Question - Analysis of Equilibrium Points in a Given Phase Portrait

\text{To find the equilibrium points, set the derivatives to zero:}

\frac{dx}{dt} = xy - 3x = 0

\frac{dy}{dt} = y^2 - y - 2 = 0

\text{Solve for x:}

x(y - 3) = 0 \Rightarrow x = 0 \text{ or } y = 3

\text{Solve for y:}

(y - 2)(y + 1) = 0 \Rightarrow y = 2 \text{ or } y = -1

\text{Combining results, we have two equilibrium points:}

(x, y) = (0, -1) \text{ and } (x, y) = (0, 2)

\text{For }(x, y) = (0, -1),\text{ by substituting into the original equations, we find that the x-nullcline is vertical while the y-nullcline is horizontal.}

\text{For }(x, y) = (0, 2),\text{ by the same substitution, the nature of the nullclines is the same.}

\text{To illustrate the solution curve starting at }(3,1), \text{we observe the phase portrait and sketch the path that the solution would take based on the arrow directions given in the phase portrait. The exact path needs numerical or analytical solutions to the differential equations but is not shown in this step-by-step format.}