Solve the problem.
68)
The table shows the population of a certain city for selected years between 1950 and 2003.
Years after 1950 Population
0 12,421
20 143,112
30 290,089
40 375,297
50 445, 052
55 471,126
Write a logistic differential equation in the form dP
dt = kP(M – P) that models the growth of the
population. [You will first need to use your calculator find the logistic regression equation that best
fits the data.]
68)
A)
dP
dt = (1.623 ×10–7)P(499,107.3 – P)
B)
dP
dt = (2.526 ×10–7)P(478,549.6 – P)
C)
dP
dt = (2.193 ×10–7)P(494,193.8 – P)
D)
dP
dt = (2.346 ×10–7)P(482,549.6 – P)
Solve the differential equation subject to the initial condition.
69)
xdy
dx + (1 + x)y =1; y(5) =2
69)
A)
y =1+10 e5– x
x
B)
y =1+9e5– x
x
C)
y =1+9e–5– x
x
D)
y =1+11e7– x
x
Solve the problem.
70)
Suppose a tank contains 100 gallons of a solution of 10 lb of salt dissolved in water, which is kept
uniform by stirring. If pure water is allowed to flow into the tank at a rate of 3 gallons per minute,
and the mixture flows out at a rate of 2 gallons per minute, find an expression for the amount of
salt in the tank after t minutes.
70)
A)
y =105
(t + 100)2
B)
y = 100 (t – 100)2
C)
y =103
(t – 100)
D)
y =105
(3t + 50)2
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
71)
dy
dx =1
x; y(1) = 1; find y(1.4)
71)
A)
1.247
B)
1.351
C)
1.274
D)
1.410
Solve the differential equation subject to the initial condition.
72)
x dy
dx +7y =x2; y(2) =12
72)
A)
y =x
9+13,312
9x7
B)
y =x2
9+10,113
9x9
C)
y =x2
9+10,113
9x7
D)
y =x2
9+13,312
9x7
Solve the problem.
73)
A wild animal preserve can support no more than 120 elephants. 34 elephants were known to be in
the preserve in 1980. Assume that the rate of growth of the population is
dP
dt =0.0007P(120 – P)
where t is time in years. Find a formula for the elephant population in terms of t. Let 1980
correspond to t = 0.
73)
A)
P =120
1 +2.53e–0.084t
B)
P =120
1 +3.53e–0.084t
C)
P =120
1 +2.53e–0.84t
D)
P =120
1 –2.53e–0.84t
Find the general solution for the differential equation.
74)
dy
dx + 2y =23
74)
A)
y =23
2+ e2x + Ce–2x
B)
y =23 + Ce2x
C)
y =23
5+ Ce2x
D)
y =23
2+ Ce–2x
Solve the problem.
75)
A population of algae consists of 2000 algae at time t = 0. Conditions will support at most 500,000
algae. Assume that the rate of growth of algae is proportional both to the number present (in
thousands) and to the difference between 500,000 and the number present (in thousands). Write a
differential equation using 0.02 for the constant of proportionality.
75)
A)
dy/dt = 0.02y(500 – y)
B)
dy/dt =2000y(500 – 0.02y)
C)
dy/dt = 0.02y(y –500)
D)
dy/dt = 0.02(500 – y)
Find the particular solution for the initial value problem.
76)
dy
dx = 4x + 21; y(0) = – 20
76)
A)
y = 2x2+ 21x – 20
B)
y = 2x2+ 21x – 10
C)
y = 4x2+ 21x – 10
D)
y = 4x2+ 21x – 20
D)
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
77)
dy
dx = xy + 2; y(0) = 0; find y(0.4)
77)
A)
0.837
B)
0.812
C)
0.828
D)
0.809
D)
Solve the differential equation subject to the initial condition.
78)
2 dy
dx – 4xy = 8x; y(0) =20
78)
A)
y = – 1 + 21ex2
B)
y = – 2 + 22ex2
C)
y = 2 + 20ex2
D)
y = – 2 + 22e–x2
D)
Find the general solution for the differential equation.
79)
dy
dx =4e3x
79)
A)
y =12e3x + C
B)
y =4
3e3x + C
C)
y =1
3e3x + C
D)
y =4e3x + C
D)
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
80)
dy
dx = 1 +y
x; y(1) = 0; find y(1.4)
80)
A)
0.428
B)
0.529
C)
0.452
D)
0.486
Find the particular solution for the initial value problem.
81)
dy
dx =ex – y; y(0) =6
81)
A)
y = ln x +6
B)
y = x +6
C)
y = – ln(ex+e6– 1)
D)
y = ln(ex+e6– 1)
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
82)
dy
dx = 2xy; y(1) = 1; find y(1.4)
82)
A)
2.591
B)
2.982
C)
2.928
D)
2.287
Solve the problem.
83)
The table shows the population of a certain city for selected years between 1950 and 2003.
Years after 1950 Population
0 7891
20 103,087
30 191,064
40 241,552
50 265,058
55 269,116
Write a logistic differential equation in the form dP
dt = kP(M – P) that models the growth of the
population. [You will first need to use your calculator find the logistic regression equation that best
fits the data.]
83)
A)
dP
dt = 0.1207 P(278,903.2 – P)
B)
dP
dt = (4.741 ×10–7)P(278,903.2 – P)
C)
dP
dt = (4.945 ×10–7)P(271,976.2 – P)
D)
dP
dt = 0.1345 P(271,976.2 – P)
Find the general solution for the differential equation.
84)
x2dy
dx + xy =4x4+7x7, x > 0
84)
A)
y =x4+x7+ C
B)
y = k x3+x6+1
x
C)
y =x3+x6+C
x
D)
y =x4+x7+C
x
Find the particular solution for the initial value problem.
85)
(4x +4)y =dy
dx ; y(0) = 1
85)
A)
y =e2x2+4x
B)
y =e4x2+4x + 1
C)
y =e2x2+4x + 1
D)
y =e4x2+4x
25
Solve the problem.
86)
Suppose an isolated island has a native population of 10,000 and a person from a visiting ship
introduces a disease which has an infection rate of 0.00005. Assume that the rate of spread of the
disease satisfies the following logistic equation:
dy
dt = k 1 –y
Ny,
where N is the size of the population and y is the number infected at time t.
How many individuals are infected after 30 days?
86)
A)
9670
B)
9970
C)
9770
D)
9870
87)
Suppose an isolated island has a native population of 9000 and a person from a visiting ship
introduces a disease which has an infection rate of 0.00004. Assume that the rate of spread of the
disease satisfies the following logistic equation:
dy
dt = k 1 –y
Ny,
where N is the size of the population and y is the number infected at time t.
Write an equation for the number of infected natives after t days.
87)
A)
y =8999
1 +9000e–0.00004t
B)
y =9000
1 +8999e–0.36t
C)
y =9000
1 +8999e–0.00004t
D)
y =8999
1 +9000e–0.36t
Find the general solution for the differential equation.
88)
dy
dx – 6x2=4
88)
A)
y =2x3+ 4x + C
B)
y = – 2x3+ 4x + C
C)
y =2x3– 2x + C
D)
y =2x3+ 2x + C
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
89)
dy
dx = – 4 + x; y(0) = 1; find y(0.5)
89)
A)
–0.900
B)
–0.940
C)
–0.540
D)
–0.880
Solve the problem.
90)
A population of ants, y, living in a colony grows at a rate
dy
dt = 0.05y – 0.1y1/2,
where t is time in weeks. The initial population is 1000 insects. Using Euler’s method with h = 1,
what is the number of ants after 2 weeks?
90)
A)
1096 ants
B)
1070 ants
C)
1063 ants
D)
1086 ants
91)
When a dead body is discovered, one of the first steps in the ensuing investigation is for a medical
examiner to determine the time of death as closely as possible. If the temperature of the medium
has been fairly constant and less than 48 hours have passed since death, Newton’s law of cooling
can be used. Newton’s law of cooling states, dT
dt = – k(T –TM), where k is a constant, T is the
temperature of the object after t hours, and TM is the (constant) temperature of the surrounding
medium. Assuming the temperature of a body at death is 98.6°F, the temperature of the
surrounding air is 69°F, and at the end of one hour the body temperature is 89°F, when will the
temperature of the body be 77°F? Round to the nearest tenth of an hour.
91)
A)
0.7 hr
B)
1.9 hr
C)
3.3 hr
D)
0.4 hr
92)
Sales (in thousands) of a certain product are declining at a rate proportional to the amount of sales,
with a decay constant of 11% per year. Write a differential equation to express the rate of sales
decline.
92)
A)
dy/dt = – 0.11y
B)
dy/dt =e–0.11t
C)
dy/dt = – 0.89y
D)
dy/dt = – 0.11t
93)
A wild animal preserve can support no more than 190 elephants. 30 elephants were known to be in
the preserve in 1980. Assume that the rate of growth of the population is
dP
dt =0.0007P(190 – P)
where t is time in years. How long will it take for the elephant population to increase from 30 to
140? [First find a formula for the elephant population in terms of t.]
93)
A)
20.3 years
B)
19.5 years
C)
22.0 years
D)
22.8 years
94)
Newton’s Law of Cooling states that the rate of change of temperature of an object is proportional
to the difference in temperature between the object and the surrounding medium. Thus, if T is the
temperature of the object after t hours and T0 is the (constant) temperature of the surrounding
medium, then
dT
dt = – k(T –T0),
where k is a constant.
A cup of coffee with a temperature of 102°F is placed in a freezer with a temperature of 0°F. After
7 minutes, the temperature of the coffee is 57.5°F. When will its temperature be 41°F? Round your
answer to the nearest minute.
94)
A)
24 minutes after being placed in the freezer
B)
14 minutes after being placed in the freezer
C)
21 minutes after being placed in the freezer
D)
11 minutes after being placed in the freezer
Find the requested equation.
95)
The system of equations
dy
dt = y – 3xy
dx
dt = – x + 5xy
describes the influence of the populations (in thousands) of two competing species on their growth
rates. Find an equation relating x and y, assuming y =4 when x = 1.
95)
A)
5y – ln y = ln x – 3x – ln 4+23
B)
5y – ln y = – 3x – ln 4+17
C)
5y – ln y = ln x – ln 4+23
D)
5y – ln y = ln x – 3x – ln 4+17
Find the particular solution for the initial value problem.
96)
x dy
dx = 4x2e2x; y(0) =16
96)
A)
y = 4xe2x – 2e2x + 18
B)
y = 4xe2x – e2x + 17
C)
y = 2xe2x + 16
D)
y = 2xe2x – e2x + 17
97)
dy
dx =x4
y; y(0) =4
97)
A)
y2=2x5
5+16
B)
y2=2x5
5+4
C)
y2=x5
5+4
D)
y2=x5
5+16
Solve the problem.
98)
A population of algae consists of 4000 algae at time t = 0. Conditions will support at most 300,000
algae. The rate of growth of algae is proportional both to the number present (in thousands) and to
the difference between 300,000 and the number present (in thousands). Given that the constant of
proportionality is 0.01, a differential equation is
dy
dt = 0.01y(300 – y).
Approximate the number of algae present when t = 2, using h = 0.5.
98)
A)
about 141 thousand
B)
about 125 thousand
C)
about 128 thousand
D)
about 133 thousand
Find the general solution for the differential equation.
99)
dy
dx = y –9
99)
A)
y = Me9x + C
B)
y =9xex+ C
C)
y = Mex+9
D)
y = Me(x +9) + C
Explanation:
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
100)
dy
dx = x + y2; y(0) = 0; find y(0.4)
100)
A)
0.060
B)
0.039
C)
0.101
D)
0.058
Find the general solution for the differential equation.
101)
2x3–2dy
dx =3
101)
A)
y =x4
4+3x + C
B)
y =x4
4–3x
2+ C
C)
y =x4
4+ C
D)
y =4x4+3x
2+ C
Solve the problem.
102)
The table shows the population of a certain city for selected years between 1950 and 2003.
Years after 1950 Population
0 12,421
20 143,112
30 290,089
40 375,297
50 445, 052
55 471,126
By using your calculator to find the logistic regression equation that best fits the data, determine
when the population of the city will first exceed 480,000.
102)
A)
In 2015
B)
In 2020
C)
In 2023
D)
In 2018
Find the particular solution for the initial value problem.
103)
dy
dx =25
x; y(1) =3
103)
A)
y =25 ln x + 3
B)
y = ln x + 1
C)
y = ln x + 3
D)
y =25 ln x + 12.5
Find the general solution for the differential equation.
104)
xdy
dx – 2y –3x = 0
104)
A)
y = – 3+ Cx
B)
y = – 3
x+ C
C)
y = – 3x2+ Cx3
D)
y = – 3x + Cx2
Find the requested equation.
105)
The system of equations
dy
dt = y – 2xy
dx
dt = – x + 3xy
describes the influence of the populations (in thousands) of two competing species on their growth
rates. Find an equation relating x and y, assuming y =8 when x = 1.
105)
A)
3y – ln y = ln x – ln 8+29
B)
3y – ln y = ln x – 2x – ln 8+26
C)
3y – ln y = ln x – 2x – ln 9+29
D)
3y – ln y = ln x – 2x +26
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
106)
dy
dx = 4x + 3; y(1) = 0; find y(1.5)
106)
A)
3.990
B)
4.181
C)
2.770
D)
3.900
Solve the problem.
107)
A rumor spreads through a community of 800 people at the rate
dN
dt = 0.1(800 – N)N1/2
where N is the number of people who have heard the rumor at time t (in hours). Use Euler’s
method with h = 0.5 to find the number who have heard the rumor after 1.5 hours, if 1 person
heard it initially.
107)
A)
about 681 people
B)
about 702 people
C)
about 711 people
D)
about 719 people
Find the general solution for the differential equation.
108)
dy
dx =8xy
108)
A)
y =e4x2+ C
B)
y = Me8x
C)
y = ln 4x2+ C
D)
y = Me4x2
Find the particular solution for the initial value problem.
109)
dy
dx =x1/2y2; y(16) =9
109)
A)
y = – 9
6x1/2 –385
B)
y = – 9
6x3/2 –360
C)
y = – 1
6x3/2 –360
D)
y = – 9
6x3/2 –385
Solve the differential equation subject to the initial condition.
110)
dy
dx – xy – x = 0; y(1) = 10
110)
A)
y = – 1 + 11e–x2
B)
y = – 1
2+11
2ex2– 1
C)
y = – 1
2+ 11e–x2
D)
y = – 1 + 11e(x2–1)/2
Find the particular solution for the initial value problem.
111)
dy
dx + 2x = 3x2; y(0) =11
111)
A)
y = x3– x2+ 11
B)
y = x3+ 2x2+ 11
C)
y = 3x3+ 2x2+ 11
D)
y = 3x3+ x2+ 11
Solve the problem.
112)
Newton’s Law of Cooling states that the rate of change of temperature of an object is proportional
to the difference in temperature between the object and the surrounding medium. Thus, if T is the
temperature of the object after t hours and T0 is the (constant) temperature of the surrounding
medium, then
dT
dt = – k(T –T0),
where k is a constant.
Suppose the air temperature surrounding a body remains at a constant 13° F, k =0.25, and the
temperature of the body at death is 98.6°F. By solving the differential equation, determine a
formula for the temperature at any time t.
112)
A)
T =85.6e0.25t+13
B)
T = 98.6e–0.25t+13
C)
T =85.6e–0.25t+13
D)
T = 98.6e–0.25t–13
113)
Richard deposits $2000 in an IRA at 10% interest compounded continuously for his retirement in 25
years. He intends to make continuous deposits at the rate of $2500 a year. How long will it take for
Richard to accumulate $90,000?
113)
A)
12.5 years
B)
13.6 years
C)
14.5 years
D)
14.1 years
114)
The table shows the population of a certain city for selected years between 1950 and 2003.
Years after 1950 Population
0 7891
20 103,087
30 191,064
40 241,552
50 265,058
55 269,116
By using your calculator to find the logistic regression equation that best fits the data, determine
when the population of the city will first exceed 271,000.
114)
A)
In 2015
B)
In 2018
C)
In 2010
D)
In 2012
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
115)
dy
dx =x
y; y(0) = 2; find y(0.3)
115)
A)
2.015
B)
2.005
C)
2.030
D)
2.024
A person’s weight depends both on the daily rate of energy intake, say C calories per day, and the daily rate of energy
consumption, typically between 12 and 20 calories per pound per day. Using an average value of 16 calories per pound
per day, a person weighing w pounds uses 16w calories per day. If C = 16w, then weight remains constant, and weight
gain or loss occurs according to whether C is greater or less than 16w.
To determine how fast a change in weight will occur, a plausible assumption is that dw/dt is proportional to the net
excess (or deficit) C – 16w in the number of calories per day.
116)
Given that 3500 calories is equivalent to one pound, a differential equation to express this
relationship is dw/dt = (C – 16w)/3500. Solve this differential equation.
116)
A)
w = C –e–0.0046M e–0.0046t
B)
w = C/16 –e–0.0065M e–0.0065t /16
C)
w = C/16 –e–0.0046M e–0.0046t /16
D)
w = C –e–0.0065M e–0.0065t
Answer Key
Testname: C10
Answer Key
Testname: C10
Answer Key
Testname: C10