\section{First Order ODE/(y - x^3) dx + (x + y^3) dy = 0}
Tags: Exact Differential Equations, Examples of First Order ODE, Examples of Exact Differential Equation, Examples of First Order ODEs

\begin{theorem}
The first order ordinary differential equation:
:$(1): \quad \paren {y - x^3} \rd x + \paren {x + y^3} \rd y = 0$
is an exact differential equation with solution:
:$4 x y - x^4 + y^4 = C$
This can also be presented as:
:$\dfrac {\d y} {\d x} + \dfrac {y - x^3} {x + y^3} = 0$
\end{theorem}

\begin{proof}
Let:
:$\map M {x, y} = y - x^3$
:$\map N {x, y} = x + y^3$
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial M} {\partial y}
      | r = 1
      | c = 
}}
{{eqn | l = \dfrac {\partial N} {\partial x}
      | r = 1
      | c = 
}}
{{end-eqn}}
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:
:$\map f {x, y} = C$
where:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial f} {\partial x}
      | r = \map M {x, y}
      | c = 
}}
{{eqn | l = \dfrac {\partial f} {\partial y}
      | r = \map N {x, y}
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = f
      | r = \map M {x, y} \rd x + \map g y
      | c = 
}}
{{eqn | r = \int \paren {y - x^3} \rd x + \map g y
      | c = 
}}
{{eqn | r = x y - \dfrac {x^4} 4 + \map g y
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = f
      | r = \int \map N {x, y} \rd y + \map h x
      | c = 
}}
{{eqn | r = \int \left({x + y^3}\right) \rd y + \map h x
      | c = 
}}
{{eqn | r = x y + \dfrac {y^4} 4 + \map h x
      | c = 
}}
{{end-eqn}}
Thus:
:$\map f {x, y} = x y - \dfrac {x^4} 4 + \dfrac {y^4} 4$
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 $C = 4 C_1$:
:$4 x y - x^4 + y^4 = C$
{{qed}}
\end{proof}

