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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
At the end of this section , you should be able to:
Solve the non-homogeneous linear DE by using
the Undetermined Coefficients – Superposition
Approach
3.4 Undetermined Coefficients
Learning Outcome
Given a non-homogeneous DE
���(�) + �����
(���) +⋯+ ��� + ��� = (�)
The general solution is
3.4 Undetermined Coefficients
Recall
� = �� + ��
�� is obtained by solving the associated homogeneous
���(�) + �����
(���) +⋯+ ��� + ��� = 0
�� is the particular solution of
���(�) + �����
(���) +⋯+ ��� + ��� = (�)
where (�) has various form.
3.4 Undetermined Coefficients
3.4 Undetermined Coefficients
Remark
���(�) + �����
(���) +⋯+ ��� + ��� = (�)
Polynomial, Exponential, Sine, Cosine or combination of these functions
Constant Coefficients
Some examples of the types of inputs (�) that are
appropriate.
� = 10
� = �� − 5�
� = 15� − 10 + 3���
� = sin 3� − 5� cos 2�
� = ��� sin � + 3�� − 1 ���
3.4 Undetermined Coefficients
The methods of undetermined coefficients is NOTAPPLICABLE to equations of the form
���(�) + �����
(���) +⋯+ ��� + ��� = (�)
if
� =1
�
� = tan �
� = sin�� �
and so on.
3.4 Undetermined Coefficients
� = ln �
Basic Idea
Making a smart guess of the general form of
by referring to kind of functions that make
up
py
).(xg
3.4 Undetermined Coefficients
Trial Particular Solutions
(�) Form of ��
1)5(any constant) &
2)5� + 7 &� + (
3)3�� − 2 &�� + (� + )
4)�+ − � + 1 &�+ + (�� + )� + ,
5)sin 4� & cos 4� + ( sin 4�
6)cos 4� & cos 4� + ( sin 4�
7)�.� &�.�
3.4 Undetermined Coefficients
(�) Form of ��
8) 9� − 2 �.� &� + ( �.�
9)���.� &�� + (� + ) �.�
10)�+� sin 4� &�+� cos 4� + (�+� sin 4�
11)5�� sin 4� &�� + (� + ) cos 4�
+ 1�� + 2� + 3 sin 4�
12)��+� cos 4� &� + ( �+� cos 4�
+ )� + , �+� sin 4�
** No function in the assumed particular solution �� is
duplicated by a function in the complementary solution ��
3.4 Undetermined Coefficients
CASE 1
No function in the assumed particular solution , �� is a
solution of the associated homogeneous DE.
Rule for Case 1
The form of �� is a linear combination of all linearly
independent functions that are generated by repeateddifferentiation of (�)
3.4 Undetermined Coefficients
CASE 2
A function in the assumed particular solution , �� is also
a solution of the associated homogeneous DE(duplicate of ��).
Rule for Case 2
If any ��4 contains terms that duplicate terms in ��, then
that ��4 must be multiplied by �� , where 5 is the
smallest positive integer that eliminates that duplication.
3.4 Undetermined Coefficients
Solve � + 4� − 2� = 2�� − 3� + 6
SolutionStep 1: Find the complementary solution
Change to auxiliary equation.
6� + 46 − 2 = 0
Solve the associated homogeneous equation
3.4 Undetermined Coefficients
Example 1
� + 4� − 2� = 0
6� + 46 − 2 = 0
6� = −2 − 6 and 6� = −2 + 6
The complementary solution is
The roots of the auxiliary equation are
�� = 7����� 8 � + 7��
��9 8 �
3.4 Undetermined Coefficients
Case 1
�� = 7��:;� + 7��
:<�
Step 2: Find the particular solution
Since the function (�) is a quadratic polynomial, lets
assume a particular solution is also in the form of quadratic polynomial
�� = &�� + (� + )
� + 4� − 2� = 2�� − 3� + 6
�� = 2&� + (
�� = 2&
Substitute into the DE
3.4 Undetermined Coefficients
Step 2.1: Assume particular solution
� + 4� − 2� = 2�� − 3� + 6
2& + 4 2&� + ( − 2 &�� + (� + ) = 2�� − 3� + 6
−2&�� + 8& − 2( � + 2& + 4( − 2) = 2�� − 3� + 6
3.4 Undetermined Coefficients
Step 2.2: Substitute into the DE
�� = &�� + (� + )
�� = 2&� + (
�� = 2&
Compare the coefficients (like terms) and constant
�� ∶ −2& = 2 ⟹ & = −1
� ∶ 8& − 2( = −3 ⟹ ( = −5
2
Constant∶ 2& + 4( − 2) = 6 ⟹ ) = −9
�� = &�� + (� + )The particular solution is
= −�� −5
2� − 9
−2&�� + 8& − 2( � + 2& + 4( − 2) = 2�� − 3� + 6
Step 2.3:
3.4 Undetermined Coefficients
Step 2.4: Particular solution
The general solution is
� = �� + ��
= 7����� 8 � + 7��
��9 8 � − �� −5
2� − 9
Step 3: General solution
3.4 Undetermined Coefficients
Find the general solution of � − � + � = 2 sin 3�
6� −6 + 1 = 0
� − � + � = 0
3.4 Undetermined Coefficients
Example 2
SolutionStep 1: Find the complementary solution
Solve the associated homogeneous equation
Change to auxiliary equation.
6� −6 + 1 = 0
= � � �⁄ � 7� cos+�� + 7� sin
+��
6 =1 ± 1 − 4
2
6� =1 + A 3
26� =
1 − A 3
2
�� = �B� 7� cosC� + 7� sin C�
3.4 Undetermined Coefficients
The roots of the auxiliary equation are
The complementary solution is
Case 3
�� = & cos 3� + ( sin 3�
� − � + � = 2 sin 3�
�� = −3& sin 3� + 3( cos 3�
�� = −9& cos 3� − 9( sin 3�
Substitute into the DE
3.4 Undetermined Coefficients
Step 2: Find the particular solution
Step 2.1: Assume particular solution
−9& cos 3� − 9( sin 3� − (−3& sin 3� + 3( cos 3�)
+ & cos 3� + ( sin 3� = 2 sin 3�
−8& − 3( cos 3� + 3& − 8( sin 3� = 2 sin 3�
� − � + � = 2 sin 3�
3.4 Undetermined Coefficients
Step 2.2: Substitute into DE
�� = & cos 3� + ( sin 3�
�� = −3& sin 3� + 3( cos 3�
�� = −9& cos 3� − 9( sin 3�
−8& − 3( cos 3� + 3& − 8( sin 3� = 2 sin 3�
cos 3� ∶ −8& − 3( = 0 ⟹ & = −3(
8
sin 3� ∶ 3& − 8( = 2 ⟹ ( = −16
73& =
6
73
�� =6
73cos 3� −
16
73sin 3�
3.4 Undetermined Coefficients
Step 2.3: Compare the coefficients (like terms) and constant
Step 2.4: Particular Solution
The particular solution is
�� = & cos 3� + ( sin 3�
The general solution is
� = �� + ��
Step 3: General solution
3.4 Undetermined Coefficients
= � � �⁄ � 7� cos+�� + 7� sin
+�� +
6
73cos 3� −
16
73sin 3�
Find the general solution of � − 2� − 3� = 4� − 5 + 6����
� − 2� − 3� = 0
3.4 Undetermined Coefficients
Example 3
SolutionStep 1: Find the complementary solution
Solve the associated homogeneous equation
Change to auxiliary equation.
6� − 26 − 3 = 0
6− 3 6 + 1 = 0
6� = 3, 6� = −1
= 7��+� + 7��
��
3.4 Undetermined Coefficients
6� − 26 − 3 = 0
The roots of the auxiliary equation are
Case 1
The complementary solution is
�� = 7��:;� + 7��
:<�
3.4 Undetermined Coefficients
Step 2: Find the particular solution
Step 2.1: Assume particular solution
� − 2� − 3� = 4� − 5 + 6����
� = the sum of two basic kinds of functions
= polynomials + (Polynomial*exponentials)
= � � + �(�)
�� = ��; + ��<
Substitute into the DE
3.4 Undetermined Coefficients
� = 4� − 5 + 6����
�� = ��; + ��<
� − 2� − 3� = 4� − 5
��; = &� + (
��; = &
��; = 0
3.4 Undetermined Coefficients
� = 4� − 5 + 6����
�� = ��; + ��<
� − 2� − 3� = 6����
��< = )���� + ,���
��< = 2)���� + )��� + 2,���
��< = 4)���� + 4)��� + 4,���
Substitute into the DE
3.4 Undetermined Coefficients
Step 2.2: Substitute into DE
� − 2� − 3� = 4� − 5
��; = &� + (
��; = &
��; = 0
0 − 2& − 3 &� + ( = 4� − 5
−3&� + −2& − 3( = 4� − 5
3.4 Undetermined Coefficients
Step 2.3: Compare the coefficients (like terms) and constant
Step 2.4: Particular Solution
The particular solution is
−3&� + −2& − 3( = 4� − 5
� ∶ −3& = 4 ⟹ & = −4
3
Constant ∶ −2& − 3( = −5 ⟹ ( =23
9
= −4
3� +
23
9
��; = &� + (
3.4 Undetermined Coefficients
Step 2.5: Substitute into DE
��< = )���� + ,���
��< = 2)���� + )��� + 2,���
��< = 4)���� + 4)��� + 4,���
� − 2� − 3� = 6����
4)���� + 4)��� + 4,��� − 2 2)���� + )��� + 2,���
− 3 )���� + ,��� = 6����
2) − 3, ��� − 3)���� = 6����
3.4 Undetermined Coefficients
Step 2.6:
Step 2.7: Particular solution
The particular solution is
2) − 3, ��� − 3)���� = 6����
��� ∶ 2) − 3, = 0
���� ∶ −3) = 6 ⟹ ) = −2
⟹ , = −4
3
= −2���� −4
3���
��< = )���� + ,���
Compare the coefficients (like terms) and constant
The general solution is
� = �� + ��
Step 3: General solution
3.4 Undetermined Coefficients
= �� + ��; + ��<
= 7��+� + 7��
�� −4
3� +
23
9− 2���� −
4
3���
3.4 Undetermined Coefficients
Example 4
SolutionStep 1: Find the complementary solution
Solve the associated homogeneous equation
Change to auxiliary equation.
Find the particular solution of � − 5� + 4� = 8��
6� − 56 + 4 = 0
� − 5� + 4� = 0
3.4 Undetermined Coefficients
The roots of the auxiliary equation are
The complementary solution is
Case 1
6� − 56 + 4 = 0
6 − 4 6 − 1 = 0
6� = 4, 6� = 1
= 7��F� + 7��
�
�� = 7��:;� + 7��
:<�
Substitute into the DE
3.4 Undetermined Coefficients
Step 2: Find the particular solution
Step 2.1: Assume particular solution
� − 5� + 4� = 8��
�� = &��
�� = &��
�� = &��
3.4 Undetermined Coefficients
�� = &��
�� = &��
�� = &��
� − 5� + 4� = 8��
&�� − 5&�� + 4&�� = 8��
0 = 8��
We have made the wrong guess of ��
Our assumption �� = &�� is already present in ��
This means that �� is a solution of the associated
homogeneous DE
3.4 Undetermined Coefficients
�� = 7��F� + 7��
�
CASE 2
A function in the assumed particular solution , �� is also
a solution of the associated homogeneous DE(duplicate of ��).
Rule for Case 2
If any ��4 contains terms that duplicate terms in ��, then
that ��4 must be multiplied by GH , where 5 is the
smallest positive integer that eliminates that duplication.
3.4 Undetermined Coefficients
Our assumption �� = &�� is already present in ��
This means that �� is a solution of the associated
homogeneous DE
3.4 Undetermined Coefficients
�� = &���
�� = &��� + &��
�� = &��� + 2&��
Substitute into the DE
Lets assume
�� = 7��F� + 7��
�
�� = 7��F� + 7��
�
3.4 Undetermined Coefficients
Step 2.2: Substitute into DE
�� = &���
�� = &��� + &��
�� = &��� + 2&��
� − 5� + 4� = 8��
&��� + 2&�� − 5 &��� + &�� + 4&��� = 8��
−3&�� = 8��
⟹ & = −8
3
3.4 Undetermined Coefficients
Step 2.3:
Step 2.4: Particular solution
The particular solution is
−3&�� = 8��
��: −3& = 8
�� = &���
�� = −8
3���
Compare the coefficients (like terms) and constant
The general solution is
� = �� + ��
Step 3: General solution
3.4 Undetermined Coefficients
= 7��F� + 7��
� −8
3���
Find the particular solution of� − 9� + 14� = 3�� − 5 sin 2� + 7��8�
3.4 Undetermined Coefficients
Example 5
� − 9� + 14� = 0
SolutionStep 1: Find the complementary solution
The complementary solution is
�� = 7���� + 7��
J�
Solve the associated homogeneous equation
3.4 Undetermined Coefficients
Step 2: Find the particular solution
Step 2.1: Assume particular solution
� − 9� + 14� = 3�� − 5 sin 2� + 7��8�
g(x)= 3�� − 5 sin 2� + 7��8�
The assumption for �� = ��; + ��< + ��M
Corresponding to 3�� ��; = &�� + (� + )
Corresponding to −5 sin 2� ��< = 1 cos 2� + 2 sin 2�
Corresponding to 7��8� ��M = 3� + N �8�
�� = 7���� + 7��
J�
No terms in this assumption duplicates a term in ��
3.4 Undetermined Coefficients
g(x)= 3�� − 5 sin 2� + 7��8�
3.4 Undetermined Coefficients
Example 6
SolutionStep 1: Find the complementary solution
The complementary solution is
Solve � − 6� + 9� = 6�� + 2 − 12�+�
� − 6� + 9� = 0
Solve the associated homogeneous equation
�� = 7��+� + 7���
+�
3.4 Undetermined Coefficients
Step 2: Find the particular solution
Step 2.1: Assume particular solution
The assumption for �� = ��; + ��<
� − 6� + 9� = 6�� + 2 − 12�+�
g(x) = 6�� + 2 − 12�+�
Corresponding to 6�� + 2 ��; = &�� + (� + )
Corresponding to −12�+� ��< = ,�+�
�� = 7��+� + 7���
+�
3.4 Undetermined Coefficients
If we multiply ��< by � , the term ��+� is still part of ��
But multiplying ��< by �� , eliminates all duplications
��< = ,���+�
g(x) = 6�� + 2 − 12�+�
��< = ,��+�
��< = ,���+�