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

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Theorem

The first order ordinary differential equation:

$(1): \quad e^y + \paren {x e^y + 2 y} \dfrac {\d y} {\d x} = 0$


is an exact differential equation with solution:

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


This can also be presented as:

$\dfrac {\d y} {\d x} + \dfrac {e^y} {x e^y + 2 y} = 0$


Proof

Let:

$\map M {x, y} = e^y$
$\map N {x, y} = x e^y + 2 y$

Then:

\(\ds \dfrac {\partial M} {\partial y}\) \(=\) \(\ds e^y\)
\(\ds \dfrac {\partial N} {\partial x}\) \(=\) \(\ds e^y\)


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:

\(\ds \dfrac {\partial f} {\partial x}\) \(=\) \(\ds \map M {x, y}\)
\(\ds \dfrac {\partial f} {\partial y}\) \(=\) \(\ds \map N {x, y}\)


Hence:

\(\ds f\) \(=\) \(\ds \int \map M {x, y} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds \int e^y \rd x + \map g y\)
\(\ds \) \(=\) \(\ds x e^y + \map g y\)

and:

\(\ds f\) \(=\) \(\ds \int \map N {x, y} \rd y + \map h x\)
\(\ds \) \(=\) \(\ds \int \paren {x e^y + 2 y} \rd y + \map h x\)
\(\ds \) \(=\) \(\ds x e^y + y^2 + \map h x\)

Thus:

$\map f {x, y} = x e^y + y^2$

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

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

$\blacksquare$


Sources