I don't know why these simple homogeneous equations are causing me so many problems, but they're kicking the teddy bear stuffing out of me.

This is the problem I'm looking at:

$\displaystyle \frac{dy}{dx} = - \frac{4x + 3y}{2x + y}$

Book Answer:

$\displaystyle \left\vert{} y + x \right\vert{} \left\vert{} y + 4x \right\vert{}^2 = c$

Here's what I'm doing:

$\displaystyle v + x \frac{dv}{dx} = - \frac{4x + 3xv}{2x + xv} = - \frac{x(4+3v)}{x(2 + v)} = - \frac{4 + 3v}{2 + v}$

$\displaystyle x \frac{dv}{dx} = - \frac{4 + 3v}{2 + v} - \frac{v(2 + v)}{2 + v} = \frac{-4 -3v}{2 + v} + \frac{-2v - v^2}{2 + v} $

$\displaystyle x \frac{dv}{dx} = \frac{- v^2 - 5v - 4}{2 + v} = - \frac{v^2 + 5v + 4}{2 + v} $

$\displaystyle x \frac{dv}{dx} = - \frac{(v + 1)(v + 4)}{2 + v}$

$\displaystyle \frac{1}{x} dx = - \frac{2 + v}{(v + 1)(v + 4)} $

$\displaystyle \frac{2 + v}{(v + 1)(v + 4)} = \frac{a}{v + 1} + \frac{b}{v + 4}$

$\displaystyle a = \frac{1}{3}$

$\displaystyle b = \frac{2}{3}$

$\displaystyle \frac{1}{x} dx = - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv$

$\displaystyle \int \frac{1}{x} dx = \int - \frac{1}{3} * \frac{1}{v + 1} - \frac{2}{3} * \frac{1}{v + 4} dv$

$\displaystyle ln \left\vert{} x \right\vert{} = - \frac{1}{3} ln \left\vert{} v \right\vert{} - \frac{2}{3} ln \left\vert{} v \right\vert{} + c = -ln \left\vert{} v \right\vert{} + c$

$\displaystyle e^{ln \left\vert{} x \right\vert{}} = e^{-ln \left\vert{} v \right\vert{} + c} = e^{-ln \left\vert{} v \right\vert{}} e^c = v^{-1} c$

$\displaystyle x = \frac{1}{v} c = c \frac{x}{y} $

...which is clearly wrong. Where am I making a mistake?