Question 2.7.1: A population P(t ) exhibits logistic growth according to the...

(a) To determine where P is increasing, we require that dP/dt > 0. If P < 0, note that (1−P/75) > 0, which makes dP/dt < 0, so we need P > 0 and
(1−P/75) > 0 to make dP/dt positive. This occurs on the interval 0 < P < 75, so for these P values, P is an increasing function of t . We note further that if P > 75 or P < 0, then dP/dt < 0 and P is a decreasing function. Finally, it is evident that both P = 0 and P = 75 are equilibrium solutions, which makes sense given the physical interpretation of the population model.

(b) Using familiar commands in Maple, we can plot the direction field for this differential equation. Note in advance the behavior we expect from our work above: two equilibrium solutions at 0 and 75, plus certain increasing and decreasing behavior. Finally, note that our analysis of the equation suggests a good range of values to select for P when plotting, say, P =−10. . . 100. As always, some experimentation with t may be necessary to get a useful plot. The plot is shown in figure 2.10.

(c) To decide where P´ is increasing most rapidly, we seek the maximum value of P. Graphically, we can observe in figure 2.10 that this appears to occur approximately halfway between P = 0 and P = 75. This is reasonable in light of the physical meaning of the logistic equation, since at this point the population has accumulated some substantial numbers to increase its growth rate, while not being close enough to the carrying capacity to have its growth slowed.

We can determine this point of greatest increase in P analytically as
well. Note that P´= 0.05P(1−P/75) = 0.05P −0.000o\overline{6}P^{2}, so that P´ is determined by a quadratic function of P. We have already observed that this quadratic function has zeros at the equilibrium solutions (P = 0 and P = 75), and furthermore, we know that every quadratic function achieves is extremum (a maximum in this case, since the function g (P) = 0.05P −0.000\overline{6}P^{2} is concave down) at the midpoint of its zeros. Hence, P is maximized precisely when P = 75/2.

(d) Our final task is to solve the given initial-value problem explicitly for P.We first solve the differential equation

\frac {dP}{dt}= 0.05P (1−P/75)

for P. Note that this equation is separable and nonlinear. Separating variables, we first write

\frac {dP}{P(1−P/75)}= 0.05dt         (2.7.2)

Because the left-hand side is a rational function of P, we may use the method of partial fractions to integrate the left-hand side of (2.7.2). Observe that

\frac {1}{P(1−P/75)}= \frac {75}{P(75−P)}

Now, letting

\frac {75}{P(75−P)}=\frac {A}{P}+\frac {B}{75−P}

it follows that A = 1 and B = 1, so that (2.7.2) may now be written as

(\frac {A}{P}+\frac {B}{75−P})dp=0.05dt            (2.7.3)

Integrating both sides of (2.7.3), we find that P must satisfy the equation

ln |P|−ln |P −75| = 0.05t +C

Using a standard property of logarithms, the left-hand side may be expressed as ln |P|/|P −75|, and hence using the definition of the natural logarithm, it follows that

|\frac {P}{P −75}|= e^{0.05t+C} = Ke^{0.05t}

where K = e^{C}. Since K is an arbitrary constant, the sign of K will absorb the ± that arises from the presence of the absolute value signs, and thus we may write

\frac {P}{P −75}=Ke^{0.05t}

Multiplying both sides by P −75 and expanding, we see that

P = PKe^{0.05t} −75Ke^{0.05t}

and gathering all terms involving P on the left,

P(1−Ke^{0.05t})=−75Ke^{0.05t}

Thus, it follows that

P=\frac {−75Ke^{0.05t}}{1−Ke^{0.05t}}

Multiplying the top and bottom of the right-hand side by −1/(Ke^{0.05t} ), it follows that

P = \frac {75}{1−Me^{−0.05t}}

where M = 1/K. In this final form, it is evident that as t →∞, P(t )→75, which fits with the given carrying capacity in the original problem. At this point, we can use the initial condition P(0) = 10 to solve for M; doing so results in the equation 10 = 75/(1−M), which yields that M =−13/2, and thus

P = \frac {75}{1+ \frac {1}{3}2 e^{−0.05t}}

A plot of this function (shown in figure 2.11), along with comparison to our work throughout this example, demonstrates that our solution is correct.