First Order ODE/(y^2 exp x y + cosine x) dx + (exp x y + x y exp x y) dy = 0

Theorem
The first order ordinary differential equation:


 * $(1): \quad \paren {y^2 e^{x y} + \cos x} \rd x + \paren {e^{x y} + x y e^{x y} } \rd y = 0$

is an exact differential equation with solution:


 * $y e^{x y} + \sin x = C$

Proof
Let:
 * $\map M {x, y} = y^2 e^{x y} + \cos x$
 * $\map N {x, y} = e^{x y} + x y e^{x y}$

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

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


 * $y e^{x y} + \sin x = C$