Modeling with Differential Equations
1. Exponential Growth and Decay models.
Definition. A quantity y(t) is said to have an exponential
growth model if it increases at a rate proportional to the
amount present. It is said to have an exponential decay model
if it decreases at a rate that is proportional to the amount
present.
In the first case, y satisfies the differential equation
dy = ky (k > 0)
dt
and in the second case it satisfies
dy =−ky (k > 0)
dt
Suppose that y satisfies an exponential growth model, and we
know that y(0) = y0 (that is we know the value of y at some time
when we begin keeping track - set to 0). Then y is the solution
of the initial value problem
dy = ky; y(0) = y
0
dt
This equation is both separable and linear. By separability we
have:
dy = kdt
y
Thus,
ln| y |= kt + C or y = Cekt
1
Substituting the initial conditions into the result gives y0 = C,
so that
y = y ekt
0
In exactly the same way we can show that the solution to the
exponential decay initial value problem
dy =−ky; y(0) = y
0
dt
is
y = y e−kt
0
Note in both cases that the proportionality constant k is still
in the final formula. Thus to get a numerical answer we will
also need to know k. This is usually accomplished by giving
one additional piece of information that lets us solve for k.
Half life and doubling time.
Suppose that a quantity y(t) is subject to exponential growth.
How long will it take to double in size? This is called the
doubling time, and it appears that this could depend on time
or on the size of y, but we will see that neither of these is
true.
If the value of y is y0 at time 0, then we want to know when the
value will be 2y0. Thus we want to solve the equation
We get
2 y = y ekt
0 0
ln2 = kt or t = ln 2
k
Thus the doubling time depends on the rate of growth k, but
not on the initial amount present or on the time when we
began to keep track of the process.
In particular, if the quantity y begins with 10 units, the time it
takes to get to 20 units is the same as the time it takes to get
from 20 to 40 or the time it takes to get from 200 to 400.
Problem. A cell of E. coli divides into two cells every 20
minutes when placed in a nutrient culture. Let y(t) be the
number of cells present at time t and suppose that the growth
of bacteria is approximated by a continuous exponential
growth model.
(a) Find an initial value problem whose solution is y(t).
(b) Find a formula for y(t).
(b) How many cells are present after 2 hrs.?
(c) How long does it take to for the number of cells to reach a
million?
(a) Clearly the initial value problem is
dy = ky; y(0) =1
dt
Now the quantity of cells doubles every twenty minute. Using
the formula for doubling time we get
20 = ln 2 or k = ln 2
k
20
so (b) y satisfies the equation
t
ln 2
y = e 20
= eln[2
t
20
t
] = 2 20
(c) After 2 hours, we have
120
y = 2 20 = 26 = 64
(d) To see when y will be 1000000, we solve
t
1000000 = 2 20
and get
ln1000000 = ln(2) t or t = 20 ln1000000 ≈ 398.63
20
ln 2
minutes.
Suppose that a quantity y(t) is subject to exponential decay.
How long will it take to half in size? This is called the half
life.
If the value of y is y0 at time 0, then we want to know when the
value will be (1/2y0. Thus we want to solve the equation
y0
= y e−kt
2 0
We get
ln 1 =−ln2 =−kt or t = ln 2
2
k
This is exactly the same formula we got for doubling
time.
Problem. Polonium 210 is a radioactive element with a half
life of 140 days. Assume that 10 milligrams of the element are
placed in a lead container and that y(t) is the number of
milligrams present t days later.
(a) Find an initial-value problem whose solution is y(t).
(b) Find a formula for y(t).
(c) How many milligrams will be present after 10 weeks?
(d) How long will it take for 70% of the original sample to
decay?
(a) Clearly the initial value problem is
dy = ky; y(0) =10
dt
The formula for half life is t = ln 2
k
so we have
140 = ln 2 or k = ln 2 ≈ 0.0050
k
140
and (b) y satisfies the equation
y =10e−0.0050t
(c) After 10 weeks or 70 days we have
y =10e−0.35 ≈ 7 milligrams.
(d) We want the time in days until the amount present is 3 mg.
Thus we need to solve the equation
3 =10e−0.0050t or .3 = e−0.0050t
The solution is ln.3 =−0.0050t or t =− ln.3 ≈ 243.2
0.0050
Problem. Assume the town of Grayrock had a population of
10000 in 1987 and 12000 in 1997. Assuming an exponential
rate of growth, in what year will the population reach 20000?
Solution. We are assuming that the population P(t) is given by a
formula
y = y ekt
0
Let us start counting in 1987 and take that as time 0. Then we
know that y0 = 10000, and so the formula for the population is
y =10000ekt
Where t is the amount of time after the year 1987. The
second piece of information is that P = 12000 in 1997, that
is when t = 10.
Thus we have 12000 =10000e10k or ln(1.2) =10k.
This leads to the result that k = ln(1.2)
10
so
ln(1.2)
t
y =10000e 10
The population will reach 20000
or
We have
ln(1.2)
t
10
when 20000 =10000e
ln(2) = ln(1.2) t
10
t =10ln(2) ≈ 38
ln(1.2)
This means that the population will reach 20000 in about
2025.
2. The logistic growth model.
Normally, populations do not grow exponentially forever, since
they must live in a system that can support no more than a
maximum of L individuals, called the carrying capacity of the
system.
The population tends to increase when it is below L , decrease
when it is above L, and stay stable when it is in equilibrium with
the system, that is when the population is L.
Logically, when the population is significantly below L, the
growth should be close to exponential. One model that meets
all these conditions is called the logistic model or the inhibited
growth model.
In this model, the population satisfies the Logistic Differential
Equation.
dy = k 1− y  y; y(0) = y (1)
0
dt  L 
Note that (assuming k to be positive) the rate of growth is
positive when y/L < 1, negative when y/L > 1, and 0 when
y = L. Moreover it is nearly exponential when y is much
smaller than L. We can rewrite the differential equation in
(1) as
dy = k ( L − y) y = k y(L − y )
dt L
L
This equation is separable and we separate the variables,
getting
Ldy = kdt
y(L − y)
We can use the method of partial fractions on the left side,
and we will see that
1 1 
∫ y + L − y dy = ∫ kdt


Thus
so
y
ln y − ln L − y = kt + C or ln
= kt +C
L− y
y
y = Aekt or
= ekt +C or
L− y
L− y
We can then get
L =1+ Ae−kt or y =
L
y
1+ Ae−kt
L − y = Ae−kt
y
Now the initial condition is that y(0) = y0. This leads to
L − y0
L
y =
, so that A =
0 1+ A
y0
This leads to the final form of the solution which is
y0 L
y=
y0 + (L − y0 )e−kt
This function has a number of forms depending on the
relationship between y0 and L.
Problem. Suppose that a population y(t) grows in accordance
with the logistic model
dy
=50 y − 0.001y 2
dt
(a) What is the carrying capacity?
(b) What is the value of k?
(c) For what value of y is the population growing most rapidly?
Solution. The standard form of the logistic equation is
dy = k y(L − y) = ky − k y2
dt L
L
Comparing this DE with the given one, we see that k = 50 and
k/L = .001. Thus means that L = 1000k = 50000. Thus we
have answered (a) and (b).
For (c), we see that the population grows most rapidly when
the derivative is a maximum. Since
dy
=50 y − 0.001y 2
dt
we must maximize 50y − 0.001y2. This maximum occurs
when
0 = d 50 y − 0.001y 2  = 50 −.002 y

dt 
or y =(50)(500)=25000.
If we substitute these numbers into the solution for the logistic
equation, we get
50000 y0
y=
y0 + (50000 − y0 )e−50t
y0 = 25000
y0 = 75000
3. Blocks and Springs
A block is attached to a vertical spring and allowed to settle
into an equilibrium position as shown. It is then set into
motion by pulling or pushing.
up
l
l - y(t)
We know by Hooke’s law that if the spring is stretched or
compressed d units from its natural position, it resists with a
force F = kd, where k is the spring constant. If m is the mass
of the block, then the weight provides a downward force of mg at all times.
Thus we see that if the equilibrium position stretches the
spring by an amount l, then we must have kl = mg.
When the block is moving and is in position y(t), the spring
provides a force of k(l − y(t)). Thus the total force on the
mass is k(l − y(t)) − mg. By Newton’s second law F = ma, so
we have the equation
my′′(t) = k (l − y(t )) − mg = kl − ky(t) − mg =−ky(t )
or y′′(t) + k y(t) = 0
m
Although this is a second order DE, it is shown in courses on
differential equations that the general solution of this equation
is
 k 
 k 
y = c cos
t + c sin
t
1
 m  2  m 




This is called simple harmonic motion.
If the block is positioned at y0 initially and released from rest,
the solution is easily determined by the conditions y(0) = y0 and
y′(0) = 0. It is
 k 
y = y cos
t
0
 m 


Two possible graphs are shown below, depending on whether
the spring is initially stretched or compressed.
y0
y0 positive
T
y0 negative
y0
The quantity
T = 2π = 2π m k
k m
is called the period of vibration. It is the time required to
complete one cycle. The frequency is 1/T or f = k m
y0 is the amplitude.
2π
Problem. Suppose a block weighs w pounds and vibrates with a
period of 3 s when it is pulled below the equilibrium position
and released. Suppose that when the process is repeated with an
additional 4 lbs of weight, then the period is 5 s.
(a) Find the spring constant
(b) Find w.
Solution. We know the weight of the block is mg and the
period of the motion is 3 = 2π m k = 2π w kg
The additional weight brings the total to w + 4, and so produces
a motion of period
5 = 2π w+ 4 kg
These equations tell us that
2
2w
3  w
4
π
 2π  = kg or k = 9 g
 
2 (w+ 4)
4
π
The other equation tells us similarly that k =
25g
2 w 4π 2 (w+ 4)
4
π
Thus
=
or 25w = 9(w+ 4)
9g
25g
so
2 w 4π 2 w π 2
9
4
π
w = ,k =
=
=
4
9g
9×32 32