Definition 3.1 (Laplace Transform) · 13 s ss. LL=−=−>0. CHAPTER 3 : LAPLACE TRANSFORM ... The step functions can be expressed into the unit step functions forms. Given a step

CHAPTER 3 : LAPLACE TRANSFORM

1

The Laplace transform of a function is defined by )(tf

( )( ) ( ) ( )0

stf t F s e f t∞ −= = ∫L dt

Example 3.1.1

Example 3.1.1:

Using definition of Laplace Transform, find if )(sF ( )f t = a , a is constant.

Solution:

( )

[ ]

0

0

0

0

0

( )

1

0 1 ( ) , 0

st

st

st

st

s s

F s e f t dt

e a dt

aes

aesa e es

a aa ss s

∞ −

∞ −

− ∞

∞−

− ∞ −

=

=

=−

⎡ ⎤= − ⎣ ⎦

⎡ ⎤= − −⎣ ⎦

= − − ∴ = >

∫∫

L

Definition 3.1 (Laplace Transform)

Remember !!!

( ) ( )1 31 and 3 , ss s

= − = −L L 0>


2

Example 3.1.2:

Find if , a is constant )(sF atetf =)(

Solution:

( )( )

( )

( )

( ) [ ] ( )

0

0

0

1 10 1 , 0 ,

st at

s a t

s a t

at

F s e e dt

e dt

es a

s a s a es a s a s a

∞ −

∞ − −

∞− −

=

=

=− −

= − − − > = > ∴ =− −

∫∫

L 1−

Example 3.1.3:

Find ( )sin atL

Solution:

( )

( )

( )

0

2 20

0

2 2 2 2

sin sin

sin cos

0 sin 0 cos 0 ,

st

st

at e at dt

e s at a ats a

e as a ss a s a

∞ −

∞−

=

⎡ ⎤= − −⎢ ⎥+⎣ ⎦

⎡ ⎤= − − − = >⎢ ⎥+ +⎣ ⎦

∫L

0

( )2 1 , 22

te ss

− = > −+

L

( ) 2

3sin 39

ts

=+

L


3

Example 3.1.4:

Find ( )( )f tL if

( )

⎪⎩

⎪⎨

⎧

<<<<<

=te

tt

tft 10,

105,050,2

4

Solution:

( )( )

( )

( )( )

( )( )

( )

5 10 4

0 5 105 4

0 10

5 4

0 10

10 45 0

10 45

2 0

2

24

2 04

2 2 , 44

st st st t

s tst

s tst

ss

ss

F s e dt e dt e e

e dt e dt

e es s

e e es s s

e e ss s s

∞− − −

∞ −−

∞−−

−− −

− −−

= + +

= +

⎡ ⎤⎡ ⎤⎢ ⎥= +⎢ ⎥

− −⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤⎡ ⎤

= − + −⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎣ ⎦

= − + >−

∫ ∫ ∫∫ ∫

dt


4

Elementary Laplace Transform

( )( ) ( )f t F s=L

( )f t

( )F s Condition on s

a

as

0s >

, 0,1,2,nt n = ... 1

!n

ns + 0s >

ate 1

s a− s a>

sin at 2 2

as a+ 0s >

cosat 2 2

ss a+ 0s >

sinh at 2 2

as a− s a>

cosh at 2 2

ss a− s a>

Note:

This table will enable us to obtain the transforms of many other functions.


5

Let f, f1, f2 be a functions whose Laplace Transforms exist for s > α and c be a constant. Then for s > α,

( ) ( ) ( )1 2 1 2f f f f± = ±L L L

( ) ( )cf c f=L L

Example 3.2.1:

Find ( )41 5 6 sin 2te t+ −L

Solution:

( ) ( ) ( ) ( )

( ) ( )

4 4

4

2

1 5 6 sin 2 1 5 6 sin 2

1 5 6 sin 2

1 5 124 4

t t

t

e t e t

e ts

s s s

+ − = + −

= + −

= + −− +

L L L L

L L

Exercises 3.2.1:

1. Show ( ) 2 2sinh aats a

=−

(

L using the linearity property.

2. Find )2sin 2 .t

( )3 cosh .te t

L

3. Find L

Definition 3.2 (Linearity Property of Laplace Transform)


6

If , ( ) ( )( )F s f t= L

then ( )( ) ( )ate f t F s a= −L

Example 3.3.1:

Find ( )2te tL

( )2 ,a f t t= =

.

Solution

( ) ( )2 2te t F s∴ = −L

( ) ( )( ) ( )where 2

1F s f t ts

= = =L L

( ) ( )( )

2

21 22

te t F ss

∴ = − =−

L

Exercises 3.3.1:

Find ( )4 coste tL .

Definition 3.3 (First Shifting Property)


7

If ( )( ) ( ) ,f t F s=L

( )( )

then

( 1) , 1,2,3,...n

n nn

d Ft f t nds

= − =L

Example 3.4.1:

Find . ( )sin 6t tL

Solution:

( ) ( ) 2

61 , sin 6 ,36

n f t t F ss

= = =+

( )

( )( ) ( )( )

( ) ( )

1

2

22

2 22 2

sin 6 ( 1)

36 0 6 2

36

12 12

36 36

dFt tds

s s

s

s s

s s

= −

+ −= −

+

−= − =

+ +

L

Exercises 3.4.1:

1. Find L . ( )tt e

(2. Find )sin 2te t t−L .

Definition 3.4 (Derivatives of the Laplace Transform)


8

Graph of

( )F t

where 0, 0 and 1,

Example 3.5.1:

( ) ( )0 , 4

Draw 41 , 4

tf t H t

t<⎧

= − = ⎨ ≥⎩

t

1

0 4

( )F t

Exercises 3.5.1:

1. Draw

2. Draw 2

3. Draw 2

Definition 3.5 (Heaviside Unit Step Function)

0, 01,

The function is defined by


9

Effect of the Unit Step Function

1) ( ) ( ) ( )3 6H t H t= − − −f t

( )F t

2) ( )0 , 0

, 0t

f t tt t

<⎧= = ⎨ ≥⎩

( )F t

3) ( ) ( )0 , 2

2, 2

tf t t H t

t t<⎧

= ⋅ − = ⎨ ≥⎩

( )F t

4) ( ) ( ) ( )0 , 2

2 22 , 2

tf t t H t

t t<⎧

= − − = ⎨ − ≥⎩

( )F t


10

Definition: Step Function A step function is a piecewise the form continuous function of

, 0 , ,

,

The step functions can be expressed into the unit step functions forms.

Given a step function , 0,

The Heaviside un it step function is

0, 0 ,

0, 0

1,

Example 3.5.2:

Express the following function rms o functions. s in te f unit step

4, 0 2, 2 40, 4

Solution:

4

4 4 2 0

4 4 2 4


11

Example 3.5.3:

Express the f e ns form. ollowing unit step functions into th step functio

1 1 2 4 Solution:

1 1 2 4

1

1 1

1 1 2

2

1, 0 2

2, 2 42, 4

Laplace Transform of H( t – a )

, 0

Theorem: Laplace Transform of Unit Step Functions

Inverse Laplace Transform of Unit Step Functions


12

Example 3.5.4:

Find ( ){ }f tL if ( ) ( ) ( )2 4f t H t H t= − − − .

Solution:

a) Using the Laplace Transform Table

( ){ } ( ) ( ){ }( ){ } ( ){ }

2 4

2 4

2 4s s

f t H t H t

H t H t

e es s

− −

= − − −

= − − −

= −

L L

L L

b) Using the Laplace Transform definition ‐ Change the unit step function into the step function form

H t 2 H t 4 0 1 2 11 2 1 3

0; 1 1; 1 0.

2

4

Hence 0, 0 21, 2 40, 4

U e Laplace transform definition sing th

· 0 · 1 · 0


13

Laplace Transform of H( t – a ). F( t – a)

Example 3.5.5:

Find ( ) ( ){ }24 4t H t− −L .

Solution

( ) ( )23

2!4, ,c f t t F ss

= = =

( ) ( ){ } ( )2 4 43

2!4 4 s st H t F s e es

− −− − = =L

( ) ( ){ }4

23

24 4set H t

s

−

∴ − − =L

Exercises 3.5.2:

1. Find . ( ) ( ){ }sin 3 3t H t− −L

2. Find . ( ) ( ){ }5 5te H t− −L

3. Find cos 2 2 2

Theorem: Second‐shift Property

Inverse Laplace Transform with Second‐shift Property


14

Example 3.5.6

Find the function whose transform is

4se2s

−

.

Solution

The numerator corresponds to ase−

where 4a = and therefore indicate

( )4H t − .

Then ( ) { } ( )2

1 F s t f t ts

= = ∴L = ( ) ( )4

12 4 4

se t H ts

−− ⎧ ⎫

∴ = − −⎨ ⎬⎩ ⎭

L

Exercises 3.5.3:

1. A function ( )f t is defined by ( )4 , 0 22 3 ,

tf t

t t 2< <⎧

= ⎨ − >⎩

Sketch the graph of the function, expressing the function ( )f t in unit

step form and determine its Laplace transforms.

2. Write the following ( )f t in terms of unit step functions and determine

the Laplace transforms.

( ) , 0 22 1 , 2

te tf t

t t

−⎧ < <= ⎨

− ≥⎩

( ){ }

( )2 12

2

1 3 2Answer:1 1

ss ef t e

s s s s

− +− ⎛ ⎞= + + −⎜ ⎟+ +⎝ ⎠

L

Note: Remember that in writing the final result ( )f t is replaced by ( )f t c− .


15

3. A function ( )f t is defined by

( )6 , 0 18 2 , 1 34 , 3

tf t t t

t

< <⎧⎪= − < <⎨⎪ >⎩

Sketch the graph and find the Laplace transform of the function.

( ){ }2 3

2 2

6 2 2 2Answer:3s s se e ef t

s s s s

− −

= − + +L−

<

4. Given

( )2

0 , 0 2, 2 5

, 5t

tf t t t

e t

⎧ < <⎪= <⎨⎪ >⎩

Find the Laplace transform of the function.

( ){ }2 10 5 5 2 5

2 2

2 5Answer:2

s s s se e e e e ef ts s s s s

− − − −⋅= + − + −

−L

s−

5. Determine the function ( )f t for which,

( ){ }2

2 2

3 4 5s se ef ts s s

− −

= − +L .

Find its inverse transform and sketch the graph of ( )f t .

( )3 , 0 1

Answer: 7 4 , 1 23 , 2

tf t t t

t t

< <⎧⎪= − < <⎨⎪ − >⎩


16

Example 3.6.1:

( ){ }6 6 ast a eδ −⋅ − =L

Example 3.6.2:

Given , 0 35, 3

( ) ( ){ } ( )( ) ( ){ } ( )

3 3

2 2

3 3 5

2 2 2

s s

s s

f t t f e e

f t t f e e

δ

δ

− −

− −

⋅ − = =

⋅ − = =

L

L

Definition 3.6 (Dirac Delta Function)

∞,0,

The Dirac Delta function is defined by

1

And

THE R M the Dirac Delta Function O E : Laplace Transform for For 0, and the inverse Laplace Transform is

For 0,


17

In many technological problems, we are dealing with forms of

mechanical vibrations or electrical oscillations and the necessity to

express such periodic functions in Laplace transforms soon arises. Let

( )f t be a periodic function of period T i.e. ( ) ( )f t f t T= + , . 0T ≠

Example 3.7.1:

Show that ( ) sin 2f t tπ= is a periodic function.

Solution:

( )( ) ( )( ) ( )

( )

sin 2

sin 2

sin 2 sin 2 = sin 2 cos 2 cos 2 sin 2cos 2 1, sin 2 0 1,2,3,...

f t t

f t T t T

f t f t T

t t Tt T t

T T T

π

π

π πTπ π π

π π

=

+ = +

= +

= +

+= = ⇒ =

π

Exercises 3.7.1:

1. Show that ( ) 2f t t= is not a periodic function.

2. Sketch the following periodic function.

( )1 , 0 1

1 , 1 2t

f tt

< <⎧= ⎨− < <⎩

( ) ( )2f t f t= +

Definition 3.7 (Periodic Function)


18

Example 3.7.2:

Sketch the following periodic function and find its Laplace transform.

( )3 , 0 20 , 2 4

tf t

t< <⎧

= ⎨ < <⎩

( ) ( )4f t f t= +

Solution:

( )F t

The expression for

( ){ } ( )

( )

4

4 0

2 4

4 0 2

2

11

1 3 01

31

sts

st sts

s

f t e f t dte

e dt e de

s e

−−

− −−

−

=−

⎡ ⎤= ⋅ +⎢ ⎥⎣ ⎦−

=+

∫

∫ ∫

L

t⋅

f (t)

Laplace Transform of Periodic Function

If ( )f t is a periodic function of period T,

then ( ){ } ( )0

1 ; 01

T stsTf t e f t dt s

e−

−= >− ∫L .


19

Exercises 3.7.2:

Sketch the following periodic function and find its Laplace transform.

( ), 0 1

1 , 1 2t t

f tt

< <⎧= ⎨ < <⎩

( ) ( )2f t f t= +


20

Definition 3.8 (Inverse Laplace Transform)

Recall that,

Notation for Laplace Transform; ( )( ) ( )f t F=L s

So, the inverse form; ( )( ) ( )1 F s f t− =L

Linearity property of Inverse Laplace Transform

If ( )( ) ( )1 F s f t− =L and ( )( ) ( )1 G s g t− =L

with α and β is a constants, then

( ) ( )( ) ( )( ) ( )( )1 1

( ) ( )

F s G s F s G s

f t g t

α β α β

α β

− −± = ±

= ±

L L 1−L

First shifting property for Inverse Laplace Transform

If ( )( ) ( )1 F s f t− =L with α as constant, then

( )( ) ( )1 atF s a e f t− − =L

or we can write as . ( )( ) ( )( )1 1atF s a e F s− −− =L L


21

Example 3.8.1:

( ) ( )2 2

4 4 3 4 sin 39 3 9 3

F s fs s

⎛ ⎞= = ⇒ =⎜ ⎟+ +⎝ ⎠t t

Example 3.8.2:

( ) ( ) 45 5

1 1 4! 14! 24

F s f t ts s

⎛ ⎞= = ⇒ =⎜ ⎟⎝ ⎠

Example 3.8.3:

( )( )4

61

F ss

=−

By shifting property, ( ) ( )1 , 1G s a G s a− = − = .

( )

( ) ( )4 4

3 3

6 3!

t

G ss sg t t f t e t

= =

⇒ = ⇒ =

Exercises 3.8.1:

Determine

1. ( )222 9s

−⎧ ⎫⎪ ⎪⎨ ⎬

− +⎪ ⎪⎩ ⎭

1L

2. ( )33

2 5s−⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭

1L


22

Example 3.8.4:

-1 -1

2 2 2 cos525 5

s s ts s

⎧ ⎫ ⎧ ⎫= =⎨ ⎬ ⎨ ⎬+ +⎩ ⎭ ⎩ ⎭L L

We can write down the corresponding function in t, provided we can

recognize it from our table of transforms.

But, what about -1

2

3 16

ss s

+⎧ ⎫⎨ ⎬− −⎩ ⎭

L ?

Solution:

( )( )2

3 1 3 16 2

s ss s s s

+ +=

− − + −3 = 1 2

2 3s s+

+ −

-1 12

3 1 1 26 2 3

⎫sL Ls s s s

−+⎧ ⎫ ⎧∴ = +⎨ ⎬ ⎨− − + −⎩ ⎭ ⎩⎬⎭

from table: = tt ee 32 2+−

The two simpler functions of 12s +

and 23s −

are called the partial

fractions of 2

3 16

ss s

+− − .

Therefore, u need to know partial fractions!!

( ) ( )2

2

3 16 2 3

3 2 33: 5 10 2

2 : 5 5 1

3 1 1 26 2 3

s A Bs s s s

1A s B s ss B Bs A A

ss s s s

+= +

− − + −− + + = +

= = ⇒ == − − = − ⇒ =

+= +

− − + −


23

Example 3.8.5:

Determine 2

5 112

ss s

− +⎧ ⎫⎨ − −⎩1L ⎬

⎭. Answer: 3 43 2t te e− +

( ) ( )2

2

3 42

5 112 3 4

4 3 5 13: 7 14 2; 4 : 7 21 3

5 1 2 312 3 45 1 2 3 2 3

12 3 4t t

s A Bs s s sA s B s ss A A s B B

ss s s s

s e es s s s

− − −

+= +

− − + −− + + = +

= − − = − ⇒ = = = ⇒ =+

= +− − + −

+⎧ ⎫ ⎧ ⎫= + = +⎨ ⎬ ⎨ ⎬− − + −⎩ ⎭ ⎩ ⎭1 1L L

Partial Fractions

There are few types of denominator that u should know:

1) A linear factor gives a partial fraction (s a+ ) As a+ where A is a

constant to be determined.

2) A repeated factor ( 2)s a+ gives ( )2A B

s a s a+

+ + .

3) Similarly ( gives )3s a+ ( ) ( )2 3A B C

s a s a s a+ +

+ + + .

4) A quadratic factor ( )2s ps q+ + gives 2

Ps Qs ps q

++ + .

5) Repeated quadratic factors ( )22s ps q+ + gives

( )22 2

Ps Q Rs Ts ps q s ps q

+ ++

+ + + + .


24

Exercises 3.8.2:

Determine

1. ( )( )2

2

4 5 61 4

s ss s

−⎧ ⎫− +⎪ ⎪⎨ ⎬

+ +⎪ ⎪⎩ ⎭

1L 3 cos 2 3sin 2e t t+ −. Answer: t−

2. ( ){ } for F s−1L ( )102

12 ++

+=

ssssF

3. ( ){ } ( ) 2

2 3 for 2

sF s F ss s

− −=

+ −1L

4. ( )32 3

4s

s−⎧ ⎫+⎪ ⎪⎨

+⎪⎩ ⎭

1L ⎬⎪ Answer:

4 25( ) 22

tf t e t t− ⎛ ⎞∴ = −⎜ ⎟⎝ ⎠


25

Example 3.8.6:

Find the inverse Laplace transform for ( )2

14s s + using Convolution

theorem.

Solution:

Let ( ) 1F ss

= and ( ) 2

14

G ss

=+ , then

( ) ( ) 11, sin 22

f t g t t= = .

( ) ( ) ( )11 , sin 22

f u g t u t u⇒ = − = − .

( ) ( ) ( )

( )

( )

[ ]

12 0

0

1 11 sin 224

cos 212 2

1 cos 2 0 cos 241 1 cos 24

t

t

t u dus s

t u

t

t

−⎧ ⎫⎪ ⎪ = −⎨ ⎬

+⎪ ⎪⎩ ⎭

− −⎡ ⎤= ⎢ ⎥−⎣ ⎦

= −⎡ ⎤⎣ ⎦

= −

∫L

Convolution Theorem

( ) ( ){ } ( ) ( )1

0

tF s G s f u g t u du− = −∫L


26

Exercises 3.8.3:

Find ( )2

22 4

s

s−⎧ ⎫⎪ ⎪⎨ ⎬

+⎪ ⎪⎩ ⎭

1L using Convolution theorem.

Find ( )22

1

1s−⎧ ⎫⎪ ⎪⎨

+⎪⎩ ⎭

1L ⎬⎪using Convolution theorem.


27

To solve a differential equation by Laplace transforms, we go through

four distinct stages.

(a) Rewrite the equation in terms of Laplace transforms.

(b) Insert the given initial conditions.

(c) Rearrange the equation algebraically to give the transform of the

solution.

(d) Determine the inverse transform to obtain the particular solution.

Transforms of Derivatives

{ }If ( ) ( ), theny t Y s=L

( ){ } 0( )y t sY s′ y= −L

{ } 20 0( ) ( )y t s Y s sy y′′ ′= − −L

( ){ } 3 20 0( )y t s Y s s y sy y′′′ ′ ′′0= − − −L

( ) ( ){ } ( )11 20 0 0

( )n nn n ny t s Y s s y s y y −− − ′= − − − −L

Definition 3.9 (Solution of Differential equations by Laplace Transforms)


28

Solution of 1st Order Differential equations

Example 3.9.1:

Solve the equation 2dy y− = 4dt , given that at 0, 1t y= = .

Solution:

(a) Rewrite the equation in Laplace transforms using the last notation

{ } { } { }2 4 2dy dyy ydt dt

⎧ ⎫ ⎧ ⎫− = ⇒ − =⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭L L L L 4L

We have

{ }

{ }

{ }

0( ) ( ) ,

( ) ( ),44

dy y t sY s ydt

y t Y s

s

⎧ ⎫ ′= = −⎨ ⎬⎩ ⎭

=

=

L L

L

L

.

Then the equation becomes

( )04( ) 2 ( )sY s y Y ss

⇒ − − = .

(b) Insert the initial condition that at 0, 1t y= = i.e. 0 1y = .

( ) 4( ) 1 2 ( )sY s Y ss

⇒ − − =

(c) Now we rearrange this to give an expression for ( )Y s

( ) ( )4 4 42 ( ) 1 ( )

2s ss Y s Y s

s s s s+ +

⇒ − = + = ∴ =−


29

(d) Finally we take inverse transforms to obtain x

( )( ) ( )4 2

42 2

s A Bs s s s

s A s B s∴ + = − +

3B

+= +

− −

i) Let 2 6 2s B= ⇒ = ∴ =

ii) Let ( )0 4 2s A A 2= ⇒ = − ∴ = −

( )4 3( )2 2

sY s 2s s s s

+∴ = =

− −−

Therefore, taking inverse transforms

{ } ( )1 1

1

2

4( ) ( )2

3 22

( ) 3 2t

sy t Y ss s

s s

y t e

− −

−

⎧ ⎫+⎪ ⎪= = ⎨ ⎬−⎪ ⎪⎩ ⎭

⎧ ⎫= −⎨ ⎬−⎩ ⎭

∴ = −

L L

L

Example 3.9.2:

Solve the equation 2 44 2 t tdx x e e

dt− = + given that at 0, 0t x= = .

Solution:

{ }2 44 2 t tdx x e edt

⎧ ⎫− = +⎨ ⎬⎩ ⎭L L

( )0

2 1( ) 4 ( )2 4

sX s x X ss s

⇒ − − = +− − .


30

0, 0t x= =

( ) 2 1( ) 0 4 ( )2 4

sX s X ss s

⇒ − − = +− −

( )

( )( ) ( )2

2 14 ( )2 4

2 1( )2 4 4

can use partial fraction can transformdirectly fromthe table

s X ss s

X ss s s

⇒ − = +− −

∴ = +− − −

Partial Fraction

( )( )2 1

2 4 2 4 2 4A B

s s s s s s1−

= + = +− − − − − −

( )( ) ( ) ( )

( )

1 12 2

2 4 4

4 2

2 1 1 1 1( ) 2 4 2 44 4

( )1

t t t

t t

x ts s s ss s

x t e e tee t e

− −⎧ ⎫ ⎧ −⎪ ⎪ ⎪= + = + +⎨ ⎬ ⎨− − − −− −⎪ ⎪ ⎪⎩ ⎭ ⎩

⎫⎪⎬⎪⎭

∴ = − + +

= + −

L L

Exercises 3.9.1:

Solve the equation 32 10 tdx x e

dt+ = given that at 0, 6t x= = .


31

Solution of 2nd Order Differential equations

The method is, in effect, the same as before, going through the same

four distinct stages.

Example 3.9.3:

Solve the equation

23

2 3 2 2 td y dy y edt dt

− + = given that ( )0y = 5 and

( )0 7y′ = .

Solution:

(a) We rewrite the equation in terms of its transforms, remembering

that

{ } ( )y Y s=L

{ } 0( ) ( )y t sY s′ = −L y

{ } 20 0( ) ( )y t s Y s sy y′′ ′= − −L

The equation becomes

⇒( ) ( )

( )

20 0 0

2( ) 3 ( ) 2 ( )3

t y ty ty⎧ ⎫⎪ ⎪⎧ ⎫ ⎛ ⎞⎧ ⎫ ⎛ ⎞⎪ ⎪⎪ ⎪ ⎨ ⎬⎜ ⎟⎨ ⎬⎜ ⎟⎨ ⎬ ⎝ ⎠⎪ ⎪⎩ ⎭⎝ ⎠⎪ ⎪⎪ ⎪⎩ ⎭ ⎩ ⎭

′′′ LLL

s Y s sy y sY s y Y ss

′− − − − + =−

(b) Insert the initial conditions. In this case 0 5y = and . 0 7y′ =

⇒ ( )2 2( ) 5 7 3 ( ) 5 2 ( )3

s Y s s sY s Y ss

− − − − + =−


32

(c) Rearrange to obtain ( )Y s

( )

( )( )

( )( )( ) ( )( )

2 23 2 ( ) 5 7 153

2 21 2 ( ) 5 7 15 5 83 3

2 5 Y(s)3 1 2 1 2

s s Y s ss

s s Y s s ss s

ss s s s s

− + − − + =−

− − = + + − = + −− −

−= +

8− − − − −

(d) Now for partial fractions

( )( )( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 5 83 1 2 1 21 1 2 3 2 4 1

3 1 2 1 2 3

ss s s s s

s s s s s s s 1

−+

− − − − −

−= + + + + = +

− − − − − − −

Therefore, taking inverse transforms

( )( )( ) ( )( ) ( ) ( )1 1

3

2 5 8 4( )3 1 2 1 2 3 1

( ) 4 t t

sy ts s s s s s s

y t e e

− −⎧ ⎫ ⎧−⎪ ⎪ ⎪= + =⎨ ⎬ ⎨− − − − − − −⎪ ⎪ ⎪⎩ ⎭ ⎩

∴ = +

L L1 ⎫⎪+ ⎬

⎪⎭

Exercises 3.9.2:

1. Solve the equation 2

2 4 24cos 2d x x tdt

− = given that at and

.

( )0x = 3

( )0 4x′ =

2. Solve the boundary value problem equation

( ) ( )3 4 cosh , 5y y t t yδ′− = − = 0.


33

REFERENCES:

1. Nagle, Saff and Snider, Fundamentals Of Differential Equations 5th Edition, Addison Wesley Longman; 2000.

2. Alan Jeffrey, Advanced Engineering Mathematics, Academic Press; 2002.

3. Abd. Wahid Md. Raji & Mohd Nor Mohamad, Persamaan Pembeza Biasa; Jabatan Matematik; UTM;Skudai 2002.

4. Glynn James et al., Advanced Modern Engineering Mathematics, Addison Wesley; 1993.

5. Boyce and di Prima, Elementary Differential Equations and Boundary Value Problems, 4th edition, John Wiley and Sons; 1986.

6. K. A. Stroud, Advanced Engineering Mathematics; MacMillan Ltd.; London; 1996.

7. Normah Maan, Halijah Osman et al., Differential Equations Module, Jabatan Matematik; UTM; Skudai 2007.

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Definition 3.1 (Laplace Transform) · 13 s ss. LL=−=−>0. CHAPTER 3 : LAPLACE TRANSFORM ... The step functions can be expressed into the unit step functions forms. Given a step