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:


 * $(1): \quad \paren {\sin x \sin y - x e^y} \rd y = \paren {e^y + \cos x \cos y} \rd x$

is an exact differential equation with solution:


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

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

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

Let:
 * $\map M {x, y} = -\paren {e^y + \cos x \cos y}$
 * $\map N {x, y} = \sin x \sin y - x e^y$

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:
 * $\map f {x, y} = C$

where:

Hence:

and:

Thus:
 * $\map f {x, y} = -\sin x \cos y - x e^y$

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


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