# Solve $xy'(\cos(2x^3)\cos(x)-\sin(2x^3)\sin(x)) -y(6x^3+x)\sin(2x^3+x)+(12x^3+2x)\sin(2x^3+x)=0$

I need to solve this differential equation. What I'm looking for is a way to simplify this equation. Can anybody give me hints/tricks to understand the following equation better:

\begin{align*} xy'(\cos(2x^3)\cos(x)-\sin(2x^3)&\sin(x))-y(6x^3+x)\sin(2x^3+x)\\&+(12x^3+2x)\sin(2x^3+x)=0 \end{align*}

Ultimately I want to separate variables and integrate to solve for $$y(x)$$

Note that : $\cos(2x^3)\cos(x)−\sin(2x^3)\sin(x) = \cos(2x^3+x)$
It seems to me that two first term of the equation: $$xy'(\cos(2x^3)\cos(x)-\sin(2x^3)\sin(x)) - y(6x^3+x)\sin(2x^3+x)$$ is somehow a part of total differential of $$xy(\cos(2x^3+x))$$ Since $$d\left(x\cos(2x^3+x)y\right)=x\cos(2x^3+x)y'+y\left(\cos(2x^3+x)-(6x^3+x)\sin(2x^3+x)\right)$$