Way to make this homogoneous ODE seperable? Is my algebra correct, turning this:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4y-3x}{2x-y}$$
Into this:
I split into the difference of the two fractions,
then factored x out of the left fraction, and factored y
out of the right fraction, getting:

$$\frac{4\frac{y}{x}}{x(1-\frac{y}{x})} - \frac{3\frac{x}{y}}{2\frac{x}{y}-1}$$
Now I can substitute $v = \frac{y}{x}, v^{-1}= \frac{x}{y}$, right?
Then separate? Then integrate?

The usual way of doing such a question is to divide the top and bottom of the fraction by $x$ yielding $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4\frac{y}{x} - 3}{2 - \frac{y}{x}}$$ and then use the substitition $v = \frac{y}{x} \implies y = vx$ so that $\frac{\mathrm{d}y}{\mathrm{d}x} = v + x\frac{\mathrm{d}v}{\mathrm{d}x}$ to get $$v + x\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{4v}{2-v} - \frac{3}{2-v}$$ can you take it from there? If not, you can continue by subtracting $v$ from both sides, giving you $$x \frac{\mathrm{d}v}{\mathrm{d}x} = - \frac{(v+3)(v-1)}{v-2} \implies \int \frac{2-v}{(v-1)(v+3)} \, \mathrm{d}v = \ln |x| + c$$ where partial fraction decomposition on the rational function of $v$ yields $$\frac{2-v}{(v-1)(v+3)} = \frac{1}{4(v-1)} - \frac{5}{4(v+3)}$$ Hence we get $$\frac{1}{4} \ln |v-1| - \frac{5}{4}\ln |v+3| = \ln |x| + c$$
