Consider the ODE $y' = (1-y)(2-y)$ with initial condition $y(0) = 3$. I have soved this and found the solution $$ y(x) = \frac{e^x - 4}{e^x - 2}$$ which I've checked on Wolfram Alpha. Consider the behaviour of $y(x)$ as $x \to \infty$. By simply taking the limit $y(x) \to 1$ as $x \to \infty$.
However, now consider $y'$ interpreted as a vector field. The 1 dimensional picture is something like $$--->-->->y=1<-<--<---<--<-y=2->-->---> $$ So I'd expect that solving the ODE for $y(0)=3$ would mean that $y(x) \to \infty$ as $x \to \infty$ (following the "flow line"). Where am I going wrong in my interpretation?
It does go to $+\infty$, but as $x$ approaches a finite value, namely $\ln(2)$.