First Order ODE/(3 x^2 - y^2) dy - 2 x y dx = 0
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Theorem
The first order ODE:
- $(1): \quad \paren {3 x^2 - y^2} \rd y - 2 x y \rd x = 0$
has the general solution:
- $\dfrac 1 y - \dfrac {x^2} {y^3} = C$
This can also be presented in the form:
- $\dfrac {\d y} {\d x} = \dfrac {2 x y} {3 x^2 - y^2}$
Proof
We note that $(1)$ is in the form:
- $\map M {x, y} \rd x + \map N {x, y} \rd y = 0$
but that $(1)$ is not exact.
So, let:
- $\map M {x, y} = -2 x y$
- $\map N {x, y} = 3 x^2 - y^2$
Let:
- $\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
Thus:
\(\ds \map P {x, y}\) | \(=\) | \(\ds -2 x - 6 x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -8 x\) |
It can be observed that:
\(\ds \frac {\map P {x, y} } {\map M {x, y} }\) | \(=\) | \(\ds \frac {-8 x} {-2 x y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 4 y\) |
Thus $\dfrac {\map P {x, y}} {\map M {x, y} }$ is a function of $y$ only.
So Integrating Factor for First Order ODE: Function of One Variable can be used:
- $\map \mu y = e^{\int -\map h y \rd y}$
Hence:
\(\ds \int -\map h y \rd y\) | \(=\) | \(\ds \int -\paren {4 / y} \rd y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds - 4 \ln y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \ln {y^{-4} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{\int -\map h y \rd y}\) | \(=\) | \(\ds \frac 1 {y^4}\) |
Thus an integrating factor for $(1)$ has been found:
- $\mu = \dfrac 1 {y^4}$
which yields, when multiplying it throughout $(1)$:
- $\paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$
which is now exact.
By First Order ODE: $\paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$, its solution is:
- $\dfrac 1 y - \dfrac {x^2} {y^3} = C$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.9$: Integrating Factors: Problem $2 \ \text{(a)}$