Example 1  Find the general solution to

given that  is a solution.

Solution

Reduction of order requires that a solution already be known.  Without this known solution we won’t be able to do reduction order. 

Once we have this first solution we will then assume that a second solution will have the form

                                                                                                                (1)

for a proper choice of v(t).  To determine the proper choice, we plug the guess into the differential equation and get a new differential equation that can be solved for v(t).

So, let’s do that for this problem.  Here is the form of the second solution as well as the derivatives that we’ll need.

Plugging these into the differential equation gives

Rearranging and simplifying gives

Note that upon simplifying the only terms remaining are those involving the derivatives of v.  The term involving v drops out.  If you’ve done all of your work correctly this should always happen.  Sometimes, as in the repeated roots case, the first derivative term will also drop out.

So, in order for (1) to be a solution then v must satisfy

                                                                                                                        (2)

This appears to be a problem.  In order to find a solution to a second order non-constant coefficient differential equation we need to solve a different second order non-constant coefficient differential equation. 

However, this isn’t the problem that it appears to be.  Because the term involving the v drops out we can actually solve (2) and we can do it with the knowledge that we already have at this point.  We will solve this by making the following change of variable.

With this change of variable (2) becomes

and this is a linear, first order differential equation that we can solve.  This also explains the name of this method.  We’ve managed to reduce a second order differential equation down to a first order differential equation.

This is a fairly simple first order differential equation so I’ll leave the details of the solving to you.  If you need a refresher on solving linear, first order differential equations go back to the second chapter and check out that section.  The solution to this differential equation is

Now, this is not quite what we were after.   We are after a solution to (2).  However, we can now find this.  Recall our change of variable.

With this we can easily solve for v(t).

This is the most general possible v(t) that we can use to get a second solution.  So, just as we did in the repeated roots section, we can choose the constants to be anything we want so choose them to clear out all the extraneous constants.  In this case we can use

Using these gives the following for v(t) and for the second solution.

Then general solution will then be,

If we had been given initial conditions we could then differentiate, apply the initial conditions and solve for the constants.

Example 2  Find the general solution to

given that  is a solution.

Solution

The form for the second solution as well as its derivatives are,

Plugging these into the differential equation gives,

Rearranging and simplifying gives the differential equation that we’ll need to solve in order to determine the correct v that we’ll need for the second solution.

Next use the variable transformation as we did in the previous example.

With this change of variable the differential equation becomes

and this is a linear, first order differential equation that we can solve.  We’ll leave the details of the solution process to you.

Now solve for v(t).

As with the first example we’ll drop the constants and use the following v(t)

Then general solution will then be,