First Order ODE/x y dy = x^2 dy + y^2 dx/Proof 1
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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
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 $(3)$:
\(\ds \frac y x\) | \(=\) | \(\ds \paren {1 - 2} \int -\dfrac 1 {y^2} y \rd y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac 1 y \rd y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \ln y + C\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds x \ln y + C x\) |
$\blacksquare$