Modeling with
Differential Equations

Learn to translate real-world phenomena into the language of \(\frac{dy}{dx}\), then build equations that capture growth, decay, and change.

AP Calculus BC

What Is a Differential Equation?

A differential equation (DE) is any equation that relates a function to one or more of its derivatives. Instead of asking "what is \(y\)?", we ask: "how does \(y\) change?"

\(y\)

The unknown function we want to find

\(\frac{dy}{dt}\)

The rate at which \(y\) changes over time

Model

The DE that encodes the real-world rule

Why Model with Differential Equations?

Many quantities in nature don't change at a constant rate: they change relative to their current value or relative to some other quantity. A DE captures this dynamic relationship.

1

Population growth: more individuals means more births: \(\frac{dP}{dt} = kP\).

2

Cooling: a hot object cools faster when the temperature gap is large: \(\frac{dT}{dt} = -k(T - T_{\text{env}})\).

3

Chemical mixing: concentration changes as solution flows in and out of a tank.

Anatomy of a Differential Equation

The Core Modeling Principle

"proportional to"

\(\frac{dy}{dt} = ky\)

"proportional to the difference"

\(\frac{dy}{dt} = k(y - A)\)

"proportional to product of \(y\) and \((L - y)\)"

\(\frac{dy}{dt} = ky(L - y)\)

Three Classic Models

Each model captures a fundamentally different behavior: unbounded growth, approach to equilibrium, and bounded growth with a carrying capacity.

Exponential vs Logistic

	Exponential	Logistic
DE	\(\frac{dy}{dt}=ky\)	\(\frac{dP}{dt}=kP(L-P)\)
Rate depends on	Current value only	Current value AND room to grow
Long-term behavior	\(y \to \infty\)	\(P \to L\) (carrying capacity)
Solution	\(y = Ce^{kt}\)	\(P = \frac{L}{1+Ae^{-kLt}}\)
Shape	J-curve	S-curve (sigmoid)

An aquarium window connects population to growth rate

Caretaker, aerator, habitat shelf, vegetation and fish establish a managed pond. Visible fish are illustrative individuals, not the model initial population or carrying capacity. Separate exact panel: Separate computed plots show a population trajectory from1000 toward8000 and a growth-rate parabola peaking at4000 with1000fish per year.

For a lake with carrying capacity 5,000 fish and 200 fish initially, observing 800 fish at 6 months determines k ≈ 0.2533 per month. The model predicts about 3,996 fish at 18 months.

A different lake: months, not years

For a lake with carrying capacity 5,000 fish and 200 fish initially, observing 800 fish at 6 months determines k ≈ 0.2533 per month. The model predicts about 3,996 fish at 18 months.

Translating Words to Equations

English Phrase	DE Translation
"rate of change of \(y\)"	\(\frac{dy}{dt}\)
"is proportional to \(y\)"	\(= ky\)
"is proportional to the difference between \(y\) and 100"	\(= k(y - 100)\)
"increases at a constant rate of 5 per hour"	\(= 5\)
"decreases at a rate proportional to \(y^2\)"	\(= -ky^2\)

How to Set Up a DE Model

1

Identify the quantity. What is changing? Call it \(y\), \(P\), \(T\), etc.

2

Identify the independent variable. Usually time \(t\). Write the derivative: \(\frac{dy}{dt}\).

3

Translate the rate description. "Proportional to" \(\to\) multiply by \(k\). "Difference" \(\to\) subtract.

4

Include signs. Growth \(\to\) positive \(k\). Decay/cooling \(\to\) negative sign or negative \(k\).

5

State the initial condition. \(y(0) = y_0\). This anchors the solution.

Example 1: Bacterial Growth

"A bacteria colony doubles every 3 hours. The rate of growth is proportional to the current population. At \(t = 0\), there are 500 bacteria."

1

Quantity: \(P\) = population. Independent variable: \(t\) (hours).

2

Rate description: "rate proportional to population" \(\to\) \(\frac{dP}{dt} = kP\).

3

Initial condition: \(P(0) = 500\).

Example 2: Newton's Cooling

"A cup of coffee at \(90°\)C is placed in a room at \(20°\)C. It cools at a rate proportional to the difference between its temperature and the room temperature."

Cooling coffee with temperature and rate shown separately

A steaming coffee cup and a cool liquid cup with condensation, each probed, with a shared room-reference probe between them.

With ambient temperature 22 °C and k > 0, T′ = −k(T − 22). A warmer liquid cools and a cooler liquid warms, both approaching ambient temperature.

Excess temperature decays toward zero

With ambient temperature 22 °C and k > 0, T′ = −k(T − 22). A warmer liquid cools and a cooler liquid warms, both approaching ambient temperature.

Example 3: Logistic Growth

"A lake can support at most 10,000 fish. The fish population grows at a rate proportional to both the current population and the remaining capacity."

1

Quantity: \(P\) = fish population. Carrying capacity \(L = 10{,}000\).

2

"proportional to both \(P\) and remaining capacity": rate \(\propto P(L - P)\).

3

The model: \(\frac{dP}{dt} = kP(10{,}000 - P)\).

Beyond the Standard Three

Not every AP problem fits neatly into exponential, cooling, or logistic. You must read carefully and translate each phrase independently.

Draining Tank

"Water flows in at 5 L/min and out at a rate of \(\frac{V}{10}\) L/min": \(\frac{dV}{dt} = 5 - \frac{V}{10}\)

Drug Dosage

"Drug is absorbed at 3 mg/hr and eliminated proportional to the amount": \(\frac{dA}{dt} = 3 - kA\)

The Role of Initial Conditions

A DE alone has infinitely many solutions (a family of curves). The initial condition \(y(t_0) = y_0\) picks exactly one curve from this family.

Common Mistakes

✕ Wrong Sign

Writing \(\frac{dT}{dt} = k(T - T_e)\) for cooling. A cooling object loses heat: the sign must be negative: \(-k(T - T_e)\).

✕ Missing Factor

Writing \(\frac{dP}{dt} = kP\) when the problem says "proportional to the product of \(P\) and \((L - P)\)." You need both factors.

✕ Forgetting Initial Condition

Setting up the DE but not stating \(y(0) = y_0\). Without it, you can't find the particular solution on the AP exam.

Summary of DE Models

Scenario	DE Form	Key Feature
Exponential growth/decay	\(\frac{dy}{dt} = ky\)	Rate \(\propto\) value
Newton's cooling/heating	\(\frac{dT}{dt} = -k(T - T_e)\)	Rate \(\propto\) difference
Logistic growth	\(\frac{dP}{dt} = kP(L - P)\)	Rate \(\propto\) value \(\times\) remaining
Mixing/flow	\(\frac{dy}{dt} = \text{in} - \text{out}\)	Balance of competing rates

Question 1 of 5

A radioactive substance decays at a rate proportional to the amount present. If \(Q\) is the amount at time \(t\), which DE models this?

"Rate proportional to the amount" gives \(\frac{dQ}{dt} = kQ\). Since the substance is decaying (decreasing), we need a negative sign: \(\frac{dQ}{dt} = -kQ\) with \(k > 0\).

Question 2 of 5

A metal rod at \(200°\)C is placed in a room at \(25°\)C. Its temperature changes at a rate proportional to the difference between its temperature and the room's. Which is the correct model?

Newton's Law of Cooling: \(\frac{dT}{dt} = -k(T - T_{\text{env}})\). Here \(T_{\text{env}} = 25\). The negative sign ensures cooling when \(T > 25\). Option (d) reverses the subtraction, giving the wrong sign for the rate when \(T > 25\).

Question 3 of 5

A population of deer in a forest grows logistically with carrying capacity \(L = 800\). Which DE models this?

Logistic growth requires the rate to be proportional to both the current population \(P\) and the remaining capacity \((L - P)\). Only \(\frac{dP}{dt} = kP(800 - P)\) includes both factors. Option (a) is exponential, option (b) is Newton's cooling form, and option (c) is not a standard model.

Question 4 of 5

Water flows into a tank at \(3\) gallons/min and out at a rate of \(\frac{V}{20}\) gallons/min, where \(V\) is the volume in the tank. Which DE models the volume?

The net rate of change is rate in minus rate out: \(\frac{dV}{dt} = 3 - \frac{V}{20}\). Options (b) and (d) reverse the signs, and (c) introduces an incorrect expression.

Question 5 of 5

If \(\frac{dy}{dt} = 0.3y\) and \(y(0) = 50\), what is the value of \(\frac{dy}{dt}\) at \(t = 0\)?

Substitute the initial condition: \(\frac{dy}{dt}\bigg|_{t=0} = 0.3 \times y(0) = 0.3 \times 50 = 15\). A common mistake is forgetting to multiply: \(0.3\) alone is the constant, not the rate at \(t=0\).

Modeling withDifferential Equations

What Is a Differential Equation?

Why Model with Differential Equations?

Anatomy of a Differential Equation

The Core Modeling Principle

Three Classic Models

Exponential vs Logistic

An aquarium window connects population to growth rate

A different lake: months, not years

Translating Words to Equations

How to Set Up a DE Model

Example 1: Bacterial Growth

Example 2: Newton's Cooling

Cooling coffee with temperature and rate shown separately

Excess temperature decays toward zero

Example 3: Logistic Growth

Beyond the Standard Three

The Role of Initial Conditions

Common Mistakes

Summary of DE Models

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Modeling with
Differential Equations