3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

Hint:

$(y^2+(y')^2)^\frac{3}{2}=y(2(y')^2+y^2+yy'')$

$y^3\left(\dfrac{(y')^2}{y^2}+1\right)^\frac{3}{2}=y^3\left(\dfrac{y''}{y}+\dfrac{2(y')^2}{y^2}+1\right)$

$\left(\dfrac{(y')^2}{y^2}+1\right)^\frac{3}{2}=\dfrac{y''}{y}+\dfrac{2(y')^2}{y^2}+1$

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

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

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

$\therefore\left(u^2+1\right)^\frac{3}{2}=u'+u^2+2u^2+1$

$u'=\left(u^2+1\right)^\frac{3}{2}-3u^2-1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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