Solve THIS ODE $ (2x-y\sin(2x))dx = (\sin^2x-2y)dy $ The following problem is in Mathematical Methods in physical science CH8 miscellaneous problems
$$
(2x-y\sin(2x))dx = (\sin^2x-2y)dy
$$
it isnt an exact equation the only difference is $-$ sign
because it not an exact equation I tried to rearrange it to produce linear first order but it did not work
I dont know what to do
I assume there an integration factor for such problem to make it exact or methods
Also I tried to use Mathematica DSolve function to find a solution
DSolve[{y'[x] == (2 x - y[x] sin (2 x))/(sin^2 x - 2 y[x])}, y[x], x]

This what it produced
DSolve::deqn: Equation or list of equations expected instead of True in the first argument {True}.
DSolve[{True}, y[x], x]

thank you

Since $d[x^2+y^2-y\sin^2x]=(2x-y\sin2x)dx+(2y-\sin^2x)dy=0$, the solution is $x^2+y^2-y\sin^2x=c$.
