First Order ODE/(y - x^3) dx + (x + y^3) dy = 0
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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:
- $\dfrac {\d y} {\d x} + \dfrac {y - x^3} {x + y^3} = 0$
Proof
Let:
- $\map M {x, y} = y - x^3$
- $\map N {x, y} = x + y^3$
Then:
\(\ds \dfrac {\partial M} {\partial y}\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \dfrac {\partial N} {\partial x}\) | \(=\) | \(\ds 1\) |
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:
\(\ds \dfrac {\partial f} {\partial x}\) | \(=\) | \(\ds \map M {x, y}\) | ||||||||||||
\(\ds \dfrac {\partial f} {\partial y}\) | \(=\) | \(\ds \map N {x, y}\) |
Hence:
\(\ds f\) | \(=\) | \(\ds \map M {x, y} \rd x + \map g y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \paren {y - x^3} \rd x + \map g y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x y - \dfrac {x^4} 4 + \map g y\) |
and:
\(\ds f\) | \(=\) | \(\ds \int \map N {x, y} \rd y + \map h x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \left({x + y^3}\right) \rd y + \map h x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x y + \dfrac {y^4} 4 + \map h x\) |
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 $C = 4 C_1$:
- $4 x y - x^4 + y^4 = C$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.8$: Exact Equations: Problem $3$