Blank B.E., Krantz S.G. Calculus: Single Variable

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where C is the constant of integration. We can determine the two constants C and k

from the two given observations: y(0) 5 1/60 and y(6000) 5 1/140. The ﬁrst data

point yields

1=60

52k  0 1 C

or C 5260. The second data point yields

1=140

52k  6000 2 60

or k 5 1/75. Thus 21/y(t) 52t/75 2 60. Solving for y(t), we obtain y(t) 5 75/

(t 1 4500). Therefore y(10500) 5 75/(10500 1 4500) 5 1/200 moles/L.

⁄ EXAMPLE 8 Let y(t) and v(t) 5 dy/dt be the height and velocity of a

projectile that has been shot straight up from the surface of Earth with initial

velocity v

. Newton’s Law of Gravitation states that

ðR 1 yÞ

;

where R is the radius of Earth, and g is the acceleration due to gravity at Earth’s

surface. Supposing that v

ﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2gR

, what is the maximum height attained by the

projectile?

Solution At

the instant the projectile reaches its maximum height and begins to

fall back to Earth, its velocity v is 0. Therefore our strategy is to ﬁnd v as a function

of y and solve for the value of y for which v 5 0. As a ﬁrst step, we use the Chain

Rule to express dv/dt in terms of dv/dy:



 v:

By equating this expression for dv/dt with the one given by Newton’s Law of

Gravitation, we obtain

v 

ðR 1 yÞ

This differential equation is separable. Following the general procedure, we obtain

vdv5



ðR 1 yÞ



5 gR

ðR 1 yÞ

1 C:

When y 5 0, we have v 5 v

. Therefore

5 gR

ðR 1 0Þ

1 C;

or C 5

2 gR. It follows that

5 gR

ðR 1 yÞ



2 gR



: ð7:6:7Þ

7.6 First Order Differential Equations—Separable Equations 595

Now we set v 5 0 in equation (7.6.7) and solve for y . We ﬁnd that the maximum

height is v

R=ð2gR 2 v

Þ. ¥

Logistic Growth Figure 6 shows the growth of a pumpkin in a controlled agricultural experiment. At

ﬁrst the growth of the pumpkin appears to be exponential. But the growth plot

eventually becomes concave down and can no longer be modeled by exponential

growth. To obtain a more accurate model, scientists often make the following

growth assumptions:

a. There is a size P

that cannot be exceeded. In population models, the constant

is called the carrying capacity;

b. P

ðtÞ-0asPðtÞ-P

. In words, as the function P(t) approaches the carrying

capacity P

, its rate of growth decreases to 0.

One way to incorporate these assumptions into a growth model is to assume that

the

rate of population increase is jointly proportional to the population P(t) and to

2P(t). The resulting differential equation

5 k  PðtÞ



2 PðtÞ



; Pð0Þ5 P

ð7:6:8Þ

is called the logistic growth equation. In this differential equation k, P

, and P

represent ﬁxed positive constants. In our discussion, we will assume that P

, P

Using the method of separation of variables, we can rewrite equation (7.6.8) as

P ðP

2 PÞ

dt 5

kdt;

P ðP

2 PÞ

dP 5

kdt: ð7:6:9Þ

⁄ EX

AMPLE 9 Use the method of partial fractions to solve equation (7.6.9).

Solution We

apply partial fractio ns and write

P ðP

2 PÞ

2 P

Before we go any further, it is a good idea to make sure that we understand the role

of each quantity in this equation. Here P

is a ﬁxed constant that we regard as

known, even though we are not given a speciﬁc numerical value. The quantity P is a

variable. We are to ﬁnd values for the unknown constants A and B so that the

equation is valid for all values of P. Putting the right side over a common

denominator and equating numerators leads to the equation

1 5 A ðP

2 PÞ1 B  P:

If we let P be 0, then we ﬁnd that A 5 1/P

. Letting P be P

results in B 5 1/P

With these values, equation (7.6.9) becomes



2 P



dt 5

kdt;

4 8 12 16

Diameter (cm)

Time

(

weeks

)

m Figure 6 Pumpkin growth

596 Chapter 7 Applications of the Integral

dt 1

2 P

dt 5 kt 1 C;





lnðPÞ2 lnð P

2 PÞ



5 kt 1 C:

We have omitted absolute values in the logarithms because P(t) is in the range

0 , P(t) , P

. We can combine the logarithms to obtain



PðtÞ

2 PðtÞ



5 P

ðkt 1 CÞ

or, on exponentiating and letting A 5 expðP

 CÞ ,

PðtÞ

2 PðtÞ

5 expðP

 kt 1 P

 CÞ5 expðP

 ktÞexpðP

 CÞ5 A  expðP

 ktÞ:

We may solve for the constant A by substituting t 5 0:

2 P

Pð0Þ

2 Pð0Þ

5 A:

Therefore

PðtÞ

2 PðtÞ

2 P

 expðP

 ktÞ:

Isolating P(t) is elem entary but somewhat tedious. The result is

PðtÞ5

 P

1 ðP

2 P

Þexpð2k  P

 tÞ

: ð7:6:10Þ ¥

The function P deﬁn

ed by equation (7.6.10) is said to be a logistic growth

function. The term sigmoidal growth is also used in biology. Notice that there are

three positive constants that determine logistic growth: the proportionality constant

k, the initial value P

of P, and the carrying capacity P

. The graph of P (t)

a. is everywhere rising,

b. is concave up for P

# P ,

c. is concave down for

, P , P

, and

d. has y 5 P

as a horizontal asymptote.

We can deduce property a by noticing that, because P

2 P

, k,andP

are all

positive, the expression (P

2 P

) exp (2 k P

t) is decreasing. Because this

expression appears in the denominator of formula (7.6.10) for P(t), we conclude

that P(t) is increasing. Property d can be expressed as

lim

t-N

PðtÞ5 P

;

which is the reason for the subscript N. This limiting value for P(t) is an immediate

consequence of formula (7.6.10):

lim

t-N

PðtÞ5 lim

t-N

 P

1 ðP

2 P

Þexpð2k  P

 tÞ

5 lim

t-N

 P

1 ðP

2 P

Þ0

5 P

7.6 First Order Differential Equations—Separable Equations 597

A proof of properties b and c is outlined in Exercise 55. A typical logistic curve is

plotted in Figure 7. In Figure 8, we have replotted the pumpkin-growth data, ad ding

the graph of the logistic function determined by P

5 3.6, P

5 46, and k 5 0.0065.

⁄ EX

AMPLE 10 Suppose that 100 children attend a day-care center. Let

P(t) be the fracti on of the day-care population with a viral infection at time t

(measured in days). Suppose that, at 9 AM on Monday (t 5 0), only 1 child was

infected. Suppose that, at 9 AM on Tuesday (t 5 1), there were 3 infected children

(including the origin al child). Assume that the rate of change of P is jointly pro-

portional to P (because P represents the fraction of the population capable of

spreading the infection) and 1 2 P (because 1 2 P represents the fraction of the

population that can still be infected). By when did the infe ction spread to half the

day-care population?

Solution Because

the rate of change of P is jointly proportional to P and 1 2 P,we

have

5 k  PðtÞ



1 2 PðtÞ



;

for some proportionality constant k. Because 1 child out of 100 was infected at time

t 5 0, we have P(0) 5 1/100. Thus the initial value problem for P(t)is

5 k  PðtÞ



1 2 PðtÞ



; Pð0Þ5 1=100:

Notice that this initial value problem is logistic growth equation (7.6.8) with P

5 1

and P

5 1/100. The solution is given by formula (7.6.10):

PðtÞ5

1=100

1=100 1 ð1 2 1=100Þexpð2k 1  tÞ

1 1 99e

2kt

: ð7:6:11Þ

We may now use the observation P(1) 5 3/100 to solve for k. Because formula

(7.6.11) gives us P(1) 5 1/(1 1 99e

), we obtain

100

1 1 99e

; or 1 1 99e

100

; or e



100

2 1



297

By applying the natural logarithm to each side of the last equation, we see that

2k 5 ln(97/297). Our ﬁnal expression for P is

PðtÞ5

1 1 99e

tlnð97=297Þ

1 1 99 exp





ð97=297Þ





1 1 99 ð97=297Þ

The solution to our problem is found by solving the equation

1 1 99 ð97=297Þ

for t. After some simpliﬁcation, we obtain 99 5 (297/97)

t 5

lnð99Þ

lnð297=97Þ

 4:1:

According to our model, about 4.1 days after the ﬁrst child was infected, that is to

say, by about 11:30 Friday morning, the ﬁftieth child became infected.

4 8 12 16

Diameter (cm)

Time

(

weeks

)

m Figure 8 Pumpkin growth



P(t)

Horizontal asymptote

Point of inflection



m Figure 7 Logistic growth curve

598 Chapter 7 Applications of the Integral

QUICK QUIZ

1. To which of the following differential equations can we apply the method

of separation of variables: a) dy=dx 5 expðxyÞ,b)dy=dx 5 expðx 1 yÞ,

c) dy=dx 5 lnðx

Þ,ord)dy=dx 5 ln ðx 1 yÞ?

2. Solve 4 dy=dx 5 3x

=ðy

1 y 1 5Þ.

3. Solve the initial value problem dy=dx 5 x=y; yð2Þ5 3.

4. If dy=dx 5 sinðx

Þ and yðπÞ5 2, then for what α, β, and C is

yðxÞ5 C 1

sinðt

Þdt?

Answers

1. b and c 2. y

1 2y

1 20y 5 x

1 C 3. y

5 x

1 5

4. α 5 π, β 5 x,andC 5 2

EXERCISES

Problems for Practice

c In Exercises 128, verify that the given function y satisﬁes

the given differential equation. In each expression for y(x),

the letter C denotes a constant. b

5 xy; yðxÞ5 Ce

5 2xy

; yðxÞ5 1=ðC 2 x

5 x 2 3y; yð xÞ5 x=3 2 1=9 1 Ce

23x

5 e

1 y; yðxÞ5 e

ðx 1 C Þ

5 x 1 y; yð xÞ5 Ce

2 x 2 1

5 x 1 xy; yðxÞ5 Ce

2 1

5 y 1 x

; yðxÞ5 Ce

2 x

2 2x 2 2

2x 2 y

x 1 y

; y 52x 1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 2C

c In each of Exercises 9218, solve the given differential

equation. b

5 6

ﬃﬃﬃﬃﬃ

10. ð4 1 y

5 x

11. ð2 1 xÞ

5 y

12.

1 3xy

5 0

13.

2 3x

y 5 0

14. x

2 y 5 3x

15. expð2x 1 3yÞ

5 1

16. cotðxÞ

5 tanðyÞ

17.

5 x secðyÞ

18.

5 y secðxÞ

c In Exercises 19236, ﬁnd the solution of the given initial

value

problem. b

19. y

ðxÞ5 2xyð1Þ5 3

20. y

ðxÞ5 2x 1 1 yð1Þ5 5

21. y

ðxÞ5 cosðxÞ yð0Þ5 2

22. y

ðxÞ5 sec

ðxÞ yðπ=4Þ5 3

23. y

ðxÞ5 x=yðxÞ yð0Þ5 1

24. y

ðxÞ5 xy

ðxÞ yð1Þ5 2

25. y

ðxÞ5 y

ðxÞsinðxÞ yð0Þ5 2

26. x

5 cos

ðyÞ yð0Þ5 1

27. y

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1 y

yð0Þ5 0

28. y

ðxÞ5 cosðxÞ=yðxÞ yð0Þ5 2

29.

5 y 1

yð0Þ5 3

30. ð1 1 x

ðxÞ5 6xyðxÞ yð0Þ5 1

31. x

ðxÞ5 yðxÞsinð1=xÞ yð2=πÞ5 3

32.

5 expðx 1 yÞ yð0Þ5 0

33. y

ðxÞ2 xy

5 xyð4x 2 yÞ yð0Þ5 1

34.

5 x 2 1 1 yx 2 yyð2Þ5 3

35.

yð1 1 x

yð1Þ5

ﬃﬃﬃ

36.

5 2x cos



yðxÞ



yð0Þ5 0

Further Theory and Practice

37. Match the functions F

(x, y) 5 x, F

(x, y) 5 y, F

(x, y) 5

xy, F

(x, y) 5 x/y, F

(x, y) 5 x

2 y

, and F

(x, y) 5 y

the graphs of the slope ﬁelds of equation (7.6.1) that are

plotted in Figures 9a through 9f.

7.6 First Order Differential Equations—Separable Equations 599

m Figure 9a

m Figure 9b

m Figure 9c

m Figure 9d

m Figure 9e

m Figure 9f

38. Match

the graphs of the solution curves in Figures 10a

through 10f to the slope ﬁelds of Figures 9a through 9f.

m Figure 10a

m Figure 10b

m Figure 10c

m Figure 10d

m Figure 10e

m Figure 10f

600 Chapter

7 Applications of the Integral

c In each of Exercises 39244, solve the given differential

equation. b

39. expðx 1 yÞ

5 x

40. y

5 x

cscðy Þ

41.

5 3x

ðy

2 yÞ

42. ð4 1 e

5 y

43. ð4 1 e

5 ye

44. ð4 1 e

5 ye

45. Solve the initial value problem y

ðxÞ5 4f

ðxÞf

ðxÞ;

f ð1Þ5 2; yð1Þ5 24. Verify your formula with f ðxÞ5 2

ﬃﬃﬃ

46. Solve the initial value problem y

ðxÞ5 xg

ðx

Þ; gð4Þ5 6;

yð2Þ5 12. Verify your formula with g(x) 5 x

2 2x 2 2.

47. Solve the initial value problem y

ðxÞ5 f

ðxÞg



f ðxÞ



;

f ð0Þ5 3; gð3Þ5 2; y ð0Þ5 6. Verify your formula with

f ðxÞ5 3 cosðxÞand gðxÞ5

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 1 x

48. Show that, when applied to the initial value problem

equation

1 2

2y 2 2

; yð0Þ5 y

;

the method of separation of variables results in the same

equation,

ðxÞ2 2yðxÞ5 x

1 2x;

when y

5 0 and when y

5 2. And yet, the solutions to the

two initial value problems clearly must differ. Explain

how this is possible.

49. According to Torricelli’s Law, if there is a hole of area a

at the bottom of a tank, then the volume V(t) and height

y(t) of water in the tank at time t are related by

52a 

ﬃﬃﬃﬃﬃﬃﬃﬃ

2gy

where g is equal to gravitational acceleration at Earth’s

surface. (Notice that the differential equation is dimen-

sionally correct—both sides bear units of L

/T where L is

length and t is time.) Use the Fundamental Theorem of

Calculus together with the Chain Rule to show that

AðyÞ

52a 

ﬃﬃﬃﬃﬃﬃﬃﬃ

2gy

50. A cubic tank ﬁlled with water has side length 1 m. At time

t 5 0, a circular hole of radius 4 cm opens up on the bot-

tom of the tank. If the proportionality constant k in

Torricelli’s Law is given by k 5 5/π m

1/2

/s, how long does

it take the tank to drain?

51. A tank in the shape of a conical frustum ﬁlled with water

is one meter deep at its central axis, has radius 2 m at its

top, and one meter at its bottom. At time t 5 0, a circular

hole of radius 1 cm opens up at the bottom. Suppose that

the proportionality constant k in Torricelli’s Law is given

by k 5 32/15 m

1/2

/s. Find an equation for the height y(t)of

water in the tank at time t. How long does it take the tank

to completely drain?

52. The escape velocity is the minimum initial velocity v

which a projectile launched straight up will never return

to Earth. Use equation (7.6.7) to ﬁnd a formula for v

terms of g and R.

53. A population P(t) satisﬁes logistic growth equation (7.6.8)

with k 5 1/2 and P

5 2P

.AtwhattimeisP(t) 5 (3/2) P

54. A population P(t ) satisﬁes the logistic growth equation

ðtÞ5 α  PðtÞ2 β  P

ðtÞ for positive constants α and β.

What is the carrying capacity?

55. By completing the square on the right side of logistic

growth equation (7.6.8), show that a solution P(t) has

exactly one inﬂection point: (t

, P(t

)) where P(t

) 5 P

/2.

56. According to the preceding exercise, the solution P(t)to

logistic growth equation (7.6.8) has one inﬂection point,

, P(t

)), where P(t

) 5 P

/2. Show that



2 P



57. Let P be the logistic growth function deﬁned by equation

(7.6.10). Let t

be the time at which P has an inﬂection

point. (Refer to the two preceding exercises.) Show that

P(t) can be expressed in terms of the parameters k, P

and t

by means of the formula

PðtÞ5

1 1 exp



2k  P

ðt 2 t



58. If a population is to be modeled by the logistic growth

function P deﬁned by equation (7.6.10), then the three

parameters P

, k, and P

must be determined. It is easiest

to solve for these constants by measuring P(t) at equally

spaced times t

2 τ, t

, and t

1 τ. Having done so, we

position the origin of the time axis so that t 5 0 corre-

sponds to the time t

, then A 5 P(2τ), P

5 P(0), and

B 5 P(τ) are the measured population values.

a. Use the equation B 5 P(τ) to express exp(2k P

τ)

in terms of B, P

, and P

b. Use the equation A 5 P(2τ) to express exp(k P

τ)

in terms of A, P

, and P

c. The product of exp(2k P

τ) and exp(k P

τ)is1.

Use this observation together with parts a and b to

show that

ðA 1 BÞP

2 2ABP

2 AB

d. Use your formula from part b to show that

expð2k  P

 tÞ5



2 A

2 P



2t=τ

7.6 First Order Differential Equations—Separable Equations 601

e. Deduce that

PðtÞ5

 P

1 ðP

2 P



2 A

2 P



2t=τ

59. A town’s population (in thousands) was 36 in 1980, 60 in

1990, and 90 in 2000. Assuming that the growth of the

population of this town is logistic, what will the popula-

tion be in the year 2020? (Use the preceding exercise.)

60. In Exercise 58, the logistic growth formula determined in

part e is obtained without solving for the growth constant

k in terms of the measured data A, P

, and B.

a. Use the formula in part d of Exercise 58 with t 5 τ to

obtain a formula for k.

b. Use the logistic growth equation to express k in terms

of the data P

, P

, and P

(0).

c. Assuming that τ is small, approximate k using part b

and the central difference approximation for the ﬁrst

derivative.

61. The plot of the pumpkin data in Figure 6 has an inﬂection

point near (9, 25). For simplicity, let us use these coordi-

nates for the inﬂection point. The initial diameter P(0) is

3.6 cm. Using Exercises 55 and 56, ﬁnd P

and k.

62. A population P satisﬁes the differential equation

ðtÞ5 10

 PðtÞ



15000 2 PðtÞ



For what value P(0) of the initial population is the initial

growth rate P

(0) greatest?

63. When sucrose is dissolved in water, glucose and fructose

are formed. The reaction rate satisﬁes

d½C



525:7 3 10

½C

:

For what value of t is the concentration of sucrose,

], equal to 1/3 its initial value?

64. The chemical reaction

COCH

1 I

- CH

COCH

I 1 HI

between acetone and iodine is governed by the differ-

ential equation

d½I



520:41 3 10

½I



(in moles per liter per second). If the initial iodine con-

centration is c

, solve for [I

] as a function of time. Sup-

pose that [I

] is plotted as a function of time. What is the

slope of the graph when [I

] 5 1/10 mole/liter? Suppose

that 1/[I

] is plotted as a function of time. What is the

slope of the graph when [I

] 5 1/10 mole/liter?

65. In the decomposition of dinitrogen pentoxide,

- 2NO

;

the reaction rate satisﬁes the equation

d½N



52k ½N



for some positive constant k.If[N

] is equal to 2.32

moles per liter when t 5 0 seconds and if [N

] is equal

to 0.37 moles per liter when t 5 3000 seconds, then what is

] as a function of time?

66. The reaction rate of

2NO 1 H

- N

O 1 H

is given by

d½N

O

5 0:44 3 10

ðh 2 ½N

OÞ

where h is the number of moles per liter of H

that are

initially present. What is the initial concentration of

O]? Solve for [N

O] as an explicit function of t.In

doing so, you will need to extract a square root. Decide

which sign, 1 or 2 to use; explain your answer.What is

lim

t-N

½N

O?

67. Suppose that φ is a function of one variable. The differ-

ential equation dy/dx 5 φ( y/x) is said to be homogeneous

of degree 0. Let w( x) 5 y(x)/x. Differentiate both sides of

the equation y(x) 5 x  w(x) with respect to x. By equating

the resulting expression for dy/dx with φ( y/x), show that

w(x) is the solution of a separable differential equation.

Illustrate this theory by solving the differential equation

dy/dx 5 2xy/(x

1 y

). For this example, φ(w) 5 2w/

(1 1 w

68. Suppose that ψ is a function of one variable and that a, b,

and c are constants. If dy/dx 5 ψ(a 1 bx 1 cy), then

show that u(x) 5 a 1 bx 1 cy(x) is the solution of a

separable equation. Illustrate by solving the equation

dy/dx 5 1 1 x 1 y.

69. The pressure P and temperature T in the outer envelope

of a white dwarf (star) are related by the differential

equation

5 C

7:5

; Pð0Þ5 0

where C is a constant. Find P as a function of T.

70. The interior temperature T

(in degrees K) of a cooling

white dwarf (star) satisﬁes the differential equation

52k



7 3 10





7=2

Here k is a constant with units degrees K per year, and t

represents time in years. Solve for T

.Ifk 5 6



K/yr and if

(0) 5 10



K, then in how many years will the star cool

to 10



71. A spherical star of radius R is in equilibrium if the pres-

sure gradient dP/dr exactly opposes gravity. In the case of

Newtonian gravitation, this amounts to

602 Chapter 7 Applications of the Integral

GmðrÞρðrÞ

; PðRÞ5 0:

Here G is the gravitational constant, r represents distance

to the center, m(r) represents the mass of that part of the

star that is within distance r of the center, and ρ(r)repre-

sents the mass density at radius r.ComputeP(r) under the

assumption that the mass density is a constant ρ.Showthat

the pressure at the center of the star is 2πGρ

/3 (in terms

of the stellar radius R)or(π/6)

1/3

2/3

4/3

(in terms of the

stellar mass M).

72. A bullet is ﬁred into a sand pile with an initial speed of

1024 feet per second. The rate at which its velocity is

decreased after entering the sand pile is 500 times the

power of the velocity if units of feet and seconds are used.

How long is it before the bullet comes to rest?

73. Refer to the preceding exercise. How far does the bullet

travel in the sand before coming to rest?

74. The standard normal probability density f satisﬁes the

differential equation

ðxÞ52xf ðxÞ; f ð0Þ5

ﬃﬃﬃﬃﬃﬃ

2π

Find f (x).

75. Suppose that, in a certain ecosystem, there is one type of

predator and one type of prey. Let x 100 denote the

number of predators and y 1000 denote the number

of prey. The Austrian mathematician A.J. Lotka

(18801949) and the Italian mathematician Vito Volterra

(18601940) proposed the following relationship, the

so-called Lotka-Volterra Equation, between the two

population sizes:

y ða 2 bxÞ

x ðcy 2 dÞ

Separate variables and solve this equation. What is the

predator-prey relationship if the initial prey population is

1500, the initial predator population is 200, and if a 5 6,

b 5 2, c 5 4, and d 5 7? Figure 11 illustrates the solution

curve of this Lotka-Volterra equation. Its potato-pancake

shape is typical.

76. Suppose that λ, A, and p are positive constants. Solve the

differential equation

 t

p21

ðA 2 yÞ:

A solution of this equation is known as a Janoschek

growth function. Such functions are often used to model

animal growth. The parameter A is called “the size of the

adult.” Why?

77. The Tractrix. The tip of a boat is at the point ( L, 0) on the

positive x-axis. Attached to the tip is a chord of length L.

A man starting at (0, 0) pulls the boat by the chord as he

walks up the y-axis. The chord is always tangent to the

path of the tip of the boat. Show that the position y(x)of

the tip of the boat is given by

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2 x

; yðLÞ5 0:

The graph of y is called the tractrix (Figure 12).

78. Actuaries use mortality tables to show the expected

number of survivors of an initial group. Let L(x) denote

the number from that group who are living at age x. Then

(x) is the instantaneous death rate and the percentage

rate, 2L

(x)/L(x) is called the force of mortality. The ﬁrst

signiﬁcant formula was derived by B. Gompertz assumed

that the force of mortality increases with age according to

an expression of the form Bg

for some constants B . 0

and g . 1. Solve for L(x) under Gompertz’s assumption.

79. M. W. Makeham emended Gompertz’s assumption for

the force of mortality to take into account accidental

deaths—refer to the preceding exercise for background.

Under Makeham’s assumption, the force of mortality has

the form M 1 Bg

for some positive constants M, B, and g

with g . 1. Solve for L(x) under this assumption. The

resulting formula is known as Makeham’s Law.

80. Du Nouy’s equation for the decreasing surface tension T

of blood serum is

Prey (in thousands)

Predators

(

in hundreds

)

2.0

2.4

2.0

1.6

1.2

2.5 3.0 3.5 4.0

m Figure 11 Solution of Lotka-Volterra equation

Man

Tractrix

Chord

(0, 0)

(

L, 0

)

Tip of boat

(x, y)

m Figure 12

7.6 First Order Differential Equations—Separable Equations 603

52k

ﬃﬃ

for some constant k. Solve this equation if T 5 τ at t 5 0.

81. Let x and y be the measures of two body parts with

relative growth rates that are proportional to a common

factor Φ(t):



5 α  ΦðtÞ and



5 β  ΦðtÞ:

Show that x and y satisfy the Huxley Allometry Equation

y 5 kx

for suitable constants k and p.

82. The shape of a Bessel horn is determined by the equation

52γ 

x 1 x

where a(x) is the bore radius at distance x from the edge

of the bell of the horn, and x

and γ are positive constants.

Solve for a as a function of x. Determine a(x) for a

trumpet with γ 5 0.7, a(0) 5 10, and a(30) 5 2. Sketch the

graph of y 5 a (x) for 0 # x # 30.

83. Verify that yðxÞ5 expð2x

expðt

Þdt, which is known

as Dawson’s Integral, is the solution of the initial value

problem

5 1 2 2xy; y





5 0.

84. For each x . 0, there is a unique y . 0 such that ye

5 x.

Denote this value of y by W(x). Show that the function W

(called Lambert’s W function) is a solution of the differ-

ential equation

xð1 1 yÞ

85. Suppose an object of mass m is propelled upwards from

the surface of the earth with initial velocity v

. Suppose

that the (downward) force of air resistance R(v) is pro-

portional to the square of the speed: R(v) 52k v

, where

k is a positive constant that carries the units of mass/

length. (This is the quadratic drag law.) Solve the initial

value problem for motion:

52kv

2 mg; vð0Þ5 v

86. Suppose that the force R of air resistance on a 0.2 kg

object thrown straight up with initial velocity 28 m/s is

given by the quadratic drag law,

RðvÞ52kv

;

where k 5 4 3 10

kg/m. Find an equation for the time τ

at which the object begins to fall.

87. Let v(t) denote the velocity at time t of an object of mass m

that is dropped from a positive height at time t 5 0. Assume

that, throughout its fall, the (upward) force of air resistance

on the object can be approximated by the formula

R(v) 51k v

,wherek is a positive constant that carries

the units of mass/length. (As in Exercises 85 and 86, this

equation is the quadratic drag law. Because drag is in the

opposite direction of motion, the sign for upward motion

is opposite to that for downward motion.) Show that

vðtÞ5

ﬃﬃﬃﬃﬃﬃﬃ

tanh



ﬃﬃﬃﬃﬃ



until impact.

88. The rate of elimination of alcohol from the bloodstream is

proportional to the amount A that is present. That is,

where k is a time constant that depends on the drug and

the individual. If k is 1/2 hour for a certain person, how

long will it take for his blood alcohol content to reduce

from 0.12% to 0.06%?

89. A column of air with cross-sectional area 1 cm

extends

indeﬁnitely upwards from sea level (i.e. to inﬁnity). The

atmospheric pressur e p at height y above sea level is the

weight of that part of this inﬁnite column of air that extends

from y to inﬁnity. Assume that the density of the air is

proportional to the pressure. (This is an approximation that

follows from Boyle’s law, assuming constant temperature.)

Show that there is a positive constant k such that

52kp:

Under this model, what is the mathematical expression

for atmospheric pressure? What physical dimensions does

K carry?

90. Suppose that α and β are positive constants. The differ-

ential equation

ðtÞ5 α  e

2βt

 PðtÞ

for a positive function P is known as the Gompertz growth

equation. (It is named for Benjamin Gompertz (17791865),

a self-educated scholar of w ide-ranging interests.) Find an

explicit formula for P(t). Use your explicit solution to show

that there is a number P

(known as the carrying capacity)

such that

lim

t-N

PðtÞ5 P

Show that the Gompertz growth equation may be written

in the form

ðtÞ5 k  PðtÞln



PðtÞ



91. In forestry, the function EðTÞ5 C  e

αT=ðT1βÞ

has been

used to model water evaporation as a function of tem-

perature T. Here α and β are constants with α . 2 and

β . 0. Show that

ðTÞ5

αβ

ðT 1 βÞ

EðTÞ

604 Chapter 7 Applications of the Integral