-AP REVIEW
DIFFERENTIAL EQUATIONS AND SLOPE FIELDS
Multiple Choice
1. (Sample Questions - Noncalculator)
The solution to the differential equation
3 4
x
4
3
(D) y  3 x 4  5
4
(A) y  3
dy x3

, where y  2  3, is
dx y 2
3 4 3
x  15
4
3
(E) y  3 x 4  15
4
(B) y  3
(C) y  3
3 4
x  15
4
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2. (Sample Questions - Noncalculator)
dy
 y sec2 x and y  5 when x  0 , then y =
If
dx
tan x
(A) e
(B) etan x  5
(C) 5etan x
(D) tan x + 5
(E) tan x  5e x
4
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3. (2003 – Noncalculator)
Shown above is a slope field for which of the following differential equations?
dy x 2
dy x 3
dy x3
dy x 2
dy x
 2

(A)
(B)
(C)
(D)
(E)



dx y
dx y 2
dx
y
dx y
dx y
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4. (Sample Questions – Noncalculator)
The slope field for a certain differential equation is shown above. Which of the following
could be a specific solution to that differential equation?
(A) y  x 2
(B) y  e x
(C) y  e  x
(D) y  cos x
(E) y  ln x
Free Response
5. (2005 – Noncalculator)
dy
2x

.
dx
y
(a) On the axes provided, sketch a slope field for the given differential equation at the twelve
points indicated.
Consider the differential equation
(b) 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) Find the particular solution y  f  x  to the given differential equation with the initial
condition f 1  1.
6. (2006 – Noncalculator)
Consider the differential equation
dy 1  y

where x  0.
dx
x
(a) On the axes provided, sketch a slope field for the given differential equation at the eight points
indicated.
(b) Find the particular solution y  f  x  to the differential equation with the initial
condition f  1 1 and state its domain.
Answers
1. E
2. C
3. E
4. E
5. (2005 – Noncalculator)
Consider the differential equation
dy
2x

.
dx
y
(a) On the axes provided, sketch a slope field for the given differential
equation at the twelve points indicated.
(b) 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) Find the particular solution y  f  x  to the given differential equation
with the initial condition f 1  1.
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(a)
1: zero slopes
1: nonzero slopes
(b) The line tangent to f at 1, 1 is y  1  2  x  1 .
Thus, f 1.1 is approximately  0.8 .
1: approximation for f 1.1
dy
2x

dx
y
y dy   2 x dx
(c)
1: equation of the tangent line
1: separates variables
1: antiderivatives
2
y
  x2  C
2
1
3
 1  C; C 
2
2
2
2
y   2x  3
1: constant of integration
1: uses initial condition
1: solves for y
Since the particular solution goes through 1, 1 ,
y must be negative.
constant of integration
Thus the particular solution is y  
6.
Consider the differential equation
Note: max 2/5 [1-1-0-0-1] if no
3  2 x2
Notes: 0/5 if no separation of variables
dy 1  y

where x  0.
dx
x
(a) On the axes provided, sketch a slope field for the given differential equation
at the eight points indicated.
(b) Find the particular solution y  f  x  to the differential equation with the
initial condition f  1 1 and state its domain.
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(a)
2: sign of slope at each point and
relative steepness of slope lines
in rows and columns
(b)
1
1 y
dy 
1
x
dx
ln 1  y  ln x  K
1 y  e
ln x  K
1 y  C x
2=C
1 y  2 x
y  2 x  1 and x  0
or
y  2 x  1 and x  0
1: separates variables
2: antiderivatives
1: constant of integration
1: uses initial condition
1: solves for y
Note: max 3/6 [1-2-0-0-0] if
no constant of integration
Note: 0/6 if no separation of
variables
1: domain