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

[Yondaime5685]

I saw this post at stackexchange:

The fact that curve length is fixed amounts to a functional constraint. So we need to use both the Euler-Lagrange equation and Lagrange multipliers. The functional to minimise is \int^a_{-a}(2\pi y-\lambda)\sqrt{1+y'^2}dx, where λ is a Lagrange multiplier. For this functional, the first integral of the Euler-Lagrange equation gives a different ODE from what I got above, which has the solution y(x)=k(\cosh\frac{x}{k}-\cosh\frac{a}{k}) for some k depending on l, as required.

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}

 

[voko]

\frac{\alpha}{2\pi}=\frac{\eta \quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}

That should be <br /> \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}]<br />

Then you square the equation: <br /> (\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]<br /> <br /> \\ (\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}<br /> = \eta^2 \frac{\quad \dot{\eta}^4 -\dot{\eta}^2 - \dot{\eta}^4 + 1+\dot{\eta}^2} {1+\dot{\eta}^2}<br /> = \eta^2 \frac{1} {1+\dot{\eta}^2}

I assume you can continue from here.

 

[Yondaime5685]

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

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