<s>[INST] Tricky second order differential equation 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?
 [/INST] 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 $u'=\dfrac{y''}{y}-\dfrac{(y')^2}{y^2}=\dfrac{y''}{y}-u^2$
$\therefore\left(u^2+1\right)^\frac{3}{2}=u'+u^2+2u^2+1$
$u'=\left(u^2+1\right)^\frac{3}{2}-3u^2-1$
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