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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.

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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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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 ?

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Hint $$\frac {v''}{v'}=(\ln v')'$$ then integrate both sides.

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Let $$y=xv$$ then we get $$xv''(x^2+1)+2v'=0$$ now substitute $$v'=u$$ and you will get $$\frac{du}{dx}=-\frac{2u}{x(x^2+1)}$$

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  • $\begingroup$ Is there any other way to solve this without letting $y=xv?$ @Dr. Sonnhard Graubner $\endgroup$ – NajmunNahar Dec 6 '19 at 16:17

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