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

Theorem
The first order ordinary differential equation:


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

is an exact differential equation with solution:


 * $x e^y + y^2 = C$

This can also be presented as:
 * $\dfrac {\mathrm d y} {\mathrm d x} + \dfrac {e^y} {x e^y + 2 y} = 0$

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

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:
 * $f \left({x, y}\right) = x e^y + y^2$

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


 * $x e^y + y^2 = C$