How find this ODE $(1-x^2)y''+2xy'-2y=-2$ Question:


  Find the ODE
  $$(1-x^2)y''+2xy'-2y=-2$$


I think we can find 
$$(1-x^2)y''+2xy'-2y=0$$
I have find a solution $y=x$,But I can't find all solution.

Thank you
By setting $y=1+f(x)$, we have that $f$ satisfies:
$$(1-x^2) f'' + 2x f' - 2f = 0, $$
that can be solved using Frobenius' method. By setting:
$$ f(z) = \sum_{j=0}^{+\infty} a_j z^j $$
we have:
$$ z\,f'(z) = \sum_{j=1}^{+\infty} j a_j\, z^j,\qquad f''(z)=\sum_{j=2}^{+\infty}j(j-1)a_j z^{j-2} $$
$$(1-z^2)f''(z) = \sum_{j=0}^{+\infty}\left((j+2)(j+1)a_{j+2}-j(j-1)a_j\right)z^j$$
so:
$$(j+2)(j+1)a_{j+2}-j(j-1)a_j+2ja_j-2a_j = 0 $$
or:
$$(j+2)(j+1)a_{j+2} = (j-1)(j-2) a_{j}\tag{1} $$
so the relation $a_{j+2}=\frac{j-1}{j+1}\cdot\frac{j-2}{j+2}a_j$ gives that the solution associated with the initial conditions $a_0=0,a_1=1$ is given by $f(x)=x$ while the solution associated with the initial conditions $a_0=1,a_1=0$ is $f(x)=x^2+1$.
That is a Legendre equation. Deriving a solution on your own will be very difficult. Here is some information about it http://en.wikipedia.org/wiki/Legendre_polynomials