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


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*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?
 A: Those ratios simplify rather nicely, so you can indeed deduce something about the radius of convergence.  But in fact there is a general theory that says there should be analytic solutions in any region (in the complex plane) where the coefficients are analytic and the leading coefficient is nonzero.  In this case the only bad points are $x = \pm 1$ where the leading coefficient is $0$, so the radius of convergence should be (at least) 1.
Also note what happens for integer values of $p$.
By the way (though this is not really connected to the power-series method), the substitution $x = \sin(z)$ produces something nice (equivalently, substitute $y(x) = u(\arcsin(x))$ in to the differential equation and simplify).
EDIT: For the ratio test, note that as $n \to \infty$, if $p$ is not an  integer
$$ \dfrac{a_{2(n+1)} x^{2(n+1)}}{a_{2n} x^{2n}} = \dfrac{((2n)^2-p^2) x^2}{(2n+2)(2n+1)} \to x^2 $$
and similarly for $a_{2(n+1)+1} x^{2(n+1)+1}/(a_{2n+1} x^{2n+1})$, so the radii of convergence of the series are $1$.  
If $p$ is an integer, one of the factors $(2i)^2 - p^2$ or $(2i+1)^2 - p^2$ is $0$, which makes either $\sum_{n} a_{2n} x^{2n}$ or $\sum_n a_{2n+1} x^{2n+1}$ a polynomial.
A: 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.
