Solve following ODE $(1-x^2)y''-xy'-2=0$ I have the differential equation $(1-x^2)y''-xy'-2=0$ with $y(0)=y'(0)=0$.
It is given that $\arcsin^2(x)$ solves it and I have to find the power series of the $\arcsin^2(x)$ in $(-1,1)$ using the differential equation.
I have seen some solutions of $(1-x^2)y''-xy'-y=0$ with power series and I have understood them. The main idea was to come to a point at the end where you have one serie and on the other $0$ so you can compare the coefficients of the left and right hand side and you have $a_n=...=0$ and you go on from there.
I have tried doing the same with my case, but the -2 disturbs me a lot, I cannot take it away in order to have only one serie in the left hand and $0$ in the right hand side.
Maybe what I am trying to do is completely wrong, I would be very happy for some help or enlightment. It is my very first day doing differential equations so I am very happy for help.
Thanks
 A: Replace $x=\sin\theta$, $u(θ)=y(x)=y(\sinθ)$ and insert to check the validity of the equation.
$$
u'(θ)=y'(x)\cosθ\\
u''(θ)=y''(x)\cos^2θ-y'(x)\sinθ=(1-x^2)y''(x)-xy'(x)=2
$$
So yes indeed, one variant of integration constants results in $u(θ)=θ^2$, that is, $y(x)=[\arcsin x]^2$.
For the power series just insert $y(x)=\sum_{n=0}^{\infty} a_nx^n$ to get in the coefficients of $x^n$
$$
(n+1)(n+2)a_{n+2}-n(n-1)a_n-na_n=2\delta_{0,n}
\\
2a_2=2\\
6a_3-a_1=0\\
12a_4-4a_2=0\\
...
$$
$a_0$ and $a_1$ obviously being the free parameters for the 2-dimensional solution family.
A: A natural way to solve this ODE is the following.
Let $z(x):=y'(x)$.
$$z'-\frac x{1-x^2}z=\frac2{1-x^2}.$$
Multiply both sides by $e^{f(x)}$ for some $f(x)$
$$\frac d{dx}(e^{f(x)}z)=\frac2{1-x^2}e^{f(x)}, \tag1$$
so that
$$f'(x)=-\frac x{1-x^2}. \tag2$$
Now solve for $f$ of the above Equation $(2)$ by simple integration. Substitute $f$ into Equation $(1)$ and solve it by integration as well. Then solve for $z$. You get the complete solution for $y$.
