Download EXAMPLE 3.1

398
Chapter 5
Integration
when x (thousand) Weenies are supplied for purchase. The current price is $2.25
per Weenie.
(a) Find the supply function p(x).
(b) At what price will 4,000 additional Weenies (x 4) be supplied?
(c) How many more Weenies will be supplied at a price of $3 per Weenie?
MARGINAL PROFIT
MARGINAL PROFIT
3
Introduction
to Differential
Equations
44. A company determines that the marginal revenue from the production of x units is
R(x) 7 3x 4x2 hundred dollars per unit, and the corresponding marginal cost
is C(x) 5 2x hundred dollars per unit. By how much does the profit change
when the level of production is raised from 5 to 9 units?
11 x
45. Repeat Problem 44 for marginal revenue R(x) and for the
x
14
2
marginal cost C(x) 2 x x .
A differential equation is an equation that involves a derivative or differential. For
example,
dy
3x2 5
dx
or
dP
kP
dt
and
dy
dx
2
3
dy
2y ex
dx
are all differential equations. Differential equations are among the most useful tools
for modeling continuous phenomena, including important situations that occur in business and economics and the social and life sciences. In this section, we introduce techniques for solving basic differential equations and examine a few practical applications.
The simplest type of differential equation has the form
dy
g(x)
dx
in which the derivative of the quantity y is given explicitly as a function of the independent variable x. Such an equation can be solved by simply finding the indefinite
integral of g(x). A complete characterization of all possible solutions of the equation
is called a general solution, and a solution that satisfies specified side conditions is
called a particular solution. This terminology is illustrated in the following examples.
EXAMPLE 3.1
Find the general solution of the differential equation
dy
x2 3x
dx
and the particular solution that satisfies y 2 when x 1.
Chapter 5 ■ Section 3
Introduction to Differential Equations
Solution
Integrating, you get
y
dy
dx dx
399
(x2 3x) dx
1
3
x3 x2 C
3
2
This is the general solution since all solutions can be expressed in this form. For the
particular solution, substitute x 1 and y 2 into the general solution:
1
3
2 (1)3 (1)2 C
3
2
1 3 1
C2 3 2 6
Thus, the required particular solution is y 1 3 3 2 1
x x .
3
2
6
EXAMPLE 3.2
The resale value of a certain industrial machine decreases over a 10-year period at a
rate that depends on the age of the machine. When the machine is x years old, the
rate at which its value is changing is 220(x 10) dollars per year. Express the value
of the machine as a function of its age and initial value. If the machine was originally worth $12,000, how much will it be worth when it is 10 years old?
Solution
dV
dx
is equal to the rate 220(x 10) at which the value of the machine is changing. Hence,
you begin with the differential equation
Let V(x) denote the value of the machine when it is x years old. The derivative
dV
220(x 10) 220x 2,200
dx
To find V, solve this differential equation by integration:
V(x) (220x 2,200) dx 110x2 2,200x C
400
Chapter 5
Integration
Notice that C is equal to V(0), the initial value of the machine. A more descriptive symbol for this constant is V0. Using this notation, you can write the general solution as
V(x) 110x2 2,200x V0
If V0 12,000, the corresponding particular solution is
V(x) 110x2 2,200x 12,000
Thus, when the machine is x 10 years old, its value is
V(10) 110(10)2 2,200(10) 12,000 $1000
V(x)
R(x)
12,000
2,200
1,000
x
10
(a) The value of the machine
V(x) = 110x2 – 2,200x + 12,000
x
10
(b) The rate of depreciation
R(x) = –220(x – 10)
FIGURE 5.2 The value of the machine and its rate of depreciation.
The negative of the rate of change of resale value of the machine
R dV
220(x 10)
dx
is called the rate of depreciation. Graphs of the resale value V of the machine and
the rate of depreciation R are shown in Figure 5.2.
SEPARABLE EQUATIONS
Many useful differential equations can be formally rewritten so that all the terms containing the independent variable appear on one side of the equation and all the terms
containing the dependent variable appear on the other. Differential equations with this
special property are said to be separable and can be solved by the following procedure involving two integrations.
Chapter 5 ■ Section 3
Introduction to Differential Equations
Separable Differential Equations
■
401
A differential equation that
can be written in the form
g(y) dy h(x) dx
Refer to Example 3.3. Store the
solution curve y (3x2 L1)^(1/3) into Y1 of the equation editor, where L1 contains
the integers {16, 12, 8,
4, 0, 4, 8, 12, 16}. Graph this
family of curves using the window [4.7, 4.7]1 by [3, 4]1
and describe what you observe.
Which of these curves passes
through the point (0, 2)?
is said to be separable. Its general solution is obtained by integrating both sides
of this equation. That is,
g(y) dy h(x) dx C
EXAMPLE 3.3
Find the general solution of the differential equation
dy 2x
2.
dx
y
Solution
dy
To separate the variables, pretend that the derivative
is actually a quotient and
dx
write
y2 dy 2x dx
Now integrate both sides of this equation to get
y2 dy 2x dx
or
1 3
y C1 x2 C2
3
where C1 and C2 are arbitrary constants
By combining constants, we can write
1 3
y x2 C3
3
where C3 C2 C1
and solve for y to get
y (3x2 3C3)1/3
or
y (3x2 C)1/3
where C 3C3
402
Chapter 5
Integration
MODELING WITH DIFFERENTIAL
EQUATIONS
The next three examples illustrate how differential equations can be used to model
situations of practical interest.
EXPONENTIAL GROWTH
AND DECAY
In Section 1 of Chapter 4, we obtained the formulas Q Q0ekt for exponential growth
and Q Q0ekt for exponential decay, and then in Section 3 of that chapter, we
showed that the rate of change of a quantity undergoing either kind of exponential
dQ
mQ for constant
change (growth or decay) is proportional to its size; that is,
dt
m. The following example shows that the converse of this result is also true.
EXAMPLE 3.4
Show that a quantity Q that changes at a rate proportional to its size satisfies Q(t) Q0emt, where Q0 Q(0).
Solution
Since Q changes at a rate proportional to its size, we have
dQ
mQ
dt
Separating the variables and integrating, we get
1
dQ Q
m dt
ln Q mt C1
and by taking exponentials on both sides
Q(t) eln Q emtC1 eC1emt
Cemt
where C eC1
Since Q(0) Q0, it follows that
Q0 Q(0) Ce0
Q0 C
so
Q(t) Q0emt
as required.
Chapter 5 ■ Section 3
LEARNING MODELS
Introduction to Differential Equations
403
In Chapter 4, we referred to the graphs of functions of the form Q(t) B Aekt as
learning curves because functions of this form often describe the relationship between
the efficiency with which an individual performs a given task and the amount of training
or experience the “learner” has had. In general, any quantity that grows at a rate
proportional to the difference between its size and a fixed upper limit is represented by
a function of this form. Here is an example involving such a learning model.
EXAMPLE 3.5
The rate at which people hear about a new increase in postal rates is proportional to
the number of people in the country who have not heard about it. Express the number of people who have heard about the increase as a function of time.
Solution
Let Q(t) denote the number of people who have heard about the increase by time t,
and let B denote the total population of the country. Then, B Q(t) people have not
heard about the increase at time t and we have
dQ
k(B Q)
dt
where k is the constant of proportionality. Separate the variables by writing
1
dQ k dt
BQ
and integrate to get
1
dQ BQ
k dt
or
ln B Q kt C
Q(t)
(Be sure you see where the minus sign came from.) This time you can drop the
absolute value sign immediately since B Q cannot be negative in this context.
Hence,
B
ln (B Q) kt C
ln (B Q) kt C
B Q ektC ekteC
B–A
t
FIGURE 5.3 A learning curve:
Q(t) B Ae
kt
.
or
Q B eCekt
404
Chapter 5
Integration
Denoting the constant eC by A and using functional notation, you can conclude that
Q(t) B Aekt
which is precisely the general equation of a learning curve. For reference, the graph
of Q is sketched in Figure 5.3.
A PRICE ADJUSTMENT MODEL
Let S(p) denote the number of units of a particular commodity supplied to the market at
a price of p dollars per unit, and let D(p) denote the corresponding number of units
demanded by the market at the same price. In static circumstances, market equilibrium
occurs at the price where demand equals supply (recall the discussion in Section 4 of
Chapter 1). However, certain economic models consider a more dynamic kind of
economy in which price, supply, and demand are assumed to vary with time. One
of these, the Evans price adjustment model,* assumes that the rate of change of price
with respect to time t is proportional to the shortage D S, so that
dp
k(D S)
dt
where k is a positive constant. Here is an example involving this model.
EXAMPLE 3.6
Suppose the price p(t) of a particular commodity varies in such a way that its rate of
change with respect to time is proportional to the shortage D S, where D(p) and
S(p) are the linear demand and supply functions D 8 2p and S 2 p.
(a) If the price is $5 when t 0 and $3 when t 2, find p(t).
(b) Determine what happens to p(t) in the “long run” (as t → ).
Solution
(a) The rate of change of p(t) is given by the separable differential equation
dp
k(D S) k[(8 2p) (2 p)]
dt
k(6 3p)
* The Evans price adjustment model and several other dynamic economic models are examined in the
text by J. E. Draper and J. S. Klingman, Mathematical Analysis with Business and Economic
Applications, Harper and Row, New York, 1967, pages 430–434.
Chapter 5 ■ Section 3
Introduction to Differential Equations
405
Separating the variables, integrating, and solving for p, you get
dp
6 3p
k dt
ln 6 3p 3kt 3C
6 3p e
1
ln 6 3p kt C1
3
1
3kt3C1
e3kte3C1 Ce3kt
1
p(t) 2 Ce3kt
3
where C e3C1
To evaluate the constant C, use the fact that p(0) 5, so that
1
1
5 p(0) 2 Ce0 2 C
3
3
C 9
p(t) 2 3e3kt
and
You still need to find k. Since p 3 when t 2, it follows that
3 p(2) 2 3e3k(2) 2 3e6k
and by solving this equation for e6k and then taking logarithms, you get
32 1
3
3
1
6k ln
1.0986
3
e6k k
Refer to Example 3.6. Graph
p(x) 2 3e1.0986x using the
window [0, 9.4]1 by [0, 7]1.
Find the smallest value of x for
which p(x) 2 .01.
1.0986
0.1831
6
Thus, the price at time t is
p(t) 2 3e6(0.1831t) 2 3e1.0986t
(b) As t increases without bound, e1.0986t approaches zero and p(t) approaches 2,
which is the price at which supply equals demand. That is, in the “long run,” the
price of p(t) approaches the equilibrium price of the commodity (see Figure 5.4).
406
Chapter 5
Integration
p (price)
p=5
p(t) = 2 + 3e –1.0986t
p=2
t (time)
FIGURE 5.4 The price p(t ) approaches the equilibrium price p 2.
P . R . O . B . L . E . M . S
5.3
In Problems 1 through 16, find the general solution of the given differential
equation.
1.
dy
3x2 5x 6
dx
2.
dP
t et
dt
3.
dy
3y
dx
4.
dy
y2
dx
5.
dy
ey
dx
6.
dy
e xy
dx
7.
dy x
dx y
8.
dy y
dx x
9.
dy
xy
dx
10.
dy y2 4
dx
xy
11.
dy
y
dx x 1
12.
dy
eyx 1
dx
13.
dy
y3
dx (2x 5)6
14.
dy
(ey 1)(x 2)9
dx
15.
dx
xt
dt
2t 1
16.
dy
tey
dt
2t 1
Chapter 5 ■ Section 3
Introduction to Differential Equations
407
In Problems 17 through 24, find the particular solution of the given differential
equation that satisfies the indicated condition.
17.
dy
e 5x; y 1 when x 0
dx
18.
dy
5x4 3x2 2; y 4 when x 1
dx
19.
dy
x
2 ; y 3 when x 2
dx y
21.
2
dy
dy
xe yx ; y 0 when x 1
y24 x ; y 2 when x 4 22.
dx
dx
23.
dy
y1
; y 2 when t 1
dt
t(y 1)
24.
dx
xtt 1 ; x 1 when t 0
dt
20.
dy
4x3y2 ; y 2 when x 1
dx
Hint: yy 11 1 y 2 1
25. Verify that the function Q B Cekt is a solution of the differential equation
dQ
k(B Q).
dt
26. Verify that the function y C1ex C2xex is a solution of the differential equad2y
dy
tion 2 2 y 0.
dx
dx
In Problems 27 through 33 write a differential equation describing the given
situation. Define all variables you introduce. (Do not try to solve the differential
equation at this time.)
GROWTH OF BACTERIA
RADIOACTIVE DECAY
INVESTMENT GROWTH
27. The number of bacteria in a culture grows at a rate that is proportional to the number
present.
28. A sample of radium decays at a rate that is proportional to its size.
29. An investment grows at a rate equal to 7% of its size.
CONCENTRATION OF DRUGS
30. The rate at which the concentration of a drug in the bloodstream decreases is
proportional to the concentration.
POPULATION GROWTH
31. The population of a certain town increases at the constant rate of 500 people per
year.
RECALL FROM MEMORY
32. When a person is asked to recall a set of facts, the rate at which the facts are recalled
is proportional to the number of relevant facts in the person’s memory that have not
yet been recalled.
408
Chapter 5
Integration
THE SPREAD OF AN EPIDEMIC
33. The rate at which an epidemic spreads through a community is jointly proportional
to the number of people who have caught the disease and the number who have not.
AGRICULTURAL PRODUCTION
34. The Mitscherlich model, a useful model of agricultural production, specifies that
the size Q(t) of a crop changes in such a way that the rate of change is proportional
to B Q(t), where B is the maximum possible size of the crop.
(a) Write this relationship as a differential equation and find its general solution.
(b) Note that this model is similar to the learning model in Example 3.5. Is this
just a coincidence or is there some meaningful analogy linking the two situations? Explain.
FLUORIDATION
35. The residents of a certain community have voted to discontinue the fluoridation of
their water supply. The local reservoir currently holds 200 million gallons of
fluoridated water that contains 1,600 pounds of fluoride. The fluoridated water is
flowing out of the reservoir at the rate of 4 million gallons per day and is being
replaced at the same rate by unfluoridated water. At all times, the remaining fluoride
is evenly distributed in the reservoir. Express the amount of fluoride in the reservoir
as a function of time.
(a) Let Q(t) be the amount of fluoride in the reservoir at time t. Explain why Q(t)
satisfies the differential equation
dQ Q
dt
50
Hint: Note that
Rate of change of fluoride
concentration of
rate of flow of
with respect to time
fluoride in water fluoridated water
Pounds
pounds per
million gallons
or
per day
million gallons
per day
(b) Solve the differential equation in part (a) to obtain Q(t). [Hint: What is Q(0)?]
DILUTION
36. A tank holds 200 gallons of brine containing 2 pounds of salt per gallon. Clear water
flows into the tank at the rate of 5 gallons per minute, and the mixture, kept uniform
by stirring, runs out at the same rate.
(a) If S(t) is the amount of salt in solution at time t, then a typical gallon of solution contains
S(t)
amount of salt
amount of fluid 200
At what rate is salt flowing out of the tank at time t?
Chapter 5 ■ Section 3
Introduction to Differential Equations
409
(b) Write a differential equation for the time rate of change of S(t). Hint:
dS
rate at which
rate at which
salt enters tank salt leaves tank
dt
(c) Solve the differential equation in part (b) to obtain S(t). [Hint: What is S(0)?]
EVANS PRICE ADJUSTMENT MODEL
37. Suppose that a particular commodity has linear demand and supply functions, D(p)
a bp and S(p) r sp, for price p and positive constants a, b, r, and s. Further
assume that price is a function of time t and that the time rate of change of price is
proportional to the shortage D S, so that
dp
k(D S)
dt
Solve this differential equation and sketch the graph of p(t). What happens to p(t)
“in the long run” (as t → )?
DOMAR DEBT MODEL
38. Let D and I denote the national debt and national income, and assume that both are
functions of time t. One of several Domar debt models assumes that the time rates
of change of D and I are both proportional to I, so that
dD
aI
dt
and
dI
bI
dt
Suppose I(0) I0 and D(0) D0.
(a) Solve both of these differential equations and express D(t) and I(t) in terms
of a, b, I0, and D0.
(b) The economist, Evsey Domar, who first studied this model, was interested in
the ratio of national debt to national income. What happens to the ratio
D(t)
as t fi ?
I(t)
39. Read an article on dynamic economic models such as the Evans model (Example
3.6 and Problem 37) and the Domar model (Problem 38). Then write a paragraph
on how such models fit in with more traditional economic methods.
ELIMINATION OF
HAZARDOUS WASTE
40. To study the degradation of certain hazardous wastes with a high toxic content,
biological researchers sometimes use the Haldane equation
dS
aS
dt
b cS S2
where a, b, and c are positive constants and S(t) is the concentration of substrate
(the substance acted on by bacteria in the waste material).* Find the general solution of the Haldane equation. Express your answer in implicit form (as an equation involving S and t).
* Michael D. LaGrega, Philip L. Buckingham, and Jeffery C. Evans, Hazardous Waste Management,
McGraw-Hill, Inc., New York, 1994, page 578.
410 Chapter 5
Integration
41. Solve the logistic equation
dP
P(k mP)
dt
by answering the following questions.
(a) Find expressions A and B so that
A
B
1
P(k mP) P k mP
(Note: A and B will involve k and m.)
(b) Evaluate
A
B
dP
P k mP
where A and B are the expressions found in part (a).
(c) Separate the variables in the given differential equation and solve, using the
result of part (b). Express P(t) in the form
P(t) C
1 De kt
where C and D are expressions involving k and m.
dQ
kQ(B Q),
dt
dQ
where k and B are positive constants, then the rate of change
is greatest when
dt
B
Q(t) . What does this result tell you about the inflection point of a logistic
2
curve? Explain.
42. Show that if a quantity Q satisfies the differential equation
4
Integration
by Parts
In this section, you will see a technique you can use to integrate certain products
f(x)g(x). The technique is called integration by parts, and as you will see, it is
a restatement of the product rule for differentiation. Here is a statement of the
technique.
Integration by Parts
■
If G is an antiderivative of g, then
f(x)g(x) dx f(x)G(x) f(x)G(x) dx