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

[jackmell]

Hi,

I derived the equation:

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

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

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

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

p=\begin{cases}\frac{i \sqrt{-y^2+y^4-4 y^2 C[1]+4 C[1]^2}}{y^2-2 C[1]}\\<br /> \\<br /> -\frac{i \sqrt{-y^2+y^4-4 y^2 C[1]+4 C[1]^2}}{y^2-2 C[1]}<br /> \end{cases}

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

Thanks,
Jack

 

[SteamKing]

What does 'C[1]' represent?

 

[jackmell]

What does 'C[1]' represent?

I should have changed that to just c. It's a constant:

p=\begin{cases}\frac{i \sqrt{-y^2+y^4-4 y^2 c+4 c^2}}{y^2-2 c}\\<br /> \\<br /> -\frac{i \sqrt{-y^2+y^4-4 y^2 c+4 c^2}}{y^2-2 c}<br /> \end{cases}

 

[fzero]

Try the substitution

$$ w = \frac{\sqrt{1+p^2}}{y}.$$

You should find that

$$ \frac{dw}{dy} = -2y w^2,$$

which is separable.

 

[JJacquelin]

Hi !

A laborious method below :

 

[jackmell]

You guys are just shear art in here. \frac{dw}{dy}=-2w^2y was tough for me to go through but I got it. Still working through Jacquelin's. I'm sure it's correct though. :)