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This is the most complicated differential equation I've encountered today:

$$y''(x)-\frac{y'^2}{y}+ \frac{\epsilon y(x)}{x^2}=0$$

It is not Euler-Cauchy, I guess the ultimate approach is to do it with variable substitution, but it is hard to make a guess on what to substitute by purely looking at the differential equation.

WolframAlpha gives the analytical solution: $$y(x)=c_2 e^{\epsilon log(x)+c_1 x}$$

Happy weekend! :)

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Just another way to do it.

Considering $$y''-\frac{y'^2}{y}+ \frac{\epsilon y}{x^2}=0$$ let $$y=e^z\implies y'=e^{z} z'\implies y''=e^{z} \left(z''+z'^2\right)$$ Replace in the original equation to get $$\frac{e^{z} \left(x^2 z''+\epsilon \right)}{x^2}=0\implies x^2 z''+\epsilon=0$$ which becomes simple.

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This equation gives $\frac{y y'' -(y')^{2}}{y^{2}}+\frac{\epsilon}{x^{2}}=0$, and thus $(\frac{y'}{y})' +\frac{\epsilon}{x^{2}}=0$. Integrating this, you have $(\frac{y'}{y})-\frac{\epsilon}{x}=C$ where $C$ is some constant. Now this is a standard linear equation, $y' - (\frac{\epsilon}{x}+C)y=0$.

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  • $\begingroup$ Clever observation! $\endgroup$ – Chee Han Mar 12 '17 at 3:36
  • $\begingroup$ Thank you for the quick respond! Your solution is correct. I think the other one is clearer and easier. :) $\endgroup$ – dellair Mar 12 '17 at 9:02

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