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Exponential Models with
Differential Equations

Learn how the equation \(\dfrac{dy}{dt} = ky\) models growth and decay in nature, finance, and science.

AP Calculus AB

What Is a Differential Equation?

A differential equation is an equation that relates a function to one or more of its derivatives. Instead of telling you the value of \(y\), it tells you how \(y\) changes.

Our goal: find the function \(y(t)\) that satisfies this relationship. This process is called solving the differential equation.

Why Exponential Models?

Many real-world quantities change at a rate proportional to their current size. This one simple idea produces the most important differential equation in calculus:

\(k > 0\)
Exponential growth: populations, investments, viral spread
\(k < 0\)
Exponential decay: radioactive isotopes, cooling, depreciation

A cell culture where more cells create more growth events

A researcher looks through a microscope beside a closed culture vessel,sample containers and an illustrative cell inset.In the model y = 100e^(0.4t), the relative growth rate stays at 0.4 while total growth y′ = 0.4y increases with the population.

A cell culture where more cells create more growth events

In the model y = 100e^(0.4t), the relative growth rate stays at 0.4 while total growth y′ = 0.4y increases with the population.In the model y = 100e^(0.4t), the relative growth rate stays at 0.4 while total growth y′ = 0.4y increases with the population.

Relative rate versus total rate

In the model y = 100e^(0.4t), the relative growth rate stays at 0.4 while total growth y′ = 0.4y increases with the population.In the model y = 100e^(0.4t), the relative growth rate stays at 0.4 while total growth y′ = 0.4y increases with the population.

Visualizing Growth and Decay

t y
\(y_0\)
\(k > 0\) (growth)
\(k < 0\) (decay)
\((0, y_0)\)

Solving \(\dfrac{dy}{dt} = ky\) by Separation

1
Separate variables: \(\dfrac{dy}{y} = k\, dt\)
2
Integrate both sides: \(\displaystyle\int \frac{1}{y}\,dy = \int k\, dt\)
3
Evaluate: \(\ln|y| = kt + C\)
4
Exponentiate: \(|y| = e^{kt+C} = e^C \cdot e^{kt}\)
5
General solution: \(y = Ae^{kt}\), where \(A = \pm e^C\)

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.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.

The General Solution

\(y_0\)
Starting amount at \(t = 0\)
\(k\)
Continuous rate constant (growth if \(k > 0\), decay if \(k < 0\))
\(t\)
Time elapsed since measurement began

Continuous vs Discrete Growth

Continuous Discrete
Model \(y = y_0 e^{kt}\) \(y = y_0 (1 + r)^t\)
Equation \(\dfrac{dy}{dt} = ky\) \(y_{n+1} = (1+r)\,y_n\)
Rate meaning Instantaneous rate Rate per period
AP Calculus focus ★ Primary Background

How to Apply the Model

1
Identify the DE: Confirm the rate of change is proportional to \(y\): \(\dfrac{dy}{dt} = ky\).
2
Write the solution: \(y(t) = y_0 e^{kt}\).
3
Plug in the initial condition to find \(y_0\). Often \(y(0) = y_0\) is given directly.
4
Use a second data point to find \(k\). Substitute, solve for \(k\) using \(\ln\).
5
Answer the question: Find \(y\) at a specific time, or find \(t\) when \(y\) reaches a target.

Example 1: Population Growth

A bacteria colony starts with 500 cells and grows at a rate proportional to its size. After 3 hours, there are 1500 cells. Find the population after 5 hours.

Example 1: Visualized

t (hours) y 0 1 2 3 4 5
\((0, 500)\)
\((3, 1500)\)
\((5, \approx 3121)\)

Example 2: Radioactive Decay

A substance has a half-life of 8 years. If the initial amount is 200 grams, how much remains after 20 years?

Example 3: Solving for Time

An investment of $2000 grows continuously at a rate of 6% per year. When will it reach $5000?

Doubling Time and Half-Life

Doubling Time
When \(k > 0\):
$$t_d = \frac{\ln 2}{k}$$
Half-Life
When \(k < 0\):
$$t_h = \frac{\ln 2}{|k|}$$

Particular vs General Solutions

t y
\((0, y_0)\)
General: \(y = Ae^{kt}\)
 Particular

Common Mistakes to Avoid

✕ Forgetting the constant
Writing \(y = e^{kt}\) instead of \(y = y_0 e^{kt}\). The initial condition \(y_0\) must be included.
✕ Wrong sign for \(k\)
Using \(k > 0\) for decay problems. If the quantity decreases, \(k\) must be negative.
✕ Mixing up rates
Plugging a percentage (like 6%) directly as \(k = 6\) instead of \(k = 0.06\).

Summary Reference Table

Concept Formula / Rule
Differential equation \(\dfrac{dy}{dt} = ky\)
General solution \(y = Ae^{kt}\)
Particular solution \(y = y_0 e^{kt}\)
Doubling time \(t_d = \dfrac{\ln 2}{k}\)
Half-life \(t_h = \dfrac{\ln 2}{|k|}\)
Finding \(k\) Use two data points and \(\ln\)
Finding time \(t\) Isolate \(e^{kt}\), take \(\ln\)

Question 1 of 5

If \(\dfrac{dy}{dt} = -0.3y\) and \(y(0) = 100\), what is \(y(t)\)?

Question 2 of 5

A population doubles every 5 years. What is the continuous growth rate \(k\)?

Question 3 of 5

A sample of 80 grams decays to 20 grams in 10 hours. What is the half-life?

Question 4 of 5

An investment grows continuously. It is worth $1000 at \(t = 0\) and $1500 at \(t = 4\). When will it be worth $3000?

Question 5 of 5

Which of the following differential equations does NOT have the solution \(y = Ce^{kt}\)?