Find the solution for Cauchy problem using transformation I'm learning differential equations for the first semester and I'm not sure how to even begin with the following problem:
Find the solution for Cauchy problem using transformation:
$\left\{\begin{matrix}
(1-t^2)x''-tx'+n^2x=0 & n \in \mathbb{Z}\\  
y(0)=1 & y'(0)=0 
\end{matrix}\right.$
Could you please help me?
 A: In fact this belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0220.pdf
Let $s=\int\dfrac{dt}{\sqrt{1-t^2}}=\sin^{-1}t$ ,
Then $\dfrac{dx}{dt}=\dfrac{dx}{ds}\dfrac{ds}{dt}=\dfrac{1}{\sqrt{1-t^2}}\dfrac{dx}{ds}$
$\dfrac{d^2x}{dt^2}=\dfrac{d}{dt}\left(\dfrac{1}{\sqrt{1-t^2}}\dfrac{dx}{ds}\right)=\dfrac{1}{\sqrt{1-t^2}}\dfrac{d}{dt}\left(\dfrac{dx}{ds}\right)+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}=\dfrac{1}{\sqrt{1-t^2}}\dfrac{d}{ds}\left(\dfrac{dx}{ds}\right)\dfrac{ds}{dt}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}=\dfrac{1}{\sqrt{1-t^2}}\dfrac{d^2x}{ds^2}\dfrac{1}{\sqrt{1-t^2}}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}=\dfrac{1}{1-t^2}\dfrac{d^2x}{ds^2}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}$
$\therefore(1-t^2)\left(\dfrac{1}{1-t^2}\dfrac{d^2x}{ds^2}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}\right)-\dfrac{t}{\sqrt{1-t^2}}\dfrac{dx}{ds}+n^2x=0$
$\dfrac{d^2x}{ds^2}+\dfrac{t}{\sqrt{1-t^2}}\dfrac{dx}{ds}-\dfrac{t}{\sqrt{1-t^2}}\dfrac{dx}{ds}+n^2x=0$
$\dfrac{d^2x}{ds^2}+n^2x=0$
$x=C_1\sin ns+C_2\cos ns$
$x=C_1\sin(n\sin^{-1}t)+C_2\cos(n\sin^{-1}t)$
$x(0)=1$ :
$C_2=1$
$\therefore x=C_1\sin(n\sin^{-1}t)+\cos(n\sin^{-1}t)$
$x'=\dfrac{C_1n\cos(n\sin^{-1}t)}{\sqrt{1-t^2}}-\dfrac{n\sin(n\sin^{-1}t)}{\sqrt{1-t^2}}$
$x'(0)=0$ :
$C_1n=0$
$C_1=0$
$\therefore x=\cos(n\sin^{-1}t)$
