AP Calculus BC, AP exams, 6–9%

Unit 7: Differential Equations

Unit 7 covers slope fields, separable equations with initial conditions, and exponential models, and BC adds Euler's method for numerical approximation and the logistic model for constrained growth. Interpreting the carrying capacity and the point of fastest growth in a logistic model is a recurring exam task.

Position: 7 of 10 units
Published weighting: 6–9% in the official framework
Topics covered: 5 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Calculus BC. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Modeling with differential equations: Translating a described rate of change into an equation and interpreting it in context.
Slope fields: Sketching and reading slope fields and matching them to equations and solution curves.
Separation of variables: Separating, integrating, and applying an initial condition to find the particular solution.
Euler's method: Stepping forward numerically from an initial condition and judging whether the estimate is high or low.
Exponential and logistic models: Growth proportional to amount versus growth limited by carrying capacity, and the fastest-growth point.

Study tips

How to practice Differential Equations.

Section-specific guidance that follows from how this material is usually tested.

Tabulate Euler's method steps rather than doing them in your head; the exam rewards the visible work.
Memorize what the carrying capacity is and where the logistic curve grows fastest, since those are asked directly.
Apply the initial condition immediately after integrating to avoid constant-of-integration errors.

Where this fits

Unit 7: Differential Equations is 7 of 10 units in the AP Calculus BC guide, and it carries a published weighting of 6–9% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus BC guide Official exam information

Sample Differential Equations questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
A population is modeled by the logistic differential equation dP/dt = 0.2P(1 - P/500). What is the carrying capacity?
1. A500
2. B250
3. C100
4. D0.2
Show answer
Answer: A. 500
In the logistic form dP/dt = kP(1 - P/L), the constant L is the carrying capacity, the population at which growth stops. Here L = 500, and 0.2 is the intrinsic growth rate rather than a population. The value 250 is where growth is fastest, not where it stops.
Key insight: The carrying capacity is the nonzero equilibrium: the P value that makes dP/dt zero.
Question 2Foundational
For the logistic model dP/dt = 0.2P(1 - P/500), at what population value is the population growing fastest?
1. A250
2. B500
3. C125
4. D1000
Show answer
Answer: A. 250
The growth rate dP/dt is a downward parabola in P with zeros at P = 0 and P = 500, so it is largest at the vertex, halfway between them at P = 250. That value is also where the solution curve has its point of inflection.
Key insight: Logistic growth peaks at half the carrying capacity, which is the inflection point of the solution.
Question 3Intermediate
Using Euler's method with two steps of size 0.25, what is the approximation of y(0.5) for dy/dx = x + y with y(0) = 1?
1. A1.625
2. B1.5
3. C1.25
4. D1.75
Show answer
Answer: A. 1.625
The first step uses slope 0 + 1 = 1, giving y(0.25) ≈ 1 + 0.25(1) = 1.25. The second step uses slope 0.25 + 1.25 = 1.5, giving y(0.5) ≈ 1.25 + 0.25(1.5) = 1.625. Taking a single step of size 0.5 instead produces 1.5.
Key insight: Each Euler step recomputes the slope at the newest point, not at the starting point.
Question 4Intermediate
The solution of dy/dx = y^2 with y(0) = 1 is evaluated at x = 1/2. What is y(1/2)?
1. A2
2. B1/2
3. C3/2
4. DThe solution is undefined at x = 1/2
Show answer
Answer: A. 2
Separating gives y^(-2) dy = dx, so -1/y = x + C. The initial condition gives C = -1, hence y = 1/(1 - x) and y(1/2) = 2. The solution does blow up, but at x = 1 rather than at x = 1/2.
Key insight: After separating, solve for y explicitly before substituting, and note where the solution's domain ends.
Question 5Advanced
A quantity satisfies dy/dt = k(100 - y) with y(0) = 20, where k is a positive constant. Which function is the solution?
1. Ay = 100 - 80e^(-kt)
2. By = 100 + 80e^(-kt)
3. Cy = 100 - 20e^(-kt)
4. Dy = 20e^(kt)
Show answer
Answer: A. y = 100 - 80e^(-kt)
Separating gives dy/(100 - y) = k dt, so -ln|100 - y| = kt + C and 100 - y = Ae^(-kt). The initial condition y(0) = 20 gives A = 80, hence y = 100 - 80e^(-kt), which rises from 20 toward the equilibrium 100. Using 20 as the coefficient confuses the initial value with the initial gap.
Key insight: In a bounded-growth model the exponential multiplies the initial distance from the equilibrium.

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Unit 7: Differential Equations