I was able to solve the following :
$(1-x^2)y''-2xy'+\lambda y = 0$ where $\lambda$ is a real constant.
The solution that I obtained is of the form $\sum_{n=0}^{\infty}C_nx^n$ with
$C_{n+2} = \frac{n(n+1)-\lambda}{(n+2)(n+1)}C_n$
Which gives me:
$y_1(x) = C_0[1-\frac{\lambda}{2}x^2+\frac{(2)(2+1)-\lambda}{(4)(3)}x^4...]$
For the even part and:
$y_2(x) = C_1[x+\frac{2-\lambda}{(3)(2)}x^3+...]$
for the odd one.
I am now required to establish the ratio of convergence using the ratio test, but I can not find a way to do it with my solution written in that form.
Consider the even and odd part as power series in $u=x^2$, $$y=y_e(x^2)+xy_o(x^2),~\text{ where }~ y_e(u)=\sum_kC_{2k}u^k,~~ y_o(u)=\sum_kC_{2k+1}u^k,$$ and compute the radius of convergence for $y_e$ and $y_o$ separately. The minimum of the radii is then the radius for the sum of both parts.