Solving the differential equation $2y'\sin x + y\cos x = y^3(x\cos x - \sin x) $ Does anyone know how to solve the following differential equation:
$$2y'\sin x  + y\cos x = y^3(x\cos x  - \sin x) $$
I tried dividing both sides by sine, then by cosine, which in either case brought me nowhere. I tried isolating $y'$ but only ended up getting a complex expression which is seemingly impossible to integrate. Are there any other approaches here? 

Hint
Substitute $u=\frac 1 {y^2}$
Then $$u'=-2 \frac {y'} {y^3}$$
$$2y'\sin x  + y\cos x = y^3(x\cos x  - \sin x)$$
$$-u'  + u\cot (x) = (x\cot(x)  - 1)$$
$$ {u'}  -u\cot (x) = (1-x\cot(x) )$$
$$.............$$
