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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

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

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