OK, so, I'm supposed to solve the differential equation $\frac{dy}{dx} = \frac{y+2x}{y-2x}$ by making the substitution $y = ux$, to make the equation separable. Then $\frac{dy}{dx} = u + x\frac{du}{dx},$ which changes the equation to $\frac{1}{x} dx = \frac{2-u}{u^2-3u-2}du,$ and then by integrating I get $\ln|x|+C = \frac{1}{2\sqrt{17}} \left[ (1-\sqrt{17}) \ln\,\left| u-\frac{3+\sqrt{17}}{2}\right| - (1+\sqrt{17}) \ln\, \left|u-\frac{3-\sqrt{17}}{2} \right| \right].$
Assuming I did that correctly, my question then becomes: if I resubstitute in for $u \ldots$ is that it? I don't think I can isolate $y$ from this, and Wolfram|Alpha seems to agree with me. I mean, is this implicit solution acceptable?
This isn't the whole story, however; I've found two linear solutions, $y = \left(\frac{3 \pm \sqrt{17}}{2}\right)x,$ to the differential equation essentially by showing that $d^2y/dx^2 \rightarrow 0$ as $x\rightarrow\pm\infty$ and then subtituting $y=Ax$. Notwithstanding the fact that it yielded correct solutions, though, I can't vouch for the validity of this method, as I already knew that those were correct and invented some fanciful method for deriving them... If this has any value to it, some basic instruction would be enormously appreciated.