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As the title said, I need to solve the differential equation:

$$y'(x)^2+y(x)y''(x)=0$$

At first I tried to separate this or use substitution $v=y'$, but it didn't work out. Then I tried my luck by using $y=e^{rx}$, from which I got:

$$r^2e^{2rx}+r^2e^{2rx}=0$$

$$r^2=0$$ $$r=0$$

so this didn't work out...I also tried for a sinusoid $y=\sin(x)$:

$$\cos^2(x)-\sin^2(x)=0$$

so this didn't work out. Any hints?

P.S. there are boundary conditions for this problem so $y=C$ is not an acceptable solution.

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    $\begingroup$ After 2+ years on the site and 180+ questions asked, this is hardly admissible. $\endgroup$
    – Did
    Feb 26, 2015 at 12:11
  • $\begingroup$ I don't understand, am I not allowed to ask this kind of questions? @Did Thank you =) $\endgroup$
    – jjepsuomi
    Feb 26, 2015 at 12:13
  • $\begingroup$ Ever read any kind of how-to-ask page on the site? Ever pondered the many comments on your previous questions referring to the same? The one who refuses to understand does not understand. $\endgroup$
    – Did
    Feb 26, 2015 at 12:14
  • $\begingroup$ I have little bit read it, and I tried to solve this myself. I don't think I have asked this myself twice. Does my question offend you in some how? Understand what? =) I always try to understand, but I'm not a professional mathematician, I don't do differential equations every single day. No one can remember 5000 physics equations as Walter Lewin stated. $\endgroup$
    – jjepsuomi
    Feb 26, 2015 at 12:16
  • $\begingroup$ I understand what you mean @Did and I try to do myself as much as possible. But sometimes when the answer isn't obvious to you, it seems to me it is more efficient in terms of optimizing time to simply ask someone than to try and try it yourself, and for what point? I will anyway get it, when somebody shows it to me and it seems I didn't spend that many seconds of anyone's time I hope. Thank you for your help. $\endgroup$
    – jjepsuomi
    Feb 26, 2015 at 12:30

2 Answers 2

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Hint:

$$\left( y\cdot y' \right)' = y'^2 + y\cdot y''$$

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    $\begingroup$ Excellent thank you! Got it =) $\endgroup$
    – jjepsuomi
    Feb 26, 2015 at 12:08
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Note that it can be written as $(yy')'=0$. So $yy'=C$. Now it is a separable first order equation.

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  • $\begingroup$ Thank you, got it and solved it! =) $\endgroup$
    – jjepsuomi
    Feb 26, 2015 at 12:09

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