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
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.9$: Integrating Factors: Problem $2 \ \text{(b)}$