Laplace Transform
Definition of Laplace Transform [IoPE2010]
Let f(t) be a function of t defined for all positive values of t, then the Laplace Transform of f(t) is defined by

L[ f (t )]  f (s)   e st f (t )dt
0
provided the integral exists. Here s is a parameter which may be real or complex number.
Laplace Transform of Some Elementary Functions
(1) L(1)  1 , s  0
(2)L(eat )  1 , s  a
s
sa
s , s0
(3)L(sinat)  a
,s0
(4)L(cosat) 
2
2
2
s a
s  a2
s ,s a
(5) L(sinhat)  a , s  a
(6)L(coshat) 
2
2
2
s a
s  a2
(7)L(t n )  n! where n  0,1,2,3,   
s n 1
Linearity Property of Laplace Transform [IoPE 2013]
If f,g,h are functions of t and a,b,c are constants, then
L[a f (t )  b g (t )  ch(t )]  a L[ f (t )]  b L[ g (t )]  c L[h(t )]. .
Examples
Find the Laplace Transforms of the following:
1) 4e5t  6t 3  3sin 4t  2cos 2t  2 [IoPE2017]
2) sin 5t  sin3 t [IoPE 2013] 3) e at [IoPE 2009]
Examples for Tutorial
Find the Laplace Transform of the following:
𝐴
𝐵
2𝐶
1) A  Bt  Ct 2 [IoPE2014]
[𝑠 + 𝑆2 + 𝑆3 ]
2) 3 𝑒 2𝑡 − 2 sin 3𝑡 [IoPE 2009]
[𝑠−2 − 𝑠2 +9]
3) sin(at + b) [IoPE 2012]
4) 4 sin4t + cos4t +4t [IoPE 2007]
5) sin2t sin5t [IoPE 2013]
6) sin2 3𝑡 [IoPE 2008]
7) sin3 3t [IoPE 2014]
8) 3cosh 5t  4sinh 5t
First Shifting Property
If L[ F (t )]  f ( s), then L[eat F (t )]  f ( s  a).
3
[cos 𝑏 (
6
𝑎
𝑠
) + sin 𝑏 (𝑠2+𝑎2 ) ]
𝑠2 +𝑎2
16
𝑠
4
[𝑠2 +16 + 𝑠2+4 + 𝑠2 ]


20s
 2

2
 ( s  9)( s  49) 
1 1
𝑠
[2 [𝑠 − 𝑠2 +36]]
162
[(𝑠2+9)(𝑠2+81)]
 3s  20 
 2

 (s  25) 
Following useful formulae are the application of this property.
n!
b
(1) L(eat t n ) 
,s  a
(2) L(eat sin b t) 
,s  a
n 1
2
2
( s  a)
( s  a)  b
sa
b
(3) L(eat cos b t) 
,s  a
(4) L(e at sinh b t) 
,s  a
( s  a)2  b2
( s  a)2  b2
sa
(5) L(eat cosh b t) 
,s  a
( s  a)2  b2
Note: Hyperbolic sine and cosine functions are given as follows:
e x  e x
e x  e x
1)sinh x 
, 2) cosh x 
2
2
Examples
Find the Laplace transform of the following functions:
1) e2t t 2
2) 𝑒 −2𝑡 (2 cos 5𝑡 − 4 sin 5𝑡) [IoPE 2012]
Examples for Tutorial
Find by using the First Shifting Theorem:
1)
 6 

4
 ( s  2) 
L ( e 2t t 3 )
2) L[ e3t (cos 4t  3sin 4t ) ] [IoPE2014]
[
𝑠+15
𝑠 2 +6𝑠+25
]
 s 1 
 2

 s  2s  3 


12s
 4

2
 s  10s  169 
3) 𝐿(𝑒 −𝑡 cosh 2𝑡) [IoPE2010]
4) L(sin 3t sinh 2t )
Second Shifting Property
0
If L[ F (t )]  f ( s) and G(t )  
 F (t  a)
when t  a
then L[ F (t )]  e-as f ( s).
when t  a
Definition of Inverse Laplace Transform
If L[f(t)] = f(s), then f(t) is called the inverse Laplace Transform of f(s) and it is denoted by
f (t )  L1[ f ( s)], where L1 is inverse Laplace Transform operator.
Linearity Property of Inverse Laplace Transform
If L1[ f ( s)]  f (t ) and L1[ g ( s)]  g (t ) then
L1[ a f ( s)  b g ( s)]  aL1[ f ( s)]  bL1[ g ( s)]
where a and b are constants.
Formulae on Inverse Laplace Transform
 s 1 a   eat , s  a
1) L1( 1 )  1, s  0
2) L1

s   cosat
4) L1 
 s2  a2 

 n
7) L1  1   t
 s n 1  n !

 1
1
5) L1 
  a sinhat
2
2
 s a 
s

 1
1
3) L1 
  a sinat
2
2
s a 

s   coshat
6) L1 
 s2  a2 
Formulae of inverse Laplace Transform using First Shifting Theorem:


at
1
  e sin b t , s  a
b
 ( s  a)2  b2 
1) L1 


at
1
  e sinh b t , s  a
b
 ( s  a)2  b2 
3) L1 





sa
  eat cos b t, s  a
2
2
 ( s  a)  b 
2) L1 
sa
  eat cosh b t, s  a
 ( s  a)2  b2 
4) L1 

at n
  e t ,s  a
n!
 ( s  a)n 1 
5) L1 
1
Inverse Laplace Transforms – Method of Partial Fractions:
We have seen that the Laplace transform of a function is a rational algebraic function. So, we express the
given function into partial fractions and then to find the inverse Laplace transform by standard formulae.
Examples: Find






2s 2  4
4s  5
1) L1 
2) L1  3s  2  3) L1 


2
 ( s  1)( s  2)( s  3) 
 s2  s  2 
 ( s  2)( s  1) 


Examples for Tutorial
Find the inverse Laplace Transform of the following functions:

1  . [IoPE 2006]
 s( s4) 
1) L1 


1  7s
. [IoPE2016]
2) L1 
 ( s  3)( s  1)( s  2) 


3s
3) L1 
 [IoPE 2016]
 s 2  2s  8 


4) L1  23s  1 
 s  4s 


1
5) L1 

2
 ( s  1)( s  1) 
 1 1 4 t 
 4  4 e 
2e3t  et  e2t 


2e4t  e2t 


[ 1  3 e2t  5 e2t ]
4 8
8
[ 1 (et  cos t  sin t )]
2


5s  3

4) L1 
 ( s  1)( s 2  2s  5) 
Laplace Transform of Derivatives
If L[ F (t )]  f ( s) then
 dn

L  n F (t )   s n f ( s )  s n-1F (0)  s n-2 F (0)    sF n-2 (0)  F n-1 (0).
 dt

Note: 1) Laplace transform of first order derivative is L[ F (t )]  s f ( s)  f (0).
2) L ( x )  x
3) L( y )  y
 dy 
4) L    L( y)  s y  y (0) where y (0) is the of y at t  0.
 dt 
 dx 
5) L    L( x)  s x  x(0) where x(0) is the of x at t  0.
 dt 
Solution of differential equations using Laplace transform
The Laplace transform can be used to solve differential equations of first order and first degree.
Steps to solve examples:
1) Take the Laplace transform on both sides of the given differential equation and apply initial conditions.
2) Collect the terms with x or y on L.H.S. and remaining terms on R.H.S.
3) Then find x or y as a function of s .
4) Resolve f(s) into partial fraction.
5) Then take inverse Laplace transform on both sides. This gives x or y as a function of t.
Examples
Solve the differential equations using Laplace transform using initial conditions.
dy
 y  3e2t , y (0)  1
1)
dt
2) 3x  2 x  e3t , x(0)  1
Examples for Tutorial
1)
dx
 x  3, x(0)  2
dt
2) y  y  sin t , y  0 when t = 0.
[ x (t )  3  e

1
 y (t )

2
t
]
et  cos t  sin t 


