First Order ODE/(y^2 - 3 x y - 2 x^2) dx = (x^2 - x y) dy

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Theorem

The first order ordinary differential equation:

$(1): \quad \paren {y^2 - 3 x y - 2 x^2} \rd x = \paren {x^2 - x y} \rd y$

is a homogeneous differential equation with solution:

$y^2 x^2 - 2 y x^3 + x^4 = C$


Proof

$(1)$ can also be rendered:

$\dfrac {\rd y} {\rd x} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$


Let:

$\map M {x, y} = y^2 - 3 x y - 2 x^2$
$\map N {x, y} = x^2 - x y$


Put $t x, t y$ for $x, y$:

\(\ds \map M {t x, t y}\) \(=\) \(\ds \paren {t y}^2 - 3 t x t y - 2 \paren {t x}^2\)
\(\ds \) \(=\) \(\ds t^2 \paren {y^2 - 3 x y - 2 x^2}\)
\(\ds \) \(=\) \(\ds t^2 \, \map M {x, y}\)
\(\ds \map N {t x, t y}\) \(=\) \(\ds \paren {t x}^2 - t x t y\)
\(\ds \) \(=\) \(\ds t^2 \paren {x^2 - x y}\)
\(\ds \) \(=\) \(\ds t^2 \, \map N {x, y}\)

Thus both $M$ and $N$ are homogeneous functions of degree $2$.

Thus, by definition, $(1)$ is a homogeneous differential equation.


By Solution to Homogeneous Differential Equation, its solution is:

$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$

where:

$\map f {x, y} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$


Hence:

\(\ds \ln x\) \(=\) \(\ds \int \frac {\d z} {\frac {z^2 - 3 z - 2} {1 - z} - z} + C_1\)
\(\ds \) \(=\) \(\ds \int \frac {1 - z} {2 z^2 - 4 z - 2} \rd z + C_1\)
\(\ds \) \(=\) \(\ds -\frac 1 4 \int \frac {4 z - 4} {2 z^2 - 4 z - 2} \rd z + C_1\)
\(\ds \) \(=\) \(\ds -\frac 1 4 \map \ln {2 z^2 - 4 z - 2} + C_1\) Primitive of Function under its Derivative


Substituting back for $z$ and tidying up, the result is obtained:

$y^2 x^2 - 2 y x^3 + x^4 = C$

$\blacksquare$


Sources

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