Modeling situations with differential equations — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Translating verbal descriptions of proportional and contextual rates of change into differential equations, identifying variables, verifying candidate solutions, and classifying differential equations by order, aligned with AP Calculus AB CED learning objectives.

You should already know: Derivative interpretation as an instantaneous rate of change. Differentiation rules for composite and exponential functions. Basic algebraic rearrangement of 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. What Is Modeling situations with differential equations?

A differential equation is any equation that contains one or more derivatives of an unknown function. Modeling with differential equations is the core process of translating verbal, contextual, or graphical descriptions of relationships between a quantity and its rate of change into a formal mathematical equation. This is always the first step in almost all differential equation problems on the AP exam: you cannot solve a differential equation if you cannot write it correctly from the given description.

According to the AP Calculus AB CED, Unit 7 (Differential Equations) makes up 6-12% of the total exam weight, and this specific topic is the gateway to all other differential equation skills in the unit, including slope fields, separation of variables, and exponential growth/decay. It appears in both multiple-choice (MCQ, as standalone questions or part of a question set) and free-response (FRQ), almost always as the opening part of a longer FRQ that leads into solving the differential equation or interpreting its solution. Synonyms for this skill include "setting up a differential equation" or "translating rate descriptions to DEs."

2. Translating Contextual Rate Descriptions to Differential Equations

The core idea of this skill is that any description of how a quantity changes automatically translates to the derivative of that quantity. The most common scenario is a rate of change proportional to the quantity itself, but you will also see rates proportional to a difference between the quantity and a fixed constant, rates proportional to a power of the quantity, and rates equal to a combination of multiple terms.

The step-by-step process for translation is: 1) Explicitly identify your dependent variable (the quantity that changes) and independent variable (the variable the quantity changes with respect to, almost always time $t$ in AP problems). 2) Translate the phrase "the rate of change of [dependent variable]" directly to the derivative of the dependent variable with respect to the independent variable. 3) Translate the relationship: "is proportional to X" always means "equals $k$ times X", where $k$ is a constant of proportionality (we use the convention that $k>0$ to make sign interpretation easier). 4) Add the correct sign: if the quantity is decreasing, the derivative will have a negative sign, even with $k>0$.

Worked Example

Problem: The number of people $N$ infected with a virus in a closed population increases at a rate proportional to the product of the number of infected people and the difference between the total population of 5000 and the number of infected people. Write the differential equation that models this situation.

1. Identify variables: Dependent variable is $N$ (number of infected people at time $t$), independent variable is $t$ (time, units unspecified here).
2. Translate the description: "rate of change of $N$" = $\frac{dN}{dt}$; "proportional to the product of $N$ and $(5000-N)$" = $k\cdot N(5000-N)$.
3. Confirm sign: The problem states the number of infected people is increasing, so the derivative is positive, which matches the expression (all terms are positive for $0<N<5000$, $k>0$).
4. Write the final differential equation: $$\frac{dN}{dt}=kN(5000-N)\text{for }k>0$$

Exam tip: Always translate the problem word-for-word, do not add extra assumptions. If a problem says "the difference between $y$ and 100", write $(y-100)$, do not swap it to $(100-y)$ unless the problem explicitly tells you the direction of change.

3. Classifying Differential Equations by Order

The order of a differential equation is defined as the order of the highest-order derivative that appears in the equation. For example, a first-order differential equation only contains first derivatives of the unknown function, while a second-order differential equation contains a second derivative. AP Calculus AB almost exclusively works with first-order differential equations, but you are expected to correctly identify the order of any DE, a common trick on MCQs.

A common student confusion is mixing up the order of the derivative with the power the derivative is raised to. For example, ${\left(\frac{dy}{dx}\right)}^4+2y=0$ is still a first-order differential equation, because the highest derivative is the first derivative — it is just raised to the 4th power. The power does not change the order of the DE. Why does order matter? The number of arbitrary constants in the general solution of a DE is equal to the order of the DE, so first-order DEs require one initial condition to find a particular solution, which is the standard AP problem.

Worked Example

Problem: What is the order of the differential equation $\frac{d^{2}y}{d{t}^2}-4t{\left(\frac{dy}{dt}\right)}^3+7y=\cos t$? Justify your answer.

1. Recall: Order is determined by the highest order of any derivative present in the equation, not the power of the derivative.
2. List all derivatives present and their orders: $\frac{dy}{dt}$ is first-order, and $\frac{d^{2}y}{d{t}^2}$ is second-order.
3. Note that $\frac{dy}{dt}$ is raised to the 3rd power, but that does not change its order as a first derivative.
4. The highest order derivative present is second-order, so the differential equation is second-order.

Exam tip: When asked for the order, circle every derivative in the DE and write its order next to it, then pick the maximum number. This eliminates the common mistake of confusing order with power.

4. Verifying Solutions to Differential Equations

To confirm a given candidate function is a solution to a differential equation, you substitute the function and all its required derivatives into the DE and check that both sides of the equation are equal for all values of the independent variable in the domain. This is a regularly tested skill on both MCQ and FRQ, often as the second part of a question after setting up the DE.

The process is straightforward: 1) Compute all derivatives of the candidate function that appear in the DE (for a first-order DE, you only need the first derivative). 2) Substitute the candidate function $y$ and its derivatives into the left-hand side and right-hand side of the DE. 3) Simplify both sides. If they are identical, the candidate is a solution; if not, it is not. This works for both general solutions (with an arbitrary constant $C$) and particular solutions (with a fixed value of $C$ from an initial condition). Never waste time solving the DE from scratch to check a solution: the candidate is already given to you.

Worked Example

Problem: Verify that $y=3e^{2x}+4x+1$ is a solution to the differential equation $\frac{dy}{dx}=2y-8x-6$.

1. Compute the first derivative of the candidate solution: $$\frac{dy}{dx}=\frac{d}{dx}\left(3e^{2x}+4x+1\right)=6e^{2x}+4$$
2. Substitute $y$ into the right-hand side of the DE and simplify: $$2y-8x-6=2\left(3e^{2x}+4x+1\right)-8x-6=6e^{2x}+8x+2-8x-6=6e^{2x}-4$$
3. Compare the left-hand side ($\frac{dy}{dx}=6e^{2x}+4$) to the simplified right-hand side. Wait, correction to step 2 arithmetic: $2-6=-4$, so RHS = $6e^{2x}-4$, which is not equal to LHS.
4. Conclusion: Since LHS ≠ RHS for all $x$, $y=3e^{2x}+4x+1$ is not a solution to the DE.

Exam tip: Double-check your differentiation and arithmetic when verifying solutions: small sign errors can lead you to incorrectly reject a valid solution or accept an invalid one.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing $\frac{dV}{dt}=k\sqrt{V}$ for a leaking tank where volume is decreasing, leaving out the negative sign. Why: Students remember "proportional" but forget to account for the direction of change described in the problem. Correct move: After writing the proportionality, explicitly ask "is the quantity increasing or decreasing?" Add a negative sign to the derivative if it is decreasing, keeping $k>0$ by convention.
• Wrong move: Classifying ${\left(\frac{d^{2}y}{d{x}^2}\right)}^3+\frac{dy}{dx}=0$ as third-order because the highest derivative is cubed. Why: Students confuse the order of the derivative (how many times you differentiate) with the power the derivative is raised to. Correct move: Circle every derivative in the DE, write down what order each is, then pick the maximum number, ignoring powers and coefficients of the derivatives.
• Wrong move: When verifying a solution, substituting only $y$ into the DE and forgetting to substitute the derivative. Why: Students rush and forget that a DE requires the derivative to be substituted. Correct move: Follow the explicit 2-step rule for verification: 1) Compute all required derivatives of the candidate solution first, 2) Substitute both $y$ and all derivatives into the DE before checking equality.
• Wrong move: Writing $\frac{dy}{dt}=k(100-y)$ for a problem that says "the rate of change of $y$ is proportional to the difference between $y$ and 100". Why: Students assume $y$ is always approaching 100 so swap the order, but the problem explicitly states the order of the difference. Correct move: Translate the phrase word-for-word: "difference between A and B" = $A-B$, unless the problem specifies it is the absolute difference.
• Wrong move: Writing $\frac{dP}{dt}=P$ when the problem says "rate of change of population $P$ is proportional to $P$", leaving out the constant of proportionality $k$. Why: Students know proportional means multiply, but forget that proportionality requires a constant of proportionality, which is not the variable itself. Correct move: Always introduce a constant of proportionality (usually $k$) when the problem says "proportional", unless it explicitly gives the constant in the problem.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following is the correct differential equation for the statement: The acceleration of a particle moving along the $x$-axis is proportional to the displacement $x$ of the particle from the origin, where acceleration is negative when $x$ is positive. Let displacement $x$ be a function of time $t$, and $k>0$ be a constant.

A) $\frac{dx}{dt}=-kx$ B) $\frac{d^{2}x}{d{t}^2}=-kx$ C) $\frac{d^{2}x}{d{t}^2}=kx$ D) $\frac{dx}{dt}=kx$

Worked Solution: First, recall that acceleration is the second derivative of displacement with respect to time, so the left-hand side of the DE must be $\frac{d^{2}x}{d{t}^2}$. This eliminates options A and D, which use the first derivative (velocity) instead of acceleration. Next, the problem states acceleration is negative when $x$ is positive, and $k>0$, so we need a negative sign to get the correct sign relationship. Option C has a positive sign, which would give positive acceleration for positive $x$, contradicting the problem statement. The correct answer is B.

Question 2 (Free Response)

Let $A(t)$ be the amount of a radioactive substance, measured in grams, remaining in a sample after $t$ days. Radioactive decay follows the rule that the rate of decay of the substance is proportional to the amount of substance remaining at time $t$.

(a) Write a differential equation that models $A(t)$, using a positive constant of proportionality $k$. (b) What is the order of this differential equation? Justify your answer. (c) Verify that $A(t)=A_0{e}^{-kt}$, where $A_0$ is the initial amount of the substance, is a solution to the differential equation you wrote in part (a).

Worked Solution: (a) The rate of change of $A(t)$ is $\frac{dA}{dt}$. Since decay means the amount is decreasing, the derivative is negative proportional to $A(t)$. For $k>0$, the differential equation is: $$\frac{dA}{dt}=-kA$$

(b) This is a first-order differential equation. The highest-order derivative present in the equation is the first derivative $\frac{dA}{dt}$, so the order equals 1 by definition.

(c) First, differentiate the candidate solution: $$\frac{dA}{dt}=\frac{d}{dt}\left(A_0{e}^{-kt}\right)=-k{A}_0{e}^{-kt}$$ Substitute $A(t)=A_0{e}^{-kt}$ into the right-hand side of the DE from (a): RHS = $-kA(t)=-k{A}_0{e}^{-kt}$. This equals the left-hand side ($\frac{dA}{dt}$) for all $t\ge 0$, so $A(t)=A_0{e}^{-kt}$ is a valid solution.

Question 3 (Application / Real-World Style)

A cup of coffee at 90°C is placed in a room kept at a constant temperature of 20°C. Newton's Law of Cooling states that the rate of change of the temperature of the coffee is proportional to the difference between the temperature of the coffee and the room temperature. Let $T(t)$ be the temperature of the coffee in degrees Celsius at time $t$ minutes, and the constant of proportionality (positive) is $k=0.05$ min⁻¹. Write the differential equation modeling this situation, and calculate the initial rate of change of the temperature of the coffee when it is first placed in the room.

Worked Solution: The difference between the coffee temperature $T(t)$ and the room temperature 20°C is $T-20$. Since the coffee is cooling, its temperature decreases over time, so the rate of change $\frac{dT}{dt}$ is negative. The differential equation is: $$\frac{dT}{dt}=-0.05(T-20)$$ At $t=0$, the initial temperature is $T(0)=90^{\circ }\text{C}$. Substitute into the DE: $$\frac{dT}{dt}{|}_{t=0}=-0.05(90-20)=-0.05(70)=-3.5^{\circ }\text{C per minute}$$ In context, this means when the coffee is first placed in the room, its temperature is decreasing at a rate of 3.5 degrees Celsius per minute.

7. Quick Reference Cheatsheet

8. What's Next

Modeling with differential equations is the foundational prerequisite for every other topic in Unit 7: you cannot draw a slope field, separate variables to solve a DE, or find a particular solution from an initial condition if you cannot first correctly write the DE from the given problem description. This topic also builds on your derivative interpretation skills from Unit 2 and connects directly to exponential growth and decay models, which are a frequent AP FRQ topic. Without mastering the skills in this chapter, you will lose easy points on the opening part of almost every differential equation FRQ, and will struggle to solve the rest of the problem even if you know separation of variables.

Next topics to study: Slope fields and differential equations Separation of variables Exponential growth and decay models Particular solutions to differential equations