Homogeneous differential eqn $(e^{3y} -y) \cos(x) {dy \over dx} = e^y \sin(x)$

Question

Solve this homogeneous differential eqn $(e^{3y} -y) \cos(x) { dy\over dx} = e^y \sin(x)$

My attempt:

$$y=vx,\ {dy\over dx} =v+x {dv\over dx} $$

$$((e^{3vx} -vx) \cos(x)) \left(v+x {dv\over dx} \right) = e^{vx} \sin(x)$$

$${dv\over dx}={-1 \over x} \left({e^{vx} \sin(x) \over (e^{3vx} -vx) \cos(x)}-v \right)$$

Now, how can I separate a variables?

“Sorry, I don’t speak English well”

Answer 1

$$(e^{3y} -y) \cos(x) {{dy}\over{dx}} = e^y \sin(x)$$

Divide both sides by $e^y$

$$\frac{(e^{3y} -y)}{e^y} \cos(x) {{dy}\over{dx}} = \sin x$$

Divide both sides by $\cos x$ and take $dx$ to the right side

$$\frac{(e^{3y} -y)}{e^y} {dy} = \frac{\sin x }{\cos x}{dx}$$ $$\int \frac{(e^{3y} -y)}{e^y} {dy} = \int \frac{\sin x }{ \cos x}{dx}$$ $$\int \left(e^{2y} - \frac{y}{e^y} \right) {dy} = \int \frac{\sin x}{\cos x}{dx}$$

Answer 2

Hint: Your equation is separable: $$\frac{e^{3y}-y}{e^y}dy=\frac{\sin(x)}{\cos(x)}dx$$