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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 $\varepsilon = 2, C = -1, 0$

enter image description here

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  • $\begingroup$ On the plot that you've attached to your question, the point $(0.4, 2.4)$ is not visible. Do you mean, for example, the point $\approx(0.075,0.325)$ where dashed orange and red lines meet? $\endgroup$ – Evgeny May 7 at 15:42
  • $\begingroup$ @Evgeny Yes, I mean the point where all the solutions are touching the parabola. The picture is for $\varepsilon=2$, because IMHO it looks better. $\endgroup$ – Lada Dudnikova May 8 at 9:13
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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.

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