First Order ODE/x y dy = x^2 dy + y^2 dx/Proof 2

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:

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

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 $x$

$\blacksquare$