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

Theorem

The first order ODE:

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

has the general 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:

\(\ds \int \map P y \rd y\)

\(=\)

\(\ds \int -\dfrac 1 y \rd y\)

\(\ds \)

\(=\)

\(\ds -\ln y\)

\(\ds \)

\(=\)

\(\ds \ln y^{-1}\)

\(\ds \leadsto \ \ \)

\(\ds e^{\int P \rd y}\)

\(=\)

\(\ds \dfrac 1 y\)

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

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.10$: Problem $4 \ \text{(b)}$