# What can be said about $(\varepsilon-x)y=y'(-x+y^2-2x^2)$ solutions?

There was an unanswered question 4 years ago. OP asked for a solution of ODE $$(\varepsilon-x)y=y'(-x+y^2-2x^2)$$

The comment to the original question proposes an implicit solution, $$2\log y + 2\epsilon\log(x + 2 x\epsilon - y^2) - (1+2\epsilon)\log(\epsilon + 2 x\epsilon - y^2) = C$$

Could you explain to me how this solution can be obtained? Are there singular solutions? If there are any orthogonal families that can be described explicitly, I would like to know about them. If it can be reduced to some special function differential equation, that would be also great!

I tried to find an integration factor, but it exists only for $$\varepsilon = -\frac{1}{4}$$

Here is a graph of solution $$\pm \sqrt{\pm \frac{\sqrt{2cx^2+cx+1}}{c}-\frac{1}{c}}$$ for $$c = -1$$

It's not separable, homogeneous, solvable for $$x$$ or $$y$$ or Lagrangian, so I'm stuck.

• An implicit solution is given by $$\frac{\frac{\log \left(-y(x)^2+2 x \epsilon +x\right)}{-2 \epsilon -1}+\frac{\log \left(-y(x)^2+2 x \epsilon +\epsilon \right)}{2 \epsilon }}{\epsilon +1}-\frac{\log (y(x))}{2 \epsilon ^3+3 \epsilon ^2+\epsilon }=c_1$$ – Dr. Sonnhard Graubner May 3 at 6:50
• I think an explicit solution isn't possible. – Dr. Sonnhard Graubner May 3 at 6:51

$$(\epsilon-x)y=y'(-x+y^2-2x^2)$$ $$(\epsilon-x)y\:dx-(-x+y^2-2x^2)dy=0 \tag 1$$ The integrating factor is $$\boxed{\mu=\frac{1}{(x+2\epsilon x-y^2)(\epsilon +2\epsilon x-y^2)\:y}}\tag 2$$ Multiplying Eq.$$(1)$$ by $$\mu$$ leads to the total differential of the sought function $$F(x,y)$$ : $$\frac{ (\epsilon-x)y\:dx-(-x+y^2-2x^2)dy }{(x+2\epsilon x-y^2)(\epsilon +2\epsilon x-y^2)\:y}=0=dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy$$
$$\int \frac{\partial F}{\partial x}dx = \int \frac{ (\epsilon-x)y\:dx }{(x+2\epsilon x-y^2)(\epsilon +2\epsilon x-y^2)\:y}=\frac{1}{1+2\epsilon}\ln\left(x+2\epsilon x-y^2 \right)-\frac{1}{2\epsilon}\ln\left(\epsilon +2\epsilon x-2y^2 \right)+f(y)$$
$$\int \frac{\partial F}{\partial y}dy=\int \frac{ -(-x+y^2-2x^2)dy }{(x+2\epsilon x-y^2)(\epsilon +2\epsilon x-y^2)\:y}=\frac{1}{1+2\epsilon}\ln\left(x+2\epsilon x-y^2 \right)-\frac{1}{2\epsilon}\ln\left(\epsilon +2\epsilon x-2y^2 \right)+\frac{1}{\epsilon(1+2\epsilon)}\ln(y)+g(x)$$ The two integrals are equal to $$F(x,y)$$, thus : $$g(x)=0\quad\text{and}\quad f(y)=\frac{1}{\epsilon(1+2\epsilon)}\ln(y)$$ $$F(x,y)=\frac{1}{1+2\epsilon}\ln\left(x+2\epsilon x-y^2 \right)-\frac{1}{2\epsilon}\ln\left(\epsilon +2\epsilon x-2y^2 \right)+\frac{1}{\epsilon(1+2\epsilon)}\ln(y)$$ Since $$dF=0$$ the function $$F$$ is constant. $$\frac{1}{1+2\epsilon}\ln\left(x+2\epsilon x-y^2 \right)-\frac{1}{2\epsilon}\ln\left(\epsilon +2\epsilon x-2y^2 \right)+\frac{1}{\epsilon(1+2\epsilon)}\ln(y)=c$$ Let $$c=\frac{C}{2\epsilon(1+2\epsilon)}$$ $$\boxed{2\epsilon\ln\left(x+2\epsilon x-y^2 \right)-(1+2\epsilon)\ln\left(\epsilon +2\epsilon x-2y^2 \right)+2\ln(y)=C}$$ I confess that the result already given in the question helped me a lot to find the integrating factor.
• It's truly wonderful. But this implicit solution goes well with positive $\varepsilon$. I can't see the solution for $\varepsilon = -0.25$ and for $(-\frac{1}{2}, 0)$ in general while graphing this implicit curve. This is somehow related with the fact that the parabolas don't intersect. – Lada Dudnikova May 5 at 17:59
• Note that the $\ln(X)$ should be written $\ln|X|$ for all $\ln$. This was implicit in my answer. Sorry for the ambiguity of typing. – JJacquelin May 5 at 18:14
• The sign in $\int \partial F/\partial x \, dx$ is reversed, and the second logarithm should be $\ln(\epsilon + 2 \epsilon x - y^2)$. – Maxim May 31 at 14:01