how to solve differential equation $y^4 = k^2 (y^2 + y'^2\csc^2\alpha)$? What's the solution of the differential equation $y^4 = k^2 (y^2 + y'^2\csc^2\alpha)$, where $y$ is a function of $x$ and $\alpha$ is a constant?
Polynomial solutions don't seem to work, because the LHS will always have higher degree than the RHS. Solutions of the form $A\cos(x\sin\alpha)+b\sin(x\sin\alpha)$ don't work either, but maybe something similar does?
This comes from the 1st integral of the Euler-Lagrange equation for the functional $\int{y^2 + y'^2\csc^2\alpha)^{1/2}}dx$, which is the arc-length of a curve $r(\theta)$ on a cone with interior angle $2\alpha$, where $y=r$ and $x=\theta$. Perhaps there's a more useful way of using the Euler-Lagrange equation, giving an ODE whose solution is obvious?
 A: Here are two possible approaches
1)Solve for $y'^2$:
$$y'^2=\frac{y^2(y^2-k^2)}{k^2\csc^2\alpha}=\frac{y^2(y^2-k^2)}{k^2\csc^2\alpha}$$
$$y'=\frac{y\sqrt{y^2-k^2}}{k\csc\alpha}$$
$$dx=k\csc\alpha\frac{dy}{y\sqrt{y^2-k^2}}$$
$$x+C=k\csc\alpha\int\frac{dy}{y\sqrt{y^2-k^2}}$$
Integral on the RHS is evaluated as follows:
$$\int\frac{dy}{y\sqrt{y^2-k^2}}=-\int\frac{d\left(\frac{k}{y}\right)}{\sqrt{1-\left(\frac{k}{y}\right)^2}}=\arccos{\frac{k}{y}}$$
Hence,
$$y=\frac{k}{\cos{\left(\frac{x+C}{k\csc\alpha}\right)}}$$
2)Alternatively, confronted with this type of equations involving radicals and even powers of $y$ and $y'$ you might consider looking for a parametric solution introducing trigonometric functions.
$$y^{2}+C\sqrt{\left(y^{2}+\frac{1}{\sin^{2}\alpha}y'^{2}\right)}=0$$
Let
$$\sin\alpha\frac{y'}{y}=\tan\psi \qquad (*)$$
$$y^{2}+\frac{Cy}{\cos\psi}=0$$
Ignoring $y=0$ which does not normally satisfy boundary conditions.
$$y=-\frac{C}{\cos\psi}$$
Now rewriting $(*)$ as follows
$$\sin\alpha\frac{1}{y}\frac{dy}{dx}=\tan\psi$$
solve for $dx$:
$$dx=\sin\alpha\frac{dy}{y\tan\psi}=-\sin\alpha\frac{\sin\psi}{\cos\psi\tan\psi}d\psi=-\sin\alpha d\psi$$
Leading to the same result
