First Order ODE/(sine x sine y - x e^y) dy = (e^y + cosine x cosine y) dx

Theorem
The first order ordinary differential equation:


 * $(1): \quad e^y + \cos x \cos y = \left({\sin x \sin y - x e^y}\right) \dfrac {\mathrm d y} {\mathrm d x}$

is an exact differential equation with solution:


 * $\sin x \cos y + x e^y = C$

This can also be presented as:
 * $\dfrac {\mathrm d y} {\mathrm d x} = \dfrac {e^y + \cos x \cos y} {\sin x \sin y - x e^y}$

Proof
First express $(1)$ in the form:
 * $(2): \quad -\left({e^y + \cos x \cos y}\right) + \left({\sin x \sin y - x e^y}\right) \dfrac {\mathrm d y} {\mathrm d x}$

Let:
 * $M \left({x, y}\right) = -\left({e^y + \cos x \cos y}\right)$
 * $N \left({x, y}\right) = \sin x \sin y - x e^y$

Then:

Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.

By Solution to Exact Differential Equation, the solution to $(1)$ is:
 * $f \left({x, y}\right) = C$

where:

Hence:

and:

Thus:
 * $f \left({x, y}\right) = -\sin x \cos y - x e^y$

and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is:


 * $\sin x \cos y + x e^y = C$