How to go about solving this first-order nonlinear differential equation?

1. Sep 23, 2012

Yondaime5685

I saw this post at stackexchange:
I ran across this post when trying to solve a homework problem. But I have no idea how he got that solution for that. When I use the Euler-Lagrange, I get this diff eq below.

Here is the simplest form I have managed to get it in:
$\frac{\alpha}{2\pi}=\frac{\eta \quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}$

Where $\alpha$ is a constant. For the simplification, $\eta = (y-y_0)$ & $λ \equiv -2\pi y_0$ is used from the quote above.

I don't know where to even begin to solve this. Any help would be good.

Thanks.

-edit
Some details that I thought I should add.

Since $h=f+\lambda g≠h(x)$, I used the special case of the E-L equation:
$\dot{y}\frac{\partial h}{\partial \dot{y}}-h=const.$

But after the substituions to trim it up the special E-L turns into this:
$\dot{\eta}\frac{\partial h}{\partial \dot{\eta}}-h=const.=\alpha$
Where $h$ now is: $h=2\pi\eta\sqrt{1+\dot{\eta}^2}$

Last edited: Sep 23, 2012
2. Sep 24, 2012

voko

That should be $$\frac{\alpha}{2\pi}=\frac{\eta \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\eta\sqrt{1+\dot{\eta}^2} = \eta [\frac{\quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}]$$

Then you square the equation: $$(\frac{\alpha}{2\pi})^2= \eta^2 [\frac{\quad \dot{\eta}^4}{1+\dot{\eta}^2} -2 \dot{\eta}^2 + 1 + \dot{\eta}^2] = \eta^2 [\frac{\quad \dot{\eta}^4}{1+\dot{\eta}^2} -\dot{\eta}^2 + 1] \\ (\frac{\alpha}{2\pi})^2= \eta^2 \frac{\quad \dot{\eta}^4 -\dot{\eta}^2(1+\dot{\eta}^2) + (1+\dot{\eta}^2)} {1+\dot{\eta}^2} = \eta^2 \frac{\quad \dot{\eta}^4 -\dot{\eta}^2 - \dot{\eta}^4 + 1+\dot{\eta}^2} {1+\dot{\eta}^2} = \eta^2 \frac{1} {1+\dot{\eta}^2}$$

I assume you can continue from here.

3. Sep 24, 2012

Yondaime5685

Thanks a lot! That made the problem a lot simpler.

I can't believe I didn't carry that $\eta$ over.

Last edited: Sep 24, 2012