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

1. Jul 23, 2013

### 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:

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

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

Thanks,
Jack

2. Jul 23, 2013

### SteamKing

Staff Emeritus
What does 'C[1]' represent?

3. Jul 23, 2013

### jackmell

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

$$p=\begin{cases}\frac{i \sqrt{-y^2+y^4-4 y^2 c+4 c^2}}{y^2-2 c}\\ \\ -\frac{i \sqrt{-y^2+y^4-4 y^2 c+4 c^2}}{y^2-2 c} \end{cases}$$

4. Jul 24, 2013

### fzero

Try the substitution

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

You should find that

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

which is separable.

5. Jul 24, 2013

### JJacquelin

Hi !

A laborious method below :

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6. Jul 24, 2013

### 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. :)

Last edited: Jul 24, 2013