Section 6-3-3

Chapter 6
Antidifferentiation

6.3 The Logistic Growth Differential Equation

6.3.3 Solution of the Logistic Growth Equation

Now we apply our antidifferential calculation to the logistic growth equation. Here is the differential equation again with variables separated and both sides ready to antidifferentiate:

An antidifferential of `(dP)/(Ptext[(]M-Ptext[)])=` some antidifferential of `k dt`.

The left-hand side has exactly the form of the left-hand side of Activity 3 - and we already knew how to antidifferentiate the right-hand side - so we are ready for the antidifferentiation step:

\frac{1}{M} \ln \frac{P}{M - P} = k t + C .

We have now solved the differential equation, in the sense that we have an equation without derivatives that relates the variables `P` and `t`. However, we don't have a completely satisfactory form of the solution yet, because we can't read off `P` as a function of `t`. The rest of the solution process consists of the necessary algebra to determine `C` from the initial condition and to solve for `P`.

First, we substitute `P=P_0` and `t=0` to satisfy the initial condition:

\frac{1}{M} \ln \frac{P_{0}}{M - P_{0}} = C .

Next, we substitute this value of `C` back into

\frac{1}{M} \ln \frac{P}{M - P} = k t + C

to get

\frac{1}{M} \ln \frac{P}{M - P} = k t + \frac{1}{M} \ln \frac{P_{0}}{M - P_{0}} .

We proceed now to a sequence of transformations of this equation that lead to a solution for `P` as a function of `t`.

$\ln \frac{P}{M - P}$	$= M k t + \ln \frac{P_{0}}{M - P_{0}}$
$\ln \frac{P}{M - P} - \ln \frac{P_{0}}{M - P_{0}}$	$= M k t$
$\ln \frac{(M - P_{0}) P}{P_{0} (M - P)}$	$= M k t$
$\frac{(M - P_{0}) P}{P_{0} (M - P)}$	$= e^{M k t}$
$(M - P_{0}) P$	$= e^{M k t} P_{0} (M - P)$
$(M - P_{0}) P$	$= P_{0} M e^{M k t} - P_{0} e^{M k t} P$
$(M - P_{0} + P_{0} e^{M k t}) P$	$= P_{0} {M e}^{M k t}$
$P$	$= \frac{P_{0} M e^{k M t}}{M - P_{0} + P_{0} e^{M k t}}$

Checkpoint 3

No wonder we couldn't guess the answer! But there it is - an explicit formula for population as a function of time, given a starting population `P_0` and an assumption of logistic growth with maximum sustainable population `M`. Is it right? For `k=0.00009`, `M=1000`, and `P_0=10` (the data for Figure 1), our formula becomes (check our arithmetic)

P = \frac{10,000 e^{0.09 t}}{990 + 10 e^{0.09 t}},

which should have a graph very similar to Figure 1.

Activity 4

Use your CAS tool to compare the graph of the symbolic solution just obtained to the graph of the numerical solution obtained earlier.
Increase `n` from `40` to `80` in the numerical solution and replot. Then increase `n` again from `80` to `160` and replot. Describe what you see.
Explain why the numerical solution turns upward later than the symbolic solution.

Comment on Activity 4

Contents for Chapter 6

Chapter 6 Antidifferentiation

6.3 The Logistic Growth Differential Equation

Chapter 6
Antidifferentiation