ode_chapter 03-05 [jan 2014]
TRANSCRIPT
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Ordinary Differential Equations[FDM 1023]
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Linear Higher-Order Differential Equations
Chapter 3
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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
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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
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Given a non-homogeneous linear DE
The general solution is
Recall
� = �� + ��
�� � �(�) + ���� � �(���) +⋯+ �� � �� + �� � � = �(�)
3.5 Variation of Parameters
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�� is obtained by solving the associated homogeneous
�� is the particular solution of
�� � �(�) + ���� � �(���) +⋯+ �� � �� + �� � � = 0
�� � �(�) + ���� � �(���) +⋯+ �� � �� + �� � � = �(�)
3.5 Variation of Parameters
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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)
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3.5 Variation of Parameters
Step 1: Find the complementary solution
Method of Solution
�� � ��� + �� � �� + �� � � = 0Solve the associated homogeneous equation
�� = ���� + ����
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3.5 Variation of Parameters
Step 2: Write the DE in standard form
��� = ��� = −��� �
� ��� = ��� = ��� �
�
� = �� ����� ��� �� = 0 ��
�(�) ��� �� = �� 0��� �(�)
��� + � � �� + � � � = �(�)
�� � ��� + �� � �� + �� � � = �(�)Standard
Form
Step 3: Compute
where
and
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3.5 Variation of Parameters
Step 4: Integrate
���
���
Step 5: Particular solution
����
�� = �� � �� � + ��(�)��(�)
Step 6: General solution
� = �� + ��
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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.
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&� − 4& + 4 = 0
&� = &� = 2
= ��(�% + ���(�%
&− 2 � = 0
3.5 Variation of Parameters
The roots of the auxiliary equation are
Case 2
The complementary solution is
�� = ��()*% + ���()*%
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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 + ��
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� ��, �� = � (�% , �(�%
= 2�(,% + (,% − 2�(,%= (,%
3.5 Variation of Parameters
��� = ��� = −��� �
� ��� = ��� = ��� �
�
Step 2.3: Compute
and
�� =? �� =?� =?
= (�% �(�%2(�% 2�(�% + (�%
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=0 �(�%(�%
1 + �� 2�(�% + (�%
= − �(,%1 + ��
3.5 Variation of Parameters
�� = 0 ���(�) ���
= (,%1 + ��
=(�% 02(�% (�%
1 + ��
�� = �� 0��� �(�)
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��� = ���
= − �(,%1 + ��(,%
= −�1 + ��
3.5 Variation of Parameters
��� = ���
=(,%
1 + ��(,%
= 11 + ��
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�� = −. �1 + �� /�
= −12 ln 1 + ��
3.5 Variation of Parameters
Step 2.4: Integrate
��� = −�1 + �� ��� = 1
1 + ��
�� = . 11 + �� /�
= 2�3���
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= ��(�% + ���(�% − 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
� = �� + ��
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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
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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 =�
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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
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� ��, ��
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�
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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
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��� = ���
3.5 Variation of Parameters
��� = ���
= −1 4⁄3
= −112
=cos 3�4 sin 3�
3= cos 3�12 sin 3�
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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�
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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�
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3.5 Variation of Parameters
Step 1: Find the complementary solution
Method of Solution
Solve the associated homogeneous equation
�� = ���� + ���� + ������ � ���� + �� � ��� + �� � �� + �� � � = 0
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3.5 Variation of Parameters
Step 2: Write the DE in standard form
Standard
Form
Step 3: Compute
and
���� + � � ��� + � � �� + @ � � = �(�)
�� � ���� + �� � ��� + �� � �� + �� � � = �(�)
��� = ��� ��� = ��
���� = ���
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3.5 Variation of Parameters
� =�� �� ����� ��� ������� ���� ����
�� =0 �� ��0 ��� ���
�(�) ���� ����
�� =�� 0 ����� 0 ������� �(�) ����
�� =�� �� 0��� ��� 0���� ���� �(�)
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3.5 Variation of Parameters
Step 4: Integrate
���
���
Step 5: Particular solution
����
�� = �� � �� � + ��(�)��(�) +��(�)��(�)Step 6: General solution
� = �� + ��
��� ��