First Order ODE/x^2 y' - 3 x y - 2 y^2 = 0

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Theorem

The first order ordinary differential equation:

$(1): \quad x^2 \dfrac {\d y} {\d x} - 3 x y - 2 y^2 = 0$


is a homogeneous differential equation with solution:

$y = C x^2 \paren {x + y}$


Proof

Let:

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


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

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


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

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


By Solution to Homogeneous Differential Equation, its solution is:

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

where:

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


\(\ds \ln x\) \(=\) \(\ds \int \frac {\d z} {\dfrac {3 z + 2 z^2} {1^2} - z} + C_1\)
\(\ds \) \(=\) \(\ds \int \frac {\d z} {2 z \paren {1 + z} } + C_1\)
\(\ds \) \(=\) \(\ds \frac 1 2 \ln \frac z {z + 1} + \ln C_2\) Primitive of $\dfrac 1 {x \paren {a x + b} }$
\(\ds \leadsto \ \ \) \(\ds \ln x^2\) \(=\) \(\ds \ln \frac z {z + 1} + \ln C_3\)
\(\ds \) \(=\) \(\ds \ln \frac {C_3 z} {z + 1}\)
\(\ds \leadsto \ \ \) \(\ds x^2\) \(=\) \(\ds \frac {C_3 y / x} {y / x + 1}\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds C x^2 \paren {x + y}\) where $C := \dfrac 1 {C_3}$

$\blacksquare$


Sources