Modeling with Differential Equations

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Quick Reference

Before You Start: Prerequisite Check

Try This First

A cup of coffee sits on a desk. Without any formula, answer three questions in order.

1. When the coffee is much hotter than the room, is it cooling fast or slowly?
2. An hour later, after it has nearly reached room temperature, is it cooling fast or slowly?
3. So the rate at which the coffee cools is large when the temperature gap is ____, and small when the gap is ____.

A Rate Is a Derivative, and a Model Equates It to the State

Many laws in science and economics describe how fast something changes, not the thing itself. A rate of change is a derivative. Modeling with a differential equation is the act of writing one sentence:

the derivative of the state = an expression built from the state.

Watch how the output (the rate) responds as the input (the current amount) moves:

• When the amount present is large, the rate $x'(t)$ is large.
• As the amount falls, the rate falls with it.
• The rate is never a single fixed number. It is a quantity that varies with the state.

Two connected representations of the same decay process appear below. Read across each row and confirm the table agrees with the words.

The picture below shows the same idea as a graph of the state over time, falling steeply at first and flattening as the amount approaches zero.

 x
100 ●
    │●
    │ ●
    │   ●
    │      ●
    │          ●
    │               ●  ●  ●  ●  ●
  0 └─────────────────────────────── t

Translate between the two: the steep part of the graph on the left matches the large rate in the top table row, and the flat part on the right matches the small rate in the bottom row. The graph and the table say the same thing.

Prerequisite Hub

Builds on (read these first):

Unlocks (what this opens next):

Quick Reference

Modeling at a Glance

Special case (growth and decay): when the rate is proportional to the amount present, the model is $x' = rx$ (growth, $r > 0$) or $x' = -rx$ (decay, $r > 0$).

The Official Definitions

Modeling a rate law as a differential equation

Setting up a differential equation translates a physical law into a relation between a state and its derivative. For a body slowing under a resistive force $F = -kv$, Newton’s second law $ma = F$ with $a = v'(t)$ gives the equation of motion

\[ m\,v'(t) = -k\,v(t). \]

The verbal rate law becomes a derivative equated to an expression in the state.

(Logan, 3rd ed., §1.1.1, pp.2-3, Eq.(1.1))

Decay (and growth) model

Processes whose rate of change is proportional to the amount present are modeled by

\[ x' = -r\,x \quad (\text{decay},\ r > 0) \qquad\text{or}\qquad x' = r\,x \quad (\text{growth},\ r > 0), \]

where $-r$ (or $r$) is the proportionality constant and $x = x(t)$ is the state.

(Logan, 3rd ed., §1.1.2 Growth-Decay Models, p.6, Eq.(1.4))

Reading note: the model is the equation itself, not its solution. A common habit from earlier courses is to jump straight to a formula for $x(t)$. Here the goal stops at the correctly written differential equation. Solving it is the next skill.

The Modeling Procedure

Goal: turn a described situation into a correct differential equation.

Step 1: Name the state and the independent variable. Choose a letter for the quantity that changes (for example $P$ for population, $v$ for velocity, $T$ for temperature) and a letter for what it changes with respect to (usually $t$ for time). State the units.

Step 2: Identify the rate. The phrase “rate of change of [state]” is the derivative of the state, written $P'(t)$ or $\frac{dP}{dt}$.

Step 3: Write the rate law in words. Find the sentence that says how fast the quantity changes. Look for “proportional to,” “increases at,” “the difference between,” or a stated physical law.

Step 4: Translate each word into symbols. “Proportional to $x$” becomes a constant times $x$. “The difference between $T$ and the room temperature $M$” becomes $(T - M)$. Keep the sign honest: cooling and decay carry a minus sign.

Step 5: Set the derivative equal to the expression. The result is the model. Stop there. Do not solve.

Worked Examples

Example 1: A population growing at a proportional rate

A bacteria culture grows so that its rate of increase is proportional to the number of bacteria present. Write a model. There are $500$ bacteria at the start.

Predict first: more bacteria means a faster increase, so the rate should rise as the count rises. The model should make $P'$ a positive multiple of $P$.

Step 1: Let $P = P(t)$ be the number of bacteria, $t$ in hours.

Step 2: The rate of increase is $P'(t)$.

Step 3: “Rate of increase is proportional to the number present” means $P' = (\text{constant}) \cdot P$.

Step 4: Growth gives a positive constant. Call it $k > 0$.

Step 5: The model is

\[ P' = kP, \qquad P(0) = 500. \]

Check against the prediction: when $P$ is large, $kP$ is large, so the rate is large. The sign is positive, matching growth. The prediction holds.

Example 2: A cooling cup of coffee

Coffee at temperature $T$ sits in a room held at $20^\circ$C. The rate at which the coffee cools is proportional to the difference between its temperature and the room temperature. Write a model.

Predict first: from the Try This First opener, a large gap drives fast cooling and a small gap drives slow cooling. The rate should depend on $(T - 20)$, and it should be negative while the coffee is hotter than the room.

Step 1: Let $T = T(t)$ be the coffee temperature in degrees Celsius, $t$ in minutes. The room temperature is the constant $M = 20$.

Step 2: The rate of change is $T'(t)$.

Step 3: “Cools at a rate proportional to the difference between its temperature and the room” gives $T' = (\text{constant}) \cdot (T - 20)$.

Step 4: While $T > 20$ the coffee is cooling, so $T'$ must be negative while $(T - 20)$ is positive. The constant is negative. Write it as $-k$ with $k > 0$.

Step 5: The model is

\[ T' = -k\,(T - 20). \]

Check against the prediction: at $T = 90$ the factor $(T - 20) = 70$ is large, so $|T'|$ is large (fast cooling). At $T = 21$ the factor is $1$, so $|T'|$ is small (slow cooling). The minus sign makes $T'$ negative while the coffee is above room temperature. Every part of the prediction holds.

Example 3: An object slowing under resistance

A small boat coasts to a stop after the engine is cut. The only force is water resistance, proportional to the speed: $F = -kv$ with $k > 0$. The boat has mass $m$. Write a model for the velocity.

Predict first: resistance opposes motion, so a faster boat feels a bigger backward force and slows more quickly. The rate of change of velocity should be negative and proportional to $v$.

Step 1: Let $v = v(t)$ be the velocity, $t$ in seconds, $m$ the constant mass.

Step 2: Acceleration is the rate of change of velocity, $a = v'(t)$.

Step 3: Newton’s second law states $ma = F$.

Step 4: Substitute $a = v'(t)$ and $F = -kv$:

\[ m\,v'(t) = -k\,v(t). \]

Step 5: This is the model, the equation of motion. Dividing by $m$ gives the equivalent decay form $v' = -\frac{k}{m}\,v$.

Check against the prediction: the right side is negative whenever $v$ is positive, so the boat slows. A larger $v$ gives a larger backward rate. The prediction holds, and this matches Logan Eq.(1.1) exactly.

Common Misconceptions

Practice Problems

Mastery Checklist

Novice (Level 1-2):

• I can read a derivative such as $P'(t)$ as a rate of change, with correct units
• I can translate “rate of change of $y$ is proportional to $y$” into $y' = ky$
• I can choose the correct sign for a growth versus a decay model

Competent (Level 3-4):

• I can write a cooling or warming model using the temperature difference $(T - M)$
• I can include an initial condition when one is given
• I can build a model from a force balance using $ma = F_{\text{net}}$ with $a = v'$

Proficient (Level 5):

• I can extend a proportional-rate model with a limiting factor (logistic growth)
• I can explain why the model is the differential equation, separate from its solution
• I can verify a model by checking its behavior at extreme values of the state

Two Ways to See a Model

One way to read $T' = -k(T - 20)$ is symbolic: a derivative on the left set equal to a function of the state on the right. Another way is dynamic: a description of how the temperature moves, fast when far from $20^\circ$C and slow when near it. Both readings describe the same equation. Which one do you find clearer, and why? Being able to switch between the symbolic form and the moving picture is the heart of working with models.

Why Build a Model Before Solving

A model is worth writing even before any solution method is known. Suppose a classmate asks why the falling-ball model in Level 4 must have a terminal velocity. Convince the classmate using only the model $m v' = mg - kv$: the rate of change $v'$ is zero exactly when $mg - kv = 0$, that is when $v = \frac{mg}{k}$. The model alone, with no formula for $v(t)$, already predicts the speed at which falling stops accelerating. Reading consequences straight off the equation is a reason modeling comes first.

Connections

Looking back:

• Introduction to Differential Equations defines the equation that a model produces.
• Exponential Growth and Decay is the first model translated from a verbal rate law.

Looking ahead:

• Direction Fields (9.2): a model $y' = f(t, y)$ is drawn as a field of slopes, a picture of every solution at once.
• Separable Equations (9.3): growth, decay, and cooling models are the first equations solved by separating variables.
• Applications of Separable Equations (9.3): each application opens by building the model on this page.

Real-world connections:

• Population biology: bacteria, fish, and animal counts modeled by growth and logistic laws.
• Thermodynamics: Newton’s law of cooling for objects approaching room temperature.
• Mechanics: motion under gravity and resistance, written as a force balance.

Resources

• Primary text (this skill): Logan, A First Course in Differential Equations, 3rd ed., §1.1.1-1.1.2, pp.2-10. Modeling the equation of motion and the decay model. PDF
• Companion text: Stewart, Calculus, Chapter 9, §9.1 Modeling with Differential Equations ($\frac{dP}{dt} = kP$, the spring-mass model). Local source: ~/math347-ingest/math_guides/MATH_347/Stewart Chapter 9/Section 1/stewart_calculus_chapter_9_section_1_modeling_with_differential_equations.tex
• Open text: OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations (translating a situation into a differential equation). OpenStax

Last updated: 2026-06-16