Ordinary Differential Equations[FCM 1023]

LAPLACE TRANSFORM

Chapter 5

Overview

Chapter 5: LAPLACE TRANSFORM

5.1. Definition and Basic Properties

5.2. Inverse Laplace Transform

5.3. Method of Solution

Learning Outcome

At the end of this section, you should be ableto:

Solve the differential equation (IVP) by usingLaplace transform

5.3 Method of Solution

If �� is continuous for � ≥ 0 , by using integration by

parts, it gives

ℒ ��(�) = �� ��

��� � ��

= �� � �(�)��� + � �� ��

�� � ��

= −� 0 + �ℒ �(�)� ��(�) = �� � − �(�) ℒ �(�) = � �

5.3 Method of Solution

Transform of a Derivatives

ℒ ��′(�) = �� ��

���′ � ��

= �� � ��(�)��� + � �� ��

��� � ��

= −�� 0 + �ℒ ��(�)= � �� � − �(0) − ��(0)

� ���(�) = ��� � − �� � − ��(�)By using the same procedure, ℒ ���′(�)� ����(�) = ��� � − ��� � − ��� � − ���(�)

5.3 Method of Solution

If �, ��, … , �"�# are continuous on $0, ∞) and are of

exponential order and if � " (�) is a piecewisecontinuous on $0, ∞) then

� �(&)(�)= �&� � − � &�' � � − � &�� �� � − ⋯ − �(&�')(�)

where � � = ℒ �(�)

5.3 Method of Solution

Theorem

ℒ �)*��) = �)+ � − �* 0 − *�(0)

ℒ �*�� = �+ � − * 0

ℒ * = + �

ℒ �,*��, = �,+ � − �)* 0 − �*� 0 − *��(0)

5.3 Method of Solution

DEApply Laplace Transform, ℒ

Transformed DE becomes algebraic

equation in +(�)

Solve transformed equation for +(�)

Apply inverse transform, ℒ�#

Solution * �of the IVP

5.3 Method of Solution

Solution:

Step 1: Transform the DE

Use the Laplace Transform to solve the initial-value problem

�*�� + 3* = 13 sin 2� , * 0 = 6

ℒ �*�� + 3* = ℒ 13sin 2�

5.3 Method of Solution

Example 1

ℒ �*�� + 3ℒ * = 13ℒ sin 2�

ℒ �*�� + 3ℒ * = 13ℒ sin 2�

�+ � − * 0 +3+ � = 13 2�) + 4

5.3 Method of Solution

�+ � − * 0 + 3+ � = 26�) + 4

ℒ �*�� + 3ℒ * = 13ℒ sin 2�

�+ � − * 0 +3+ � = 13 2�) + 4

5.3 Method of Solution

�+ � − * 0 + 3+ � = 26�) + 4

�+ � − 6 + 3+ � = 26�) + 4

�+ � − 6 + 3+ � = 26�) + 4

� + 3 + � = 6 + 26�) + 4

+ � = 6� + 3 + 26

� + 3 (�) + 4)

= 6�) + 50� + 3 (�) + 4)

5.3 Method of Solution

+(�) = 6�) + 50� + 3 (�) + 4)

We are looking for * � = ℒ�# +(�)

6�) + 50� + 3 (�) + 4) = 7

� + 3 + 8� + 9�) + 4

= 8� + 3 + −2� + 6

�) + 4

Step 2: Compute inverse Laplace transform

5.3 Method of Solution

* � = ℒ�# +(�) = ℒ�# 6�) + 50� + 3 (�) + 4)

= ℒ�# 8� + 3 + −2� + 6

�) + 4

= 8ℒ�# 1� + 3 − 2ℒ�# �

�) + 4 + 6ℒ�# 1�) + 4

* � = 8��,� − 2 cos 2� + 3 sin 2�

5.3 Method of Solution

= ℒ�# 8� + 3 − 2�

�) + 4 + 6�) + 4

Use the Laplace Transform to solve the initial-value problem

*�� − 3*� + 2* = ��=�, * 0 = 1, *� 0 = 5

ℒ *�� − 3ℒ *� + 2ℒ * = ℒ ��=�

5.3 Method of Solution

Example 2

Solution:

Step 1: Transform the DE

ℒ *�� − 3*� + 2* = ℒ ��=�

ℒ *�� − 3ℒ *� + 2ℒ * = ℒ ��=�

�)+ � − �* 0 − *�(0) −3 �+ � − * 0 +2+(�) = 1� + 4

5.3 Method of Solution

�)+ � − �* 0 − *�(0) − 3�+ � + 3* 0 + 2+ � = 1� + 4

ℒ *�� − 3ℒ *� + 2ℒ * = ℒ ��=�

�)+ � − �* 0 − *�(0) −3 �+ � − * 0 +2+(�) = 1� + 4

5.3 Method of Solution

�)+ � − �* 0 − *�(0) − 3�+ � + 3* 0 + 2+ � = 1� + 4

�)+ � − � 1 − 5 − 3�+ � + 3 1 + 2+ � = 1� + 4

�)+ � − 3�+ � + 2+ � − � − 2 = 1� + 4

�)+ � − 3�+ � + 2+ � − � − 2 = 1� + 4

�) − 3� + 2 + � = 1� + 4 + � + 2

+ � = 1(� + 4)(�) − 3� + 2) + � + 2

(�) − 3� + 2)= �) + 6� + 9

(� + 4)(�) − 3� + 2)= �) + 6� + 9

(� + 4)(� − 1) � − 2

5.3 Method of Solution

+(�) = �) + 6� + 9(� + 4)(� − 1) � − 2

�) + 6� + 9(� + 4)(� − 1) � − 2 = −#?

@� − 1 +

)@?

� − 2 +#

,�(� + 4)

We are looking for * � = ℒ�# +(�)

Step 2: Compute inverse Laplace transform

5.3 Method of Solution

* � = ℒ�# −#?@

� − 1 +)@?

� − 2 +#

,�(� + 4)

* � = − 165 ℒ�# 1

� − 1 + 256 ℒ�# 1

� − 2 + 130 ℒ�# 1

� + 4

= − 165 �� + 25

6 �)� + 130 ��=�

5.3 Method of Solution

Use the Laplace Transform to solve the following IVPs

2)*�� − 7*� + 10* = 4�,�, * 0 = 1, *� 0 = −2

5.3 Method of Solution

Exercise

3)*�� + 9* = ��, * 0 = 0, *� 0 = 0

1)*� + 2* = 0, * 0 = 3