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:


 * $(1): \quad \left({y^2 e^{x y} + \cos x}\right) \mathrm d x + \left({e^{x y} + x y e^{x y} }\right) \mathrm d y = 0$

is an exact differential equation with solution:


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

Proof
Let:
 * $M \left({x, y}\right) = y^2 e^{x y} + \cos x$
 * $N \left({x, y}\right) = e^{x y} + x y e^{x y}$

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

where:

Hence:

and:

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

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


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