Fixed points for 1-D ODE I'm doing some independent work, and have managed to come across the following interesting 1-D autonomous ODE:
$\dot{x} = x(1-x) \log^2\left[\frac{x}{1-x}\right]$.
For the fixed points, i.e., where $\dot{x} = 0$, I know that the only valid one should be for $x = 1/2$ because of the Log function. However, technically, $x = 1$ and $x = 0$ also satisfy $\dot{x} = 0$. Even though the log function blows up at these points, can one not make some type of argument where since a linear function goes to zero "faster" than the log function, these other points should be fixed points too? 
Just wondering!
Thanks.

Writing
\begin{equation}
 f(x) = x(1-x) \log^2 \frac{x}{1-x},
\end{equation}
the limits $\lim_{x\to 0} f(x)$ and $\lim{x \to 1} f(x)$ both exist and yield 0. As you noticed, the singularity in the log is 'overpowered' by the fact that $x(1-x)$, which is polynomial, has roots at $x = 0$ and $x = 1$. So, $x = 0$ and $x = 1$ are both valid equilibria of the ODE $\dot{x} = f(x)$.
Fun fact: the solution to the ODE is given by (corrected):
\begin{equation}
 x(t) = \frac{e^{\frac{1}{t_0-t}}}{1 + e^{\frac{1}{t_0-t}}}
\end{equation}
