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 $(3)$:

\(\ds \frac y x\) \(=\) \(\ds \paren {1 - 2} \int -\dfrac 1 {y^2} y \rd y\)
\(\ds \) \(=\) \(\ds \int \dfrac 1 y \rd y\)
\(\ds \) \(=\) \(\ds \ln y + C\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds x \ln y + C x\)

$\blacksquare$