Common point of solutions of $(\varepsilon-x)y=y'(-x+y^2-2x^2)$ Let there be the same equation as here.
$(\varepsilon-x)y=y'(-x+y^2-2x^2)$
@JJacquelin found the integrating factor
$$\boxed{\mu=\frac{1}{(x+2\epsilon x-y^2)(\epsilon +2\epsilon x-y^2)\:y}}\tag 2$$
The implicit answer is 
$$\boxed{2\epsilon\ln\left(|x+2\epsilon x-y^2| \right)-(1+2\epsilon)\ln\left(|\epsilon +2\epsilon x-2y^2| \right)+2\ln(|y|)=C}$$ for $\varepsilon \neq 0 , -\frac{1}{2}$
I have graphed the parabolas where $\mu$ is undefined, took $\varepsilon = 6.5$ and found that all the solutions intersect in one special point $(0.4, 2.4)$. That holds for all positive $\varepsilon$, though the point is moving along the parabola $y^2 = (2\varepsilon+1)x$. 
What is this special point where all the solutions meet, where does it come from?
In the picture you see two solutions for $\varepsilon = 2, C = -1, 0$


The conditions of the existence and uniqueness theorem do not hold when $-x + y^2 -2 x^2 = 0$. Substituting this into the ODE gives
$$(x, y) \in
 \left\{ (-1/2, 0), (0, 0),
 \left( \epsilon, -\sqrt {\smash[b] {\epsilon (1 + 2 \epsilon)}} \right),
 \left( \epsilon, \sqrt {\smash[b] {\epsilon (1 + 2 \epsilon)}} \right) \right\}.$$
Or, since the ODE is $f(x, y) dx = g(x, y) dy$, one can analyze the equilibrium points of the system $(\dot x, \dot y) = (g(x, y), f(x, y))$.
There is a mismatch with the plot because of a typo in the first integral.
