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

3.2. Reduction of Order

Learning Outcome

At the end of this section you should be able to

know:

Find the second solution, given a non-trivial

solution of a linear second order differential

equation

Given the linear second order DE

�� � ��� + �� � �� + �� � � = 0and �� is a non-trivial solution on an interval .

We need to search for a solution �� such that ��, �� isa linearly independent set of solution on .

3.2. Reduction of Order

What?

I

I

Remark

1y

2yIf is a linearly independent set of solutions, then

cannot be written as a multiple of .

{ }21

, yy

Cy

yCyy ≠⇒≠

1

2

12

1

2

y

yIis non-constant on

)()(

)(

1

2 xuxy

xy= )()()(

12xyxuxy =⇒

3.2. Reduction of Order

Solution:

Step 1

Given �� = � is a solution of ��� − � = 0 on the interval

−∞,∞ use reduction of order to find a second solution ��

�� = �(�)��= �(�)�

��� = �� + � ������ = �� + � �� + � ��� + ���

= �� + 2� �� + � ���

3.2. Reduction of Order

Example

Step 2

Step 3

��� − � = 0� ��� + 2�� = 0

3.2. Reduction of Order

�� + 2� �� + � ��� − �� = 0��� + 2�� = 0 � ≠ 0

Step 4

Use substitution � = ���� + 2� = 0

The new equation is now a linear first order equation in �(Solve by using linear equation or separable variable)

�� + 2� = 0���� = −2�

� 1� �� = �−2��ln� = −2� + ln �

�� = ����

�(�) = ����� ��

= ����� + ��

3.2. Reduction of Order

Step 5

� = ���� ln� − ln � = −2�

� = ���� = �

−2�� = ����

= � ��� −2 + ��

We know that

= �(�)� = ����� + �� � = ���� + ���

The general solution is � = ��� + ����

3.2. Reduction of Order

Step 6

�� = �(�)��

�� � ��� + �� � �� + �� � � = 0Divide by ��(�) ��� + � � �� + � � � = 0Define � = � � ��(�)It follows that ��= ���� + ����

��� = ����� + 2����� + �����

� ���� + ���� + ��� + ����� + 2��� + ��� �� = 0

Substitute

into DE

= 0We now have ����� + 2��� + ��� �� = 0

Std. Form

3.2. Reduction of Order

General Method

Use substitution � = ��, ����� + 2��� + ��� �� = 0���� + 2��� + ��� � = 0

��� = − 2���

�� + � ��ln� = −2 ln �� −���� + �

ln ���� = −���� + ����� = ���� !" ⇒ � = $%&' ()*

+%,

3.2. Reduction of Order

We know that � = �� and integrate to get the �

� = ��� !" ��� ��

But �� = ���

�� = ����� ! " ��� ���� = ����� ! " ��� ��

Reduction of order formula

3.2. Reduction of Order

��� + � � �� + � � � = 0

Solution:

Step 1: Change to standard form

The function �� = �� is a solution of ����� − 3��� + 4� = 0. Find the general solution of the DE on the interval 0,∞

��� − 3� �� +

4�� � = 0

��� + � � �� + � � � = 0 Std. Form

3.2. Reduction of Order

Example

Step 2: Use the reduction of order formula

�� = ����� ! " ��� ��

= ����� �/ " �� � ��

3.2. Reduction of Order

= ���� / " �0 ��

= ����/ 12 �0 ��

�� = ����/ 12 �0 ��

= ���1� ��

= ��ln�The general solution is given by � = ���� + ����

� = ���� + ���� ln �

3.2. Reduction of Order

1.��� − 4�� + 4� = 0, �� = ��

The indicated function ��(�) is a solution of the given

differential equation. Find the second solution andwrite the general solution of the given DE.

2.��� + 16� = 0, �� = cos 4�3.����� − ��� + 2� = 0, �� = � sin(ln �)

Answer : �� = ��� , �� = sin 4�, �� = � cos(ln �)

3.2. Reduction of Order

Exercise 3.2