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.



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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, John B, cansomeonehelpmeout, Dando18, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 5
    $\begingroup$ 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

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


Not the answer you're looking for? Browse other questions tagged or ask your own question.