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.

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

  • $\begingroup$ 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. $\endgroup$ – Lada Dudnikova May 5 at 17:59
  • $\begingroup$ Check out another question that appeared thanks to you! math.stackexchange.com/q/3214859/477927 $\endgroup$ – Lada Dudnikova May 5 at 18:05
  • $\begingroup$ 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. $\endgroup$ – JJacquelin May 5 at 18:14
  • $\begingroup$ Oh, I see, it's me, who lost the solutions! There are "capital omegas", mirrored in x-axis and I lost the solutions inside parabolas in my original post. $\endgroup$ – Lada Dudnikova May 5 at 18:42
  • $\begingroup$ The sign in $\int \partial F/\partial x \, dx$ is reversed, and the second logarithm should be $\ln(\epsilon + 2 \epsilon x - y^2)$. $\endgroup$ – Maxim May 31 at 14:01

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