It looks like you're trying to solve the ODE via Frobenius' method. It's not immediately clear where the solution is to be centered, but I think we can safely assume it's to be taken about
. Dividing through by
, we get
and we observe that
is a regular singular point.
Now, taking a solution of the form
and substituting into the ODE gives
The indicial polynomial admits only one root (of multiplicity two) at
, so in fact a regular power series solution will exist. So in fact the ODE above reduces to
leaving us with the recurrence
We can solve this by recursively substituting the right hand side:
and so one of the solutions will be
Now the literature on the Frobenius method when the indicial polynomial has repeated roots suggests a second solution will take the form
but it's not clear to me *why* this works.