First Order ODE/y dx + x dy + 3 x^3 y^4 dy

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Theorem

The first order ODE:

$(1): \quad y \rd x + x \rd y + 3 x^3 y^4 \rd y = 0$

has the general solution:

$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$


This can also be presented in the form:

$\dfrac {\rd y} {\rd x} + \dfrac y {3 x^3 y^4 + x} = 0$


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:

$\map M {x, y} = y$
$\map N {x, y} = 3 x^3 y^4 + x = x \paren {3 x^2 y^4 + 1}$

Let:

$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

Thus:

\(\ds \map P {x, y}\) \(=\) \(\ds 1 - \paren {9 x^2 y^4 + 1}\)
\(\ds \) \(=\) \(\ds -9 x^2 y^4\)


It can be observed that:

\(\ds \frac {\map P {x, y} } {\map N {x, y} y - \map M {x, y} x}\) \(=\) \(\ds \frac {-9 x^2 y^4} {\paren {3 x^3 y^4 + x} y - y x}\)
\(\ds \) \(=\) \(\ds -\frac {-9 x^2 y^4} {3 x^3 y^5}\)
\(\ds \) \(=\) \(\ds -\frac 3 {x y}\)

Thus $\dfrac {\map P {x, y} } {\map M {x, y} }$ is a function of $x y$.

So Integrating Factor for First Order ODE: Function of Product of Variables can be used.

Let $z = x y$.

Then:

$\map \mu {x y} = \map \mu z = e^{\int \map g z \rd z}$

Hence:

\(\ds \int \map g z \rd z\) \(=\) \(\ds \int -\frac 3 z \rd z\)
\(\ds \) \(=\) \(\ds -3 \ln z\)
\(\ds \) \(=\) \(\ds \map \ln {z^{-3} }\)
\(\ds \leadsto \ \ \) \(\ds e^{\int \map g z \rd z}\) \(=\) \(\ds \frac 1 {z^3}\)
\(\ds \) \(=\) \(\ds \frac 1 {x^3 y^3}\)

Thus an integrating factor for $(1)$ has been found:

$\mu = \dfrac 1 {x^3 y^3}$

which yields, when multiplying it throughout $(1)$:

$\dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$

which is now exact.


By First Order ODE: $\dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$, its solution is:

$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$

$\blacksquare$


Sources