Hi,(adsbygoogle = window.adsbygoogle || []).push({});

I derived the equation:

[tex]1+(y')^2-y y''-2y\left(1+(y')^2\right)^{3/2}=0[/tex]

Letting [itex]y'=p[/itex] and [itex]y''=p\frac{dp}{dy}[/itex], I obtain:

[tex]\frac{dp}{dy}=\frac{1+p^2-2y(1+p^2)^{3/2}}{yp}[/tex]

I believe it's tractable in p because Mathematica gives a relatively simple answer:

[tex]p=\begin{cases}\frac{i \sqrt{-y^2+y^4-4 y^2 C[1]+4 C[1]^2}}{y^2-2 C[1]}\\

\\

-\frac{i \sqrt{-y^2+y^4-4 y^2 C[1]+4 C[1]^2}}{y^2-2 C[1]}

\end{cases}[/tex]

I don't see how to get to that answer and I was wondering if someone here could help me with this.

Thanks,

Jack

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Non-linear second order from calculus of variation I can't solve

**Physics Forums | Science Articles, Homework Help, Discussion**