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? 
 A: HINT.
This is a Bernoulli differential equation, that is an ODE of the form
$$y'(x) + p(x) y(x) = q(x) y^n(x)$$
Where $p(x)$ and $q(x)$ are functions.
You can easily transform yours into a Bernoulli type, and the next step will be to use this substitution:
$$z = y^{1-n}$$
In your case, $n = 3$.
After this substitution, your ODE will turn into the form
$$z'(x) + (1-n)p(x) z(x) = (1-n)q(x)$$
Which is then a simple first order linear differential equation, which has the form
$$z'(x) + Q(x)z(x) = f(x)$$
it's straightforward to connect the coefficient to your terms.
Remember then, that the solution of the above ODE is given by
$$\large z(x) = \frac{1}{e^{\int Q(x)\ dx}}\left(\int e^{\int Q(x)\ dx}\ f(x)\ dx + C\right)$$
A: Hint
Let $y=\frac 1 {\sqrt z}$, replace and simplify to get
$$z'\sin (x) -z \cos (x)=\sin (x)-x \cos (x)$$ which looks much more pleasant to work.
I am sure that you can take it from here.
A: Hint
Substitute $u=\frac 1 {y^2}$
Then $$u'=-2 \frac {y'} {y^3}$$
$$2y'\sin x  + y\cos x = y^3(x\cos x  - \sin x)$$
$$-u'  + u\cot (x) = (x\cot(x)  - 1)$$
$$ {u'}  -u\cot (x) = (1-x\cot(x) )$$
$$.............$$
