Find a solution to the ODE?

Question

$dy/dx = x\cos2x/3y^2$ So far I've rearranged.

$dy3y^2 = x\cos2x dx$

Then do I just solve for $y$? If so how do I do that? I'm just a little confused on the next steps.

Thanks for any help.

Answer 1

Suppose that we write it as : $y'(x) = \frac{xcos(2x)}{y^(x)} $

Hint : First multiply both sides with : $3y^2(x)$ and then integrate both sides with respect to x. You will then find $y(x)$.