1)
1 1 1 5 1
Or, y = . – + .
3 s  2 s 1 3 s 1
Taking inverse Laplace transformation on both sides, we get
1 5
y = e– 2t – e– t + et [Answer]
3 3

Exercises:
1. Define Laplace transformation with examples.
2. Find the Laplace transform of
3t4 – 2t3 + 4e–3t – 2sin 5t + 3 cos 2t
72 12 4 10 3s
[Ans: 5 – 4 + – 2 + 2 ]
s s s  3 s  25 s 4

32
 s2  a2
3. Prove that L[t cos at] = .
(s 2  a 2 ) 2
8s  8
4. Find the Laplace transform of (tetsin4t). [Ans: ]
( s  2s  17) 2
2

48
5. Show that (i) L[sin32t] = ;
s  40s 2  144
4

2s  21
(ii) L[e3t(2cos 5t – 3sin 5t)] = .
s  6s  34
2

t
 : 0ta
6. If f(t) is defined by f(t) =  a
1 : t  a
1  e  as
Find the Laplace transformation of f(t). [Ans: ]
as 2
7. Find the Laplace transformation of f(t), where
cos t : 0  t  2 s(1  e 2s )
f(t) =  [Ans: ]
0 : t  2 1 s2
8. Find the Laplace transform of the periodic function f(t) with
period 2c and is defined by
t : 0  t  c 1  e  cs
f(t) =  [Ans: 2 ]
2c  t : c  t  2c s (1  e cs )
1  et  s 1
9. Calculate the value of (i) L[ ] [Ans: ln  ]
t  s 
e  at  e bt sa
(ii) L[ ] [Ans: ln  ]
t  sb
10. Find the inverse Laplace transforms of
s 1 2 sinh t
(i) ln( ) [Ans: ]
s 1 t

33
 1 1 2 1 1
(ii) [Ans: t + cos 2t – ]
s ( s  4)
3 2
8 16 16
11. Using convolution theorem evaluate
3 3 3 3
(i) L1 [ 2 ] [Ans: t + e– 2t – ]
s ( s  2) 2 4 4
1
(ii) L1 [ 2 ] [Ans: te– t + 2e– t + t – 2]
s ( s  1) 2

12. Using Majumdar-Shahidul method, find the inverse Laplace

s2  4
transform of 2
( s  1)( s 2  4) 2
5 4 1
[Ans  sin t  sin 2t  t cos 2t ]
9 9 3
 8 
13. Using Majumdar-Shahidul method evalute L1  3
.
[( s  1)  1] 
2

[Ans: 3e t sin t  3te t cos t  t 2 e t sin t. ]

14. Using Laplace transformation solve the equation
d2y
2
+ y = t cos 2t, given that y(0) = 0, y/(0) = 0.
dt
5 5 1 sin 2t
[Ans: sin t + sin 2t + ( – t cos 2t)]
9 18 3 2
15. Solve the given boundary value problems using the method of
Laplace transformation:
(i) y// – 3y/ + 2y = e3t; y(0) =0, y/(0) = 0
1 1
[Ans: y = et – e2t + e3t]
2 2
(ii) y// + 2y/ + y = 0; y(0) = 0, y(1) = 2 [Hints: let y/(0) = c and
calculate c = 2e from the result] [Ans: y(t) = 2te1 – t]
// / /
(iii) y + 3y + 2y = 0; y(0) = 1, y (0) = 2
[Ans: y = – 3e– 2t + 4e– t]

34