First Order ODE/(y - x^3) dx + (x + y^3) dy = 0

Theorem
The first order ordinary differential equation:


 * $(1): \quad y - x^3 + \left({x + y^3}\right) \dfrac {\mathrm d y} {\mathrm d x} = 0$

is an exact differential equation with solution:


 * $4 x y - x^4 + y^4 = C$

This can also be presented as:
 * $\dfrac {\mathrm d y} {\mathrm d x} + \dfrac {y - x^3} {x + y^3} = 0$

Proof
Let:
 * $M \left({x, y}\right) = y - x^3$
 * $N \left({x, y}\right) = x + y^3$

Then:

Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.

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

where:

Hence:

and:

Thus:
 * $f \left({x, y}\right) = x y - \dfrac {x^4} 4 + \dfrac {y^4} 4$

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


 * $x y - \dfrac {x^4} 4 + \dfrac {y^4} 4 = C_1$

which can be simplified by multiplying through by $4$ and setting $C = 4 C_1$:
 * $4 x y - x^4 + y^4 = C$