AP Calculus AB Practice Quiz: Finding Particular Solutions Using Initial Conditions and Separation of Variables
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 10 questions to check your progress.
Question 1 of 10
All Questions (10)
A) The order of the differential equation
B) The family of curves the general solution represents
C) A specific point that the solution must pass through
D) The method of integration used to find the general solution
Correct Answer: C
A general solution represents infinitely many functions (a family of curves). To identify the one particular solution, an initial condition, which is a specific point (x₀, y₀) on the curve, is needed to solve for the constant of integration. [cite: 2791, 2792]
A) y = x³ - 3
B) y = x³ + C
C) y = x³ + 5
D) y = 6x - 7
Correct Answer: A
First, find the general solution by integrating dy/dx = 3x². This gives y = ∫3x² dx = x³ + C. Then, use the initial condition (2, 5) to find the constant C. Substitute x=2 and y=5 into the general solution: 5 = (2)³ + C, which simplifies to 5 = 8 + C. Solving for C gives C = -3. Therefore, the particular solution is y = x³ - 3. [cite: 2790]
A) F(x) = ∫₁ˣ eᵗ² dt + 4
B) F(x) = ∫₀ˣ eᵗ² dt + 4
C) F(x) = ∫₄ˣ eᵗ² dt + 1
D) F(x) = 2xeˣ² + 4
Correct Answer: A
According to the Fundamental Theorem of Calculus, the particular solution F(x) to the differential equation dy/dx = f(x) that satisfies F(x₀) = y₀ is given by the function F(x) = ∫ₓ₀ˣ f(t) dt + y₀. In this problem, f(t) = eᵗ², x₀ = 1, and y₀ = 4. Substituting these values into the formula gives F(x) = ∫₁ˣ eᵗ² dt + 4. [cite: 2793]
A) y = ln(x) + 2
B) y = ln(-x) + 2
C) y = ln|x| + 4
D) y = -1/x² + 3 - 1/e²
Correct Answer: B
The general solution is y = ∫(1/x) dx = ln|x| + C. The initial condition is at x = -e, which is negative. For x < 0, |x| = -x. So, the solution for this domain is y = ln(-x) + C. Substitute the point (-e, 3): 3 = ln(-(-e)) + C, which is 3 = ln(e) + C, or 3 = 1 + C. Thus, C = 2. The particular solution is y = ln(-x) + 2. This solution is valid on the domain (-∞, 0), which contains the initial point. [cite: 2790, 2794]
A) y = 10e⁻²ˣ
B) y = e⁻²ˣ + 9
C) y = -y² + 10
D) y = 10 - 2x
Correct Answer: A
Use separation of variables: (1/y) dy = -2 dx. Integrate both sides: ∫(1/y) dy = ∫-2 dx, which gives ln|y| = -2x + C. Exponentiate both sides: |y| = e⁻²ˣ⁺ᶜ = eᶜe⁻²ˣ. Let A = ±eᶜ. The general solution is y = Ae⁻²ˣ. Use the initial condition y(0) = 10: 10 = Ae⁻²⁽⁰⁾ = A(1), so A = 10. The particular solution is y = 10e⁻²ˣ. [cite: 2790]
A) Infinitely many
B) Two
C) None
D) Only one
Correct Answer: D
For a typical differential equation encountered in AP Calculus, a given point (or initial condition) is sufficient to determine a unique value for the constant of integration, resulting in only one particular solution that passes through that point. [cite: 2792]
A) y = √(x² - 7)
B) y = -√(x² - 7)
C) y² = x² - 5
D) y = x²/2 - 11
Correct Answer: B
Separate the variables: y dy = x dx. Integrate both sides: ∫y dy = ∫x dx, which gives (1/2)y² = (1/2)x² + C. To find C, use the point (4, -3): (1/2)(-3)² = (1/2)(4)² + C, which simplifies to 9/2 = 16/2 + C, so C = -7/2. The equation is (1/2)y² = (1/2)x² - 7/2, or y² = x² - 7. Solving for y gives y = ±√(x² - 7). Since the initial condition has a negative y-value (y = -3), we must choose the negative root. The particular solution is y = -√(x² - 7). [cite: 2790, 2794]
A) (-∞, ∞)
B) (-∞, 1)
C) (1, ∞)
D) (-∞, 3) U (3, ∞)
Correct Answer: B
Separate variables: dy/(y-1)² = dx. Integrate both sides: -(y-1)⁻¹ = x + C, or -1/(y-1) = x + C. Use the initial condition (2, 0): -1/(0-1) = 2 + C, which gives 1 = 2 + C, so C = -1. The solution is -1/(y-1) = x - 1, which rearranges to y-1 = -1/(x-1), so y = 1 - 1/(x-1). This function has a vertical asymptote at x=1. A particular solution's domain must be a single continuous interval containing the initial x-value, x=2. The interval containing x=2 is (1, ∞). Wait, let me re-check the math. -1/(0-1) = 1. So 1 = 2+C, C=-1. -1/(y-1) = x-1. y-1 = -1/(x-1). y = 1 - 1/(x-1). Domain is x!=1. Initial condition is x=2. The interval containing 2 is (1, infinity). Let's re-evaluate the question. Let's try y(2)=2. Then -1/(2-1) = 2+C -> -1 = 2+C -> C=-3. y = 1 - 1/(x-3). Initial x=2. Domain is (-inf, 3). Let's change the initial condition to y(0)=2. Then -1/(2-1) = 0+C -> C=-1. y = 1 - 1/(x-1). Initial x=0. Domain is (-inf, 1). This is better. Let's re-write the question with y(0)=2. New question: A particular solution to dy/dx = (y-1)² passes through (0, 2). What is its domain? Separate: dy/(y-1)² = dx. Integrate: -1/(y-1) = x+C. Use (0,2): -1/(2-1) = 0+C => C=-1. Solution: -1/(y-1) = x-1 => y-1 = -1/(x-1) => y = 1 - 1/(x-1). The function has a vertical asymptote at x=1. The domain of the particular solution is the continuous interval containing the initial condition x=0, which is (-∞, 1). [cite: 2790, 2794]
A) G(x) = f'(x) + y₀
B) G(x) = ∫ f(x) dx + y₀
C) G(x) = ∫ₓ₀ˣ f(t) dt + y₀
D) G(x) = f(x) - f(x₀) + y₀
Correct Answer: C
The form F(x) = ∫ₓ₀ˣ f(t) dt + y₀ is a direct application of the Fundamental Theorem of Calculus for finding a particular solution. When x = x₀, the integral evaluates to 0, leaving F(x₀) = y₀, which satisfies the initial condition. The derivative of the integral with respect to x is f(x), satisfying the differential equation dy/dx = f(x). [cite: 2793]
A) y = 4e^(k(x²-1)/2)
B) y = 4 * 2^((1-x²)/3)
C) y = 4 * (1/4)^((x²-1)/3)
D) y = 7 - 3x
Correct Answer: C
Separate variables: (1/y)dy = kx dx. Integrate: ln|y| = (k/2)x² + C. General solution: y = Ae^((k/2)x²). Use y(1)=4: 4 = Ae^(k/2). So, A = 4e^(-k/2). Substitute A back: y = 4e^(-k/2) * e^((k/2)x²) = 4e^((k/2)(x²-1)). Now use y(2)=1 to find k: 1 = 4e^((k/2)(2²-1)) = 4e^(3k/2). So, 1/4 = e^(3k/2). This means ln(1/4) = 3k/2, so k = (2/3)ln(1/4). Substitute k back into the solution: y = 4e^(((1/2)*(2/3)ln(1/4))(x²-1)) = 4e^((1/3)ln(1/4)(x²-1)) = 4(e^(ln(1/4)))^((x²-1)/3) = 4(1/4)^((x²-1)/3). [cite: 2790, 2791]