Differential equation problem!!!

Solve this initial following

[(e^x cosy+2y)/siny]dy/dx + e^x = 0

The equation gives and exact differential form.

(e^x cos(y) + 2y) dy + e^x sin(y) dx = 0.

∂/∂y e^x sin(y) = e^x cos(y) = ∂/∂x (e^x cos(y) + 2y).

So all solutions are defined implicitly by f(x,y) = C where

∂f/∂x = e^x sin(y) and ∂f/∂y = e^x cos(y) + 2y.

Integrate the latter with respect to y.

f(x, y) = ∫ e^x cos(y) + 2y dy = e^x sin(y) + y² + g(x).

Then taking the partial with respect to x

∂f/∂x = e^x sin(y) = e^x sin(y) + g '(x) ==> g'(x) = 0.

So the solutions are defined by

e^x sin(y) + y² = C.

You can't back y out of this equation, but you could express x as a function of y by

x = ln[(C - y²) csc(y)].