First note that
is a regular singular point; in particular
is a pole of order 1 for
.
We seek a solution of the form
where
is to be determined. Differentiating, we have
and substituting into the ODE gives
The indicial polynomial,
, has roots at
and
. Because these roots are separated by an integer, we have to be a bit more careful, but we'll get back to this later.
When
, we have the recurrence
valid for
. When
, with
, we find
and so on, with a general pattern of
Similarly, when
for
, we find
and so on, with the general pattern
So the first indicial root admits the solution
which you can recognize as the power series for
and
.
To be more precise, the second series actually converges to
, which doesn't satisfy the ODE. However, remember that the indicial equation had two roots that differed by a constant. When
, we may seek a second solution of the form
where
. Substituting this into the ODE, you'll find that
, and so we're left with
Expanding
, you'll see that all the terms
with
in the expansion of this new solutions are already accounted for, so this new solution really only adds one fundamental solution of the form
. Adding this to
, we end up with just
.
This means the general solution for the ODE is
