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