I got stuck on this differential equation. Most of the equation I solved has degree of highest derivative 1. But in this equation, degree of $y''$ is 2. Can anyone help me or give me some hint so I can solve it? I really appreciate.

Solve this differential equation: $$\frac{y''^2 - y'y''}{y'^2} = {1\over x^2}$$


closed as off-topic by 6005, user230715, Claude Leibovici, tired, user223391 Nov 8 '15 at 18:23

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  • 2
    $\begingroup$ The left hand side is the derivative of a simple expression. $\endgroup$ – Julián Aguirre Nov 4 '15 at 3:08
  • $\begingroup$ Take $Y = \log(y') \implies Y'(Y'-1) = \frac{1}{x^2}$. Solve for $Y'$, integrate to get $Y$, exponentiate to get $y' = e^{Y}$ and integrate again to get $y$. $\endgroup$ – Winther Nov 4 '15 at 3:30
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    $\begingroup$ @JuliánAguirre: Can you please explain clearer? Which function are you mentioning? $\endgroup$ – le duc quang Nov 4 '15 at 6:26
  • $\begingroup$ When I wrote my comment I wasn't wearing my glasses and I missed two $'$s (I thought the left hand side was $-(y'/y)'$.) $\endgroup$ – Julián Aguirre Nov 4 '15 at 10:55

Note that $$ \frac{y''^2 - y'y''}{y'^2} = \left(\frac{y''}{y'}\right)^2 - \frac{y''}{y'} $$

Let $u = \frac{y''}{y'}$, this becomes $$u^2 - u =\frac{1}{x^2}$$ $$ \left( u - \frac{1}{2} \right)^2 = \frac{1}{x^2} + \frac{1}{4}$$ $$ u = \frac{1}{2} \pm \sqrt{\frac{1}{x^2} + \frac{1}{4}}$$

Since $$\frac{y''}{y'} = \frac{d}{dx}\ln(y')$$

This gives $$y' = \exp\left( \int u(x)\,dx \right)$$

  • $\begingroup$ Thanks a lot for your solution. $\endgroup$ – le duc quang Nov 4 '15 at 17:55

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