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!
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!