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

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:

It can be observed that:

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:

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