Finding the general solution of a 2nd order ODE?

Question

SO here's a problem that I'm not having much progress with:

Using substitution $u=cosx$, how can I find the general solution of $sinx(d^2y/dx^2)-cosx(dy/dx)+2ysin^3x=0$

Thank you so much for helping!

My workings so far:

$(d^2y/dx^2)-(cosx/sinx)(dy/dx)+2ysin^2x=0$

$(du/dx)=-sinx$

$(dy/du)=(dy/dx)(dx/du)=(-1/sinx)(dy/dx)$

$=>(d^2y/dx^2)+u(dy/du)+2ysin^2x=0$

$=>(d^2y/dx^2)+u(dy/du)+2y(1-u^2)=0$

How do I go on further? Is this even the best way to go about the problem??

Answer 1

You have to replace both the first and second derivatives.

With $u=\cos x$ it holds that $$ \frac{dy}{dx}=-\frac{dy}{du}\sin x $$ and $$ \begin{aligned} \frac{d^2y}{dx^2} & =\frac{d}{dx}\Bigl(-\frac{dy}{du}\sin x\Bigr)\\ & =\frac{d^2y}{du^2}\sin^2x -\frac{dy}{du}\cos x\\ & = (1-u^2)\frac{d^2y}{du^2}-u\frac{dy}{du}. \end{aligned} $$ Thus, $$ \begin{aligned} \sin x \frac{d^2y}{dx^2}&-\cos x \frac{dy}{dx}+2y\sin^3 x\\ &=\sin x\Bigl((1-u^2)\frac{d^2y}{du^2}-u\frac{dy}{du}+u\frac{dy}{du}+2y(1-u^2)\Bigr)\\ & = \sin x(1-u^2)\Bigr(\frac{d^2y}{du^2}+2y\Bigr). \end{aligned} $$ Can you continue from here?