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

Theorem

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

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

Let $(1)$ be rearranged as:

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

Hence:

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

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:

$\map P y = \dfrac 2 y$

$\map Q y = \dfrac {e^y} {y^2}$

Thus:

$\ds \int \map P y \rd y$

$=$

$\ds \int \dfrac 2 y \rd y$

$\ds $

$=$

$\ds 2 \ln y$

$\ds $

$=$

$\ds \ln y^2$

$\ds \leadsto \ \ $

$\ds e^{\int P \rd y}$

$=$

$\ds y^2$

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

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

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

$\blacksquare$