Unit 7: Differential Equations — AP Calculus AB Unit Overview

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: This unit overview covers all seven required sub-topics for AP Calculus AB Unit 7 (Differential Equations): modeling situations, solution verification, slope field sketching, reasoning with slope fields, separation of variables, general/particular solutions, and exponential models.

You should already know: How to compute derivatives of all elementary functions; How to evaluate indefinite integrals; Basic algebraic manipulation to rearrange equations.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. Why This Matters

Differential equations are the core bridge between the rate of change of a quantity (studied with derivatives) and the quantity itself, making them the foundation of almost all real-world applications of calculus. On the AP Calculus AB exam, this unit counts for 6–12% of your total score, and it appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a standalone multi-part FRQ or a question that ties integration to applied contexts. Beyond the exam, differential equations let us model everything from population growth to cooling of objects to compound interest, turning abstract calculus into usable problem-solving. This unit ties together the two central big ideas of AP Calculus: change (derivatives) and the accumulation of change (integration), since solving a differential equation is essentially integrating to recover the original function from its known rate of change. All sub-topics in this unit build toward being able to set up, interpret, and solve these real-world rate problems, a key skill tested heavily on the exam.

2. Concept Map

The seven sub-topics of Unit 7 build sequentially, each relying on mastery of the previous to move from conceptual understanding to applied problem-solving. We start with Modeling situations with differential equations, the conceptual foundation: you learn to translate a verbal description of a quantity’s rate of change into a mathematical equation relating the derivative of the quantity to the quantity itself and its independent variable. This sets up every other problem in the unit.

Next, Verifying solutions for differential equations reinforces what a differential equation actually means: you check if a proposed function satisfies the equation by plugging it and its derivative back in, which solidifies the core relationship between derivative and solution. Next, we turn to graphical methods for understanding solutions before we learn to solve algebraically: first you learn Sketching slope fields, where you draw small line segments representing the slope of a solution at a grid of points to visualize how solutions behave. That is followed by Reasoning using slope fields, where you use that visualization to answer questions about long-term behavior of solutions, match differential equations to their graphs, or sketch a particular solution through a given point.

Next, we introduce the only algebraic solution method required for AP Calculus AB: separation of variables. First, you find General solutions via separation of variables, where you rearrange the equation to separate dependent and independent variables on opposite sides, integrate both sides, and get a family of solutions with one unknown constant of integration. You then use the given starting (initial) condition to solve for that constant and get Particular solutions with initial conditions, the unique solution that matches the problem context. Finally, you apply this entire process to the most common applied case on the exam: Exponential models with differential equations, which describe growth or decay proportional to the current size of the quantity.

3. A Guided Tour

We will walk through a single exam-style problem to show how multiple core sub-topics connect to solve it end-to-end:

Problem: A biologist is studying the growth of a bacteria culture. At time $t$ hours, the number of bacteria $B$ is changing at a rate proportional to the square root of the current number of bacteria. The initial number of bacteria is 100, and after 2 hours the number of bacteria is 144.

Step 1: Apply Modeling situations with differential equations: Translate the description: "rate of change of $B$ is proportional to $B$ " becomes: $\frac{d B}{d t} = k B, k > 0$ That is the full differential equation model for the problem.

Step 2: Apply Verifying solutions for differential equations: A proposed general solution is $B (t) = (\frac{k t}{2} + C)^{2}$ . To verify, first compute the derivative: $\frac{d B}{d t} = 2 (\frac{k t}{2} + C) \cdot \frac{k}{2} = k (\frac{k t}{2} + C)$ Take the square root of $B (t)$ : $B = \frac{k t}{2} + C$ , so $\frac{d B}{d t} = k B$ , which matches our differential equation. The proposed solution is valid.

Step 3: Apply General solutions via separation of variables and Particular solutions with initial conditions: Separate variables to derive the solution ourselves: $\frac{d B}{B} = k d t$ Integrate both sides: $2 B = k t + C ⟹ B = \frac{k t}{2} + \frac{C}{2} ⟹ B (t) = (\frac{k t}{2} + C^{'})^{2}$ Apply initial condition $B (0) = 100$ : $100 = (C^{'})^{2} ⟹ C^{'} = 10$ (we take the positive root since population is positive). Next, use $B (2) = 144$ : $144 = (\frac{2 k}{2} + 10)^{2} ⟹ 12 = k + 10 ⟹ k = 2$ Our final particular solution is $B (t) = (t + 10)^{2}$ . Using Reasoning with slope fields, we can confirm this makes sense: slope increases as $t$ and $B$ increase, so the growth rate speeds up as the population grows, matching the problem description.

4. Common Cross-Cutting Pitfalls

Wrong move: When separating variables, you leave the constant of integration on only one side of the equation, then solve for the initial condition with the constant only on one side. Why: Students forget that integrating both sides produces a constant on each side, which must be combined to get the correct general solution. Skipping this leads to wrong values for $C$ . Correct move: After integrating both sides, immediately combine the two constants of integration into a single constant on one side before proceeding.
Wrong move: When modeling exponential decay, you write $\frac{d P}{d t} = k P$ and keep $k$ positive, leading to a growing solution instead of a decaying one. Why: Students memorize the exponential model form but forget the sign corresponds to the direction of change. Correct move: Always set the sign of the proportionality constant based on context: write $\frac{d P}{d t} = - k P$ (with $k > 0$ ) for decay, and explicitly confirm the sign matches the problem description.
Wrong move: When matching a differential equation to a slope field, you check the slope at one grid point and select the first matching option, without checking a second point. Why: Students rush to save time on MCQ, but two different differential equations can have the same slope at one point but different slopes at another. Correct move: After identifying a candidate, always check a second grid point (usually where $x = 0$ or $y = 0$ ) to confirm the slope matches before selecting.
Wrong move: When verifying a solution, you plug only the proposed function into the differential equation and forget to substitute the derivative. Why: Students confuse solutions to algebraic equations (which only require plugging in the variable) with solutions to differential equations (which require the derivative of the function to also satisfy the relationship). Correct move: When verifying, always compute the derivative of the proposed solution first, then substitute both the derivative and the original function into the differential equation.
Wrong move: After integrating $\frac{1}{y}$ you drop the absolute value around $ln (y)$ , leading to an incorrect sign on the particular solution. Why: Students drop the absolute value out of convenience, forgetting that it accounts for negative values of $y$ . Correct move: Keep the absolute value when integrating reciprocal functions, and use the initial condition to find the correct sign before removing the absolute value.

5. Quick Check: When To Use Which Sub-Topic

For each problem prompt below, identify which sub-topic you would use to answer it:

"The rate of change of the value of a car is proportional to the current value of the car. Write an equation describing this relationship."
"Is $y = 2 e^{3 x} + 5$ a solution to $\frac{d y}{d x} = 3 y - 15$ ?"
"What happens to $y (x)$ as $x \to \infty$ for solutions to $\frac{d y}{d x} = y (3 - y)$ if $y (0) = 1$ ?"
"Find the unique solution to $\frac{d y}{d x} = x y^{3}$ that passes through $(1, 1)$ ."

Answers:

Modeling situations with differential equations
Verifying solutions for differential equations
Reasoning using slope fields
Particular solutions with initial conditions (via separation of variables)

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following differential equations matches a slope field with these properties: (1) slope is 0 everywhere along the line $y = 3$ , (2) slope is negative when $y > 3$ , (3) slope is positive when $y < 3$ , (4) slope does not depend on $x$ ?

A) $\frac{d y}{d x} = x (3 - y)$ B) $\frac{d y}{d x} = y - 3$ C) $\frac{d y}{d x} = (3 - y) x^{2}$ D) $\frac{d y}{d x} = 3 - y$

Worked Solution: First, property 4 states that slope does not depend on $x$ , so any differential equation with an $x$ term can be eliminated immediately. This removes options A and C. Next, check property 1: slope is 0 when $y = 3$ . For both remaining options B and D, slope is 0 at $y = 3$ , so we check property 2: slope is negative when $y > 3$ . For option B, $\frac{d y}{d x} = y - 3$ , which is positive when $y > 3$ , contradicting property 2. For option D, $\frac{d y}{d x} = 3 - y$ , which is negative when $y > 3$ , positive when $y < 3$ , 0 at $y = 3$ , and independent of $x$ , matching all four properties. Correct answer: D.

Question 2 (Free Response)

Consider the differential equation $\frac{d y}{d x} = 3 x^{2} y$ . (a) For the grid points $x = - 1, 0, 1$ and $y = - 1, 0, 1$ , calculate the slope at each grid point for the slope field. (b) Find the general solution of the differential equation using separation of variables. (c) Find the particular solution that passes through the point $(0, - 2)$ and state its domain.

Worked Solution: (a) For any $(x, y)$ , $\frac{d y}{d x} = 3 x^{2} y$ . Calculate slopes:

$x = - 1$ : $y = - 1 \to 3 (- 1)^{2} (- 1) = - 3$ ; $y = 0 \to 0$ ; $y = 1 \to 3$
$x = 0$ : all $y$ have slope $0$
$x = 1$ : $y = - 1 \to 3 (1)^{2} (- 1) = - 3$ ; $y = 0 \to 0$ ; $y = 1 \to 3$ Draw small line segments with these slopes at each grid point.

(b) Separate variables for $y \neq = 0$ : $\frac{d y}{y} = 3 x^{2} d x$ Integrate both sides: $ln ∣ y ∣ = x^{3} + C ⟹ ∣ y ∣ = e^{C} e^{x^{3}} ⟹ y = A e^{x^{3}}$ where $A = \pm e^{C}$ is an arbitrary constant. This is the general solution.

(c) Substitute $x = 0, y = - 2$ : $- 2 = A e^{0} ⟹ A = - 2$ The particular solution is $y = - 2 e^{x^{3}}$ , defined for all real numbers, so domain is $(- \infty, \infty)$ .

Question 3 (Application / Real-World Style)

A 1000 mL mug of coffee brewed at 95°C is set on a counter to cool in a room with constant temperature 22°C. The rate of change of the coffee temperature $T$ is proportional to the difference between $T$ and the room temperature, with a proportionality constant of $k = 0.03$ per minute, where $t$ is time in minutes after the coffee is brewed. (a) Write the differential equation that models $T (t)$ . (b) Find the particular solution for $T (t)$ with the given initial condition. (c) What is the temperature of the coffee after 30 minutes? Give your answer to one decimal place.

Worked Solution: (a) The rate of change is negative when $T > 22$ (temperature cools), so the differential equation is: $\frac{d T}{d t} = - 0.03 (T - 22)$

(b) Separate and integrate: $\frac{d T}{T - 22} = - 0.03 d t ⟹ ln ∣ T - 22∣ = - 0.03 t + C$ Exponentiate and substitute initial condition $T (0) = 95$ : $T (t) - 22 = 73 e^{- 0.03 t} ⟹ T (t) = 22 + 73 e^{- 0.03 t}$

(c) Substitute $t = 30$ : $T (30) = 22 + 73 e^{- 0.03 (30)} = 22 + 73 e^{- 0.9} \approx 22 + 73 (0.4066) \approx 51. 7^{\circ} C$ In context, after 30 minutes of cooling, the coffee is approximately 51.7°C, which is a drinkable warm temperature.

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Differential Equation Definition	$\frac{d y}{d x} = f (x, y)$	Relates derivative of $y$ to $x$ and $y$ ; only first-order DEs are tested on AP Calculus AB
Separation of Variables	1. Rearrange to $g (y) d y = f (x) d x$ 2. Integrate both sides	Only works for separable DEs, the only type solved on AB
Constant of Integration	Combine into one constant $C$	Always combine constants from both sides after integration
Verifying Solutions	1. Compute $\frac{d y}{d x}$ for proposed $y (x)$ 2. Substitute into DE to check equality	Always substitute both $y$ and $\frac{d y}{d x}$
Slope Field Construction	Slope at $(x, y)$ equals the value of $\frac{d y}{d x}$ at that point	Draw small line segments at each grid point
General Solution	Family of solutions with one unknown constant (for first-order DE)	All solutions to the DE are included in this family
Particular Solution	Substitute initial condition $y (x_{0}) = y_{0}$ to solve for $C$	Unique solution that satisfies both the DE and the initial condition
Exponential Growth/Decay Model	$\frac{d P}{d t} = k P$ , $P (t) = P_{0} e^{k t}$	$P_{0} = P (0)$ (initial value); $k > 0$ = growth, $k < 0$ = decay

8. What's Next

This unit ties together all the core big ideas of AP Calculus AB, connecting derivatives and integration to solve applied problems. Mastery of the sub-topics in this unit is required for nearly all multi-part FRQ questions on the AP exam, and it sets the foundation for further study of calculus in AP Calculus BC, where you will learn more complex solution methods for non-separable differential equations. To master each sub-topic individually, use the in-depth study guides linked below: