Laplace Transform

Laplace Transform: Let F (t ) be a function defined for t  0, then the Laplace transform of F (t )
is denoted by L{F (t )}, or f ( s ) and is defined as follows:

L{F (t )} = f ( s) =  e− st F (t )dt
0
where s is real or complex.


Note: If the integral 0
e − st F (t )dt converges for some value of s, then L{F (t )} is said to exist
otherwise it does not exist.

The Linearity Property of Laplace Transform: A Laplace transform L{F (t )} is said to be

linear if for every pair of functions F1 (t ) and F2 (t ) satisfy the following condition:
L{c1 F1 (t ) + c 2 F2 (t )} = c1 L{F1 (t )} + c 2 L{F2 (t )}
where c1 and c 2 are any constant.

First Translation or Shifting Property: If L{F (t )} = f ( s ), then L{e at F (t )} = f ( s − a).

 F (t − a) if t  a
Second Translation or Shifting Property: If L{F (t )} = f ( s ) and G (t ) =  then
 0 if t  a
L{G(t )} = e − as f ( s)

1 s
Change of Scale Property: If L{F (t )} = f ( s ), then L{F (at )} = f  .
a a

Laplace Transform of Derivatives: If L{F (t )} = f ( s ), then

L{F n (t )} = s n f ( s) − s n −1 F (0) − s n − 2 F (0) −  − sF n − 2 (0) − F n −1 (0)

Some Relations on Laplace Transform and Inverse Laplace Transform:

n!  n! 
1. L{t n } = n +1  t n = L−1  n +1 
s s 
1 1 
2. L{1} =  1 = L−1  
s s
1  1 
3. L{e at } =  e at = L−1  
s−a s − a
1  1 
4. L{e − at } =  e − at = L−1  
s+a s + a
 s  s 
5. L{cos at} =  cos at = L−1  2 2 
s + a2 2
s + a 
a  a 
6. L{sin at} = 2  sin at = L−1  2 2 
s + a2 s + a 
 n !  at n
7.L e at t n  =
n!
 L−1  n +1 
=e t
(s − a) ( )
n +1
 s − a 

−1 
 at
8.L e at sin bt =
b b
 L   = e sin bt
( s − a ) + b2  ( s − a ) + b 
2 2 2

s−a 
−1  s−a  at
9.L e at cos bt =  L   = e cos bt
( s − a ) − b2  ( s − a ) − b 
2 2 2

−1 
  at
10.L e at sinh bt =
b b
 L   = e sinh bt
( s − a ) − b2  ( s − a ) − b 
2 2 2

s−a −1 
 s−a  at
11.L e at cosh bt =  L   = e cosh bt
( s − a ) − b2  ( s − a ) − b 
2 2 2

Theorem:
1. If the function F (t ) is piecewise condition over any close interval 0  t  n and of
exponential  for t>n then its Laplace transformation f(s)=L{F(t)} exist for all s   .
Proof : since F(t) is integrable in the close interval, its F(t) is exist.
Now

L{F (t )} =  e − st F (t )dt  F (t )  Me t 
0

     e− ( s − )t  M
0 0 0 0
t − st  t
− s − a  = s − a ; s  
− st − st − st
 L{F (t )} = e F ( t ) dt  e F (t ) dt  e Me dt = M e e dt = M
 
Hence the Laplace transformation exist for s   .
2. State and prove linear property of Laplace Transformation.
3. State and prove first transformation or shifting property of Laplace Transformation
Or, If L{F (t )} = f ( s) then, L{e at F (t )} = f ( s − a).
Question: Find the Laplace transformation by using first shifting method.
(n + 1)
( i ) L{t n eat } = n +1 ( )
ii L{e at sin bt} ( iii ) L{e 2t (3sin 4t − 4sin 4t} ( iv ) L{t 3e5t }
(s − a)
3! 6
Solution ( iv ) let F (t ) = t 3 then L{F (t )} = L{t 3 } = = = f (s)
s4 s4
we knowthe first sheifting method ,
L{e F (t )} = f ( s − a ) then
at

6
L{e5t t 3 } = f ( s − 5) = .(ans )
( s − 5)
4
 4. State and prove second transformation or shifting property of Laplace Transformation
 F (t − a),; t  a
Or, If L{F (t )} = f ( s ) , G (t ) =  then L{G (t )} = e − sa f ( s ).
 0 ;t  a
Question: Find the Laplace transformation by using 2nd shifting method.
  2  2
( t − 1)2 when t  1 ( t − 1)3 when t  2 cos  t − 3  when t  3
( i ) F (t ) =  ( ii ) F (t ) =  ( iii )( ii ) F (t ) =   
0 when 0  t  1 0 when 0  t  2 0 2
when 0  t 
 3
t / a when 0  t  a
( iv ) F (t ) = 
1 when t  a

5. State and prove change of scale property of Laplace Transformation.

1 s
Or, If L{ f (t )} = f ( s ) then showthat L{F (at )} = f ( ).
a a
 sin t  −1 1  sin at  −1 a
Questions If L   = tan then showthat L   = tan .
 t  s  t  s
6. L{F (t )} = f ( s ) then showthat L{F (t )} = sf ( s ) − F (0).
7. L{F (t )} = f ( s) then showthat L{F (t )} = s 2 f (s ) − sF (0) − F ' (0).
8. L{F (t )} = f ( s ) then showthat L{F (t )} = s 3 f (s ) − s 2 F (0) − sF ' (0) − F  ( 0 ) .

9. L{F (t )} = f ( s) then showthat ( i ) L  F (u)du = f (ss) .(ii ) L  sinu u du  = 1s tan
0
t

0
t
−1 1
s
. ( iii ) L{t n F (t )} = ( −1) f n ( s )
n

Theorem:
(n + 1)
(i) Show that L t n  = when n  −1 and s  0.
s n +1
Show that L t n  = n +1 when n = 0,1, 2,3.... and s  0.
n!
(ii)
s
Solution: By the definition of Laplace transformation

L{F (t )} = f ( s ) =  e − st F (t )dt
0

Given function F (t ) = t n , then


L{t n } = f ( s) =  e − st t n dt.......(1)
0

1
Now putting st = y , then sdt = dy dt = dy equation (1)we get ,
s
n
y
Also t n = n ; t = 0 then y = 0 and t =  then y = 
s
the equation (1) we get ,
 y n dy 1  1 (n + 1) n ! 
L{t n } = f ( s) =  e − y = n +1  e − y y ( ) dy = n +1 (n + 1) = = n +1 ;[ (n) =  e − x x n −1dx; n  0]
n +1 −1
n n +1
0 s s s 0 s s s 0

Problem:
 (i) show that
1 a 1  −1  
(i ) L 1 = ;(ii ) L a = ; ( iii ) L t = 2 ; ( iv ) L t 2  = .
s s s   s
(ii) If s>a then show that
(i ) L e at  = (ii ) L eiat  =
1 1 a s
( iii ) L sin at = 2 2 ( iv ) L cos at = 2 2
s−a s − ia s +a s +a
a s
( v ) L sinh at = 2 2 ( vi ) L cosh at = 2 2
s −a s −a
(iii) Find the Laplace transformation of the following expression.
( i ) 4e5t + 6t 3 − 3cos 4t + 4sin 5t ( ii ) 3coh5t − 4sin 5t (iii ) e 4t + 4t 3 − 2sin 3t + 3cos 5t
.
( iv ) 3t 4 − 2t 3 − 4e−3t − 2sin 5t + 3cos 2t.
(iv) Find the Laplace transformation of the following function.
t , 0  t  2
( i ) F (t ) = t sin at ( ii ) t cos at ( iii ) F (t ) =  .
3, t  2

Question: Using Laplace transformation of 1st derivatives to show that.

1 1 1
( i ) L{1} = , s  0; ( ii ) L{t} = 2 , s  0; ( iii ) L{e at } = , s  0.
s s s−a

Question: Using Laplace transformation of 2nd derivatives to show that.

a a
( i ) L{sin at} = , s  0; ( ii ) L{sinh at} = 2 .
s +a 2
2
s − a2

Question: show that

s2 − a2 6as 2 − 2a 3 2 s 3 − 6a 2 s
2 ( ) 
iii L t 2 sin at = 3 ( ) 
iv L t 2 cos at =
2as
( i ) L t sin at = 2 ( ) 
ii L t cos at =
(s + a2 ) (s + a2 ) (s + a2 ) (s + a2 )
2 2 2 2 3

Question:

sin at 
1. show that 
0
t
dt = ;
2


2. Prove by use of Laplace transformation  e − x dx =
2
;
0
2

e
− st
F (t )dt
Theorem: If F(t) has period T>0 then L{F (t )} = 0
.
1 − e sT
 sin t , 0  t  
Question: Find the Laplace transformation of the function F ( t ) = 
0 ,   t  2

Question:Prove that
 
s2 − a2 2as
( i )  te− st cos at dt = 2 ( )
. ii te − st sin at dt = .
( s2 + a2 ) ( s2 + a2 )
2
0 0

Inverse Laplace Transform

Inverse Laplace Transform: If L{F (t )} = f ( s ), then F (t ) is called an inverse Laplace transform

of f (s) and it is denoted by
F (t ) = L−1  f ( s)
where L−1 is called inverse Laplace transformation operator.

The Linearity Property of Inverse Laplace Transform: If L{F1 (t )} = f 1 ( s), L{F2 (t )} = f 2 ( s)

and c1 , c 2 are constants, then
L−1{c1 f 1 ( s) + c 2 f 2 ( s)} = c1 L−1{ f 1 ( s)} + c 2 L−1{ f 2 ( s)} = c1 F1 (t ) + c 2 F2 (t )

First Translation or Shifting Property: If L−1{ f ( s)} = F (t ), then L−1{ f ( s − a)} = e at F (t ).

Second Translation or Shifting Property: If L−1{ f ( s)} = F (t ), then

 F (t − a), t  a
L−1 {e − as f ( s)} = 
 0 , ta

1 t
Change of Scale Property: If L−1{ f ( s)} = F (t ), then L−1 { f (as )} = F  .
a a

Inverse Laplace Transform of Derivatives: If L−1{ f ( s)} = F (t ), then L−1 { f n ( s)} = (−1) n t n F (t ).

Inverse Laplace Transform of Integrals: If L−1{ f ( s)} = F (t ), then L−1 s



f (u )du =
F (t )
t
.

Convolution Property: If L−1 { f ( s)} = F (t ) and L−1 {g ( s)} = G(t ), then

t
L−1{ f ( s) g ( s)} =  F (u )G (t − u ) du
0
Some Relations on Laplace Transform and Inverse Laplace Transform:
 n!  n! 
1. L{t n } = n +1
 t n = L−1  n +1 
s s 
1 1 
2. L{1} =  1 = L−1  
s s
1  1 
3. L{e at } =  e at = L−1  
s−a s − a
1  1 
4. L{e − at } =  e − at = L−1  
s+a s + a
s  s 
5. L{cos at} =  cos at = L−1  2 2 
s + a2 2
s + a 
a  a 
6. L{sin at} = 2  sin at = L−1  2 2 
s + a2 s + a 

 n !  at n
7.L e at t n  =
n!
 L−1  n +1 
=e t
(s − a)  ( s − a ) 
n +1


−1 
 at
8.L e at sin bt =
b b
 L   = e sin bt
( s − a ) + b2  ( s − a ) + b 
2 2 2

s−a −1 
 s−a  at
9.L e at cos bt =  L   = e cos bt
( s − a ) − b2  ( s − a ) − b 
2 2 2

−1 
  at
10.L e at sinh bt =
b b
 L   = e sinh bt
( s − a ) − b2  ( s − a ) − b 
2 2 2

s−a −1 
 s−a  at
11.L e at cosh bt =  L   = e cosh bt
( s − a ) − b2  ( s − a ) − b 
2 2 2

Theorem:
1. If L{F1 (t )} = f 1 ( s), L{F2 (t )} = f 2 ( s) and c1 , c 2 are constants, then show that
L−1{c1 f 1 ( s) + c 2 f 2 ( s)} = c1 L−1{ f 1 ( s)} + c 2 L−1{ f 2 ( s)} = c1 F1 (t ) + c 2 F2 (t )

Question:

1. Prove that

−1 
 at  s − a  at
−1 
( i ) L−1 
1  sin at −1  1  tn b
2
= . ( ii ) L  n+1  = . ( iii ) L   = e sin bt . ( iv ) L   = e cos bt
s + a   s  ( n + 1)  ( s − a ) + b   ( s − a ) + b 
2 2 2 2 2
a

Theorem: state and prove first transformation or shifting property of inverse Laplace
transformation
Or If L−1{ f ( s)} = F (t ), then show that L−1{ f ( s − a)} = e at F (t ).
Question: Find the inverse Laplace transformation of

 3s − 8 4 s − 24  6s − 4 
( i ) L−1 
1   1   1  −1 
2 ( ) 2 (
ii L−1  2 iii ) L−1  3/2  ( iv ) L−1  2 − 2  (v ) L  2 
s + a  s + a  s   s + 4 s − 16   s − 4s + 20 
2

 2s − 11  4s + 12   
−1  −1  −1  7 s + 12  −1  s 2 + 2s + 3 
( vi ) L   ( vii ) L  2  ( viii ) L  2  ( ix ) L  2 
 ( s + 2 )( s − 3)   s + 8s + 16   s +9  (
 ( s + 2s + 2 ) ( s + 2s + 5 )
2
) 
s +1  −1  
( x ) L−1  2
4  −1  15  −1  1
 ( xi ) L  2  ( xii ) L  2  ( xiii ) L  3 2 
 4s + 16   s + 4s + 13   s + 6s + 25   s ( s + 4) 

−1  2s 2 − 4 
( ) 
xiv L 
 ( s + 1)( s − 2 )( s − 3) 

Theorem: state and prove 2nd transformation or shifting property of inverse Laplace
transformation
Or, If L−1{ f ( s)} = F (t ), then show that
 F (t − a), t  a
L−1 {e − as f ( s)} = 
 0 , ta

Theorem; If L−1{ f ( s)} = F (t ), then show that L−1 { f n ( s)} = (−1) n t n F (t ).

Question: Find inverse transformation of derivatives
   
−1  s  −1  s +1 
( )  2 2 2 ( )  2
i L ii L 
 ( s + a )   ( s + 2s + 2 ) 
2 2

 1  t 2
Question: Show that L−1  3 2  = + cos t − 1.
 s ( s + 1)  2

Convolution: If F(t) and G(t) be two functions then the convolution of two function F(t) and
G(t) denoted by F*G is define by the relation
If L−1 { f ( s)} = F (t ) and L−1 {g ( s)} = G(t ), then
t
L−1{ f ( s) g ( s)} =  F (u )G (t − u ) du = F * G
0
Theorem: State and prove convolution theorem.
Question: Using convolution theorem to show that
 
 
( i ) L−1  2 2 2  =
s t sin at 1
; ( ii ) L−1  2 2
= te −t + 2e −t + t − 2
 ( s + a )  2 a  s ( s + 1) 
Question: Using convolution theorem find the value of
   
 1  −1  1  −1  s  
−1  3  −1  1 
(i ) L −1
 ( ii ) L   ( iii ) L   ( iv ) L   (v) L  2 2 
 ( s + 1)( s − 2 )   ( s − 2 ) ( s + 1)   ( s + 4 )   s ( s + 2 )   s ( s + 4) 
2 2 3 2

   
( vi ) L−1   −1 
s 1
 ( vi ) L  2 2
 ( s + a )   s ( s + 1) 
2 2 2

Theorem: State and prove Heaviside’s expansion theorem.

Question: Using Heaviside’s expansion formula evaluate
   2s 2 − 4   3s + 1 
3s −1  −1 
(i ) L −1
 ( ii ) L   ( iii ) L  
 ( s + 1)( s − 3)   ( s + 1)( s − 2 )( s − 3)   ( s − 1) ( s + 1) 
2

Question: prove that

  
  
( i )  e− x dx = ( ii )  cos x 2 dx = ( iii )  sin x 2dx =
2
;using Laplace transformation.
0
2 0 2 2 0 2 2
Question: Solve the following differential equation using Laplace transformation.
( i ) Y  + Y = 0, Y (0) = 0, Y (0) = 1 ( ii ) Y  + Y = 1, Y (0) = 2, Y (0) = 0 ( iii ) Y  + Y = t , Y (0) = 1, Y (0) = −2.
( iv ) Y  − Y = t , Y (0) = 2, Y (0) = −3. ( v ) Y  + 4Y = 9t , Y (0) = 0, Y (0) = 7 ( vi ) y '' − 2 y ' − 8 y = 0, y(0) = 3, y ' (0) = 6.
( vii ) y '' − 2 y ' − 6 y = 0, y(0) = 1, y ' (0) = 0; ( viii ) Y  − 3Y  + 2Y = 4e2t , Y (0) = −3, Y (0) = 5.
( ix ) Y  − 3Y  + 2Y = 2e−t , Y (0) = 2, Y (0) = −1. ( x ) Y  + Y = 8cos t , Y (0) = 1, Y (0) = −1.
( xi ) Y  + 25Y = 10 cos 5t , Y (0) = 2, Y (0) = 0. ( xii ) Y  + 2Y  + 5Y = e −t sin t , Y (0) = 0, Y (0) = 1.
( xiii ) Y  + 2Y  + 5Y = e−t sin t , ( xiv ) Y  − 3Y  + 2Y = 3et , Y (0) = 0, Y (0) = 1, Y  ( 0 ) = 2.
( xv ) Y  − 3Y  + 3Y  − Y = t 2et , Y (0) = 1, Y (0) = 0, Y  ( 0 ) = −2.
( xvi ) Y  − 3Y  + 3Y  − Y = t 2et ,
( xvii ) Y  + 2Y ' + Y = 3te−t , Y (0) = 4, Y (0) = 2. ( xviii ) y '' − 6 y ' + 9 y = t 2 33t , y (0) = 2, y ' (0) = 6
( xix ) y '' + 2 y ' + y = 4sin t , y(0) = −2, y ' (0) = 1; ( xx ) x '' ( t ) + x ( t ) = 6 cos 2t; x ( 0 ) = 3, x ' ( 0 ) = 1

( xxi ) Y  + a 2Y = F ( t ) , Y (0) = 1, Y (0) = −2. ( xxii ) Y  + 9Y = cos 2t , Y (0) = 1, Y ( ) = 1.
2
d 2x dx
( xxiii ) 2
+ 3 + 2 x = t + sin t subject x ( 0 ) = 1 and x ' ( 0 ) = −2 and verify the result .
dt dt
Question: Solve the following differential equation using Laplace transformation.
( i ) ty + y + ty = 0, y ( 0 ) = 1, y ( 0 ) = c ( ii ) ty + y + 4ty = 0, y ( 0 ) = 3, y ( 0 ) = 0.
( iii ) y + y + 4ty = 0, y ( 0 ) = 3, y ( 0 ) = 0
Md.Nurul Alam
Lecturer(Mathematics)
Barishal Engineering College, Barishal