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 of 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 nonconstant coefficient differential equation we need to solve a
different second order nonconstant 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.
