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I have the following Differential equation which I need to solve:

$$\frac{\mathrm{d^2} y}{\mathrm{d} x^2} = \left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2$$

I know how to solve a second order linear differential equations but this is something strange equation which I have seen before while practicing the Differential equation Chapter. Please tell me how to go about solving this Differential equation problem. Any intial hint would do for me.

Thanks

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closed as off-topic by Saad, John B, cansomeonehelpmeout, Dando18, GNUSupporter 8964民主女神 地下教會 Apr 26 '18 at 16:33

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    $\begingroup$ let $v = \frac {dy}{dt}.$ Now you have a first order differential equation. $\frac {dv}{dt} = v^2$ $\endgroup$ – Doug M Apr 26 '18 at 3:05
  • $\begingroup$ You should probably see how to solve non-linear differential equations and then come back to this one $\endgroup$ – Francisco José Letterio Apr 26 '18 at 3:15
  • $\begingroup$ @DougM I understood your point. After solving, this gives $\frac{-1}{v}=t+c$ for the Differential equation $\frac{dv}{dt}=v^2$. Should we again substitute, the value of p into the Differential equation? $\endgroup$ – RAHUl JHa Apr 26 '18 at 3:28
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Put $\frac{dy}{dx}$ = $v$, after differentiating w.r.to $x$ you get $\frac{dv}{dx}$ = $v^2$

then $\frac{-1}{v}$ = $x+C$

or, $v$ = $\frac{-1}{x+C}$

so $\frac{dy}{dx}$ = $\frac{-1}{x+C}$

$y$ = $-ln(x+C)$ + K

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