AP Calculus AB: Differential Equations

Differential equations bring the entire course together: derivatives describe how something changes, and solving the equation recovers the original function. This free, complete lesson covers general and particular solutions, slope fields, separable differential equations, exponential growth and decay, and Newton's Law of Cooling. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.

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1. General and Particular Solutions

A differential equation (d.e.) is any equation involving a derivative. A solution is any function that satisfies it. Since antidifferentiation always introduces an arbitrary constant, a d.e. has infinitely many solutions, one for each value of C, called the general solution. Applying a specific initial condition pins down a single particular solution.

A particular solution must not only satisfy the differential equation and the initial condition, it must also be differentiable on an interval that contains the initial point. Vertical tangents or asymptotes can restrict the domain of the solution, even when the algebraic formula looks defined everywhere.

Worked Example: Given dy/dx = 6x² − 4x with general solution y = 2x³ − 2x² + C, find the particular solution satisfying y(1) = 5.

y(1) = 2(1) − 2(1) + C = C = 5, so y = 2x³ − 2x² + 5

1. Given dy/dx = 3x² + 2 with general solution y = x³ + 2x + C, find the particular solution satisfying y(0) = 4.

Explanation: y(0) = 0 + 0 + C = C = 4, so y = x³ + 2x + 4.

2. A particular solution to a differential equation must satisfy which conditions?

Explanation: All three conditions are required for a function to genuinely qualify as the particular solution.

3. Verify that x² + y² = r² is a solution of the d.e. dy/dx = −x/y, using implicit differentiation.

Explanation: Differentiating both sides with respect to x: 2x + 2y(dy/dx) = 0, which rearranges to dy/dx = −x/y, matching the given d.e.

4. Why might a particular solution have a restricted domain even though its algebraic formula looks defined everywhere?

Explanation: Even when a formula is algebraically defined at a point, that point may involve a vertical tangent or an undefined derivative, disqualifying it from the actual domain of the particular solution.

2. Slope Fields

A slope field represents a d.e. graphically: at many points, a short line segment is drawn with the slope given by the d.e. at that point. Following segments tangentially from an initial point traces out a solution curve.

Worked Example: The d.e. dy/dx = y tells us the slope at any point equals its y-coordinate. What is the slope at any point where y = 4?

The slope equals 4 at every such point, regardless of x.

1. For the d.e. dy/dx = x − y, what is the slope of the solution curve at the point (3, 1)?

Explanation: Slope = x − y = 3 − 1 = 2.

2. A slope field shows horizontal segments (slope 0) along the entire x-axis. This is consistent with which d.e.?

Explanation: For dy/dx = y, the slope equals y itself, so along the entire x-axis (where y = 0), the slope is 0 for every x, producing horizontal segments there.

3. Match the d.e. dy/dx = 2x to its general solution family.

Explanation: y = ∫2x dx = x² + C.

4. If a slope field has vertical segments (undefined slope) along the line x = 0, which d.e. below is most consistent with this?

Explanation: dy/dx = 1/x is undefined at x = 0, matching a slope field with vertical (undefined-slope) behavior along that line.

5. When sketching a solution curve on a slope field starting from a given initial point, what should you do?

Explanation: The segments approximate the solution curve's tangent direction at each point, so tracing tangentially from segment to segment reconstructs the actual solution curve.

3. Separable Differential Equations

A first-order d.e. is separable if it can be written as dy/dx = f(x)/g(y), with all y-terms and all x-terms isolated to opposite sides. Solve by separating, then integrating both sides independently.

Worked Example: Solve dy/dx = −x/y given y(0) = 2.

Separate: y dy = −x dx. Integrate: y²/2 = −x²/2 + k, so x² + y² = C.

Using y(0) = 2: 0 + 4 = C, so x² + y² = 4 (with y > 0)

1. Find the general solution of du/dv = e^v−u.

Explanation: Rewrite as e^u du = e^v dv. Integrating: e^u = e^v + C, so u = ln(e^v + C).

2. If ds/dt = √(st) and s = 1 when t = 0, separate the variables.

Explanation: Since √(st) = √s · √t, dividing both sides by √s gives ds/√s = √t dt.

3. Using ds/√s = √t dt, find the general solution after integrating.

Explanation: ∫s^−1/2 ds = ∫t^1/2 dt gives 2s^1/2 = 23t^3/2 + C.

4. Solve dy/dx = 6x²y, given y(0) = 5.

Explanation: dy/y = 6x² dx gives ln|y| = 2x³ + C, so y = Ce^2x³. Using y(0) = 5: C = 5.

5. A first-order differential equation is called "separable" if it can be written in which form?

Explanation: This is the defining form of a separable d.e., all y-dependence isolated on one side, all x-dependence on the other.

4. Exponential Growth and Decay

If a quantity y changes at a rate proportional to the amount present, dy/dt = ky, then y = ce^kt, where c is the initial amount. If k > 0, this is exponential growth; if k < 0, exponential decay. The half-life is the time required for a decaying quantity to reduce by half.

Worked Example: A population grows at a rate proportional to its size, at 3% per year. How long will it take to double?

dP/dt = 0.03P, so P = P₀e^0.03t. Setting 2P₀ = P₀e^0.03t:

ln 2 = 0.03t, so t = ln 20.03 ≈ 23.1 years

1. Bacteria in a culture increase at a rate proportional to the number present. If the number doubles in 5 hours, how many times the original amount will be present in 15 hours?

Explanation: Doubling every 5 hours means 15 hours is exactly 3 doubling periods: 2³ = 8 times the original amount.

2. A radioactive substance has a half-life of 20 years. What fraction of the original amount remains after 60 years?

Explanation: 60 years is exactly 3 half-lives, so the remaining fraction is (1/2)³ = 1/8.

3. At a yearly rate of 4% compounded continuously, set up the equation to find how long it takes for an investment to double.

Explanation: A = Pe^0.04t; setting A = 2P and dividing by P gives 2 = e^0.04t.

4. Using 2 = e^0.04t from the previous problem, solve for t (to the nearest year).

Explanation: t = ln 2 / 0.04 ≈ 17.3, so about 17 years.

5. If a quantity satisfies dy/dt = ky with k < 0, what does this tell us about the quantity's behavior over time?

Explanation: A negative proportionality constant k means the quantity shrinks over time, exponential decay.

6. A fossil contains 40% of its original carbon-14. Using Q = Q₀e^−0.00012t, set up the equation to find its age t.

Explanation: 0.40Q₀ = Q₀e^−0.00012t; dividing both sides by Q₀ gives 0.40 = e^−0.00012t.

5. Newton's Law of Cooling

Newton's Law of Cooling states that a hot object cools at a rate proportional to the difference between its own temperature and that of its surroundings. This is a case of restricted growth: T(t) = A + ce^−kt, where A is the (constant) ambient temperature the object approaches, and c is determined by the initial condition.

Worked Example: A cup of coffee at 190°F is placed in a 70°F room. Using T(t) = 70 + ce^−kt, find c.

T(0) = 190: 70 + c = 190, so c = 120

1. Continuing the coffee example, T(t) = 70 + 120e^−kt. Using T(10) = 150, find e^−10k.

Explanation: 150 = 70 + 120e^−10k, so 120e^−10k = 80, giving e^−10k = 80/120 = 2/3.

2. A cold drink at 40°F is placed in a 75°F room. Using Newton's Law of Warming, T(t) = 75 − ce^−kt, find c using T(0) = 40.

Explanation: 40 = 75 − c, so c = 35.

3. In the restricted growth model f'(t) = k[A − f(t)] with f(t) increasing, what does A represent as t → ∞?

Explanation: As the exponential term decays to 0, f(t) approaches A, making A a horizontal asymptote, the limiting value.

4. A falling object's velocity approaches a limiting value L due to air resistance, following V(t) = L(1 − e^−kt). What is V(0)?

Explanation: V(0) = L(1 − e⁰) = L(1 − 1) = 0, meaning the object starts at rest, as expected.

5. A leaking tire's pressure P(t) satisfies P(t) = O + ce^−kt, where O is the outside pressure. What happens to P(t) as t → ∞?

Explanation: As t → ∞, the exponential term vanishes, so the inside pressure equalizes with the outside pressure O.

Common Mistakes to Avoid

Forgetting the constant of integration when solving a separable d.e. Every general solution needs it, and it's exactly what an initial condition is used to pin down.
Separating variables incorrectly. Every y-term (including dy) must end up on one side, and every x-term (including dx) on the other, double-check that no mixed terms remain before integrating.
Confusing the sign of k in growth versus decay. A positive k means growth (y = ce^kt increasing); a negative k means decay. Setting up the wrong sign flips the entire behavior of the model.
Mixing up half-life calculations with the growth/decay constant k. Half-life and k are related but distinct, always set up the exponential equation explicitly rather than trying to shortcut with memorized half-life arithmetic alone.
Forgetting that Newton's Law of Cooling approaches the ambient temperature, not zero. The limiting value in T(t) = A + ce^−kt is A (the surrounding temperature), not 0, a very common point of confusion.

Frequently Asked Questions

How do I know if a differential equation is separable?

Check whether you can algebraically rearrange it so every y (including dy) is on one side and every x (including dx) is on the other. If a term genuinely mixes x and y in a way that can't be factored apart, the equation isn't separable by this method.

What's the difference between a slope field and an actual solution curve?

A slope field shows the direction (slope) a solution curve would have at many sample points, it's a picture of the differential equation itself. A solution curve is one specific function that is actually tangent to those slope segments everywhere along its path.

Why do exponential growth and Newton's Law of Cooling use such similar formulas?

Both come from the same family of first-order linear differential equations, just with different setups: unrestricted growth (dy/dt = ky) has no limiting value, while restricted growth (like cooling) has a rate proportional to the distance from a fixed target value, producing the extra additive constant A in the solution.

Where can I practice more problems like these?

The Differential Equations quizzes in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.