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