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```Differential Equations
Big Ideas
Slope fields – draw a slope field, sketch a particular solution
Separation of variables – separable differential equations
General solution Particular solution
Growth decay problems
Antidifferentiation by substitution
Antidifferentiation by parts
Winplot graphing program is very good for drawing slope fields.
http://math.exeter.edu/rparris/winplot.html
Domain Issues
The Domain of Solutions to Differential Equations by Larry Riddle (AP Central)
http://apcentral.collegeboard.com/apc/members/repository/ap07_calculus_DE_domain_fin.pdf
Definition: The solution of a differential equation is a differentiable function on an open interval that contains the
initial x-value.
For all parts of the domain, the derivative of the explicit solution does not contradict the original differential
equation
The derivative exists for all values in its domain
Mike Koehler
6-1
Differential Equations
Mike Koehler
6-2
Differential Equations
Remember the Domain!
1.
Find a solution y = f ( x) to the differential equation
dy 1
2.
= satisfying y (−1) =
dx x
2.
Find a solution y = f ( x) to the differential equation
dy 1 2
= y satisfying y (1) = 1 .
dx 3
3.
Find a solution y = f ( x) to the differential equation =
4.
Find a solution y = f ( x) to the differential equation = ( y + 2 ) 2 satisfying y (0) = −1 .
dy
dx
Mike Koehler
dy
dx
6-3
1 − y 2 satisfying y (1) = 0 .
3
Differential Equations
Mike Koehler
6-4
Differential Equations
AP Multiple Choice Questions
2008 AB Multiple Choice
22 23 27
2008 BC Multiple Choice
27
2003 AB Multiple Choice
12.
The rate of change of the volume, V , of water in a tank with respect to time, t , is directly proportional to
the square root of the volume. Which of the following is a differential equation that describes this
relationship?
A)
B) V (t ) = k V
V (t ) = k t
C)
E)
19.
dV
=k t
dt
dV
=k V
dt
D)
dV
k
=
dt
V
A curve has slope 2 x + 3 at each point ( x, y ) on the curve. Which of the following is an equation for this
curve if it passes through the point (1, 2 ) ?
A)
=
y 5x − 3
y x2 + 1
B) =
D)
y = x 2 + 3x − 2
E)
y x 2 + 3x
C) =
y = 2 x 2 + 3x − 3
2003 BC Multiple Choice
14.
Shown to the right is a slope field for which of the
following differential equations?
A)
dy x
=
dx y
B)
dy x 2
=
dx y 2
D)
dy x 2
=
dx
y
E)
dy x 3
=
dx y 2
Mike Koehler
C)
6-5
dy x 3
=
dx
y
Differential Equations
1998 AB Multiple Choice
21.
dy
= ky and k is a nonzero constant, then y could be
dt
If
A)
84.
2e kty
B)
2e kt
C)
e kt + 3
D)
kty + 5
1 2 1
ky +
2
2
E)
dy
= ky , where k is a constant and t is measured in years. If
dx
the population doubles every 10 years, then the value of k is
A)
0.069
B)
.0200
C) 0.301
D) 3.322
E) 5.000
Population y grows according to the equation
1998 BC Multiple Choice
8.
π
dy
when x
and if y 0=
, what is the value of y when x = 0 ?
= sin( x) cos 2 ( x) =
2
dx
1
1
−
A) -1
B)
C) 0
D)
E) 1
3
3
If
1993 AB Multiple Choice
33.
42.
dy
=
2 y 2 and if y =
−1 when x =
1, then when x =
2, y =
dx
1
2
−
B)
C) 0
D)
A) −
3
3
If
1
3
E)
2
3
A puppy weighs 2.0 pounds at birth and 3.5 pounds two months later. If the weight of the puppy during its
first 6 months is increasing at a rate proportional to its weight, then how much will the puppy weigh when it
is 3 months old?
A)
4.2 pounds
B) 4.6 pounds
C) 4.8 pounds
D)
5.6 pounds
E) 6.5 pounds
1993 BC Multiple Choice
13.
38.
dy
= x 2 y, then y could be
dx
x3
 x
B)
A) 3ln  
3
e
+7
3
 
If
C)
2e
x3
3
D) 3e 2 x
E)
x3
+1
3
During a certain epidemic, the number of people that are infected at any time increases at a rate
proportional to the number of people that are infected at that time. If 1,000 people are infected when the
epidemic is first discovered, and 1,200 are infected 7 days later, how many people are infected 12 days after
the epidemic is first discovered?
A)
343
B)
1,343
C) 1,367
D) 1,400
E) 2,057
Mike Koehler
6-6
Differential Equations
1988 BC Multiple Choice
39.
43.
dy
If
=
y sec 2=
x and y 5=
when x 0,=
then y
dx
A) e tan( x ) + 4
B) e tan( x ) + 5
D) tan( x) + 5
E) tan( x) + 5e x
C) 5e tan( x )
Bacteria in a certain culture increase at a rate proportional to the number present. If the number of bacteria
doubles in three hours, in how many hours will the number of bacteria triple?
 27 
9
3ln 3
2 ln 3
ln 3
ln  
E)
A)
B)
C)
D) ln  
ln 2
ln 2
ln 2
 2 
2
1985 BC Multiple Choice
At each point ( x, y ) on a certain curve, the slope of the curve is 3x 2 y . If the curve contains the point
( 0,8) , then its equation is
44.
A)
y = 8e x
D)
y= ln ( x + 1) + 8
3
B)
=
y x3 + 8
E)
2
y=
x3 + 8
C) =
y ex + 7
3
Integration by substitution Multiple Choice
2003
2ab
∫
1
0
e −4 x dx =
A)
−e −4
4
B)
−4e −4
C)
e −4 − 1
D)
1 e −4
−
4 4
2
2003
2 x + 1, ∫ 2 x + 1dx is equivalent to
Using the substitution u =
0
11ab
1 12
1 2
A)
B)
u du
u du
∫
−
1
2
2
2 ∫0
D)
∫
2
0
u du
E)
∫
5
1
4 − 4e −4
1 5
u du
2 ∫1
u du
1998 What is the average value=
of y x 2 x 3 + 1 on the interval [ 0, 2] ?
27ab
26
52
26
52
B)
C)
D)
A)
9
9
3
3
Mike Koehler
C)
E)
6-7
E)
24
Differential Equations
1998
If f is a continuous function and if F '( x) = f ( x) for all real numbers x , then
82ab
1
1
2 F (3) − 2 F (1)
A)
B)
F (3) − F (1)
2
2
2 F (6) − 2 F (2)
C)
D) F (6) − F (2)
1993
7bc
e−t + C
B)
e
e tan x
∫ cos2 x dx =
A)
0
B)
B)
∫
1
0
t
D)
2e 2 + C
E)
et + C
1
C)
e −1
D)
e
E)
e +1
1
e
4
C)
e −1
D)
e
E)
4 ( e − 1)
4
1
( e − 1)
4
x
∫
3x 2 + 5
dx =
3
1
3x 2 + 5) 2 + C
(
9
1
1
3x 2 + 5) 2 + C
(
3
D)
1985
32ab
t
e2 + C
+C
x 3 e x dx =
A)
1985
30ab
t
2
C)
−
π
4
0
A)
1988
7ab
f (2 x) dx =
t
1
2
e
dt =
2∫
A)
1997
18ab
3
1
1
1
F (6) − F (2)
2
2
E)
1997
6ab
∫
3
1
3x 2 + 5) 2 + C
(
4
E)
1
3
3x 2 + 5) 2 + C
(
2
C)
1
1
3x 2 + 5) 2 + C
(
12
C)
1
ln cos ( 2 x ) + C
2
tan ( 2x ) dx =
∫
A)
−2 ln cos ( 2x ) + C
D)
2 ln cos ( 2x ) + C
∫
B)
π 3
0
A)
Mike Koehler
1
B) − ln cos ( 2 x ) + C
2
1
E)
sec ( 2 x ) tan ( 2 x ) + C
2
sin ( 3x ) dx =
-2
B)
−
2
3
C)
0
D)
6-8
2
3
E)
2
Differential Equations
1985
3bc
∫
x +1
dx =
x + 2x
2
2
1
ln 8 − ln 3
A)
1985
40bc
1973
21ab
∫ ( x + 1) e
1
x2 + 2 x
0
3ln 2
2
E)
3ln 2 + 2
2
∫
12
∫
2x
1 − x2
0
1−
B)
e3 − 1
2
C)
∫
2
1
1− u2
du
2u
B)
1 3
ln
2 4
C)
e4 − e
2
D)
e3 − 1
E)
e4 − e
E)
2− 3
dx =
3
2
π
D)
6
π
6
−1
x 4 − x 2 dx =
(4 − x )
2 32
3
−
D)
∫
C)
dx =
e3
2
A)
1969
43ab
D)
ln 8
2
A)
1973
20bc
C)
 x
1−  
4
x
 2  dx =
If the substitution u = is made, the integral ∫
2
2
x
2
2
2 1− u
4 1− u
B) ∫
A)
du
∫1 u du
2
u
2
2
2 1− u
4 1− u
E)
D)
du
∫1 4u
∫ 2 2u du
A)
1973
27ab
ln 8 − ln 3
2
B)
+C
x2 ( 4 − x2 )
3
B) − ( 4 − x 2 )
32
32
(4 − x )
−
C)
x2 ( 4 − x2 )
3
32
+C
2 32
+C
E)
3
+C
sin ( 2 x + 3) dx =
A)
D)
Mike Koehler
1
cos ( 2 x + 3) + C
2
1
− cos ( 2 x + 3) + C
2
B) cos ( 2 x + 3) + C
E)
C)
− cos ( 2 x + 3) + C
1
− cos ( 2 x + 3) + C
5
6-9
Differential Equations
Integration by parts Multiple Choice
2003
23bc
∫ x sin(6 x)dx =
− x cos(6 x) + sin(6 x) + C
A)
x
1
− cos(6 x) + sin(6 x) + C
6
6
6 x cos(6 x) − sin(6 x) + C
C)
E)
1993
43ab
1997
25ab
A)
x f ( x) − ∫ x f ′( x) dx
B)
D)
x f ( x) − ∫ f ′( x) dx
E)
∫
x
1
− cos ( 2 x ) + sin ( 2 x ) + C
2
4
x
1
cos ( 2 x ) − sin ( 2 x ) + C
2
4
−2 x cos ( 2 x ) − 4sin ( 2 x ) + C
E)
∫
x2
x2
f ( x) − ∫
f ′( x) dx
2
2
x2
f ( x) dx =
2 ∫
C)
x f ( x) −
x2
f ( x) + C
2
x sin ( 2 x ) dx =
C)
1988
16bc
D)
x
1
− cos(6 x) + sin(6 x) + C
6
36
x
1
cos(6 x) + sin(6 x) + C
6
36
∫ x f ( x) dx =
A)
1997
84bc
B)
B)
x
1
− cos ( 2 x ) − sin ( 2 x ) + C
2
4
D)
−2 x cos ( 2 x ) + sin ( 2 x ) + C
x 2 sin x dx =
A)
− x 2 cos x − 2 x sin x − 2 cos x + C
B)
C)
− x 2 cos x + 2 x sin x + 2 cos x + C
D)
E)
2 x cos x + C
∫ xe
A)
D)
2x
− x 2 cos x + 2 x sin x − 2 cos x + C
x3
cos x + C
3
dx =
xe 2 x e 2 x
−
+C
2
4
xe 2 x e 2 x
+
+C
2
2
B)
E)
xe 2 x e 2 x
−
+C
2
2
x 2 e2 x
+C
4
1985 If
− f ( x) cos x + ∫ 3 x 2 cos x dx, then f ( x) could be
∫ f ( x) sin x dx =
21bc
A)
B)
C) − x 3
D) sin x
3x 2
x3
(insightful)
Mike Koehler
6 - 10
C)
xe 2 x e 2 x
+
+C
2
4
E)
cos x
Differential Equations
AP Free Response Questions
2012 AB5
The rate at which a baby bird gains weight is proportional
to the difference between its adult weight and its current
weight. At time t = 0 , when the bird is first weighed, its
weight is 20 grams. If B (t ) is the weight of the bird, in
grams, at time t days after it is first weighed, then
dB 1
=
(100 − B ) .
dt 5
Let y = B (t ) be the solution to the differential equation
above with initial condition B (0) = 20 .
a) Is the bird gaining weight faster when it weighs 40 grams or when it weighs 70 grams? Explain your
reasoning.
b)
d 2B
d 2B
Find
in
terms
of
B.
Use
to explain why the graph of B cannot resemble the graph above.
dt 2
dt 2
c) Use separation of variables to find y = B (t ) , the particular solution to the differential equation with initial
condition B (0) = 20 .
2011 AB5
At the beginning of 2010, a landfill contained 1400 tons of solid waste. The increasing function W models the total
amount of solid waste stored at the landfill. Planners estimate that W will satisfy the differential equation
dW
1
=
(W − 300 ) for the next 20 years. W is measured in tons, and t is measured in years from the start of 2010.
dt
25
a)
b)
c)
Use the line tangent to the graph of W at t = 0 to approximate the amount of solid waste that the landfill
1

contains at the end of the first 3 months of 2010  time t =  .
4

2
2
dW
dW
in terms of W . Use
Find
to determine whether your answer in part (a) is an underestimate or
2
dt 2
dt
1
an overestimate of the amount of solid waste that the landfill contains at time t = .
4
dW
1
Find the particular solution W = W (t ) to the differential equation=
(W − 300 ) with initial condition
dt
25
W (0) = 1400 .
Mike Koehler
6 - 11
Differential Equations
2010 AB6
dy
d2y
y 3 1 + 3 x 2 y 2 . Let y = f ( x) be a particular
Solutions to the differential equation = xy 3 also satisfy =
dx
dx 2
dy
solution to the differential equation
= xy 3 with f (1) = 2 .
dx
a) Write an equation for the line tangent to the graph
of y f=
=
( x) at x 1 .
b) Use the tangent line equation from part (a) to approximate f (1.1) . Given that f ( x) > 0 for 1 < x < 1.1 , is the
(
c)
)
approximation for f (1.1) greater than or less than f (1.1) ? Explain your reasoning.
Find the particular solution y = f ( x) with initial condition f (1) = 2 .
2008 AB5
Consider the differential equation
dy y − 1
= 2 , where x ≠ 0 .
dx
x
a)
b)
On the axes provided, sketch a slope field for the given differential equation at the nine points indicated.
Find the particular solution y = f ( x) to the differential equation with the initial condition f (2) = 0 .
c)
For the particular solution y = f ( x) described in part (b), find lim f ( x)
x →∞
2006 AB 5
Consider the differential equation
a)
b)
dy 1 + y
, where x ≠ 0 .
=
dx
x
On the axis provided, sketch a slope field for the given differential equation at the eight points indicated.
Find the particular solution y = f ( x) to the differential equation with the initial condition f (−1) =
1 and state
its domain.
Mike Koehler
6 - 12
Differential Equations
2005 AB 6
Consider the differential equation
a)
b)
dy
2x
= − .
dx
y
On the axis provided, sketch a slope field for the given differential equation at the twelve points indicated.
Let y = f ( x) be the particular solution to the differential equation with the initial condition f (1) = −1 . Write
an equation for the line tangent to the graph of f at (1, −1) and use it to approximate f (1.1) .
c)
2004
Find the particular solution y = f ( x) to the given differential equation with the initial condition f (1) = −1 .
AB6
dy
Consider the differential equation = x 2 ( y − 1) .
dx
a)
b)
c)
On the axis provided, sketch a slope field for the given differential equation at the twelve points indicated.
While the slope field in part (a) is drawn for only twelve points, it is defined at every point in the xy-plane.
Describe all points in the xy-plane for which the slopes are positive.
Find the particular solution y = f ( x) to the given differential equation with the initial condition f (0) = 3.
Mike Koehler
6 - 13
Differential Equations
2003
AB5
A coffeepot has the shape of a cylinder with radius 5 inches. Let h be the depth of the coffee in the pot, measured
in inches, where h is a function of time t , measured in seconds. The volume V of coffee in the pot is changing at
the rate of −5π h cubic inches per second.
(The volume V of a cylinder with radius r and height h is V = π r 2 h .)
a)
dh
h
Show that
.
= −
dt
5
b)
dh
h
Given that h = 17 at time t = 0 solve the differential equation
for h as a function of t .
= −
dt
5
c) At what time t is the coffeepot empty?
2000
AB6
Consider the differential equation
dy 3 x 2
=
.
dx e 2 y
a)
Find a solution y = f ( x) to the differential equation satisfying f (0) =
b)
Find the domain and range of the function f found in part (a).
1998
1
.
2
AB4
Let f be a function with f (1) = 4 such that for all points ( x, y ) on the graph of f the slope is given by
a)
b)
c)
d)
3x 2 + 1
.
2y
Find the slope of the graph of f at the point where x = 1 .
Write an equation for the line tangent to the graph of f at x = 1 and use it to approximate f (1.2) .
Find f ( x) by solving the separable differential equation
dy 3 x 2 + 1
with initial condition f (1) = 4
=
dx
2y
Use your solution from part c to find f (1.2) .
1993 AB 6
Let P (t ) represent the number of wolves in a population at time t years, when t ≥ 0 . The population P (t ) is
increasing at a rate directly proportional to 800 − P(t ) , where the constant of proportionality is k .
a) If P (0) = 500, find P(t ) in term of t and k .
b)
c)
If P (2) = 700, find k .
Find lim P (t ) .
Mike Koehler
t →∞
6 - 14
Differential Equations
1989
AB6
Oil is being pumped continuously from a certain oil well at a rate proportional to the amount of oil left in the well;
dy
that is,
= ky , where y is the amount of oil left in the well at any time t . Initially there were 1,000,000 gallons
dt
of oil in the well, and 6 years later there were 500,000 gallons remaining. It will no longer be profitable to pump oil
when there are fewer than 50,000 gallons remaining.
a) Write an equation for y , the amount of oil remaining in the well at any time t .
b) At what rate is the amount of oil in the well decreasing when there are 600,000 gallons of oil remaining?
c) In order not to lose money, at what time t should oil no longer be pumped from the well?
2000 BC6
dy
2
Consider the differential equation given by = x ( y − 1) .
dx
a)
b)
c)
d)
On the axis provided, sketch a slope field for the differential equation at the eleven points indicated.
Use the slope field for the given differential equation to explain why a solution could not have the graph
shown in the figure on the right above.
Find the particular solution y = f ( x) to the given differential equation with the initial condition f (0) = −1 .
Find the range of the solution fount in part (c).
Mike Koehler
6 - 15
Differential Equations
Mike Koehler
6 - 16
Differential Equations
Textbook Problems
Calculus, Finney, Demanna, Waits, Kennedy; Prentice Hal, l2012
Section
7.1
7.2
7.3
QQ p 353
7.4
QQ p376
7.R
Questions
35-40 59 60
69 79
33 34 35
1234
24 25
12
39-42 60
Handout Problems
Mike Koehler
6 - 17
Differential Equations
Mike Koehler
6 - 18
Differential Equations
AP Calculus
Chapter 6 Section 1 Slope Fields
A slope field is a plot of short line segments with slope f ( x, y ) for a lattice of points in the plane.
Slope fields enable us to graph solution curves without solving the differential equation.
Let
dy
= f ( x, y=
) xy + y . Sketch the slope field at the points indicated on the axis provided.
dx
For the point (1,1), the slope is equal to 1 ⋅1 + 1 =2 . Draw a short line segment with slope 2 through the point (1,1).
Repeat for each of the lattice points in the graph below.
Draw a possible graph for the function f with the given slope field that goes through the point (0,1).
Mike Koehler
6 - 19
Differential Equations
Mike Koehler
6 - 20
Differential Equations
AP Calculus
Chapter 6 Section 1 Slope Fields
Draw the slope field for each of the following differential equations.
dy
dy
2.
1.
= 2y
= x +1
dx
dx
3.
dy
= x+ y
dx
4.
dy
= 2x
dx
5.
dy
= y −1
dx
6.
dy
y
= −
dx
x
Mike Koehler
6 - 21
Differential Equations
Match the following differential equations to their slope fields.
dy
= x + .5 y
dx
dy 2
=
dx y
i.
ii.
dy
= −x
dx
dy
= y−x
dx
dy
= .5 y
dx
iii.
iv.
v.
4
-4
-3
-2
4
3
3
2
2
1
1
1
-1
2
3
4
5
-4
-3
-2
-1
-1
-2
-2
-3
-3
-4
A.
-3
-2
3
2
2
1
1
1
2
3
4
5
-4
-3
-2
1
-1
-1
-1
-2
-2
-3
-3
-4
C.
3
4
5
4
3
-1
2
-4
B.
4
-4
1
-1
D.
2
3
4
5
-4
4
3
2
1
-4
-3
-2
1
-1
2
3
4
5
-1
-2
-3
E.
Mike Koehler
-4
6 - 22
Differential Equations
AP Calculus
Chapter 6 Section 2
Match the following integrals to one of the following types:
A)
1
∫ u du
B) ∫ u n du
C)
∫ e du
u
D) Other
Identify u for each integral.
dx
u=
2.
∫ x +1
dx
u=
u=
4.
ln x
2
∫ 2 x dx
u=
u=
6.
∫
dx
u=
8.
∫
2
u=
10.
∫ 3+ e
u=
12.
∫x
u=
1.
∫ −2e
3.
1
2
5.
∫x
1
7.
∫
x2
9.
∫ tan x sec
11.
∫ ( x + 1)
13.
2 3
∫ e x dx
15.
∫
17.
∫ cos
19.
∫ 1 + cos x dx
21.
∫ x cos(1 − x
−2 x
dx
∫ x ln x
dx
ln x
x +5
3
1
2
1
1+ x
x dx
dx
dx
2 tan 2 x
dx
2
2x
sin x
sin −1 x
2
) dx
23.
∫
25.
ex
∫ (2e x − 5)3 dx
Mike Koehler
1 − x2
dx
2x
x2 − 1
−2 x + 2
ex
2
−2 x
e− x
dx
u=
dx
u=
dx
u=
3x
dx
−4
u=
14.
e x + e− x
∫ e x − e− x dx
u=
u=
16.
∫ cos x(e
u=
u=
18.
− ∫ sin x cos 4 x dx
u=
20.
∫
u=
22.
∫
u=
24.
∫ ln(cos x) tan x dx
u=
u=
26.
2e x
∫ (e x + 3) dx
u=
6 - 23
−x
2
sin x
) dx
ln 6 x
dx
x
1+1 x
x2
u=
u=
dx
u=
Differential Equations
AP Calculus
Chapter 6 Section 2
Match the following integrals to one of the following types:
A)
1
∫ u du
B) ∫ u n du
C)
∫ e du
u
D) Other
Identify u for each integral.
dx
u = −2 x
2.A
∫ x +1
dx
u= x + 1
u = ln x 2
4.B
ln x
2
∫ 2 x dx
u = ln x
u = ln x
6.B
∫
=
u x3 + 5
8.C
∫
u = tan x
10.A
∫ 3+ e
u= x + 1
12.A
∫x
u=x
14.A
e x + e− x
∫ e x − e− x dx
u= e x − e − x
u= 1+ x
16.C
∫ cos x(e
u = sin x
u = tan 2 x
18.B
− ∫ sin x cos 4 x dx
u = 1 + cos x
20.B
∫
u = 1 − x2
22.B
∫
u = sin −1 x
24.B
∫ ln(cos x) tan x dx
u = ln(cos x)
=
u 2e x − 5
26.A
2e x
∫ (e x + 3) dx
u= e x + 3
1.C
∫ −2e
3.A
1
2
5.B
∫x
1
7.B
∫
x2
9.B
∫ tan x sec
11.B
∫ ( x + 1)
13.B
2 3
∫ e x dx
15.B
∫
17.B
∫ cos
19.A
∫ 1 + cos x dx
21.D
∫ x cos(1 − x
−2 x
dx
∫ x ln x
dx
ln x
dx
x +5
3
1
2
1
1+ x
2
x dx
dx
dx
2 tan 2 x
dx
2
2x
sin x
sin −1 x
2
) dx
23.B
∫
25.B
ex
∫ (2e x − 5)3 dx
Mike Koehler
1 − x2
dx
6 - 24
2x
x2 − 1
−2 x + 2
e
x2 − 2 x
e− x
dx
=
u x2 − 1
dx
=
u x2 − 2x
dx
u= 3 + e − x
3x
dx
−4
=
u x2 − 4
−x
2
sin x
) dx
ln 6 x
dx
x
1+1 x
x2
u = cos x
u = ln x
dx
u= 1+1 x
Differential Equations
AP Calculus
Chapter 6 Section 2
Part 1: Calculate du for the given function.
1. u =
1 − x2
2. u =
sin x
3. u =
x3 − 2
4. u =
2 x4 + 8x
5. u =
cos( x 2 )
6. u =
tan x
Part 2: Write the integral in terms of u and du . Then evaluate the integrals. Final answer should be in terms of x .
1.
∫ ( x − 7) dx
4.
∫ x( x + 1)
7.
u=
x−7
2.
∫ 2x
dx u =
x +1
5.
∫ sin(2 x − 4)dx
3
∫
9
x +1
(x
2
+ 2x)
10.
∫
13.
∫ x cos( x
2x − 4
u=
3.
∫ ( x + 1)
6.
∫
−2
dx u =
x +1
x3
(x
4
+ 1)
4
dx u =
x4 + 1
( )
1
x
dx u =
x2 + 2x
dx u =
sin x dx u =
x
8. ∫
8x + 5
9. ∫
3
x
(8x + 5)
3
4 x − 1dx u =
4x −1
2
x 2 + 1dx u =
x2 + 1
x2
)dx u
∫x
11.
4x −1
4 x − 1dx u =
∫ (1 + sin x )
12.
9
cos xdx u = 1 + sin x
=
=
15. ∫ sec 2 ( x) tan( x)dx u tan( x)
14. ∫ sin 2 ( x) cos( x)dx u sin(
x)
Part 3: Evaluate the indefinite integrals.
1.
∫ ( 4 x + 3)
4.
∫ sin ( x − 7 ) dx
7.
∫ ( x + 9)
10.
1
∫ ( 3x
2
2
4
dx
dx
+ 1)( x3 + x ) dx
2
13.
∫ ( 3x + 9 )
16.
∫x
19.
∫ sec ( 4 x + 9 ) dx
22.
∫ (1 + sin 2 x )
10
2
dx
sin ( x3 ) dx
2
cos ( 2 x )
Mike Koehler
2
dx
2.
∫ x (x
5.
∫x
x 2 − 4dx
8.
∫
x
11.
2
3
+ 1) dx
3
x2 + 9
∫
dx
5x4 + 2 x
(x
+x
5
)
2 3
dx
1
3.
∫
6.
∫ ( 2 x + 1) ( x
9.
∫
dx
x−7
( 4x
3
+ 3x 2 )
12.
∫ x (x
15.
∫ x ( x + 1)
2
∫ sin ( x ) cos ( x ) dx
18.
∫x
20.
∫ sec ( x ) tan ( x ) dx
21.
∫ sin 4 x
23.
∫ cos x ( 3sin x − 1) dx
24.
∫
2
4
6 - 25
dx
4
17.
5
2
+ 1) dx
3
∫ x ( 3x + 9 )
dx
3
2 x2 + x
14.
10
+ x ) dx
2
2
7
2
sin ( x 3 + 1) dx
cos x
x
dx
cos 4 x + 1 dx
dx
Differential Equations
AP Calculus
Chapter 6 Section 2 Answers
Part 1:
1. du 2=
2. du cos
3. du 3 x 2 dx
xdx
=
xdx
4. du =
5. du =
−2 x sin( x 2 )dx
6. du =
sec 2 xdx
(8x3 + 8) dx
Part 2:
1.
∫ u du = ( x − 7 )
4.
∫ ( u − 1) u du = ∫ u
3
1
4
9
4
+C
10
2.
− u 9 du =
1
2
du =
2
3
(x
2
+ 1) + C
3
2
3.
∫u
−2
du =
−1
+C
x +1
1
1
11
10
( x + 1) − ( x + 1) + C
11
10
−3
1 −4
−1
u du = ( x 4 + 1) + C
4∫
12
1 1 u −5
1
1 
5
−1
−2 
−2
−3
8.
 du = ∫ u − 5u du =  −1( 8 x + 5 ) + ( 8 x + 5 )  + C
3 
∫
8 u  8 
64
64 
2

5.
7.
∫u
1 1
−1
du =
+C
3
2
∫
2
2 u
4 ( x + 2x)
1
∫ sin udu =− 2 cos(2 x − 4) + C
6.
9. 2 ∫ sin udu =
−2 cos x +C
3
10.
1 12
1
u du =
( 4 x − 1)
∫
4
6
13.
1
1
cos(u )du =
sin ( x 2 ) + C
2∫
2
2
+C
Part 3:
1
5
1.
( 4 x + 3) + C
20
4. − cos ( x − 7 ) + C
7. −
10.
13.
15.
16.
19.
22.
1
( x + 9)
+C
3
1 3
x + x) + C
(
3
1
11
( 3x + 9 ) + C
33
1
2
1
6
10
9
( x + 1) + ( x + 1) + ( x + 1) + C
10
9
6
1
− cos ( x3 ) + C
3
1
tan ( 4 x + 9 ) + C
4
1
−1
− (1 + sin 2 x ) + C
2
Mike Koehler
3
1
5
3 
1  u − 1  12
1
1 2
2
2
2
u 2 − u 2 du =

 u du =
 ( 4 x − 1) − ( 4 x − 1)  + C
∫
∫
4  4 
16
16  5
3

1
10
12. ∫ u 9 du = (1 + sin x ) + C
10
1
2
2
3
1
14. ∫ u du =
sin ( x) + C
15. =
3
∫ u du 2 tan x + C
11.
4
1 3
( x + 1) + C
12
3
1 2
x − 4) 2 + C
5.
(
3
2.
8.
x2 + 9 + C
11. −
14.
1
1
+C
5
2 ( x + x 2 )2
3. 2 x − 7 + C
4
1 2
x + x) + C
(
4
−1
1
9. − ( 4 x 3 + 3 x 2 ) + C
6
6.
12.
5
1 3
x + 1) + C
(
15
1
1
12
11
( 3x + 9 ) − ( 3x + 9 ) + C
108
11
1 6
sin ( x ) + C
6
1 5
20.
tan ( x ) + C
5
1
2
23.
( 3sin x − 1) + C
6
17.
6 - 26
1
18. − cos ( x3 + 1) + C
3
3
1
21. − ( cos 4 x + 1) 2 + C
6
24. 2sin x + C
Differential Equations
AP Calculus
Chapter 6 Section 4
1.
Consider the differential equation
dy 3 − x
.
=
dx
y
Let y = f ( x) be the particular solution to the given differential equation for 1 < x < 5 such that the line
y = −2 is tangent to the graph of f . Find the x-coordinate of the point of tangency, and determine whether f has a
local maximum, local minimum, or neither at this point. Justify your answer.
a.
Let y = g ( x) be the particular solution to the given differential equation for −2 < x < 8 , with the initial
condition g (6) = −4 . Find y = g ( x) .
b.
2.
Let f be the function satisfying f ′( x) = x f ( x) for all real numbers x, where f (3) = 25 .
a.
Find f ′′(3).
b.
Write an expression for y = f ( x) by solving the differential equation
dy
= x y with the initial condition
dx
f (3) = 25
dy
2
Consider the differential equation = ( y − 1) cos (π x ) .
dx
a.
On the axes provided, sketch a slope field for
the given differential equation at the nine points
indicated.
3.
b.
There is a horizontal line with equation y = c that satisfies this differential equation. Find the value of c .
c.
Find the particular solution y = f ( x) to the differential equation with the initial condition f (1) = 0 . Show
Mike Koehler
6 - 27
Differential Equations
AP Calculus
Chapter 6 Section 2 Answers
dy
a. = 0=
when x 3.
dx
Use second derivative test or first derivative test to justify that the function has a minimum at x = 3 .
1.
b.
y=
− 6 x − x 2 + 16
2.
a.
=
f ′′( x)
b.
11 
1 2
1
y =  x2 +  =
x + 11
4
4
16


f ( x) +
2
3.
x2
19
. Show the work that leads to this answer. f ′′(3) =
.
2
2
(
)
2
a.
b.
The line y = 1 satisfies the differential equation so c = 1 .
c.
y = 1−
Mike Koehler
π
for - ∞ < x < ∞
sin (π x ) + π
6 - 28
Differential Equations
AP Calculus
Chapter 6
1.
Shown on the right is a slope field for which of the
following differential equations?
dy
= xy
dx
dy
= xy + x
dx
A)
D)
B)
E)
dy
= xy − y
dx
dy
3
= ( x + 1)
dx
C)
dy
= xy + y
dx
x
dx =
−4
2.
∫x
3.
∫x
4.
∫ xe
5.
Let R be the region between y = e −2 x and the x-axis for 1 ≤ x ≤ 4 . Find the area of R . Solve analytically.
6.
Solve the differential equation
2
2
cos ( x 3 )dx =
x2
dx =
dy x 2
with initial condition y (3) = −2 .
=
dx
y
7.
The rate of change of the volume, V , of water in a tank with respect to time, t , is directly proportional to
the square root of the volume. Write a differential equation that describes this relationship.
8.
The slope of the line tangent to the curve y = f ( x) is given by
point ( 2,1) , find the positive value of x when y = −1 .
9.
Solve the differential equation
Mike Koehler
dy
= xy 2 . If the curve passes through the
dx
dy
= −2 xy and y (1) = 4 .
dx
6 - 29
Differential Equations
dy
= ky , where k is a
dt
constant and y is the number of bacteria present. The initial population is 1000 and the population triples during the
first 5 days.
10.
At any time t ≥ 0 , in days, the rate of growth of a bacteria population is given by
a.
Write an expression for y at any time t ≥ 0.
b.
What will the population of bacteria be in 12 days?
c.
When will there be 6000 bacteria?
11.
Consider the differential equation
dy
= x 4 ( y − 2)
dx
a.
On the axes provided, sketch a slope field for
the given differential equation.
b.
While the slope field in part (a) is drawn at
only twelve points, it is defined at every point in the
xy-plane. Describe all points in the xy-plane for which
the slopes are negative.
c.
Find the particular solution y = f ( x) to the given differential equation with the initial condition f (0) = 0 .
1.
C
1
ln x 2 − 4 + C
2.
2
1
3.
sin ( x 3 ) + C
3
1 x2
4.
e +C
2
1 −2 −8
1  e6 − 1 
5.
e −e ) =
(


2
2  e8 
6.
7.
2 x3
y=
−
− 14
3
dV
=k V
dt
Mike Koehler
8.
8=2 2
9.
y = 4e1− x
2
ln 3
t
10.=
a. y 1000
=
e5
1000e0.2197 t
b. 13967 bacteria
5ln(6)
c. t = 8.1546 days
=
ln(3)
11. a. Slope field.
b. Slopes are negative at points
( x, y ) where x ≠ 0 and y < 2.
1
c.
6 - 30
y= 2 − 2e 5
x5
Differential Equations
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