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.
Best Answer
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 .