General solution of $(1-x^2)y''-2y=0$ about $x_0=0$?

Question

I've expanded this differential equation as a series to obtain the recurrence relation $$a_{n+2}=\frac{a_n(n^2-n+2)}{n^2+3n+2}.$$ I don't know how to find $a_n$ in terms of $a_0$ and $a_1$ so that I can get the general solution. How do I do this?

Answer 1

This is not an answer but it is too long for a comment.

As commented by Winther, finding something nicer than the general recurence seem to be almost impossible.

Playing with a CAS and starting from what Winther wrote, what I obtained is the following monster for $a_{2n}$ $$\frac{ 4^{n-1} \cosh \left(\frac{\sqrt{7} \pi }{2}\right) \Gamma \left(\frac{1}{4} \left(1-i \sqrt{7}\right)\right) \Gamma \left(\frac{1}{4} \left(1+i \sqrt{7}\right)\right) \Gamma \left(n-\frac{i \sqrt{7}}{4}-\frac{1}{4}\right) \Gamma \left(n+\frac{i \sqrt{7}}{4}-\frac{1}{4}\right)}{\pi ^2 \Gamma (2 n+1)}a_0$$ $a_{2n+1}$ is as nice.

As Winther said, this kind of beauty just reflects the fact that the solution of the differential equation involves ugly hypergeomatric functions $$y=c_1 \, _2F_1\left(-\frac{1}{4}-\frac{i \sqrt{7}}{4},-\frac{1}{4}+\frac{i \sqrt{7}}{4};\frac{1}{2};x^2\right)+i c_2 x \, _2F_1\left(\frac{1}{4}-\frac{i \sqrt{7}}{4},\frac{1}{4}+\frac{i \sqrt{7}}{4};\frac{3}{2};x^2\right)$$