First Order ODE/y dx + x dy + 3 x^3 y^4 dy
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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:
\(\ds \map P {x, y}\) | \(=\) | \(\ds 1 - \paren {9 x^2 y^4 + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -9 x^2 y^4\) |
It can be observed that:
\(\ds \frac {\map P {x, y} } {\map N {x, y} y - \map M {x, y} x}\) | \(=\) | \(\ds \frac {-9 x^2 y^4} {\paren {3 x^3 y^4 + x} y - y x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {-9 x^2 y^4} {3 x^3 y^5}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 3 {x y}\) |
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:
\(\ds \int \map g z \rd z\) | \(=\) | \(\ds \int -\frac 3 z \rd z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -3 \ln z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \ln {z^{-3} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{\int \map g z \rd z}\) | \(=\) | \(\ds \frac 1 {z^3}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {x^3 y^3}\) |
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:
- $-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.9$: Integrating Factors: Problem $2 \ \text{(c)}$