|
Example 1 Determine
a series solution for the following differential equation about  .

Solution
Notice that in this case p(x)=1 and so every point is an ordinary point. We will be looking for a solution in the
form,

We will need to plug this into our differential equation
so we’ll need to find a couple of derivatives.

Recall from the power series review section on power series that we can
start these at n=0 if we need to,
however it’s almost always best to start them where we have here. If it turns out that it would have been easier
to start them at n=0 we can easily
fix that up when the time comes around.
So, plug these into our differential equation. Doing this gives,

The next step is to combine everything into a single
series. To do this requires that we
get both series starting at the same point and that the exponent on the x be the same in both series.
We will always start this by getting the exponent on the x to be the same. It is usually best to get the exponent to
be an n. The second series already has the proper
exponent and the first series will need to be shifted down by 2 in order to
get the exponent up to an n. If you don’t recall how to do this take a
quick look at the first review section
where we did several of these types of problems.
Shifting the first power series gives us,

Notice that in the process of the shift we also got both
series starting at the same place.
This won’t always happen, but when it does we’ll take it. We can now add up the two series. This gives,

Now recalling the fact
from the power series review section we know that if we have a power series
that is zero for all x (as this is)
then all the coefficients must have been zero to start with. This gives us the following,

This is called the recurrence
relation and notice that we included the values of n for which it must be true.
We will always want to include the values of n for which the recurrence relation is true since they won’t
always start at n = 0 as it did in
this case.
Now let’s recall what we were after in the first
place. We wanted to find a series
solution to the differential equation.
In order to do this we needed to determine the values of the an’s. We are almost to the point where we can do
that. The recurrence relation has two
different an’s in it so
we can’t just solve this for an
and get a formula that will work for all n. We can however, use this to determine what
all but two of the an’s
are.
To do this we first solve the recurrence relation for the an that has the largest
subscript. Doing this gives,

Now, at this point we just need to start plugging in some
value of n and see what happens,
Notice that at each step we always plugged back in the
previous answer so that when the subscript was even we could always write the
an in terms of a0 and when the coefficient
was odd we could always write the an
in terms of a1. Also notice that, in this case, we were
able to find a general formula for an’s
with even coefficients and an’s
with odd coefficients. This won’t
always be possible to do.
There’s one more thing to notice here. The formulas that we developed were only
for k=1,2,… however, in this case
again, the will also work for k=0. Again, this is something that won’t always
work, but does here.
Do not get excited about the fact that we don’t know what a0 and a1 are. As you
will see, we actually need these to be in the problem to get the correct
solution.
Now that we’ve got formulas for the an’s let’s get a solution. The first thing that we’ll do is write out
the solution with a couple of the an’s
plugged in.

The next step is to collect all the terms with the same
coefficient in them and then factor out that coefficient.

In the last step we also used the fact that we knew what
the general formula was to write both portions as a power series. This is also our solution. We are done.
|