First Order ODE/(1 - x y) y' = y^2

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Theorem

The first order ODE:

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

has the general solution:

$x y = \ln y + C$


Proof

Let $(1)$ be rearranged as:

\(\ds \frac {\d x} {\d y}\) \(=\) \(\ds \frac {1 - x y} {y^2}\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d y} + \frac x y\) \(=\) \(\ds \frac 1 {y^2}\)


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:

$\map P y = \dfrac 1 y$
$\map Q y = \dfrac 1 {y^2}$

Thus:

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

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

\(\ds \map {\dfrac \d {\d y} } {x y}\) \(=\) \(\ds \dfrac 1 y\)
\(\ds \leadsto \ \ \) \(\ds x y\) \(=\) \(\ds \int \frac {\d y} y\)
\(\ds \) \(=\) \(\ds \ln y + C\)

$\blacksquare$


Sources

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