First Order ODE/y - x y' = y' y^2 exp y

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Theorem

The first order ODE:

$(1): \quad y - x y' = y' y^2 e^y$

has the solution:

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


Proof

Let $(1)$ be rearranged as:

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

Hence:

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


It can be seen that $(2)$ is a linear first order ODE in the form:

$\dfrac {\d x} {\d y} + \map P y x = \map Q y$

where:

$\map P y = -\dfrac 1 y$
$\map Q y = y e^y$

Thus:

\(\displaystyle \int \map P y \rd y\) \(=\) \(\displaystyle \int -\dfrac 1 y \rd y\)
\(\displaystyle \) \(=\) \(\displaystyle -\ln y\)
\(\displaystyle \) \(=\) \(\displaystyle \ln y^{-1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle e^{\int P \rd y}\) \(=\) \(\displaystyle \dfrac 1 y\)

Thus from Solution by Integrating Factor, $(2)$ can be rewritten as:

$\map {\dfrac {\d} {\d y} } {\dfrac x y} = e^y$

and the general solution is:

$\dfrac x y = e^y + C$

or:

$x = y e^y + C y$

$\blacksquare$


Sources