566
CHAPTER 7
.
.
FirstOrder Differential Equations
72
we return to the topic of differential equations and present additional examples. However,
a more thorough examination of this vast ﬁeld will need to wait for a course focused on
this topic.
100
200
300
400
4
3
2
1
Number of bacteria
(millions per ml)
Time (hours)
FIGURE 7.1
Growth of bacteria
Number of
Time
Bacteria
(hours)
(millions per ml)
0
1.2
0.5
2.5
1
5.1
1.5
11.0
2
23.0
2.5
45.0
3
91.0
3.5
180.0
4
350.0
7.1
MODELING WITH DIFFERENTIAL EQUATIONS
Growth and Decay Problems
In this age, we are all keenly aware of how infection by microorganisms such as
Escherichia
coli
(
E. coli
) causes disease. Many organisms (such as
E. coli
) produce a toxin that can
cause sickness or even death. Some bacteria can reproduce in our bodies at a surprisingly
fast rate, overwhelming our bodies’ natural defenses with the sheer volume of toxin they
are producing. The table shown in the margin indicates the number of
E. coli
bacteria
(in millions of bacteria per ml) in a laboratory culture measured at halfhour intervals
during the course of an experiment. We have plotted the number of bacteria per milliliter
versus time in Figure 7.1. What would you say the graph most resembles? If you said, “an
exponential,” you guessed right. Careful analysis of experimental data has shown that many
populations grow at a rate proportional to their current level. This is quite easily observed in
bacterial cultures, where the bacteria reproduce by binary ﬁssion (i.e., each cell reproduces
by dividing into two cells). In this case, the rate at which the bacterial culture grows is
directly proportional to the current population (until such time as resources become scarce
or overcrowding becomes a limiting factor). If we let
y
(
t
) represent the number of bacteria
in a culture at time
t
, then the rate of change of the population with respect to time is
y
±
(
t
).
Thus, since
y
±
(
t
)is proportional to
y
(
t
), we have
y
±
(
t
)
=
ky
(
t
)
,
(1.1)
for some constant of proportionality
k
(the
growth constant
). Since equation (1.1) involves
the derivative of an unknown function, we call it a
differential equation.
Our aim is to
solve
the differential equation, that is, ﬁnd the
function y
(
t
). Assuming that
y
(
t
)
>
0 (this
is a reasonable assumption, since
y
(
t
) represents a population), we have
y
±
(
t
)
y
(
t
)
=
k
.
(1.2)
Integrating both sides of equation (1.2) with respect to
t
,we obtain
±
y
±
(
t
)
y
(
t
)
dt
=
±
kdt
.
(1.3)
Substituting
y
=
y
(
t
)in the integral on the lefthand side, we have
dy
=
y
±
(
t
)
and so,
(1.3) becomes
±
1
y
=
±
.
Evaluating these integrals, we obtain
ln

y
+
c
1
=
kt
+
c
2
,
where
c
1
and
c
2
are constants of integration. Subtracting
c
1
from both sides yields
ln

y
=
+
(
c
2
−
c
1
)
=
+
c
,