First Order ODE/x dy = (y + x^2 + 9 y^2) dx
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Theorem
The first order ODE:
- $(1): \quad x \rd y = \paren {y + x^2 + 9 y^2} \rd x$
has the general solution:
- $\map \arctan {\dfrac {3 y} x} = 3 x + C$
Proof 1
Divide both sides of $(1)$ by $x^2 \rd x$ to get:
- $\dfrac 1 x \dfrac {\d y} {\d x} = \dfrac 1 x \paren {\dfrac y x } + 1 + 9 \paren {\dfrac y x}^2$
Now apply the substitution:
- $y = u x$
This implies then that:
- $\dfrac {\d y} {\d x} = u + x \dfrac {\d u} {\d x}$
Now substitute everything into $(1)$ to get:
\(\ds \dfrac 1 x \paren {u + x \dfrac {\d u} {\d x} }\) | \(=\) | \(\ds \dfrac 1 x u + 1 + 9 u^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\d u} {\d x}\) | \(=\) | \(\ds 1 + 9 u^2\) |
Now it becomes Solution to Separable Differential Equation and we end up with:
\(\ds \frac {\d u} {9 u^2 + 1}\) | \(=\) | \(\ds \d x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 3 \tan^{-1} \paren {3 u}\) | \(=\) | \(\ds x + K\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tan^{-1} \paren {3 u}\) | \(=\) | \(\ds 3 x + C\) |
Substitute back for $u$:
- $\tan^{-1} \paren {\dfrac {3 y} x} = 3 x + C$
$\blacksquare$
Proof 2
Let $z = \map \arctan {3y / x}$.
Then:
- $\dfrac {\partial z} {\partial x} = \dfrac 1 {1 + \paren {3 y / x}^2} \dfrac {-3 y} {x^2} = \dfrac {-3 y} {x^2 + 9 y^2}$
- $\dfrac {\partial z} {\partial y} = \dfrac 1 {1 + \paren {3 y / x}^2} \dfrac 3 x = \dfrac 3 {x^2 + 9 y^2}$
So:
- $\d z = \dfrac {3 x \rd y - 3 y \rd x} {x^2 + 9 y^2}$
Multiplying $(1)$ by $3$ and manipulating:
- $\dfrac {3 x \rd y - 3 y \rd x} {x^2 + 9 y^2} = 3 \rd x$
From Differential of Arctangent of Quotient:
- $\map \d {\map \arctan {\dfrac {3 y} x} } = 3 \rd x$
and so
- $\map \arctan {\dfrac {3 y} x} = 3 x + C$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.9$: Integrating Factors: Problem $4 \ \text{(e)}$