First Order ODE/x dy = (y + x^2 + 9 y^2) dx/Proof 1

Theorem

$(1): \quad x \rd y = \paren {y + x^2 + 9 y^2} \rd x$

$\map \arctan {\dfrac {3 y} x} = 3 x + C$

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 \ \ $

$\ds \dfrac {\d u} {\d x}$

$=$

$\ds 1 + 9 u^2$

$\ds \frac {\d u} {9 u^2 + 1}$

$=$

$\ds \d x$

$\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 $u$:

$\tan^{-1} \paren {\dfrac {3 y} x} = 3 x + C$

$\blacksquare$