hints on solving $\sin^2 x {d^2y \over dx^2} = 2 y$

How to solve this differentiation equation? $$\sin^2 x {d^2y \over dx^2} = 2 y$$ I don't know how to begin. Can it be any simpler than this?

-

$$y \left( x \right) ={\frac {c_{{1}} \left( \cos \left( 2\,x \right) +1 \right) }{\sin \left( 2\,x \right) }}+{\frac {c_{{2}} \left( x\cos \left( 2\,x \right) -\sin \left( 2\,x \right) +x \right) }{\sin \left( 2\,x \right) }}$$
I think it is better to exhibit the solution as $$y\left( x \right) = {c_1}\cot x + {c_2}\left( {x\cot x - 1} \right)$$ – Pedro Tamaroff Jul 27 '12 at 19:06
@PeterTamaroff how did you get that? solution according to book is $$y \tan (x) = c_1(\tan x - x) + c_2$$ same as yours – Santosh Linkha Jul 27 '12 at 19:09
I just rewrote Robert's solution using $\cot(2x)+\csc(2x)=\cot(x)$ – Pedro Tamaroff Jul 27 '12 at 19:11