We explore a technique for reducing a second order nonhomgeneous linear differential equation to first order when we know a nontrivial solution to the complementary homogeneous equation.

### Reduction of Order

In this section we give a method for finding the general solution of

if we know a nontrivial solution of the complementary equation The method is called*reduction of order*because it reduces the task of solving (eq:5.6.1) to solving a first order equation. Unlike the method of undetermined coefficients, it does not require , , and to be constants, or to be of any special form.

By now you shouldn’t be surprised that we look for solutions of (eq:5.6.1) in the form

where is to be determined so that satisfies (eq:5.6.1). Substituting (eq:5.6.3) and into (eq:5.6.1) yields Collecting the coefficients of , , and yields However, the coefficient of is zero, since satisfies (eq:5.6.2). Therefore (eq:5.6.4) reduces to with (It isn’t worthwhile to memorize the formulas for and !) Since (eq:5.6.5) is a linear first order equation in , we can solve it for by variation of parameters as in Module linearFirstOrderDiffEq, integrate the solution to obtain , and then obtain from (eq:5.6.3).- (a)
- Find the general solution of given that is a solution of the complementary equation
- (b)
- As a byproduct of item:5.6.1a, find a fundamental set of solutions of (eq:5.6.7).

item:5.6.1b By letting in (eq:5.6.10), we see that is a solution of (eq:5.6.6). By letting and , we see that is also a solution of (eq:5.6.6). Since the difference of two solutions of (eq:5.6.6) is a solution of (eq:5.6.7), is a solution of (eq:5.6.7). Since is nonconstant and we already know that is a solution of (eq:5.6.6), Theorem thmtype:5.1.6 implies that is a fundamental set of solutions of (eq:5.6.7).

Although (eq:5.6.10) is a correct form for the general solution of (eq:5.6.6), it’s silly to leave the arbitrary coefficient of as where is an arbitrary constant. Moreover, it’s sensible to make the subscripts of the coefficients of and consistent with the subscripts of the functions themselves. Therefore we rewrite (eq:5.6.10) as by simply renaming the arbitrary constants. We’ll also do this in the next two examples, and in the answers to the exercises.

- (a)
- Find the general solution of given that is a solution of the complementary equation As a byproduct of this result, find a fundamental set of solutions of (eq:5.6.11).
- (b)
- Solve the initial value problem

As we explained above, we rename the constants in (eq:5.6.15) and rewrite it as

item:5.6.2b Differentiating (eq:5.6.16) yields

Setting in (eq:5.6.16) and (eq:5.6.17) and imposing the initial conditions and yieldsUsing reduction of order to find the general solution of a homogeneous linear second order equation leads to a homogeneous linear first order equation in that can be solved by separation of variables. The next example illustrates this.

### Text Source

Trench, William F., ”Elementary Differential Equations” (2013). Faculty Authored and Edited Books & CDs. 8. (CC-BY-NC-SA)