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
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.7$: Homogeneous Equations: Problem $1 \ \text{(b)}$