Assuming we're looking for a power series solution centered around
, take
Substituting into the ODE yields
The first series starts with a constant term; the second series starts at
; the last starts at
. So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a
term. We have
Re-index the first sum to have it start at
(to match the the other two sums):
So now the ODE is
Consolidate into one series starting
:
![\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle2a_2%2B%286a_3-a_0%29x%2B%5Csum_%7Bn%5Cge1%7D%5Cbigg%5B%28n%2B3%29%28n%2B2%29a_%7Bn%2B3%7D%2B%28n-1%29a_n%5Cbigg%5Dx%5E%7Bn%2B1%7D%3D0)
Suppose we're given initial conditions
and
(which follow from setting
in the power series representations for
and
, respectively). From the above equation it follows that
Let's first consider what happens when
, i.e.
. The recurrence relation tells us that
and so on, so that
except for when
.
Now let's consider
, or
. We know that
, and from the recurrence it follows that
for all
.
Finally, take
, or
. We have a solution for
in terms of
, so the next few terms (
) according to the recurrence would be
and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for
as
So the series solution to the ODE is given by
Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at
and
overlaid with the series solution (orange) with
and
. (Note the rapid convergence.)
