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

Theorem
The first order ODE:
 * $(1): \quad y \, \mathrm d x + x \, \mathrm d y + 3 x^3 y^4 \mathrm d y = 0$

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

This can also be presented in the form:
 * $\dfrac {\mathrm d y} {\mathrm d x} + \dfrac y {3 x^3 y^4 + x} = 0$

Proof
We note that $(1)$ is in the form:
 * $M \left({x, y}\right) \mathrm d x + N \left({x, y}\right) \mathrm d y = 0$

but that $(1)$ is not exact.

So, let:
 * $M \left({x, y}\right) = y$
 * $N \left({x, y}\right) = 3 x^3 y^4 + x = x \left({3 x^2 y^4 + 1}\right)$

Let:
 * $P \left({x, y}\right) = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

Thus:

It can be observed that:

Thus $\dfrac {P \left({x, y}\right)} {M \left({x, y}\right)}$ 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:
 * $\mu \left({x y}\right) = \mu \left({z}\right) = e^{\int g \left({z}\right) \mathrm d z}$

Hence:

Thus an integrating factor for $(1)$ has been found:
 * $\mu = \dfrac 1 {x^3 y^3}$

which yields, when multiplying it throughout $(1)$:
 * $\dfrac 1 {x^3 y^2} \, \mathrm d x + \dfrac 1 {x^2 y^3} \, \mathrm d y + 3 y \, \mathrm d y = 0$

which is now exact.

Let $M$ and $N$ be redefined as:


 * $M \left({x, y}\right) = \dfrac 1 {x^3 y^2}$
 * $N \left({x, y}\right) = \dfrac 1 {x^2 y^3} + 3 y$

By Solution to Exact Differential Equation, the solution to $(1)$ is:
 * $f \left({x, y}\right) = C$

where:

Hence:

and:

Thus:
 * $f \left({x, y}\right) = -\dfrac 2 {x^2 y^2} + \dfrac {3 y^2} 2$

and by Solution to Exact Differential Equation, the solution to $(1)$ is:


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