AP Calculus AB, AP exams, 6–12%

Unit 7: Differential Equations

Unit 7 covers modeling with differential equations at the AB level: interpreting and sketching slope fields, verifying solutions, and solving separable equations with an initial condition. Exponential growth and decay is the standard applied model. Correct treatment of the constant of integration and the domain of the solution is expected.

Position: 7 of 8 units
Published weighting: 6–12% 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 AB. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Modeling situations with differential equations: Translating a described rate of change into an equation relating a function and its derivative.
Verifying solutions: Confirming that a proposed function satisfies a given differential equation and initial condition.
Slope fields: Sketching and interpreting slope fields, and matching them to differential equations.
Separation of variables: Separating and integrating both sides, then applying the initial condition to determine the constant.
Exponential growth and decay: Recognizing the model in which the rate is proportional to the amount, and using it in context.

Study tips

How to practice Differential Equations.

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

Apply the initial condition immediately after integrating, before simplifying, which avoids sign and constant errors.
Practice sketching a particular solution through a given point on a slope field.
State the domain of the solution when the problem context requires it; the rubric sometimes awards a point for it.

Where this fits

Unit 7: Differential Equations is 7 of 8 units in the AP Calculus AB guide, and it carries a published weighting of 6–12% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus AB 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 P satisfies dP/dt = 0.05P with P(0) = 200. Which function is the particular solution?
1. AP = 200e^(0.05t)
2. BP = 200 + 0.05t
3. CP = 0.05e^(200t)
4. DP = 200e^(t/0.05)
Show answer
Answer: A. P = 200e^(0.05t)
When the rate of change is proportional to the amount, the solution is exponential: P = P(0)e^(kt) with k = 0.05, so P = 200e^(0.05t). A linear model would require dP/dt to be constant rather than proportional to P, and swapping the constants puts 200 in the exponent where it does not belong.
Key insight: dP/dt = kP always integrates to P = P(0)e^(kt); the initial value is the coefficient, not the exponent.
Question 2Foundational
Which constant function is a solution of the differential equation dy/dx = y - 2?
1. Ay = 0
2. By = 2
3. Cy = -2
4. DThere is no constant solution
Show answer
Answer: B. y = 2
A constant function has dy/dx = 0, so the equation requires y - 2 = 0 and therefore y = 2. Checking confirms it: the derivative of the constant 2 is 0, and the right side is 2 - 2 = 0. Constant solutions like this are the equilibrium solutions that slope fields show as horizontal lines.
Key insight: Equilibrium solutions come from setting the derivative expression equal to zero.
Question 3Intermediate
What is the particular solution of dy/dx = 2xy with y(0) = 3?
1. Ay = 3e^(x^2)
2. By = 3e^(2x)
3. Cy = 3 + x^2
4. Dy = e^(x^2) + 2
Show answer
Answer: A. y = 3e^(x^2)
Separating variables gives (1/y) dy = 2x dx, so ln|y| = x^2 + C and y = Ae^(x^2). Applying y(0) = 3 gives A = 3, so y = 3e^(x^2). Integrating 2x as 2x rather than x^2 produces the tempting exponent 2x.
Key insight: Separate, integrate both sides, then apply the initial condition before simplifying further.
Question 4Intermediate
A radioactive sample decays according to dA/dt = -kA, and 80 grams decays to 40 grams in 10 years. What is the value of k?
1. Aln(2)/10
2. Bln(2)/20
3. C10 ln(2)
4. D1/10
Show answer
Answer: A. ln(2)/10
The solution is A = 80e^(-kt), and A(10) = 40 gives e^(-10k) = 1/2. Taking natural logarithms gives -10k = -ln(2), so k = ln(2)/10 ≈ 0.0693 per year. The half-life divides the logarithm; it does not multiply it.
Key insight: For a half-life h, the decay constant is k = ln(2)/h.
Question 5Advanced
The volume V of water in a leaking tank, in liters, satisfies dV/dt = -0.4√V with V = 100 liters at time t = 0 minutes. At what time is the tank empty?
1. At = 25 minutes
2. Bt = 50 minutes
3. Ct = 100 minutes
4. Dt = 250 minutes
Show answer
Answer: B. t = 50 minutes
Separating gives V^(-1/2) dV = -0.4 dt, so 2√V = -0.4t + C. The initial condition V(0) = 100 gives C = 20, hence √V = 10 - 0.2t and V = 0 when t = 50 minutes. Dividing 100 by 0.4 to get 250 ignores the square root entirely.
Key insight: Antidifferentiating V^(-1/2) produces 2√V; the factor of 2 controls the whole answer.

Continue through AP Calculus AB

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Unit 7: Differential Equations