# 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.

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 .
The next step is (where $$A$$ is an arbitrary constant) $$\begin{eqnarray*} \ln(v')=-2 \ln(x) + \ln(x^2+1) + \ln(A) \\ \frac{dv}{dx} = A\frac{ x^2+1}{x^2}. \end{eqnarray*}$$ Should be easy from here ?
Hint $$\frac {v''}{v'}=(\ln v')'$$ then integrate both sides.
Let $$y=xv$$ then we get $$xv''(x^2+1)+2v'=0$$ now substitute $$v'=u$$ and you will get $$\frac{du}{dx}=-\frac{2u}{x(x^2+1)}$$
• Is there any other way to solve this without letting $y=xv?$ @Dr. Sonnhard Graubner – NajmunNahar Dec 6 '19 at 16:17