First Order ODE/Cosine (x + y) dx = sine (x + y) dx + x sine (x + y) dy

Theorem
The first order ordinary differential equation:
 * $\map \cos {x + y} \rd x = \map \sin {x + y} \rd x + x \map \sin {x + y} \rd y$

is an exact differential equation with solution:
 * $x \map \cos {x + y} = C$

Proof
Express it in the form:
 * $\paren {x \map \sin {x + y} - \map \cos {x + y} } \rd x + x \map \sin {x + y} \rd y$

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

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

and by Solution to Exact Differential Equation, the solution to $(1)$ is:
 * $-x \map \cos {x + y} = C_1$

or setting $C = -C_1$:
 * $x \map \cos {x + y} = C$