# 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 $$\varepsilon = 2, C = -1, 0$$ • 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? – Evgeny May 7 at 15:42
• @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. – Lada Dudnikova May 8 at 9:13

## 1 Answer

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.