FINAL EXIST EXAM 2023-2024 WITH
VERIFIED ANSWERS
AP® CALCULUS AB
2010 SCORING GUIDELINES (Form B)
Question 5
x+1
Consider the differential equation dy = .
dx y
(a) On the axes provided, sketch a slope field for the given differential equation at the
twelve points indicated, and for − < <1 x 1, sketch the solution curve that passes
through the point (0, −1 .)
(Note: Use the axes provided in the exam booklet.)
(b) While the slope field in part (a) is drawn at only twelve points, it is defined at every
point in the xy-plane for which y ≠ 0. Describe all points in the xy-plane, y ≠ 0, for
dy which
=−1.
dx
(c) Find the particular solution y = f x( ) to the given differential equation with the initial
condition f ( )0 =−2.
(a) ⎧ 1 : zero slopes
⎪
3 : ⎨ 1 : nonzero slopes
⎪ 1 : solution curve through 0,( −1)
⎩
, x+ 1
(b) − =1 ⇒ y =− −x 1
y
dy
=−1 for all (x, y) with y =− −x 1 and y ≠ 0
dx 1 : description
⎧ 1 : separates variables ⎪
(
c)
∫y dy = ∫(x+1) dx 1 : antiderivatives
⎪
C 5 : ⎨ 1 : constant of integration
⎪ 1 : uses initial condition
0C⇒C=2 ⎪
y = x + +2x 4
2 2 ⎩ 1 : solves for y
Since the solution goes through (0, 2 ,− ) y must be
Note: max 2 5 [1-1-0-0-0] if no constant
negative. Therefore y =− x2 + +2x 4. of integration
Note: 0 5 if no separation of variables
© 2010 The College Board.
Visit the College Board on the Web: www.collegeboard.com.
© 2010 The College Board.
Visit the College Board on the Web: www.collegeboard.com.
VERIFIED ANSWERS
AP® CALCULUS AB
2010 SCORING GUIDELINES (Form B)
Question 5
x+1
Consider the differential equation dy = .
dx y
(a) On the axes provided, sketch a slope field for the given differential equation at the
twelve points indicated, and for − < <1 x 1, sketch the solution curve that passes
through the point (0, −1 .)
(Note: Use the axes provided in the exam booklet.)
(b) While the slope field in part (a) is drawn at only twelve points, it is defined at every
point in the xy-plane for which y ≠ 0. Describe all points in the xy-plane, y ≠ 0, for
dy which
=−1.
dx
(c) Find the particular solution y = f x( ) to the given differential equation with the initial
condition f ( )0 =−2.
(a) ⎧ 1 : zero slopes
⎪
3 : ⎨ 1 : nonzero slopes
⎪ 1 : solution curve through 0,( −1)
⎩
, x+ 1
(b) − =1 ⇒ y =− −x 1
y
dy
=−1 for all (x, y) with y =− −x 1 and y ≠ 0
dx 1 : description
⎧ 1 : separates variables ⎪
(
c)
∫y dy = ∫(x+1) dx 1 : antiderivatives
⎪
C 5 : ⎨ 1 : constant of integration
⎪ 1 : uses initial condition
0C⇒C=2 ⎪
y = x + +2x 4
2 2 ⎩ 1 : solves for y
Since the solution goes through (0, 2 ,− ) y must be
Note: max 2 5 [1-1-0-0-0] if no constant
negative. Therefore y =− x2 + +2x 4. of integration
Note: 0 5 if no separation of variables
© 2010 The College Board.
Visit the College Board on the Web: www.collegeboard.com.
© 2010 The College Board.
Visit the College Board on the Web: www.collegeboard.com.
