If N(t) is the population at time t  0, K is the

carrying capacity (maximum sustainable population—a positive constant) and r is

the rate of maximum growth (also a positive constant), then we have Equation (1), in

which 0 < N(t)  K.

dN/dt = (rN (K-N))/K (1)

(a) By making the change of variable x(t) = N(t)/K, show that Equation (1) may

be transformed into Equation (2).

dx/dt = rx (1-x) (2)

(b) Without solving Equation (2), show that the function x(t) is always non-decreasing

for t>=0.

(c) Find the general solution of Equation (2).

(d) Given rate r = 5 and an initial population such that x(0) = 0.02, compute the

particular solution of Equation (2) for x(t).

(e) Use appropriate computer software to plot the solution of the previous part for

0>=t>=2.

(f) For the particular solution found in part (d), algebraically determine the time

when the population reaches half carrying capacity.