Answer
h
$y′=y(3-y)$
Work Step by Step
The directional field for question 19 shows a field with an equilibrium solution at $y=3$ and another at $y=0$; it converges as $t →∞$, for $y_0\gt0$, to $y=3$. For $y_0\lt0$, $y→-∞$. Because there are two equilibrium solutions, it can be inferred that the differential equation is not of the form $\frac{dy}{dt}=(a+by)^n$, this is because equations in this form can have a single equilibrium solution. This rules out answers (a), (b), (c), (f), (g), (i)and (j). All remaining answer choices are of the form $\frac{dy}{dt}=(a+by)(c+dy)$.
Answer choice (d) does not have an equilibrium solution at $y=3$, so it is ruled out.
The answer is either (e) or (h).
Since the field is divergent for $y_0\lt0$, this must mean that $y′\lt0$ for this space; this is only true for answer choice (h).
