By "ck" I'm assuming you're referring to the series solution's coefficients, i.e.
.
Differentiating gives
Substituting into the ODE yields
By extracting the first term (
) of the third series, you can consolidate the second and third series:
Next, since the first series starts with a constant term while the other two start with a quadratic term, remove the first two terms:
Now shift the index so that the series begins at the same starting index by replacing
with
.
So now the series can be joined into one:
which means the ODE is equivalent to
It follows that
and
. Using this, you solve the recurrence
with the initial values
and
.
Because
, it follows that
; in general,
for all
.
Also, because the RHS of the recurrence vanishes when
, it follows that
; or generally,
for
, leaving you with only one not-necessarily-zero term of
.
So the hard part is finding a solution for
when
is a multiple of 3. By finding the first few of these terms, you'll start to notice a pattern. Starting with
, you find
and so on, following a general pattern of
valid for
.
So the series solution is
So the two solutions are
and
.
To solve the IVP, notice that when
, you have
so the solution to the IVP is the same as that to the general solution with the unknowns replaced accordingly.