<s>[INST] Solve $(x^2+1)y''-2xy'+2y=0$ 
Solve $$(x^2+1)y''-2xy'+2y=0$$

Seems I can't use Euler Differential method. I tried it 
\begin{align}
(x^2+1)y''-2xy'+2y&=0\\
\text{Let }y&=xv\\
(x^2+1)(xv''+2v')-2x(xv'+v)+2(xv)&=0\\
x(x^2+1)v''+2v'&=0\\
\frac{v''}{v'}&=-\frac{2}{x(x^2+1)}\\
\frac{v''}{v'}&=-\frac{2}{x}+\frac{2x}{x^2+1}
\end{align}
Can I integrate both side and treat LHS as $\int\frac{1}{v'}dv'?$ Any help will be appreciated.
Edit: Actually letting $y=xv$ made the work easy. But I suddenly think Is there any intuitive reason behind it$?$ Because I just take it as a guess without any investigate.
 [/INST] Hint: Put $x^2+1 = t$ and differentiate and put back in your ODE
Think backward, let $y=x^n$ maybe one of solution of your ODE. Then it must satisfy the ODE,
\begin{align}
y'&=nx^{n-1}\\
y''&=n(n-1)x^{n-2}
\end{align}
\begin{align}
(x^2+1)n(n-1)x^{n-2}-2xnx^{n-1}+2x^n&=0\\
(n^2-3n+2)x^n+n(n-1)x^{n-2}&=0\\
(n-1)(n-2)x^n+n(n-1)x^{n-2}&=0\\
(n-1)((n-2)x^n+nx^{n-2})&=0
\end{align}
Yes we luckily got that $y=x$ is one of the solution of your ODE. The rest of your work is simply Variation of parameters method .
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