(1) Some diseases spread through a (finite) population P at a rate proportional to the number of
people, N(t), infected by time t and the number, P − N(t), not yet infected. Thus N ′(t) = kN(P − N)
and, for some positive c and k,
(2) A rumor (or fad or new religious cult) often spreads through a population P according to the
formula in (1), where N(t) is the number of people who have heard the rumor (acquired the fad,
converted to the cult), and P − N(t) is the number who have not.
(3) Bacteria in a culture on a Petri dish grow at a rate proportional to the product of the existing
population and the difference between the maximum sustainable population and the existing
population. (Replace bacteria on a Petri dish by fish in a small lake, ants confined to a small
receptacle, fruit flies supplied with only a limited amount of food, yeast cells, and so on.)
BC ONLY
(4) Advertisers sometimes assume that sales of a particular product depend on the number of TV
commercials for the product and that the rate of increase in sales is proportional both to the existing
sales and to the additional sales conjectured as possible.
(5) In an autocatalytic reaction a substance changes into a new one at a rate proportional to the
product of the amount of the new substance present and the amount of the original substance still
unchanged.
EXAMPLE 20
Because of limited food and space, a squirrel population cannot exceed 1000. It grows at a rate
proportional both to the existing population and to the attainable additional population. If there
were 100 squirrels 2 years ago, and 1 year ago the population was 400, about how many
squirrels are there now?
SOLUTION: Let P be the squirrel population at time t. It is given thatwith P(0) = 100 and P(1) = 400. We seek P(2).
We will find the general solution for the given d.e. (3) by separating the variables: