TITLE: Solving a differential equation
QUESTION [9 upvotes]: 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.

REPLY [3 votes]: These solutions are the same as ${\displaystyle u = {3 \pm \sqrt{17} \over 2}}$, which are the constant solutions to ${\displaystyle x{du \over dx} = {-u^2 + 3u + 2 \over u - 2}}$. They dissappear when you write ${\displaystyle {1 \over x} dx = {2 - u \over u^2 - 3u - 2}du}$ because they correspond to when the denominator of the right-hand side is identically zero. 
In this sort of situation you have to consider such constant values of $u$ separately and check to see if they correspond to solutions of the original differential equation, and they often do.