In this project we explore how two populations develop when onepreys on the other. Clearly if there are no predators, the preypopulation will keep growing, whereas if there are no prey, thepredators will go extinct. Suppose x and y denote the populationsof the prey and predators respectively.

If y = 0, we will assume that

dx/dt = ax, a > 0.

If y does not equal 0, it is natural to assume that the numberof encounters between predators and prey is jointly proportional tox and y. If we further assume a proportion of these encountersleads to the prey being eaten, we have

dx/dt = dx − bxy, a, b > 0.

Similarly, we have

dy/dt = −cy + dxy, c, d > 0.

This system of equations is called Volterra’s predator-preyequations.

Part a) Solve this system of equations to findsolutions in the form g(y) = f(x). You will see that we cannotexplicitly find y in terms of x, so our solutions are implicit. Wecan still, however, study these solutions.

Part b) Suppose g(y) = C1, where C1 is aconstant. Determine how many solutions there are to this equationby using calculus techniques. Note this may well depend on thevalue of C1. Do the same thing for f(x) = C2, where C2 is aconstant.

Part c) Hence determine the shape of thetrajectories in the x, y-plane (do a sketch!), and theirdirections.

Part d) Clearly the system has a rest point atx = c/d and y = a/b. By making the substitutions x = c/d + X and y= a/b + Y , assuming X and Y are small enough that we can neglectany second order terms in X and Y , show that near the rest point,trajectories are approximately ellipses.

Part e) Finally, sketch graphs of x(t) and y(t)against t on the same axes. To help, show that d^2 * y / dt^2 >0 whenever dx/dt > 0 are think about what this means in terms ofthe shapes of the graphs.