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CHAPTER 5
Laplace TransformFocus of Attention
What is a Laplace Transforms?
How to use the standard Laplace
Transforms Tables?
What are the properties of Laplace
transforms? Linearity?
What are the First Shift Theorem?
Multiplication by ttheorem?
What are the Laplace Transforms of
first and second order derivatives?
How to solve differential equations
using Laplace Transforms?
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5.1 Introduction
Laplace transforms will allow us to transform
a Differential Equation into an algebraic
equation.
5.1.1Definition and Notation
Let )(tfy = , the Laplace transform ofy is
defined by
L{ }
=0
)()( dttfetf st
The transformed function is called )(sF , thus
L ( ){ } ( )sFtf =
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Revision: Integrating improper integrals
( ) lim ( )
N
Na a
f x dx f x dx+
=
( )a
f x dx
+
is convergent if the limit is finite
and ( )a
f x dx+ is divergent, if the limit is non-
finite.
5.1.2Laplace Transforms of Some Simple
FunctionsThe definition of the Laplace transform is
L{ }
=0
)()( dttfetfst
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Finding Laplace transforms of some functions
using basic principles.
Example 5.1 (a): Laplace transform of
( ) 1f t =
If 1)( == tfy , then L{ }
=0
1 dtest
=
s
edte
stst
sss
est
1)10(
1
0
==
L {1} = s1
Since: = dxxfkdxxkf )()( then
L{ })(tkf = k L{ })(tf
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Example 5.1 (b):Laplace transform of the
form ( )f t k=
Find L{ }5 .
L{ }5 = 5 L{ }1 = s5
Example 5.1 (c):Laplace transform of( )f t t=
If ttfy == )( , then
L{ } =)(tf L{ } =0
dttetst
= dtteI st - use integration by parts with
s
evdtedv
dtdutu
stst
==
==
[ ]12
2
+=
+=
=
sts
e
s
e
s
te
dts
e
s
teI
st
stst
stst
[ ]22
02
1101
ssst
s
e st=
=
+
L{ } 21
st =
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Example 5.1 (d): Laplace transform of
( ) sinf t at=
Find L{ }atsin .
Therefore, L{ }atsin =0
sin dtate st .
Using integration by parts twice:
( )
+=
0220
cossinsin ataatsas
edtate
stst
L{ }atsin 22 asa
+=
Important note: must know how to use the
Tables of Laplace Transforms.
Example 5.2: Using tables to find Laplace
Transforms
Find the Laplace Transform of each of the
following function:
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(a)2( )f t t= (b) ttf 5cos)( =
Solution
(a)2( )f t t=
Looking at the table, we find
L{ }nt = 1
!, 1, 2, 3...
n
nn
s +=
So, L ( )2t = 2 1 32! 2
s s+ =
(b) ttf 5cos)( =
From the table, we find L { atcos } = 22 ass
+
Therefore,L
{cos5t
} = 2 25
s
s +
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Structured Examples: Finding Laplace
Transforms
Using the standard tables of Laplace
Transforms
Question 1
Find the Laplace Transform
of each of the following
function:
(a) 3( ) 5f t t=
(b) ( ) 6 sin 2f t t=
Prompts/
Questions
What is a
Laplace
transform?
How does the
function
compares to
the standard
function?
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How to find Laplace Transforms of
2( ) ; ( ) sin 2 ; ( ) cos
t tf t te f t t t f t e t= = = ?
We will need:
5.2 Properties of Laplace Transforms(1) Linearity
Let L{ })(1 tf and L{ })(2 tf exist with and
constants, then
L{ } =+ )()( 21 tftf L{ })(1 tf + L{ })(2 tf
(2) First Shift Theorem
Let L ( ){ } ( )sFtf = with a constant, then
L } )()( asFtfeat =
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Example 1.11 :Using tables and linearity
law
(a) Find the Laplace Transform of each of the
following function
(1) 2( ) 2 4 1f t t t= + (2) 2( ) 2 sin 3 tf t t e=
Solution
(1)2
( ) 2 4 1f t t t= +
L 2(2 4 1)t t + = 2 L ( )2t 4 L ( )t + L (1) =
3 2
2! 1! 1
2 4s s s
+ = 3
4 4s s
s
+.
(2)2
( ) 2 sin 3t
f t t e=
L ( )2
2sin3
t
t e = 2 L (sin 3t) L ( )2t
e
= 2 23 1
23 2s s
+ .
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Example 1.12: Using tables and the First
Shift Theorem
(a) Find the Laplace Transform of each of the
following function
(1) 2( ) sin 3tf t e t= (2) 2( ) tf t t e=
Solution
(1) L ( )2 sin3te t = 2 23
( 2) 3s +
(2) L ( )2 tt e = 32!
( 1)s
+
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Question 13
Find the Laplace Transform
of each of the following
function:
(a)2( ) cosf t t=
(b)2
( ) (1 2 )f t t=
(c)2 2( ) (2 )tf t e=
Prompts/
Questions
What is a
Laplace
transform?
How does the
function
compares to
the standard
function?
o Which
formula do
you need tochange the
expression
into the
standard
form? Which
theorems must
be used?
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1.3.1 Transform of a Derivative
We can find the Laplace transforms of the
first and second derivative. Thus, we can use
these transforms to convert the differential
equations into an algebraic form.
(1) Transform of the First Derivative
L [ ] stf = )( L [ ] )0()( ftf
If )(tfy = then L [ ] sy = L [ ] )0(yy
OR L [ ]y = ( ) (0)sY s y
Example 1.12 (a): Laplace transform of the
first derivative
Find the Laplace Transform of ty 6= with the
initial condition that 5)0( =y .
Solution:
Take Laplace transform of both sides
L [ ]y = L [ ]t6
Use Table of Laplace Transform
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L [ ] sy = L [ ] )0(yy
L [ ] 26
6s
t =
s L [ ] )0(yy = 26
s
s L [ ] 5y = 26
s .
L [ ]y = 2 31 6 6 5
5s s s s
+ = +
(2) Transform of the Second Derivative
L [ ] 2)( stf = L [ ] )0()0()( fsftf
If)(tfy =
thenL
[ ] 2sy = L
[ ] )0()0( ysyy
ORL [ ]2
( ) (0) (0)y s Y s sy y =
Example 1.12(b): Laplace transforms of
the second derivative
Find the Laplace Transform of 2 3 0y y y + =
with the initial condition that (0) 2y = and
(0) 1y =
Solution
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From the tables: L [ ] 2sy = L [ ] )0()0( ysyy and L
[ ] sy = L [ ] )0(yy
L ( )2 3y y y + = L (0)
We know that L [ ]y = Y(s) and (0) 2y = and
(0) 1y =
2
( ) (0) (0) 2( ( ) (0)) 3 ( ) 0s Y s sy y sY s y Y s + =2
2
( )[ 2 3] (2) (1) 2(2) 0
2 5( )
2 3
Y s s s s
sY s
s s
+ =+
=+
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Question 14
Find the Laplace transform
of each expression and
substitute the given initial
conditions
(2) 2)0(,35 = yyy
4) 3)0(,2)0(,32 ==++ yyyyy
Prompts/
Questions
What is theLaplace
Transform of
first and
second order
derivatives?
o What doyou do
with the
initial
values?
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1.3.2 Inverse Transforms
IfL [ ] )()( sFtf = , then )(tf is called the Inverse
Laplace Transform of )(sF and is written as
L1 [ ])(sF )(tf=
We find the inverse Laplace
Transform by reading the same tables but
in reverse.
We have chosen a few functions to
illustrate how it is done.
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L1 [ ])(sF )(tf=
)(sF )(tf
s
aa
1
!
+ns
n ...3,2,1, =ntn
as1 at
e22
as
a
+
atsin
22 as
s
+
atcos
22 as
a
atsinh
22 as
s
atcosh
Example 1.13:Finding the inverse Laplace
Transforms
Find ( )f t for the following Laplace
transforms:
(a) 164)(
2 +=
ssF (b) 2( )
4sF s
s=
(c) 21
( )9 1
F ss
=+
Solution:
(a) 222 44
16
4)(+
=+
=ss
sF
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Use the Table of Laplace transform: ttf 4sin)( =
(b) 2 2 2( ) 4 2s s
F ss s
= =
( ) cosh 2f t t=
(c) 21
( )9 1
F ss
=+
Rearrange or modify so that it looks
like a function in the table
make sure that you still have the
original function:
2 22 2
1
1 1 1 13( ) sin
19 1 3 319 39 3
F s t
s s s
= = = =
+ + +
Review: Changing expressions into
standard forms some algebraic
manipulations
(1) Completing the square
(2) Partial Fractions
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Example 1.14(a):Completing the square
Findy ifL [ ]y 21
4 20
s
s s
=
+ +
Solution:
1) change 21
4 20
s
s s
+ + to an expression in a
form similar to those that can be found in
the tables.
Check denominator: standard? factorise?
use method of completing the square?
( )( ) ( )
2 2
2 2 2
4 20 4 4 20 4
2 16 2 4
s s s s
s s
+ + = + + + = + + = + +
Check numerator: modify if necessary
2 2 2
1 1
4 20 ( 2) 4
s s
s s s
=+ + + +
2 2 2 2 2 2 2 2
1 ( 2) 1 2 ( 2) 3
( 2) 4 ( 2) 4 ( 2) 4 ( 2) 4
s s s
s s s s
+ += =
+ + + + + + + +
2 2 2 2 2 2 2 2( 2) 3 ( 2) 3 4
( 2) 4 ( 2) 4 ( 2) 4 4 ( 2) 4s s
s s s s+ + = + + + + + + + +
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2) Findy
[ ]y 2 2 2 2( 2) 3 4
( )( 2) 4 4 ( 2) 4
sF s
s s
+ = = + + + +
2 2
2
3cos 4 sin 4
4
3cos 4 sin 4
4
t t
t
y e t e t
e t t
=
=
Example 1.14 (b):Using Partial Fractions
Findy if [ ]y 2 122 8s s=
Solution:
1) modify to be comparable to standard
forms in the tables:
Check denominator:
( ) ( )2 2 8 4 2s s s s = +
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Check expression: suitable for conversion
to partial fractions
12
( 4)( 2) 4 2
A B
s s s s= +
+ +
12 ( 2) ( 4)
( 4)( 2) ( 4)( 2)
A s B s
s s s s
+ + =
+ +
Solve forA andB : Compare numerator:
12 2 4As A Bs B= + +
0
2 4 12
A B
A B
+ = =
Solve simultaneously:
6 12 2B B= =
2A =12 2 2
( 4)( 2) 4 2s s s s= + +
2) Findy
[ ]y = 2 2
4 2s s
+
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( )4 2 4 2( ) 2 2 2t t t t y t e e e e = =
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REVISION:
Other Standard forms of partial fractions
(a) Expressions are of the form( )
( )
P s
Q swhere
( )P s and ( )Q s are polynomials ins and the
degree of( )P s
is less than the degree of( )Q s
.Example 1.15 (a): ( )P s is a constant and ( )Q s
has linear factors
3
( 1)( 2) ( 1) ( 2)
A B
s s s s= ++ +
Example 1.15(b): ( )P s is a constant and
( )Q s has linear factors with some factors
repeated
2 2
3
( 1) ( 2) ( 1) ( 1) ( 2)
A B C
s s s s s= + +
+ + +
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Example 1.15(c): ( )P s is a constant and ( )Q s
has linear and quadratic factors
(i) 2 23
( 4)( 3) ( 4) ( 3)
As B C
s s s s
+= +
+ +
(ii) ( ) ( )2 22 1
( 3)( 1 4)( 3) ( 1 4)s As B C
ss s s += + + +
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1.3.3 Solving Differential Equations using
Laplace Transform
The method converts the differential
equations into algebraic expressions in
terms ofs.
The expressions have to bemanipulated such that the function
( )y f t= can be obtained from the inverse
Laplace Transforms.
Example 1.16: Solve the differential equation
2 3 0y y y + = with the initial conditions
(0) 0, (0) 2y y= =
Solution:
1. Take Laplace transforms of both sides
[ ]2 3y y y + = [0]67
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[ y ]+ [ ]2y [ ]3y = 0Use the formula of Laplace Transforms of
first and second order derivatives:
2s [ ] (0) (0)y sy y + 2s [ ] 2 (0)y y 3[ ] 0y =
2. Solve for [ ]y
Given: (0) 0; (0) 2y y= = ; substitute the
initial conditions
2s [ ] 2y + 2s [ ]y 3 [ ] 0y =
[ ]y 2( 2 3) 2s s+ =
[ ]y ( )22 2
( 3)( 1)2 3 s ss s= = + +
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3. Convert into standard forms
Use partial fractions
2
( 3)( 1) 3 1
A B
s s s s= +
+ +
2 ( 1) ( 3)
( 3)( 1) ( 3)( 1)
A s B s
s s s s
+ +=
+ +
Compare numerator:
2 3As A Bs B= + +
Thus,
0
3 2
A B
A B
+ =
+ =
Solving simultaneously:
14 2
2B B= = and 1
2A =
[ ]y = ( )1 1
2( 3) 2 1s s +
+
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4. Find the inverse Laplace Transforms:
use tables
31
2
t ty e e =
Example 1.17:Solve the differential equation
4 3 0y y + = wherey is a function oft, ify
and y are both zero at 0t = .
Solution:
5. Take Laplace transforms of both sides
[ ]4 3y y + = [0] [ ]y + [ ]4y [ ]3 = 0
2s [ ] (0) (0)y sy y + 4 [ ]y 3s
= 0
6. Solve for [ ]y
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( )2 4s + [ ]y = 3s
since (0) 0; (0) 0y y= =
[ ]y = ( ) ( )2 2 23 3
4 2s s s s=
+ +
7. Convert into standard forms
Use partial fractions
( ) ( )2 2 2 23
2 2
A Bs C
ss s s
+= +
+ +
( )
( ) ( )
( )
2 2
2 2 2 2
23
2 2
A s Bs C s
s s s s
+ + +=
+ +
Compare numerator:
2 23 4As A Bs Cs= + + +
Thus,
0; 4 3; 0A B A C+ = = =
3 3;
4 4A B= =
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[ ]y = ( )2 23 3
4 4 2
s
s s
+
8. Find the inverse Laplace Transform:
use tables
( )3 3 3
(1) cos 2 1 cos 24 4 4
y t t= =
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Question 15
Solve the following initial
value problems
(a)6 10 0;
(0) (0) 3
y y y
y y
+ == =
(b)4 8 cos 2 ;
(0) 2, (0) 1
y y y t
y y
+ + == =
(c)2 24 4 ;
(0) (0) 0
ty y y t e
y y
+ + == =
Prompts/
Questions
What is the
Laplace
Transforms or
equations?
What is the
Laplace
Transform of first
and second orderderivatives?
o What do you
do with the
initial values?
How does your
expressioncompare to the
standard forms?
o Which
algebraic
manipulations
do you need?
How do you find
the inverse
Laplace
Transforms?