Find solution of $(1-x^2)y''-xy'+p^2y=0, p \in \mathbb{R}$ The following differential equation is given:
$$(1-x^2)y''-xy'+p^2y=0, p \in \mathbb{R}$$


*

*Find the general solution of the differential equation at the interval $(-1,1)$ (with the method of power series).

*Are there solutions of the differential equation that are polynomials?


That's what I have tried:
We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence $R>0$.
$$y'(x)= \sum_{n=1}^{\infty} n a_n x^{n-1}= \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n$$
$$y''(x)= \sum_{n=1}^{\infty} (n+1)n a_{n+1}x^{n-1}= \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n$$
$$-x^2y''(x)= \sum_{n=0}^{\infty} -n(n-1)a_n x^n$$
We have:
$$\sum_{n=0}^{\infty} \left[ (n+2)(n+1) a_{n+2}-n(n-1) a_n-na_n+p^2a_n\right]x^n=0, \forall x \in (-R,R)$$
It has to hold:
$$(n+2)(n+1) a_{n+2}+ \left[ -n(n-1)-n+p^2\right] a_n=0 \\ \Rightarrow (n+2)(n+1)a_{n+2}+ \left[ -n(n-1+1)+p^2 \right]a_n=0 \\ \Rightarrow a_{n+2}=\frac{n^2-p^2}{(n+2)(n+1)} a_n \forall n=0,1,2, \dots$$
For $n=0: a_2=-\frac{p^2}{2}a_0$
For $n=1: a_3= \frac{1-p^2}{2 \cdot 3} a_1$
For $n=2: a_4= \frac{(2^2-p^2)(-p^2)}{2 \cdot 3 \cdot  4 } a_0$
For $n=3: a_5=\frac{(3^2-p^2)(1-p^2)}{2 \cdot 3 \cdot 4 \cdot 5}a_1$
For $n=4: a_6=\frac{(4^2-p^2)(2^2-p^2)(-p^2)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}a_0$
For $n=5: a_7=\frac{(5^2-p^2)(3^2-p^2)(1-p^2)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}$
So:
$$a_{2n}=\frac{\prod_{i=0}^{n-1} (2i)^2-p^2}{(2n)!} a_0$$
$$a_{2n+1}=\frac{\prod_{i=0}^{n-1} (2i+1)^2-p^2}{(2n+1)!} a_1$$
So the solution $y$ can be written as follows:
$$y(x)=\sum_{n=0}^{\infty} a_{2n}x^{2n}+ \sum_{n=0}^{\infty}a_{2n+1}x^{2n+1}$$
Is it right so far? If so, then do we have to find for  the power series $\sum_{n=0}^{\infty} a_{2n}x^{2n}$ and $\sum_{n=0}^{\infty}a_{2n+1}x^{2n+1}$ the radius of convergence and show that they define functions that are infinitely many times differentiable?
If so, then do we have to take the ratio
$$\left| \frac{a_{2(n+1)} x^{2(n+1)}}{a_{2n} x^{2n}}\right|$$
? If yes, could we deduce something from the latter about the radius of convergence?

The Equation can be rewritten as( D^2-(x/(1-x^2))×D+x^2/1-x^2)y=0.
This equation can easily be solved using one integral method .
As P+XQ=0 then solution will assume type of y=xv.
substitute dy/dx by xdv/dx+v ,d^2y/dx^2 by xd^2/dx^2+2dv/dx and y by xv
This equation will be converted to linear diffrential equation with constant coefficient or other forms which r very easy to evaluate.
