First Order ODE/(x y - 1) dx + (x^2 - x y) dy = 0

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Theorem

The first order ODE:

$(1): \quad \paren {x y - 1} \rd x + \paren {x^2 - x y} \rd y = 0$

has the general solution:

$x y - \ln x - \dfrac {y^2} 2 + C$


This can also be presented in the form:

$\dfrac {\d y} {\d x} + \dfrac {x y - 1} {x^2 - x y} = 0$


Proof

We note that $(1)$ is in the form:

$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$

but that $(1)$ is not exact.

So, let:

$\map M {x, y} = x y - 1$
$\map N {x, y} = x^2 - x y = x \paren {x - y}$

Let:

$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

Thus:

\(\ds \map P {x, y}\) \(=\) \(\ds x - \paren {2 x - y}\)
\(\ds \) \(=\) \(\ds -\paren {x - y}\)


It can be observed that:

\(\ds \frac {\map P {x, y} } {\map N {x, y} }\) \(=\) \(\ds \frac {-\paren {x - y} } {x \paren {x - y} }\)
\(\ds \) \(=\) \(\ds -\frac 1 x\)

Thus $\dfrac {\map P {x, y} } {\map N {x, y}}$ is a function of $x$ only.

So Integrating Factor for First Order ODE: Function of One Variable can be used:

$\map \mu y = e^{\int \map g x \rd x}$

Hence:

\(\ds \int \map g x \rd x\) \(=\) \(\ds \int -\paren {1 / x} \rd x\)
\(\ds \) \(=\) \(\ds -\ln x\)
\(\ds \) \(=\) \(\ds \map \ln {x^{-1} }\)
\(\ds \leadsto \ \ \) \(\ds e^{\int \map g x \rd x}\) \(=\) \(\ds \frac 1 x\)

Thus an integrating factor for $(1)$ has been found:

$\mu = \dfrac 1 x$

which yields, when multiplying it throughout $(1)$:

$\paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$

which is now exact.


By First Order ODE: $\paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$, its solution is:

$x y - \ln x - \dfrac {y^2} 2 + C$

$\blacksquare$


Sources