So I have the following equation, where we have y as a function of x: $$(y^2+(y')^2)^{3/2}=y(2(y')^2+y^2+yy'')$$

This is a second order autonomous equation. I make the following substitutions : $y\rightarrow p=y', x\rightarrow s=y(x) $

We have $y''=\frac{d}{dx}y'=\frac{dp}{ds}\frac{ds}{dx}=p'p$

So our equation now is $$(s^2+p^2)^{3/2}=s(2p^2+s^2+sp'p)$$

This can be turned into $\frac{dp}{ds}=\frac{(s^2+p^2)^{3/2}-2sp^2-s^3}{s^2p}\rightarrow ((s^2+p^2)^{3/2}-2sp^2-s^3)ds-(s^2p)dp=0$

This is an exact equation, which doesn't seem very friendly. Can anyone suggest a better way of tackling the initial equation? If not can someone tell me an integrating factor for the above exact equation? Can the exact equation be solved easily?

  • $\begingroup$ i think there is no solution in the known elementary functions $\endgroup$ – Dr. Sonnhard Graubner Mar 30 '16 at 20:11
  • $\begingroup$ You would think so right? But I happen to have the answer(just the answer, not the solution). And the answer is y(x)=c1(2x+c2) $\endgroup$ – prometheus21 Mar 30 '16 at 20:21
  • $\begingroup$ You might have made a typo in the original ODE; there is no linear function which solves the ODE in your question, unfortunately (other than the trivial solution y = 0, obviously). $\endgroup$ – Frits Veerman Mar 30 '16 at 20:34
  • $\begingroup$ You are correct. There is a typo. I hate it when this happens... It's y(2y'^2+...) $\endgroup$ – prometheus21 Mar 30 '16 at 20:39
  • $\begingroup$ I will make the computations again $\endgroup$ – prometheus21 Mar 30 '16 at 20:43





This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0511.pdf.

Let $u=\dfrac{y'}{y}$ ,

Then $u'=\dfrac{y''}{y}-\dfrac{(y')^2}{y^2}=\dfrac{y''}{y}-u^2$




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