How to solve for $y(x)$ if $\sin^2(x)y'/\sqrt{(\sin(x)y')^2 + 1} = C$?

Question

I would like to solve for $y(x)$ given that (for some constant C):

$$ \frac{\sin^2(x)y'}{\sqrt{(\sin(x)y')^2 + 1}} = C$$

I thought one thing I could do something like:

\begin{align*} &\quad \frac{\sin^2(x)y'}{\sqrt{(\sin(x)y')^2 + 1}} = C \\ &\equiv \sin^2(x)y'= C\sqrt{(\sin(x)y')^2 + 1} \\ &\equiv \sin^4(x)y'^2 = C\left(\sin^2(x)y'^2 + 1\right) \\ &\equiv (y')^2 = \frac{C}{\sin^4(x) - C\sin^2(x)} \\ &\implies y' = \pm \sqrt{\frac{C}{\sin^4(x) - C\sin^2(x)}} \end{align*}

Does the last step make sense? It doesn't really make sense to me. Do I have two ODEs for the same function? Two ODEs defining different parts of the same function? Is there a way to avoid the whole $\pm$ deal with a nice substitution?

Answer 1

Take a look at your original equation. If $C > 0$ and you use the positive square root, this implies that $y' > 0$.

You introduced the possible negative $y'$ when you squared it.

This is another example of the need to check for extraneous solutions when expressions have been squared.

Note that the solution runs into trouble when $\sin^2(x) = C$, since the denominator is $\sin^2(x)(\sin^2(x)-C)$.

The solution, slightly simplified, is $ y' = \frac1{\sin(x)} \sqrt{\frac{C}{\sin^2(x) - C}} $.

Incidently, Wolfy says that (the second and third lines are my simplification)

$\begin{array}\\ \int \frac1{\sin(x)}\sqrt{\frac{C}{\sin^2(x) - C}} dx &= (-C)^{-3/2}(C \sqrt{\frac{-C}{C - sin^2(x)}} \sqrt{\sin^2(x) - C} \tanh^{-1}\left(\frac{ \sqrt{-2C} cos(x)}{\sqrt{-2 C - cos(2 x) + 1}}\right)\\ &= (-C)^{-3/2}(C^{3/2} \tanh^{-1}\left(\frac{ \sqrt{-2C} cos(x)}{\sqrt{-2 C - cos(2 x) + 1}}\right)\\ &= (-1)^{-3/2}( \tanh^{-1}\left(\frac{ \sqrt{-2C} cos(x)}{\sqrt{-2 C - cos(2 x) + 1}}\right)\\ \end{array} $

So it seems that that integral can be done, though I don't trust Wolfy's result because of the complex values introduced.

I am too lazy to check it or try to integrate it myself.