AP Calc BC Differential Equations and Integral — Questions and Answers
Question 1: If dy/dx = xy and y(0) = 2, what is y(2)?
- 2e² (Correct answer)
- e²
- 2e⁴
- 4e²
Correct answer: 2e²
Separating gives dy/y = x dx, so ln|y| = x²/2 + C. Since y(0) = 2, y = 2e^(x²/2), and y(2) = 2e².
The equation dy/dx = xy is separable, so move all y terms to one side: (1/y) dy = x dx. Integrating gives ln|y| = x²/2 + C, so y = Ae^(x²/2). The initial condition y(0) = 2 gives A = 2. Evaluating at x = 2 gives the exponent 4/2 = 2, so y(2) = 2e². A common error is to forget the 1/2 from integrating x, which gives e⁴ in place of e².
Question 2: A population satisfies dP/dt = 0.2P(1 − P/500). For what value of P is the population growing most rapidly?
- 100
- 500
- 250 (Correct answer)
- 400
Correct answer: 250
The growth rate 0.2P(1 − P/500) is a downward-opening parabola in P. Its maximum is halfway between the zeros at P = 0 and P = 500, which is P = 250.
In a logistic model dP/dt = kP(1 − P/L), the rate of change is a quadratic function of P with zeros at 0 and L. Its vertex is at P = L/2, so the population grows fastest when it is half the carrying capacity. Here L = 500, so the maximum growth rate occurs at P = 250, where dP/dt = 0.2(250)(0.5) = 25. This is also the inflection point of the solution curve. At P = 500 the population is at equilibrium and the growth rate is 0.
Question 3: Euler's method with two steps of size 0.5 is used to approximate y(1) for the differential equation dy/dx = x + y with y(0) = 1. What is the approximation?
- 2.0
- 2.25
- 3.0
- 2.5 (Correct answer)
Correct answer: 2.5
At (0, 1) the slope is 1, so y(0.5) ≈ 1 + 0.5(1) = 1.5. At (0.5, 1.5) the slope is 2, so y(1) ≈ 1.5 + 0.5(2) = 2.5.
Euler's method uses y_new = y_old + (dy/dx at the old point) × Δx. In step 1 the slope at (0, 1) is 0 + 1 = 1, so y rises by 0.5 to 1.5. In step 2 the slope must be recomputed at (0.5, 1.5), which gives 0.5 + 1.5 = 2, so y rises by 1 to 2.5. Reusing the first slope for both steps would give 2.0, a common mistake.
Question 4: What is the area of the region enclosed by the graphs of y = x² and y = 2x?
- 8/3
- 4/3 (Correct answer)
- 4
- 2/3
Correct answer: 4/3
The curves meet where x² = 2x, at x = 0 and x = 2, and 2x is the upper curve on that interval. The area is ∫₀² (2x − x²) dx = 4 − 8/3 = 4/3.
Setting x² = 2x gives x(x − 2) = 0, so the boundaries are x = 0 and x = 2. On (0, 2) the line y = 2x lies above the parabola. The area is the integral of top minus bottom: ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 4/3. Integrating only 2x gives 4, and integrating only x² gives 8/3, which are common wrong answers.
Question 5: The region bounded by y = x², the x-axis, and the line x = 2 is revolved about the y-axis. What is the volume of the solid generated?
- 8π (Correct answer)
- 32π/5
- 16π
- 4π
Correct answer: 8π
Using washers in y, the outer radius is 2 and the inner radius is √y, for 0 ≤ y ≤ 4. The volume is π∫₀⁴ (4 − y) dy = π(16 − 8) = 8π.
Each horizontal slice at height y is a washer whose outer radius is the line x = 2 and whose inner radius is the curve x = √y. The area of a slice is π(4 − y), and y runs from 0 to 4, so V = π[4y − y²/2]₀⁴ = π(16 − 8) = 8π. Shells give the same result: 2π∫₀² x·x² dx = 2π(4) = 8π. The value 32π/5 comes from revolving the region about the x-axis, where V = π∫₀² x⁴ dx.
Question 6: What is the length of the curve y = x^(3/2) from x = 0 to x = 4?
- (8/27)(10√10)
- (2/3)(10√10 − 1)
- (4/9)(10√10 − 1)
- (8/27)(10√10 − 1) (Correct answer)
Correct answer: (8/27)(10√10 − 1)
Here y′ = (3/2)x^(1/2), so the integrand is √(1 + 9x/4). Integrating gives (8/27)(1 + 9x/4)^(3/2), and evaluating from 0 to 4 gives (8/27)(10√10 − 1).
The arc length formula is L = ∫ₐᵇ √(1 + (y′)²) dx. With y′ = (3/2)x^(1/2), (y′)² = 9x/4, so L = ∫₀⁴ (1 + 9x/4)^(1/2) dx. The antiderivative is (4/9)·(2/3)(1 + 9x/4)^(3/2) = (8/27)(1 + 9x/4)^(3/2). At x = 4 the base is 10, and at x = 0 it is 1, so L = (8/27)(10^(3/2) − 1) = (8/27)(10√10 − 1). Forgetting the 4/9 factor from the inner function leads to the wrong coefficient.
Question 7: Consider dy/dx = y² with y(0) = 1. On which interval is the particular solution defined, given that the interval must contain x = 0?
- All real numbers
- x < 1 (Correct answer)
- x > 1
- −1 < x < 1
Correct answer: x < 1
Separating gives −1/y = x + C, and y(0) = 1 gives C = −1. So y = 1/(1 − x), which has a vertical asymptote at x = 1, and the solution through x = 0 lives on x < 1.
Separate to get dy/y² = dx, so −1/y = x + C. Using y(0) = 1 gives −1 = C, so −1/y = x − 1 and y = 1/(1 − x). The solution is undefined at x = 1, where it blows up. A solution must be defined and differentiable on a single interval containing the initial x-value, which is 0. So the domain is x < 1. This shows that a solution can fail to exist for all x even when the right side of the differential equation looks harmless.
Question 8: A population satisfies dP/dt = 0.03P − 0.0006P² with P(0) = 10. What is lim (t→∞) P(t)?
- 20
- 50 (Correct answer)
- 100
- 500
Correct answer: 50
Factor the right side as 0.03P(1 − P/50), which is a logistic equation with carrying capacity 50. Since P(0) = 10 > 0, the population approaches 50.
Set dP/dt = 0 and factor: P(0.03 − 0.0006P) = 0, so the equilibria are P = 0 and P = 0.03/0.0006 = 50. Equivalently, dP/dt = 0.03P(1 − P/50). For 0 < P < 50 the derivative is positive, so the population increases, and it levels off at 50. The equilibrium at 0 is unstable and the one at 50 is stable, so any positive starting value tends to 50.
Question 9: What is the area of the region enclosed by the curves x = y² and x = y + 2?
- 7/2
- 9
- 9/2 (Correct answer)
- 11/2
Correct answer: 9/2
The curves intersect where y² = y + 2, at y = −1 and y = 2, and x = y + 2 is the right-hand curve. The area is ∫₋₁² (y + 2 − y²) dy = 9/2.
Solving y² = y + 2 gives (y − 2)(y + 1) = 0, so the curves meet at y = −1 and y = 2. Between those values the line x = y + 2 is to the right of the parabola x = y². The area is ∫₋₁² (y + 2 − y²) dy = [y²/2 + 2y − y³/3] from −1 to 2. At y = 2 this is 2 + 4 − 8/3 = 10/3, and at y = −1 it is 1/2 − 2 + 1/3 = −7/6. The difference is 10/3 + 7/6 = 27/6 = 9/2.
Question 10: Consider dy/dx = y(2 − y) with y(0) = 1. What is lim (x→∞) y?
Correct answer: 2
The equilibria are y = 0 and y = 2. For 0 < y < 2 the slope y(2 − y) is positive, so a solution starting at y = 1 increases toward y = 2.
Setting y(2 − y) = 0 gives the horizontal equilibrium lines y = 0 and y = 2. On a slope field, solution curves cannot cross these lines. For 0 < y < 2 both factors are positive, so dy/dx > 0 and the solution is increasing. It is bounded above by y = 2, so it must level off there. The equation is a logistic model with carrying capacity 2, so the limit is 2. The solution stays at y = 1 only if the slope there is 0, but the slope is 1 · (2 − 1) = 1.
