First Order ODE/(y + y cosine x y) dx + (x + x cosine x y) dy = 0

Theorem
The first order ordinary differential equation:


 * $(1): \quad \paren {y + y \cos x y} \rd x + \paren {x + x \cos x y} \rd y = 0$

is an exact differential equation with solution:


 * $x y + \sin x y = C$

This can also be presented as:
 * $\dfrac {\d y} {\d x} + \dfrac {y + y \cos x y} {x + x \cos x y} = 0$

Proof
Let:
 * $\map M {x, y} = y + y \cos x y$
 * $\map N {x, y} = x + x \cos x y$

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

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


 * $x y + \sin x y = C$