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:


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

is an exact differential equation with solution:


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

Proof
Let $(1)$ be expressed as:
 * $\paren {e^x \sin y - y \sin x y} \rd x + \paren {e^x \cos y - x \sin x y} \rd y = 0$

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

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

where:

Hence:

and:

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

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


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