4 laplace transforms (1)

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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?

4 laplace transforms (1)

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