Exam
Name___________________________________
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Find the general solution for the differential equation.
1)
dy
dx =4x3y –7x6y
1)
A)
y = Me(x4–x7)
B)
y = – x4–x7
C)
y = – 1
x4–x7
D)
y =e(x4–x7)+ C
Solve the problem.
2)
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
Use the logistic regression function on your calculator to determine the logistic equation that best
fits the data.
2)
A)
P =278,715.3
1 + 25.311e–0.1374t
B)
P =254,180.3
1 + 26.118e–0.1402t
C)
P =266,076.8
1 + 23.128e–0.1215t
D)
P =271,976.2
1 + 24.361e–0.1345t
3)
Assume that the rate of change of population of a certain city is given by dy
dt = 6000e0.06t, where y
is the population at time t, in years. The population was 100,000 in 1980 (t = 0 in 1980). Predict the
population in 2010.
3)
A)
600,965
B)
624,965
C)
614,965
D)
604,965
4)
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 dish of lasagna baked at 375°F is taken out of the oven into a kitchen that is 68°F. After 7 minutes,
the temperature of the lasagna is 310°F. What will its temperature be 11 minutes after it was taken
out of the oven? Round your answer to the nearest degree.
4)
A)
288°F
B)
264°F
C)
273°F
D)
279°F
Solve the differential equation subject to the initial condition.
5)
dy
dx +10xy –e–5x2= 0; y(0) =4
5)
A)
y = xe–5x2+4
B)
y = (x +4)e–5x2
C)
y = (x +4)e–5x
D)
y = xe–5x +4
Find the particular solution for the initial value problem.
6)
x dy
dx –7y x = 0; y(0) = 1
6)
A)
y =e14x1/2
B)
y =e7x–1/2 + 1
C)
y =e14x–1/2
D)
y =e14x1/2 + 1
Solve the problem.
7)
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
Use the logistic regression function on your calculator to determine the logistic equation that best
fits the data.
7)
A)
P =494,193.8
1 + 24.126e–0.1084t
B)
P =478,549.6
1 + 21.095e–0.1209t
C)
P =499,107.3
1 + 23.521e–0.0981t
D)
P =482,549.6
1 + 22.095e–0.1132t
8)
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 remain uninfected after 5 days?
8)
A)
9788
B)
9588
C)
9988
D)
12
9)
An influenza epidemic spreads at a rate proportional to the product of the number of people
infected and the number not yet infected. Assume that 80 are infected at the beginning of the
epidemic in a community of 10,000 people and 300 are infected 10 days later. Write an equation for
the number of people infected, y, after t days.
9)
A)
y =10,000
1 +124e–0.13t
B)
y =10,000
1 +9999e–0.13t
C)
y =10,000
1 +124e–0.15t
D)
y =10,000
1 +9999e–0.15t
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
10)
dy
dx = x + y2; y(0) = 0; find y(0.5)
10)
A)
0.141
B)
0.100
C)
0.098
D)
0.061
Find the general solution for the differential equation.
11)
dy
dx =12xy –12x2y
11)
A)
y = ke(6x2–4x3)
B)
y = ln(6x2–4x3) + C
C)
y =6x2–4x3+ C
D)
y = ke(12x –6x2)
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.
12)
Assuming C is constant, a differential equation to express this relationship is dw/dt = k(C – 16w),
where k is the constant of proportionality. The units of dw/dt are pounds per day, and the units of
C – 16w are calories per day. What units must k have?
12)
A)
pounds calories/day2
B)
pounds/day
C)
pounds/calorie
D)
calorie/pounds
Solve the differential equation subject to the initial condition.
13)
dy
dx +5y = 3; y(0) = 1
13)
A)
y =3
5e–5x +2
5
B)
y =2
5e–5x +3
5
C)
y =3
5e5x +2
5
D)
y =2
5e5x +3
5
Solve the problem.
14)
The rate of change in the concentration of a drug with respect to time in a user’s blood is given by
dC
dt = – kC + D(t),
where D(t) is dosage at time t and k is the rate at which the drug leaves the bloodstream. If D(t) is a
constant D, and C(0) =C0, solve the differential equation to find a formula for C(t), the
concentration at time t.
14)
A)
C(t) =
D + (kC0– D)e–kt
k
B)
C(t) =
D + (kC0– D)ekt
k
C)
C(t) =D(1 –e–kt)
k
D)
C(t) =
D + (kC0– 1)e–kt
k
Find the general solution for the differential equation.
15)
dy
dx +5y =10
15)
A)
y =1
2+ Ce5x
B)
y =10 + Ce–5x
C)
y =2+ Ce–5x
D)
y =2+ Ce–10x
Solve the problem.
16)
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 0.00006. 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 natives who remain uninfected after t days.
16)
A)
y =80,991,000
8999 +e0.54t
B)
y =80,991,000
8999 +e–0.54t
C)
y =9000
8999 +e0.54t
D)
y =80,991,000
8999 +e0.00006t
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
17)
dy
dx = 1 – e–x; y(0) = 0; find y(0.4)
17)
A)
0.072
B)
0.054
C)
0.042
D)
0.051
18)
dy
dx = x2– x; y(0) = 0; find y(0.3)
18)
A)
–0.055
B)
–0.046
C)
–0.025
D)
–0.034
Find the general solution for the differential equation.
19)
dy
dx =y2(4 –ex)
19)
A)
y =33
4x –ex+ C
B)
y =4x –ex+ C
C)
y =1
ex–4x + C
D)
y =k
4x –ex
Solve the problem.
20)
Suppose a tank contains 100 gallons of a solution of 10 lb of salt dissolved in water, which is kept
uniform by stirring. 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. How long will it take for the amount of
salt in the mixture to be reduced to 4.5 lb?
20)
A)
Approx. 69 min
B)
Approx. 49 min
C)
Approx. 35 min
D)
Approx. 89 min
21)
Sally plans to make continuous deposits to a savings account. She wants to accumulate $80,000 in
20 years. If the account earns 9% interest compounded continuously and she makes no initial
deposit, what yearly deposit will be required? Round your answer to the nearest dollar.
21)
A)
$1626
B)
$1326
C)
$1526
D)
$1426
Find the general solution for the differential equation.
22)
x2+7xy =7x dy
dx
22)
A)
y =x3
28 + C
B)
y = – x
7+ Cex
C)
y =x – 1
7+ Cex
D)
y = – x + 1
7+ Cex
23)
x dy
dx + 2y =5x2+6x
23)
A)
y =5x3+6x2+C
x
B)
y =5x4
4+2x3+ C
C)
y =5x2+6x +C
x2
D)
y =5x2
4+2x +C
x2
Solve the problem.
24)
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 104°F is placed in a freezer with a temperature of 0°F. After 7
minutes, the temperature of the coffee is 60.7°F. What will its temperature be 11 minutes after it is
placed in the freezer? Round your answer to the nearest degree.
24)
A)
45°F
B)
37°F
C)
41°F
D)
35°F
Find the general solution for the differential equation.
25)
ydy
dx =x2–4x
25)
A)
y2=2
3x3–4x2+ C
B)
y =1
3x3–4x + C
C)
y =2
3x3–4x2+ C
D)
y2=1
3x3–4x2+ C
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.
26)
Assume C is constant and write a differential equation to express this relationship. Use k to
represent the constant of proportionality.
26)
A)
dw
dt = C(k – 16w)
B)
dw
dt = k(C – 16w)
C)
dw
dt = k(16w – C)
D)
dw
dt = k(C + 16w)
Find the particular solution for the initial value problem.
27)
dy
dx = 4xe2x; y(0) =13
27)
A)
y = 2xe2x – e2x + 14
B)
y = 2xe2x + 13
C)
y = 4xe2x – 2e2x + 15
D)
y = 4xe2x – e2x + 14
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
28)
dy
dx = x2y; y(0) = 1; find y(0.5)
28)
A)
1.014
B)
1.003
C)
1.042
D)
1.030
Find the particular solution for the initial value problem.
29)
dy
dx =2–x2
3y +8; y(0) =2
29)
A)
3
2y2+8y =2x –1
3x3+ 2
B)
3
2y2+8y =2x –1
3x3+ 22
C)
3
2y2+8y =2x –x3+ 22
D)
3
2y2+8y =2x –1
2x3+47
2
Solve the problem.
30)
If a population is changed by either immigration or emigration, a model for the population is
dy
dt = ky + f(t),
where y is the population at time t and f(t) is some function of t that describes the net effect of the
emigration/immigration. Assume that k = 0.02 and y(0) = 10,000. Solve this differential equation
for y, given that f(t) = – 13t.
30)
A)
y = – 650t + 32,500 – 22,500e–0.02t
B)
y =650t + 32,500 – 22,500e–0.02t
C)
y =650t + 32,500 – 22,500e0.02t
D)
y = – 650t – 32,500 – 22,500e0.02t
Find the general solution for the differential equation.
31)
dy
dx =y
x5
31)
A)
y = – x4
4+ C
B)
y = – 1
4 ln 1
x4+ C
C)
y = Me–1/(6x6)
D)
y = Me–1/(4x4)
32)
y – x dy
dx =12x3, x > 0
32)
A)
y =6x2+ Cx
B)
y = – 6x2+ C
C)
y = – 12x3+ Cx
D)
y = – 6x3+ Cx
Solve the problem.
33)
Suppose a tank contains 100 gallons of a solution of 10 lb of salt dissolved in water, which is kept
uniform by stirring. 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. How much salt is present in the tank
after 40 minutes?
33)
A)
Approx. 5.1 lb
B)
Approx. 6.9 lb
C)
Approx. 8.3 lb
D)
Approx. 5.9 lb
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
34)
dy
dx – y = 3ex; y(0) = 80; find y(0.4)
34)
A)
119.718
B)
119.140
C)
120.018
D)
118.736
Solve the problem.
35)
If a population is changed by either immigration or emigration, a model for the population is
dy
dt = ky + f(t),
where y is the population at time t and f(t) is some function of t that describes the net effect of the
emigration/immigration. Assume that k = 0.02 and y(0) = 10,000. Solve this differential equation
for y, given that f(t) =4t.
35)
A)
y =200t + 10,000 + 20,000e–0.02t
B)
y =200t – 10,000 + 20,000e–0.02t
C)
y = – 200t – 10,000 + 20,000e0.02t
D)
y = – 200t – 10,000 + 20,000e–0.02t
36)
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 the
limiting size of the population.
36)
A)
271,976
B)
266,078
C)
278, 715
D)
254,180
37)
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 71°F, and at the end of one hour the body temperature is 88°F, what is the
temperature of the body after 4 hours? Round to the nearest tenth of a degree.
37)
A)
71.5°F
B)
75°F
C)
4°F
D)
88°F
38)
The amount of a radioactive substance decreases exponentially, with a decay constant of 3% per
month. Find a general solution to the differential equation which expresses the rate of change.
38)
A)
y = Me–0.03t
B)
y = – 0.03t
C)
y = Me–0.97t
D)
y = – Me0.03t
Solve the differential equation subject to the initial condition.
39)
dy
dx + y = 2ex; y(0) =16
39)
A)
y = 4e2+ 20e–x
B)
y = ex+ 15e–x
C)
y =16ex
D)
y = 2ex+ 13e–x
Find the particular solution for the initial value problem.
40)
x5dy
dx = y; y(1) = 1
40)
A)
y =e–1/(4x4) + 1
B)
y =e–1/x4+ 1
C)
y =e–1/(4x4) + 1/4
D)
y =e–1/(6x6) + 1/6
Solve the problem.
41)
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.
Assume that the temperature of a body at death is 98.6°F, the temperature of the surrounding air is
60°F, and at the end of one hour the body temperature is 88°F. Use Newton’s Law of Cooling to
determine the number of hours that have elapsed when the temperature of the body is 71°F.
41)
A)
4.7 hours
B)
5.1 hours
C)
3.9 hours
D)
6.2 hours
Find the general solution for the differential equation.
42)
dy
dx =y2e7x
42)
A)
y = – 1
7e7x + C
B)
y = – e7x
7+ C
C)
y =1
e7x
7
+ C
D)
y = – 1
e7x
7
+ C
43)
6dy
dx –12xy – x = 0
43)
A)
y = – 1
12 + Cex2/2
B)
y = – 1
12 + Cex2
C)
y = – 1
12 + Ce–x2
D)
y =1
12 + Ce–x2/2
Find the particular solution for the initial value problem.
44)
dy
dx =(x +3)3ey; y(1) = 0
44)
A)
y = – ln 257 –(x +3)2
2
B)
y = – ln 257 –(x +3)4
4
C)
y =e65 –(x +3)4/4
D)
y = – ln 65 –(x +3)4
4
45)
dy
dx = 6x2– 4x + 9; y(1) =13
45)
A)
y = 6x3– 4x2+ 9x + 2
B)
y = 2x3– 2x2+ 9x – 4
C)
y = 2x3– 4x2+ 9x + 6
D)
y = 2x3– 2x2+ 9x + 4
46)
dy
dx =y2
x; y(e) =3
46)
A)
y =3
ln |x|
B)
y = – 3
3 ln |x| –4
C)
y =3 ln |x|
D)
y =3
4 ln |x| –3
Solve the differential equation subject to the initial condition.
47)
dy
dx +3x2y –6xe–x3= 0; y(0) =300
47)
A)
y =6x2ex3+300ex3
B)
y =3xe–x3+300e–x3
C)
y =3x2e–x2+300e–x2
D)
y =3x2e–x3+300e–x3
Find the general solution for the differential equation.
48)
dy
dx =15x2
48)
A)
y =x3
3+ C
B)
y =15x3+ C
C)
y = x3+ C
D)
y =5x3+ C
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
49)
dy
dx =ey+e–x; y(1) = 2; find y(1.3)
49)
A)
12.278
B)
12.704
C)
11.907
D)
12.908
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.
50)
Write a differential equation using the fact that 3500 calories is equivalent to one pound.
50)
A)
dw/dt = (16w – C)/3500
B)
dw/dt = (C – 16w)/3500
C)
dw/dt = 3500(C – 16w)
D)
dw/dt = C(3500 – 16w)
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
51)
dy
dx = e–y+ x; y(0) = 0; find y(0.4)
51)
A)
0.408
B)
0.342
C)
0.502
D)
0.406
52)
dy
dx = x2; y(0) = 2; find y(0.4)
52)
A)
2.020
B)
2.005
C)
2.014
D)
2.026
Find the general solution for the differential equation.
53)
(y6– y) dy
dx = x
53)
A)
7y7– 2y2= 2x2+ C
B)
y7–7y2=7x + C
C)
2y7–7y2=7x2+ C
D)
7y7– 2y = 2x2+ C
Solve the problem.
54)
Sales (in thousands) of a certain product are declining at a rate proportional to the amount of sales,
with a decay constant of 14% per year. By writing and solving a differential equation, determine
how much time will pass before sales become 30% of their original value.
54)
A)
9.3 years
B)
8.1 years
C)
8.6 years
D)
10.1 years
Find the general solution for the differential equation.
55)
dy
dx + 2xy =13x
55)
A)
y =13
5+ Cex2
B)
y =13
2+ Ce–x2
C)
y =13
2+ 2x + Ce–x2
D)
y =13 + Cex2
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
56)
dy
dx = x +y; y(0) = 1; find y(0.4)
56)
A)
1.603
B)
1.491
C)
1.501
D)
1.402
Solve the differential equation subject to the initial condition.
57)
dy
dx + xy =4x; y(0) =1
57)
A)
y = – 3ex2/2 +4
B)
y =4ex2/2 – 3
C)
y = – 3e–x2/2 +4
D)
y =4e–x2/2 – 3
Solve the problem.
58)
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 the
limiting size of the population.
58)
A)
478,550
B)
482,550
C)
499,107
D)
494,194
59)
Suppose a rumor starts among 2 people in an office building. That is, y0=2. Suppose 400 people
work in the building and 40 people have heard the rumor in 3 days. Write an equation for the
number who have heard the rumor in t days. Assume that the rate of spread of the rumor satisfies
the logistic equation:
dy
dt = k 1 –y
Ny,
where N is the size of the population and y is the number who have heard the rumor after t days.
59)
A)
y =400
1 +199e–0.914t
B)
y =400
1 +399e–0.914t
C)
y =400
1 +199e–1.032t
D)
y =400
1 +399e–1.032t
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
60)
dy
dx = x2+ y2; y(0) = 1; find y(0.4)
60)
A)
1.689
B)
1.573
C)
1.569
D)
1.462
Find the general solution for the differential equation.
61)
xdy
dx +2xy –x2= 0
61)
A)
y =x
2–1
4+ Ce–2x
B)
y =x
2+1
2+ Ce2x
C)
y =x
2–1
2+ Ce–2x
D)
y =x
2–1
4+ Ce2x
Use Euler’s method to approximate the indicated function value to three decimal places using h = 0.1.
62)
dy
dx = 4x + 3; y(1) = 2; find y(1.4)
62)
A)
3.630
B)
5.040
C)
2.770
D)
3.080
63)
dy
dx = 1 + y; y(0) = 2; find y(0.5)
63)
A)
3.864
B)
4.561
C)
3.427
D)
3.832
Solve the problem.
64)
The population of a country is predicted to increase from 21.5 million in 2000 to 49.5 million in
2050. Assuming the unlimited growth model dy/dt = ky fits this population growth, express the
population y as a function of the year t. Let 2000 correspond to t = 0.
64)
A)
y =49.5
1 +e–0.01468t
B)
y =49.5e0.01668t
C)
y =21.5e0.01668t
D)
y =21.5e0.01468t
Find the general solution for the differential equation.
65)
dy
dx =18x2– 14x
65)
A)
y =18x3– 7x2+ C
B)
y =18x3– 14x2+ C
C)
y =6x3– 7x2+ C
D)
y =6x3– 14x2+ C
Solve the problem.
66)
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 much will he have
accumulated in 18 years? Round your answer to the nearest dollar.
66)
A)
$140,340
B)
$145,340
C)
$151,340
D)
$138,340
Find the general solution for the differential equation.
67)
dy
dx = x – 15
67)
A)
y =x2
2– x + C
B)
y = 2x2– 15 + C
C)
y = x3– 15x + C
D)
y =x2
2– 15x + C