First Order ODE/exp x sine y dx + exp x cos y dy = y sine x y dx + x sine x y dy

Theorem
The first order ordinary differential equation:


 * $(1): \quad e^x \sin y \, \mathrm d x + e^x \cos y \, \mathrm d y = y \sin x y \, \mathrm d x + x \sin x y \, \mathrm d y$

is an exact differential equation with solution:


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

Proof
Let $(1)$ be expressed as:
 * $\left({e^x \sin y - y \sin x y}\right) \, \mathrm d x + \left({e^x \cos y - x \sin x y}\right) \, \mathrm d y = 0$

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

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

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


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