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 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

Constant Coefficients

3.4 Undetermined Coefficients

Remark

���(�) + �����

(���) +⋯+ ��� + ��� = (�)

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�+�

��< = ,��+�

��< = ,���+�