First Order ODE/x y dy = x^2 dy + y^2 dx
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Theorem
The first order ODE:
- $(1): \quad x y \rd y = x^2 \rd y + y^2 \rd x$
has the general solution:
- $y = x \ln y + C x$
Proof 1
Let $(1)$ be rearranged as:
- $(2): \quad \dfrac {\d x} {\d y} - \dfrac x y = -\dfrac {x^2} {y^2}$
It can be seen that $(2)$ is in the form:
- $\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$
where:
- $\map P y = -\dfrac 1 y$
- $\map Q y = -\dfrac 1 {y^2}$
- $n = 2$
and so is an example of Bernoulli's equation.
By Solution to Bernoulli's Equation it has the general solution:
- $(3): \quad \ds \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \map \mu y \rd y + C$
where:
- $\map \mu y = e^{\paren {1 - n} \int \map P y \rd y}$
Thus $\map \mu y$ is evaluated:
| \(\ds \paren {1 - n} \int \map P y \rd y\) | \(=\) | \(\ds \paren {1 - 2} \int -\dfrac 1 y \rd y\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {-\ln y}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \ln y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \mu y\) | \(=\) | \(\ds e^{\ln y}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds y\) |
and so substituting into $(3)$:
| \(\ds \frac y x\) | \(=\) | \(\ds \paren {1 - 2} \int -\dfrac 1 {y^2} y \rd y\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int \dfrac 1 y \rd y\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \ln y + C\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds x \ln y + C x\) |
$\blacksquare$
Proof 2
Let $(1)$ be rearranged as:
- $(2): \quad y^2 \rd x = \paren {x y - x^2} \rd y$
Let:
- $\map M {x, y} = y^2$
- $\map N {x, y} = x y - x^2$
Put $t x, t y$ for $x, y$:
| \(\ds \map M {t x, t y}\) | \(=\) | \(\ds \paren {t y}^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds t^2 \paren {y^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds t^2 \map M {x, y}\) |
| \(\ds \map N {t x, t y}\) | \(=\) | \(\ds t x t y - \paren {t x}^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds t^2 \paren {x y - x^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds t^2 \map N {x, y}\) |
Thus both $M$ and $N$ are homogeneous functions of degree $2$.
Thus, by definition, $(1)$ is a homogeneous differential equation:
- $\dfrac {\d x} {\d y} = \dfrac {x^2 - x y} {y^2}$
By Solution to Homogeneous Differential Equation, its solution is:
- $\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
- $\map f {y, x} = \dfrac {x y - x^2} {y^2}$
Hence:
| \(\ds \ln y\) | \(=\) | \(\ds \int \frac {\d z} {\frac {z - z^2} 1 - z} + C\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\int \frac {\d z} {z^2} + C\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 z + C\) | Primitive of Power | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \ln y\) | \(=\) | \(\ds \frac y x + C\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds x \ln y + C x\) | multiplying through by $x$ |
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $9$