I have the following differential equation:
$$2\theta'' = h \sin\theta$$
After multiplying both sides by $\frac{\mathrm{d}\theta}{\mathrm{d}x}$, I obtain:
$$\frac{\mathrm{d}}{\mathrm{d}x} \left[\left(\frac{\mathrm{d}\theta}{\mathrm{d}x}\right)^2+h\cos\theta\right]=0$$
Therefore,
$\left(\frac{\mathrm{d}\theta}{\mathrm{d}x}\right)^2+h\cos\theta=c$
where $c$ is a constant that I need to determine via energy minimization.
I can take the square root of the expression to obtain:
$\mathrm{d}x=\frac{\mathrm{d}\theta}{\sqrt{c-h \cos\theta}}$.
Mathematica gives me a solution in terms of Jacobi Amplitudes. I am trying to express the right hand side as an elliptic integral of the first kind, but so far I have not succeeded. Can anyone give me some ideas on how to proceed?
Edit: I have already found a solution without the use of elliptic integrals in order to find the constant $c$. In that case, the constant $c$ is known and I have to solve the following differential equation:
$\left(\frac{\mathrm{d}\theta}{\mathrm{d}x}\right)^2=c-h\cos\theta$.
Could I find an expression for $\theta(x)$ without the use of elliptic integrals in that case?
I think an easy way to do it would be to write
$\cos\theta=1-2\sin\left(\theta/2\right)^2$
Then,
$c-h\cos\theta=c-h\left(1-2\sin\left(\theta/2\right)^2\right)=c-h+2h\sin\left(\theta/2\right)^2=c-h\left(1+\frac{2h}{c-h}\sin\left(\theta/2\right)^2\right)$
Consequently,
$\sqrt{c-h}\;\mathrm{d}x=\frac{\mathrm{d}\theta}{\sqrt{1+\frac{2h}{c-h}\sin\left(\theta/2\right)^2}}$
Then if we integrate, the right hand side is the elliptic integral of the first kind with modulus $k\equiv -2h/(c-h)$: $F\left(\theta/2,\frac{2h}{h-c}\right)$.
As a result, $ \theta(x)=2\; \text{am}\left(\sqrt{c-h}\;x,\frac{2h}{h-c}\right)$
Edit: the $\sqrt{c-h}$ is legit since $c-h\cos\theta>0$ ,$\forall \theta$. Consequently, $c>h$.