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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?

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up vote 3 down vote accepted

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

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Many thanks for this! – Harry Macpherson Jun 4 '12 at 12:33
(+1) nice answer. – Mhenni Benghorbal Oct 8 '13 at 0:01

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