Why do some of the solutions of $y'=(1-y)(2-y)$ have a vertical asymptote?

Question

I wanted to sketch the family of solutions of the differential equation $y'=(1-y)(2-y)$. I first did it without solving the equation, by sketching a "vector field", something like this:

So from that I successfully sketched the solutions that lie in between $y=1$ and $y=2$ (have $1

Now of course, as soon as I solved the equation, I saw the error in my solutions. And after seeing the StreamPlot, I recognize that the gradients from the exponential do not quite fit with the gradients from the differential equation. Note that part of my question requires that this is done by hand, I'm only using these computer generated plots to help make my question more clear.

My problem was the vertical asymptotes, the horizontal ones were so easy to see that I expected a vertical asymptote would be the same, however it's not. So now the question is, provided that $y'$ does not go to $\infty$ as $x$ or $y \rightarrow a$, where $a$ is some real number, how could I have known that there is a vertical asymptote without solving the equation? Since the method I'm using here are usually employed where the solutions cannot be optained explicitly.

Answer 1

One has $y'=(y-1)(y-2)>0$ when $y>2$, hence in the half plane $y>2$ the solutions are horizontally translated copies of a single monotonically increasing curve $\gamma:\>x\mapsto \eta(x)$. If the right hand side of the ODE would be essentially linear in $y$ then the increase would be exponential in $x$, and the model function $\eta(\cdot)$ would live for $-\infty

Consider as model example the IVP $${dy\over dt}=y^2,\qquad y(0)=y_0>0\ .$$ Its solution is given by $$y(t)={1\over{1\over y_0}-t}\qquad\left(-\infty