1550078481-Ordinary_Differential_Equations__Roberts_

(jair2018) #1
Applications of the Initial Value Problem y' = f(x, y); y(c) = d 131

where Bis a positive constant. The term -By represents the process of


forgetting and since B > 0 it produces a decrease in attainment. Solve

numerically and graph the learning curve of the initial value problem

I = .06(50 - y) _ 02.

y 1 + .05t. y, y(O) =^0


on the interval [O, 50]. When does y(t) = 25? When does y(t) = 37?

Compare your results with the results for example 1 and exercise l.

What is the maximum value of y(t)?

3.3 Population Models


A central figure in the history of population growth modelling is Thomas
Robert Malthus (1766-1834). Malthus was the second of eight children of an
English country gentleman. He graduated from Cambridge University and
in 1788 was ordained as a minister in the Church of England. In the first
edition of his essay on population, which was published in 1798, Malthus
noted that the population of Europe doubled at regular intervals. Further
research indicated that the rate of increase of the population was proportional
to the present population. So, Malthus formulated what we presently call the
Malthusian population model:

(1) dP =kP
dt

where k > 0.

In this model P(t) represents the population at time t and the positive con-
stant of proportionality k represents the "growth rate" per individual. If at
some time to we determine that the population has size Po, then the solu-
tion at any time t of the differential equation ( 1) which satisfies the initial
condition P(to) = Po, is

(2) P(t) = Poek(t-to).

Verify that (2) satisfies the initial value problem

(3) dP = kP.

dt '

P(to) =Po


for any constant k. Observe that k need not be positive. If we let b represent
the "birth rate" per individual and d represent the "death rate" per individual,
then the constant k = b - d represents the "growth rate" per individual of
the population. A graph of the solution (2) of the initial value problem (3)


for k > 0, k = 0, and k < 0 is shown in Figure 3.9. Observe for k > 0

the population grows exponentially and is unbounded- that is, as t -t oo,
P(t) -t oo. Fork = 0 the population maintains the constant value Po. And

for k < 0 the population decreases exponentially to zero.
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