Solve a nonlinear differential equation of the first order

This is not an answer to the question but a complement to the Robert Israel's answer. It was not possible to edit it in the comments section.

$$y(x) = \sqrt {(c-c^2 )(2\,x-1)}-cx+c \tag 1$$ is not the complete set of solutions. One must not forget $$y(x) = -\sqrt {(c-c^2 )(2\,x-1)}-cx+c \tag 2$$ Among them two are trivial :

$y(x)=0\quad$ corresponding to $c=0$ ,

$y(x)=1-x\quad$ corresponding to $c=1$ .

The map drawn below shows the curves corresponding to Eqs.$(1)$ and $(2)$. The small numbers written on the curves are the values of $c$.

The envelops of the set of curves are also solutions. They are four of them :

Two already given by WA :

$$y(x) = \frac12$$ $$y(x) =\frac12-x$$ The third and fourth are discutable (not given by WA) : $$x(y)=\frac12$$ $$x(y)=0$$ In fact these solutions results from the transformation of the ODE : $$x(2x-1)\left(\frac{dy}{dx}\right)^2-(2x-1)(2y-1)\frac{dy}{dx}+y(2y-1)=0.$$ into : $$x(2x-1)-(2x-1)(2y-1)\frac{dx}{dy}+y(2y-1)\left(\frac{dx}{dy}\right)^2=0.$$ which avoid to forget the solutions to which $\frac{dy}{dx}$ is infinite, i.e. the vertical lines $\frac{dx}{dy}=0$ at $x=\frac12$ and $x=0$ .

This problem is about isotomic transversals:

In general, each point of the plane is on a pair of isotomic transversals with respect to some given triangle $ABC$. So is defined a field of directions in the plane. Write down the differential equation linked to this field and integrate it.