ODE_Chapter 03-05 [Jan 2014]

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Ordinary Differential Equations[FDM 1023]

Linear Higher-Order Differential Equations

Chapter 3

Overview

Chapter 3: Linear Higher-Order Differential Equations

3.1. Definitions and Theorems

3.2. Reduction of Order

3.3. Homogeneous Linear Equations with

Constant Coefficients

3.4. Undetermined Coefficients

3.5. Variation of Parameters

3.6. Cauchy-Euler Equations

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

Solve the non-homogeneous linear DE byusing the Variation of Parameters method

3.5 Variation of Parameters

Learning Outcome

Given a non-homogeneous linear DE

The general solution is

Recall

� = �� + ��

�� � �(�) + ���� � �(���) +⋯+ �� � �� + �� � � = �(�)

3.5 Variation of Parameters

�� is obtained by solving the associated homogeneous

�� is the particular solution of

�� � �(�) + ���� � �(���) +⋯+ �� � �� + �� � � = 0

�� � �(�) + ���� � �(���) +⋯+ �� � �� + �� � � = �(�)

3.5 Variation of Parameters

The assumption for particular solution is

�� = �� � �� � + �� � �� � + ��(�)��(�)where �� � ,�� (�) and ��(�) obtained form the

solutions of the associated homogeneous equation

3.5 Variation of Parameters

Particular Solution

�� = �� � �� � + �� � �� � (second order)

(third order)

3.5 Variation of Parameters

Step 1: Find the complementary solution

Method of Solution

�� � ��� + �� � �� + �� � � = 0Solve the associated homogeneous equation

�� = ���� + ����

3.5 Variation of Parameters

Step 2: Write the DE in standard form

��� = ��� = −��� �

� ��� = ��� = ��� �

� = �� ����� ��� �� = 0 ��

�(�) ��� �� = �� 0��� �(�)

��� + � � �� + � � � = �(�)

�� � ��� + �� � �� + �� � � = �(�)Standard

Form

Step 3: Compute

where

and

3.5 Variation of Parameters

Step 4: Integrate

���

���

Step 5: Particular solution

����

�� = �� � �� � + ��(�)��(�)

Step 6: General solution

� = �� + ��

Solve ��� − 4�� + 4� = !"#�$%"

&� − 4& + 4 = 0

��� − 4�� + 4� = 0

3.5 Variation of Parameters

Example 1

SolutionStep 1: Find the complementary solution

Solve the associated homogeneous equation

Change to auxiliary equation.

&� − 4& + 4 = 0

&� = &� = 2

= ��(�% + ���(�%

&− 2 � = 0

3.5 Variation of Parameters

The roots of the auxiliary equation are

Case 2

The complementary solution is

�� = ��()*% + ���()*%

Step 2: Find the particular solution

3.5 Variation of Parameters

Step 2.1: Identify �� and ���� = ��(�% + ���(�%

�� = (�% �� = �(�%

Step 2.2: DE in standard form and identify � ���� − 4�� + 4� = (�%

1 + ��

� � = (�%1 + ��

� ��, �� = � (�% , �(�%

= 2�(,% + (,% − 2�(,%= (,%

3.5 Variation of Parameters

��� = ��� = −��� �

� ��� = ��� = ��� �

Step 2.3: Compute

and

�� =? �� =?� =?

= (�% �(�%2(�% 2�(�% + (�%

=0 �(�%(�%

1 + �� 2�(�% + (�%

= − �(,%1 + ��

3.5 Variation of Parameters

�� = 0 ���(�) ���

= (,%1 + ��

=(�% 02(�% (�%

1 + ��

�� = �� 0��� �(�)

��� = ���

= − �(,%1 + ��(,%

= −�1 + ��

3.5 Variation of Parameters

��� = ���

=(,%

1 + ��(,%

= 11 + ��

�� = −. �1 + �� /�

= −12 ln 1 + ��

3.5 Variation of Parameters

Step 2.4: Integrate

��� = −�1 + �� ��� = 1

1 + ��

�� = . 11 + �� /�

= 2�3���

= ��(�% + ���(�% − 12 (�% ln 1 + �� + �(�% 2�3�� �

= −12 ln 1 + �� (�% + 2�3�� � �(�%

Step 2.5: Particular solution

3.5 Variation of Parameters

�� = �� � �� � + ��(�)��(�)

Step 3: General solution

The general solution is

� = �� + ��

3.5 Variation of Parameters

Example 2

SolutionStep 1: Find the complementary solution

Solve the associated homogeneous equation

Change to auxiliary equation.

Solve 4��� + 36� = csc 3�

4��� + 36� = 0

4&� + 36 = 0

3.5 Variation of Parameters

The roots of the auxiliary equation are

Case 3

The complementary solution is

4&� + 36 = 0

& = ±39&� = 39, &� = −39

= �� cos 3� + ��sin3��� = (<% �� cos=� + �� sin =�

Step 2: Find the particular solution

3.5 Variation of Parameters

Step 2.1: Identify �� and ��

Step 2.2: DE in standard form and identify � �

�� = �� cos 3� + ��sin3��� = cos 3� �� = sin 3�

��� + 9� = csc 3�4

4��� + 36� = csc 3�

� � = csc 3�4

� ��, ��

3.5 Variation of Parameters

��� = ��� = −��� �

� ��� = ��� = ��� �

Step 2.3: Compute

and

�� =? �� =?� =?= � cos 3� , sin 3�

= 3= cos 3� sin 3�

−3 sin 3� 3 cos 3�

3.5 Variation of Parameters

�� = 0 ���(�) ��� �� = �� 0

��� �(�)

=0 sin 3�csc 3�4 3 cos 3�

= −14 = cos 3�

4 sin 3�

=cos 3� 0

−3 sin 3� csc 3�4

��� = ���

3.5 Variation of Parameters

��� = ���

= −1 4⁄3

= −112

=cos 3�4 sin 3�

3= cos 3�12 sin 3�

3.5 Variation of Parameters

Step 2.4: Integrate

��� = −112 ��� = cos 3�

12 sin 3�

�� = −. 112/�

= −12�

�� = . cos 3�12 sin 3� /�

= 136 ln sin 3�

Step 2.5: Particular solution

3.5 Variation of Parameters

�� = �� � �� � + ��(�)��(�)

Step 3: General solution

The general solution is

� = �� + ��

�� = − 112� cos 3� +

136 ln sin 3� sin 3�

= �� cos 3� + ��sin3� − 112 � cos 3� +

136 sin 3� ln sin 3�

3.5 Variation of Parameters

Step 1: Find the complementary solution

Method of Solution

Solve the associated homogeneous equation

�� = ���� + ���� + ������ � ���� + �� � ��� + �� � �� + �� � � = 0

3.5 Variation of Parameters

Step 2: Write the DE in standard form

Standard

Form

Step 3: Compute

and

���� + � � ��� + � � �� + @ � � = �(�)

�� � ���� + �� � ��� + �� � �� + �� � � = �(�)

��� = ��� ��� = ��

���� = ���

3.5 Variation of Parameters

� =�� �� ����� ��� ������� ���� ����

�� =0 �� ��0 ��� ���

�(�) ���� ����

�� =�� 0 ����� 0 ������� �(�) ����

�� =�� �� 0��� ��� 0���� ���� �(�)

3.5 Variation of Parameters

Step 4: Integrate

���

���

Step 5: Particular solution

����

�� = �� � �� � + ��(�)��(�) +��(�)��(�)Step 6: General solution

� = �� + ��

��� ��

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