Unit 7.1 – Modeling Situations with Differential Equations

AP® Calculus AB & BC | Introduction to Differential Equations

Why This Matters: Differential equations are equations that contain derivatives! They're one of the most powerful tools in mathematics because they describe how things change. From population growth to radioactive decay, from cooling coffee to moving planets—differential equations model the real world. Unit 7 introduces you to this fundamental concept that connects calculus to physics, biology, economics, and engineering. Master this and you'll see how calculus explains the universe!

🎯 What is a Differential Equation?

DEFINITION

A differential equation is an equation that relates a function to one or more of its derivatives.

General Form:

\[ F\left(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}, \ldots\right) = 0 \]

Examples:

\(\frac{dy}{dx} = 3x^2\) — relates \(y\) to its first derivative
\(\frac{dy}{dx} = ky\) — exponential growth/decay model
\(\frac{d^2y}{dx^2} + y = 0\) — simple harmonic motion
\(\frac{dy}{dt} = k(A - y)\) — Newton's Law of Cooling

📝 Key Idea: Unlike regular equations that give you a NUMBER as a solution, differential equations have FUNCTIONS as solutions!

📚 Order and Basic Terminology

Order of a Differential Equation:

The order is the highest derivative that appears in the equation.

First-order: \(\frac{dy}{dx} = f(x, y)\)
Second-order: \(\frac{d^2y}{dx^2} = f(x, y, y')\)

AP® Calculus AB & BC focus: Primarily first-order differential equations!

✅ Solutions to Differential Equations

Types of Solutions

1. General Solution:

Contains arbitrary constants (usually \(C\))

\[ \text{Example: } \frac{dy}{dx} = 2x \quad \Rightarrow \quad y = x^2 + C \]

2. Particular Solution:

Specific solution when initial condition is given

\[ \text{Example: } \frac{dy}{dx} = 2x, \, y(0) = 3 \quad \Rightarrow \quad y = x^2 + 3 \]

Verifying Solutions:

To check if \(y = f(x)\) is a solution to a differential equation:

Find the derivative(s) of \(y\)
Substitute \(y\) and its derivative(s) into the differential equation
Verify both sides are equal

🎲 Initial Value Problems (IVPs)

INITIAL VALUE PROBLEM

A differential equation together with an initial condition that specifies the value of the solution at a particular point.

Standard Form:

\[ \frac{dy}{dx} = f(x, y), \quad y(x_0) = y_0 \]

where \((x_0, y_0)\) is the initial condition

🌍 Modeling Real-World Situations

General Modeling Strategy:

Identify variables: What quantities are changing?
Find the rate of change: What does the problem say about how things change?
Translate to mathematical equation: Express rate as derivative
Write the differential equation: Relate derivative to other quantities
Include initial conditions: What are the starting values?

Key Translation Phrases:

"The rate of change of \(y\)" → \(\frac{dy}{dt}\)
"increases/grows/expands" → positive rate
"decreases/decays/shrinks" → negative rate
"proportional to \(y\)" → multiply by \(ky\)
"proportional to difference" → \(k(A - y)\) or \(k(y - A)\)
"at a rate of" → equals some expression

📊 Common Differential Equation Models

Model 1: Exponential Growth and Decay

\[ \frac{dy}{dt} = ky \]

where:

\(y\) = quantity (population, mass, money, etc.)
\(k\) = growth rate constant
\(k > 0\): exponential growth
\(k < 0\): exponential decay

Solution: \(y = y_0 e^{kt}\)

Applications:

Population growth (bacteria, animals)
Radioactive decay
Compound interest
Epidemics (early stages)

Model 2: Newton's Law of Cooling/Heating

\[ \frac{dT}{dt} = k(T - T_s) \quad \text{or} \quad \frac{dT}{dt} = k(T_s - T) \]

where:

\(T\) = temperature of object
\(T_s\) = surrounding/ambient temperature
\(k\) = cooling/heating constant (negative for cooling)

Applications:

Cooling coffee
Forensics (time of death)
Heating/cooling systems

Model 3: Logistic Growth

\[ \frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right) \]

where:

\(P\) = population
\(k\) = growth rate
\(M\) = carrying capacity (maximum population)

Logistic growth starts exponentially but slows as population approaches carrying capacity!

Model 4: Limited Growth/Decay (Approach to Limit)

\[ \frac{dy}{dt} = k(L - y) \]

where:

\(y\) = current value
\(L\) = limiting value
\(k > 0\): rate constant

Applications:

Learning curves
Drug concentration in bloodstream
Sales approaching saturation

🔧 Setting Up Models from Word Problems

Example 1: Population Growth

Problem: A bacteria population grows at a rate proportional to its current size. If there are 1000 bacteria initially, write a differential equation to model this situation.

Solution:

Identify variable: Let \(P(t)\) = population at time \(t\)
Rate of change: "grows at a rate" → \(\frac{dP}{dt}\)
"proportional to current size": → \(\frac{dP}{dt} = kP\)
Initial condition: \(P(0) = 1000\)

\[ \frac{dP}{dt} = kP, \quad P(0) = 1000 \]

Example 2: Cooling Problem

Problem: A cup of coffee at 90°C is placed in a room at 20°C. The temperature decreases at a rate proportional to the difference between the coffee's temperature and room temperature. Set up the differential equation.

Solution:

Variable: Let \(T(t)\) = temperature of coffee at time \(t\)
Rate of change: \(\frac{dT}{dt}\)
"proportional to difference": → \(\frac{dT}{dt} = k(T - 20)\)
Since cooling, \(k < 0\) (or write \(\frac{dT}{dt} = -k(T - 20)\) with \(k > 0\))
Initial condition: \(T(0) = 90\)

\[ \frac{dT}{dt} = k(T - 20), \quad T(0) = 90 \]

Example 3: Tank Problem

Problem: A tank contains 100 gallons of brine (salt water). Pure water flows in at 5 gal/min, and the mixture flows out at 5 gal/min. If the tank initially contains 20 pounds of salt, write a differential equation for the amount of salt \(S(t)\) at time \(t\).

Solution:

Variable: \(S(t)\) = pounds of salt at time \(t\)
Rate of change: \(\frac{dS}{dt} = (\text{rate in}) - (\text{rate out})\)
Rate in: 0 (pure water, no salt)
Rate out: Concentration × flow rate = \(\frac{S(t)}{100} \cdot 5 = \frac{S}{20}\) lb/min
Initial condition: \(S(0) = 20\)

\[ \frac{dS}{dt} = -\frac{S}{20}, \quad S(0) = 20 \]

✓ Verifying Solutions

Example: Verify that \(y = 3e^{2x}\) is a solution to \(\frac{dy}{dx} = 2y\)

Solution:

Find derivative: \(\frac{dy}{dx} = 3 \cdot 2e^{2x} = 6e^{2x}\)
Find \(2y\): \(2y = 2(3e^{2x}) = 6e^{2x}\)
Compare: \(\frac{dy}{dx} = 6e^{2x} = 2y\) ✓

Yes, it's a solution!

📖 Essential Vocabulary

Differential Equations Terminology
Term	Definition
Differential Equation	Equation involving derivatives
Order	Highest derivative in equation
Solution	Function that satisfies the equation
General Solution	Family of solutions with constant(s)
Particular Solution	Specific solution (no constants)
Initial Condition	Given value at a specific point
IVP	Initial Value Problem

💡 Tips and Common Mistakes

✅ Success Tips:

Read carefully: Word choice matters! "Proportional to" means multiply
Define variables clearly: Always state what each symbol represents
Check units: Make sure rates have correct units
Sign matters: Growth is positive, decay is negative
Initial conditions: Don't forget to include them!
Verify solutions: Plug back in to check your work

❌ Common Mistakes:

Mistake 1: Confusing \(\frac{dy}{dx}\) with \(y\)
Mistake 2: Wrong sign for decay/cooling
Mistake 3: Forgetting that "proportional to" requires a constant \(k\)
Mistake 4: Not stating initial conditions
Mistake 5: Mixing up dependent and independent variables
Mistake 6: Not checking if solution satisfies initial condition

📝 Practice Problems

Set Up Differential Equations:

A population decreases at a rate of 5% per year. Write a DE for \(P(t)\).
An object at temperature 80°F is placed in a freezer at 0°F. The rate of cooling is proportional to the temperature difference. Set up the DE.
Money in an account grows at 4% annual interest compounded continuously. Write the DE.
The rate of change of velocity is proportional to velocity. Set up the DE for \(v(t)\).

Answers:

\(\frac{dP}{dt} = -0.05P\)
\(\frac{dT}{dt} = k(T - 0) = kT\) with \(k < 0\), or \(\frac{dT}{dt} = -k(T - 0)\) with \(k > 0\)
\(\frac{dA}{dt} = 0.04A\)
\(\frac{dv}{dt} = kv\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Clear variable definitions: State what each variable represents
Correct notation: Use \(\frac{dy}{dt}\) or \(y'\) consistently
Initial conditions: Include them when given
Units (if applicable): Show understanding of context
Justification: Explain why you set up equation that way
Verification: Check your solution works

⚡ Quick Reference Guide

COMMON MODELS SUMMARY

Model	Differential Equation	Application
Exponential Growth	\(\frac{dy}{dt} = ky\), \(k > 0\)	Population, bacteria
Exponential Decay	\(\frac{dy}{dt} = -ky\), \(k > 0\)	Radioactive decay
Newton's Law	\(\frac{dT}{dt} = k(T - T_s)\)	Cooling/heating
Logistic Growth	\(\frac{dP}{dt} = kP(1 - \frac{P}{M})\)	Limited population
Limited Growth	\(\frac{dy}{dt} = k(L - y)\)	Learning curves

Master Modeling with Differential Equations! A differential equation relates a function to its derivatives. The key to modeling is translating words into mathematics: "rate of change" becomes a derivative, "proportional to" means multiply by a constant. Common models include exponential growth/decay (\(\frac{dy}{dt} = ky\)), Newton's Law of Cooling (\(\frac{dT}{dt} = k(T - T_s)\)), and logistic growth (\(\frac{dP}{dt} = kP(1-\frac{P}{M})\)). An Initial Value Problem (IVP) includes an initial condition that determines a particular solution. To verify a solution, substitute it into the DE and check both sides are equal. Always define your variables clearly, pay attention to signs (growth = positive, decay = negative), and include initial conditions. Practice translating word problems into differential equations—this skill is essential for AP® success. Next, you'll learn to solve these equations and create slope fields to visualize solutions! 🎯✨