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

Theorem
The first order ordinary differential equation:


 * $(1): \quad \left({y + y \cos x y}\right) \mathrm d x + \left({x + x \cos x y}\right) \mathrm d y = 0$

is an exact differential equation with solution:


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

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

Proof
Let:
 * $M \left({x, y}\right) = y + y \cos x y$
 * $N \left({x, y}\right) = x + x \cos x y$

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:
 * $f \left({x, y}\right) = C$

where:

Hence:

and:

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

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


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