Solve a nonlinear differential equation of the first order I am trying to solve a nonlinear differential equation of the first order that comes from a geometric problem ; $$x(2x-1)y'^2-(2x-1)(2y-1)y'+y(2y-1)=0.$$
edit1   I am looking for human methods to solve the equation 
edit2  the geometric problem was discussed on this french forum http://www.les-mathematiques.net/phorum/read.php?8,1779080,1779080 
We can see the differential equation here  http://www.les-mathematiques.net/phorum/read.php?8,1779080,1780782#msg-1780782
edit 3 I do not trust formal computer programs: look at Wolfram's answer when asked to calculate the cubic root of -1 https://www.wolframalpha.com/input/?i=%7B%5Csqrt%5B3%5D%7B-1%7D%7D%29+.
 A: 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 $x=0$ .

A: 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.
A: Maple finds the solutions
$$y(x) = \frac12,\ y(x) =\frac12-x,\ y(x) =  \sqrt {(c-c^2 )(2\,x-1)}-cx+c $$
A: Not a complete solution, but you can try a polynomial solution approach:
$$y(x)=\sum_{n=0}^{m}a_nx^n$$
Plug $y(x)=a_mx^m$ in the equation to find the degree of the polynomial:
$$2m^2a_m^2-4a_m^2m+2a_m^2=0$$
$$(m-1)^2=0 \implies m=1$$
Then,
$$y(x)=a_1x+a_2$$
Plug that solution in the equation to find $a_1,a_2$
$$x(2x-1)y'^2-(2x-1)(2y-1)y'+y(2y-1)=0.$$
$$x(2x-1)a_1^2-(2x-1)(2a_1x+2a_2-1)a_1+(a_1x+a_2)(2a_1x+2a_2-1)=0.$$
Edit:
It's a polynomial in x write it like this
$$\alpha x+\beta=0$$
Since this must be true for all $x$
$$\implies \begin{cases}\alpha=0 \\ \beta=0
\end{cases}
$$
$$\implies \begin{cases}2a_1a_2-a_1+2a_2^2-a_2 =0\\ a_1(a_1+1)=0
\end{cases}
$$
The solutions of the system are:
$$S=\{(0,0),(0,\frac 1 2),(-1,\frac 1 2 ),(-1,1)\}$$
These are also solution of the DE:
$y(x)=0$. And  $y(x)=-x+1$. 
