Series solution of $(1-x)^3y’’-6x^2y-6xy=0$ For the differential equation $(1-x)^3y’’-6x^2y-6xy=0$, I considered a series solution about the ordinary point $x=0$. Using a series solution, I got a very nice recurrence relation of simply
$$a_{n+3}=a_n.$$
However, I am unable to progress further. I am unsure of what restrictions to place on $a_0$, $a_1$ and $a_2$. I know that I should have 2 linearly independent solutions, but I don’t know how to invoke my restrictions. Could someone lead me in the right direction? Thanks!
 A: You are absolutely correct in searching for two linearly independent solutions. But are these two solutions unique? The answer is no; for example, each can be multiplied by an arbitrary constant. If you want a unique solution you generally need input from boundary conditions. How does this relate to your problem with determining the first coefficients?
As for your recurrence relation, I think the simplicity might be too good to be true, unfortunately. I ended up with a different, more complicated relation. Suppose we look for solutions in the form of power series about $x=0$.
\begin{equation}
y(x) = \sum_{n=0}^\infty a_n x^n
\end{equation}
This is my result for the recurrence relation: for $n = 0,1,2,...$
\begin{equation}
a_{n+2}(n+2)(n+1)-3a_{n+1}(n+1)n+3a_n n(n-1) -a_{n-1}(n^2-3n+8)-6a_{n-2} = 0
\end{equation}
Here $a_n = 0$ if $n < 0$.
One reason to be suspicious of your recurrence relation is that it leaves 3 coefficients undetermined. How many should there be for a 2nd order ODE? Hope this helps!
