Beginning in the next lecture we examine a method for determining these solutions when the coefficients of the differential equation in are constants

Azərbaycan Dövlət Neft və Sənaye Universiteti

Differential Equations

Reduction of Order

In the preceding lecture we saw that the general solution of a homogeneous linear second-order differential equation

is a linear combination

where and are solutions that constitute a linearly independent set on some interval .

 

Beginning in the next lecture we examine a method for determining these solutions when the coefficients of the differential equation in are constants.

 

This method yields only a single solution of the DE. It turns out that we can construct a second solution of a homogeneous equation provided that we know a nontrivial solution of the DE.

 

The basic idea described in this lecture is that equation can be reduced to a linear first-order DE by means of a substitution involving the known solution .

 

A second solution of (1) is apparent after this first-order differential equation is solved.

 

Suppose that denotes a nontrivial solution of (1) and that is defined on an interval .

 

We seek a second solution so that the set consisting of and is linearly independent on .

 

Recall that if and are linearly independent, then their quotient is nonconstant on — that is

or

 

The function can be found by substituting into the given differential equation.

 

This method is called reduction of order because we must solve a linear first-order differential equation to find .

 

Example 1. A Second Solution by Reduction of Order

Given that is a solution of

on the interval use reduction of order to find a second solution .

 

Solution

If then the Product Rule gives

 

And so

 

Since the last equation requires

 

Solution (continued)

If we make the substitution this linear second-order equation in becomes

which is a linear first-order equation in .

 

Using the integrating factor we can write .

 

Integrating again then yields

 

Thus

 

Solution (continued)

After integrating we get or .

 

Solution (continued)

By picking and we obtain the desired second solution

 

Because for every the solutions are linearly independent on .

 

Since we have shown that and are linearly independent solutions of a linear second-order equation the expression in is actually the general solution of on .

 

Reduction of Order:

General Case

Suppose we divide by to put equation in the standard form

where and are continuous on some interval .

 

Let us suppose further that is a known solution of on and that for every in the interval.

 

If we define it follows that

 

And so

 

This implies that we must have

or

where we have let .

 

Observe that the last equation in is both linear and separable.

 

Separating variables and integrating we obtain

 

or

 

We solve the last equation for use and integrate again.

 

We thus obtain

 

By choosing and we find from that a second solution of equation is

 

Homework

It makes a good review of differentiation to verify that the function defined in satisfies equation and that are linearly independent on any interval on which is not zero.

 

Example 2. A Second Solution by Formula

The function is a solution of

 

Find the general solution of the differential equation on the interval .

 

Solution

Build the standard form of the differential equation

 

Solution

We then find from

 

The general solution on the interval is given by

 

Remark

Reduction of order can be used to find the general solution of a nonhomogeneous equation

whenever a solution of the associated homogeneous equation is known.

 

Thank you for attention