Modeling with Differential Equations1. 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 modelif it decreases at a rate that is proportional to the amount present.

In the first case, y satisfies the differential equation

and in the second case it satisfies

( 0)dy ky kdt

= >

( 0)dy ky kdt

=− >

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

; (0)0

dy ky y ydt

= =

This equation is both separable and linear. By separability we have: dy kdt

y=

Thus,1

ln| | or kty kt C y Ce= + =

Substituting the initial conditions into the result gives y0 = C, so that

0kty y e=

In exactly the same way we can show that the solution to the exponential decay initial value problem

; (0)0

dy ky y ydt

=− =

is

0kty y e−=

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

20 0

kty y e=We get

ln 2ln2 or kt tk

= =

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 getfrom 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

; (0) 1dy ky ydt

= =

Now the quantity of cells doubles every twenty minute. Using the formula for doubling time we get

ln 2 ln 220 or 20

kk

= =

20ln 2 20ln[2 ]20 2

t tty e e= = =

so (b) y satisfies the equation

(c) After 2 hours, we have

12020 62 2 64y= = =

(d) To see when y will be 1000000, we solve

and get

ln(2) ln1000000ln1000000 or 20 398.6320 ln 2

t t= = ≈

201000000 2

t

=

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

002

y kty e−=We get

1 ln 2ln ln2 or 2

kt tk

=− =− =

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

; (0) 10dy ky ydt

= =

The formula for half life is ln 2tk

=

0.005010 ty e−=

and (b) y satisfies the equation

(c) After 10 weeks or 70 days we have

so we haveln 2 ln 2140 or 0.0050

140k

k= = ≈

0.3510 7 milligrams.y e−= ≈

(d) We want the time in days until the amount present is 3 mg. Thus we need to solve the equation

0.0050 0.00503 10 or .3t te e− −= =

The solution is orln.3 0.0050t=− ln.3 243.20.0050

t =− ≈

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

0kty y e=

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

10000 kty e=

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.

1012000 10000 ke=Thus we have orln(1.2) 10 .k=

This leads to the result that ln(1.2)10

k =

so ln(1.2)

1010000t

y e=

The population will reach 20000 whenln(1.2)

1020000 10000t

e=

or ln(1.2)ln(2)10

t=

We have 10ln(2) 38ln(1.2)

t = ≈

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.

1 ; (0) (1)0

ydy k y y yLdt

= − =

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 kL y y y L ydt L L

= − = −

This equation is separable and we separate the variables, getting

( )Ldy kdt

y L y=

−

1 1dy kdt

y L y

+ =∫ ∫ −

We can use the method of partial fractions on the left side, and we will see that

Thusln ln or ln yy L y kt C kt C

L y− − = + = +

−

so y kt CeL y

+=−

or y ktAeL y

=−

or L y ktAey− −=

We can then get

1 or 1

L LktAe ykty Ae

−= + =−+

0, so that 0 1 0

L yLy AA y

−= =+

Now the initial condition is that y(0) = y0. This leads to

This leads to the final form of the solution which is

0( )0 0

y Lykty L y e

=−+ −

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

(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?

250 0.001dy

y ydt

= −

2( )dy k ky L y ky ydt L L

= − = −

Solution. The standard form of the logistic equation is

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).

250 0.001dy

y ydt

= −

For (c), we see that the population grows most rapidly when the derivative is a maximum. Since

we must maximize 50y − 0.001y2. This maximum occurs when 20 50 0.001 50 .002d y y y

dt = − = −

or y =(50)(500)=25000.

50000 050(50000 )0 0

yyty y e

=−+ −

If we substitute these numbers into the solution for the logistic equation, we get

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.

upl

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′′ = − − = − − =−

( ) ( ) 0ky t y tm

′′ + =or

Although this is a second order DE, it is shown in courses on differential equations that the general solution of this equation is

This is called simple harmonic motion.

1 2cos sink ky c t c t

m m

= +

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

0cos ky y t

m

=

y0 negative

y0 positive

Two possible graphs are shown below, depending on whether the spring is initially stretched or compressed.

The quantity

is called the period of vibration. It is the time required to complete one cycle. The frequency is 1/T ory0 is the amplitude.

2 2T m kk mπ π= =

y0

y0

T

2k mfπ

=

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 2 2m k w kgπ π= =

The additional weight brings the total to w + 4, and so produces a motion of period 5 2 4w kgπ= +

These equations tell us that2 23 4 or

2 9w wkkg g

ππ

= =

The other equation tells us similarly that24 ( 4)25

wkg

π +=

Thus2 24 4 ( 4) or 25 9( 4)

9 25w w w w

g gπ π += = +

so 2 2 29 4 4,4 9 9 32 32

w ww kg

π π π= = = =×