Laplace and Inverse Laplace Transformation – Complete Notes with Solved Examples (HK Das Reference)

Laplace Transformation 885




13
Laplace Transformation


13.1 INTRODUCTION
Laplace transforms help in solving the differential equations with boundary values without
finding the general solution and the values of the arbitrary constants.
13.2 LAPLACE TRANSFORM
Definition. Let f (t) be function defined for all positive values of t, then
st
F s e f t dt
0
provided the integral exists, is called the Laplace Transform of f (t). It is denoted as
st
L f t F s e f t dt
0


13.3 IMPORTANT FORMULAE
1 n!
1. L(1)  2. L tn  , when n  0, 1, 2, 3...
s s n 1
1 s
3. L e at
sa 4. L (cosh at) = s2  a2
sa s  a2
2

a a
5. L (sinh at) = 2 (s2  a2 ) 6. L (sin at) = s0
s  a2 s2  a2
s
7. L (cos at) = 2 s0
s  a2
1
1. L 1 =
s
 
  e  st  1 1  1 1

 st
Proof. L(1) = 1 e dt       st    0  1 
0
  s 0 s  e 0 s s
1
Hence L(1) = Proved.
s
n!
2. L tn where n and s are positive.
sn 1


Proof. L(tn) =  0
e  st t n dt

x dx
Putting st = x  t  dt =
s s




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, 886 Laplace Transformation
n
  x dx 1 
Thus, we have L(tn) =
 0
e x  
s s
 L(tn) =
s n 1  0
e x . x n dx
  
 L (t n ) 
n 1 n n!  n 1   0
e x .x n dx 
 L t    Proved.
s n 1 s n 1  and n +1  n ! 
 
at 1
3. L e , where s > a
s a
 
Proof. L (eat) =  0
e  st .e at dt   0
e  st  at .dt


   sa t  e sa t  1  1 
 
s a t
= e .dt  e .dt    –  
0 0   s  a  0 s  a  e sa t 
0
1 1
= 0 1  Proved.
sa sa
s
4. L cosh at
2
s a2
 e at  e  at   e at  e at 
Proof. L cosh at = L    cosh at  
2  2 
 
1 at 1  at 1  1 1   at 1 
= L e  L e =  
2  s  a s  a  L e  s  a 
2 2  
1 s  a s  a s
=  2  Proved.
2  s  a 2  s 2  a 2
a
5. L sinh at
s a2
2

 1 at  at 
Proof. L (sinh at) = L  e  e 
2 
1 at  at 1 1 1  1 s a s a
= [ L( e )  L(e )]    
2 2  s  a s  a  2  s 2  a 2 
a
= 2 Proved.
s  a2
a
6. L sin at
s a2
2

 eiat  e iat   eiat  e iat 
Proof. L (sin at)  L    sin at  
 2i   2i 
1  iat  1
=
iat
L e e   L eiat  L e  iat 
2i   2i  
1 1 1  1 s  ia  s  ia 1 2ia a
=    =  Proved.
2i  s  ia s  ia  2i s 2  a 2 2i s2  a2 s  a2
2

s
7. L (cos at ) 2
s a2
 eiat  e – iat   eiat  e – iat 
Proof. L (cos at )  L    cos at  
2  2 
 




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, Laplace Transformation 887

1 1 1 1 1  1 s  ia  s – ia
 [L(eiat  e – iat )]  [L(eiat )  L(e – iat )]    
2 2 2  s – ia s  ia  2 s2  a 2
s
 Proved.
s  a22

Example 1. Find the Laplace transform of f(t) defined as
t
 , when 0  t  k
f (t )   k
1, when t  k
k   e st  
k t  1  e st  k e  st
  dt    
 st  st
Solution. L [f (t)] = e dt  1.e dt   t   
0 k k k  s  0 0 s   s 
  k
  ks  st k  ks  ks  sk  ks
1 ke e  e 1  ke e 1 e
    2       2  2
k   s  s 0  s k  s s s  s
 
e  sk 1 e ks 1 1 e ks 1
     2 [ e  ks  1] Ans.
s k s2 k s2 s ks
Example 2. From the first principle, find the Laplace transform of (1 + cos 2 t).
Solution. Laplace transform of (1 + cos 2 t)
   e 2it  e 2it 
  e  st 1  cos 2t dt   e  st 1   dt
0 0
 2 

1  2e  st e 
 s  2i t  s  2i t
1  e
 2e  st  e  s  2 i t  e  s  2i t dt  
 
2 0   2  s
 
 s  2i s  2i  0


1  2 1 1 
 0  0 1  0 1 
2   s   s  2i  s  2i 
1 2 1 1  1 2 2s 
     2
2  s s  2i s  2i  2  s s  4 
 
1 s 2s 2  4
  2  Ans.
s s  4 s s2  4
13.4 PROPERTIES OF LAPLACE TRANSFORMS
(1) L[af1 t bf 2 t ]  a L  f1 t   b L[ f 2 t ]

Proof. L[af1 t  bf 2 t ]   0
e  st [af1 t  bf 2 t ] dt
 
a  0
e  st f1 t dt  b  0
e  st f 2 t dt

 a L [ f1 t ]  b L [ f 2 t ] Proved.
(2) First Shifting Theorem. If L f (t) = F (s), then
L [e at f (t)] = F (s – a)
 
Proof. L [e at f (t)] =  e  st ·eat f (t) dt =
0  0
e  s  a t f t dt

= 0 e  rt f t dt where r = s – a
= F (r) = F (s – a) Proved.




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, b s a
4. L (eat sin bt) = 2 2 5. L (eat cos bt) = 2
sa b sa  b2




888 Laplace Transformation

With the help of this property, we can have the following important results :
n!  n n! 
 L t  s n 1 
at n
(1) L e t  n 1
sa  
sa b
(2) L e at cosh bt  2
(3) L e at sinh bt  2
sa b 2
s  a  b2
at b sa
(4) L e sin bt 
at
2 (5) L e cos bt  2
s  a  b2 s  a  b2
Example 3. Find the Laplace transform of cas2 t.
Solution. cos 2 t = 2 cos2 t – 1
1
 cos2t = [cos 2 t+ 1]
2
1  1
L (cos2 t) = L  cos 2t  1   2  L cos 2t  L 1 
2 

1 s 1 1  s 1
= 2 2    2   Ans.
 s  2
2
s 2 s  4 s

1

Example 4. Find the Laplace Transform of t 2 .

Solution. We know that L t n  n  1
s n 1
1 1
 1
1  1
Put n =  , L t 1/ 2
 1/2 2 1  2  , where   Ans.
2 s s s 2
Example 5. Find the Laplace Transform of t sin at.
 eiat  e iat  1
   L t.e  L t.e 
iat  iat
Solution. L t sin at  L  t 
 2i  2i
1  1  s  ia  s  ia 2 
2
1 1
     
2i  s  ia 2 s  ia  2i  s  ia s  ia 
2 2 2

2 2 2 2
1 s  2ias  a  s  2ias  a
 2
2i s2  a2
1 4ias 2as
 
2i s  a
2 2 2
s  a2
2 2 Ans.
Example 6. Find the Laplace Transform of t 2 cos at.

 eiat  e iat  1
Solution.    L t .e  L t e 
2 iat 2  iat
L t 2 cos at  L  t 2 
 2  2
1  2! 2! 
3 3
s  ia  s  ia
    
2  s  ia 3 3
s  ia 
3
s  ia s  ia
3





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