Exponential Models
with Differential Equations

Master the relationship between differential equations and exponential growth and decay, the foundation of modeling in calculus.

AP Calculus BC

What Is a Differential Equation?

A differential equation is any equation that relates a function to one or more of its derivatives. It tells us how a quantity changes rather than what the quantity is directly.

Solving a differential equation means finding the function $y(t)$ that satisfies the relationship. In this unit, we focus on equations whose solutions are exponential functions.

The Exponential Growth and Decay Model

The simplest and most important differential equation in applied mathematics is:

Growth ($k > 0$)

The quantity increases over time: populations, investments, viral spread.

Decay ($k < 0$)

The quantity decreases over time: radioactive decay, cooling, depreciation.

Visualizing Growth vs Decay

Solving $\dfrac{dy}{dt} = ky$ via Separation

1

Separate variables: $\dfrac{dy}{y} = k\,dt$

2

Integrate both sides: $\displaystyle\int \frac{1}{y}\,dy = \int k\,dt$ gives $\ln|y| = kt + C$

3

Exponentiate: $|y| = e^{kt+C} = e^C \cdot e^{kt}$

4

Apply initial condition: At $t = 0$, $y = y_0$, so $e^C = y_0$

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.

The Exponential Growth/Decay Theorem

The key insight: the exponential function is the only function that is proportional to its own derivative.

Growth Rate vs Growth Factor

	Growth Rate $k$	Growth Factor $e^k$
Definition	Continuous rate of change	Multiplier per unit time
Appears in	$\dfrac{dy}{dt} = ky$	$y = y_0 \cdot (e^k)^t$
Sign meaning	$k > 0$: growth; $k < 0$: decay	$e^k > 1$: growth; $e^k < 1$: decay
Example	$k = 0.06$ (6% per year)	$e^{0.06} \approx 1.0618$

Continuous vs Discrete Models

Continuous

$y = y_0 e^{kt}$
Rate is applied at every instant. Uses $\dfrac{dy}{dt} = ky$.

Discrete

$y = y_0(1 + r)^t$
Rate is applied at fixed intervals. No differential equation needed.

How to Solve an Exponential Model Problem

1

Identify the DE: Confirm the equation has the form $\dfrac{dy}{dt} = ky$. If not, it may require a different technique.

2

Write the general solution: $y(t) = y_0 e^{kt}$.

3

Use initial conditions to find $y_0$: plug in $t = 0$.

4

Use a second condition to find $k$: plug in the known $(t, y)$ pair and solve for $k$.

5

Answer the question: Find $y$ at a specific $t$, or find $t$ for a specific $y$.

Example 1: Population Growth

A bacteria colony grows at a rate proportional to its size. At $t = 0$ there are 500 bacteria. After 3 hours, there are 1500. Find the population at $t = 5$ hours.

1

Model: $\dfrac{dy}{dt} = ky$ with $y(0) = 500$, so $y = 500e^{kt}$.

2

Find $k$: $1500 = 500e^{3k}$ gives $e^{3k} = 3$, so $k = \dfrac{\ln 3}{3} \approx 0.3662$.

3

Answer: $y(5) = 500e^{5 \cdot \frac{\ln 3}{3}} = 500 \cdot 3^{5/3} \approx $ 3120 bacteria.

Example 2: Radioactive Decay

A substance has a half-life of 8 years. If you start with 100 grams, how much remains after 20 years?

1

Model: $y = 100e^{kt}$. Half-life means $y(8) = 50$.

2

Find $k$: $50 = 100e^{8k}$ gives $e^{8k} = \tfrac{1}{2}$, so $k = \dfrac{-\ln 2}{8} \approx -0.08664$.

3

Answer: $y(20) = 100e^{20 \cdot \frac{-\ln 2}{8}} = 100 \cdot 2^{-20/8} = 100 \cdot 2^{-2.5} \approx $ 17.68 grams.

Example 3: Continuously Compounded Interest

You invest \$2000 at a continuously compounded annual rate of 4.5%. How long until the investment doubles?

1

Model: $A = 2000\,e^{0.045t}$. We want $A = 4000$.

2

Solve: $4000 = 2000\,e^{0.045t}$ gives $e^{0.045t} = 2$, so $0.045t = \ln 2$.

3

Answer: $t = \dfrac{\ln 2}{0.045} \approx $ 15.4 years.

Particular Solutions and Initial Conditions

Doubling Time and Half-Life

Doubling Time

$$t_d = \frac{\ln 2}{k}$$ Time for quantity to double. Requires $k > 0$.

Half-Life

$$t_{1/2} = \frac{\ln 2}{|k|}$$ Time for quantity to halve. Uses $|k|$ since $k < 0$.

Common Mistakes to Avoid

✕ Wrong sign for $k$

Decay requires $k < 0$. If the problem says "decreasing," make sure your $k$ is negative, not positive.

✕ Using $(1+r)^t$ instead of $e^{kt}$

Differential equations model continuous change. The discrete formula gives a different (wrong) answer.

✕ Forgetting the constant $C$

After integration, omitting $+C$ means you lose the general family of solutions and cannot apply initial conditions.

Summary of Key Formulas

Concept	Formula	When to Use
General DE	$\dfrac{dy}{dt} = ky$	Rate proportional to amount
General Solution	$y = y_0 e^{kt}$	Always, for this DE type
Doubling Time	$t_d = \dfrac{\ln 2}{k}$	When $k > 0$, find time to double
Half-Life	$t_{1/2} = \dfrac{\ln 2}{\|k\|}$	When $k < 0$, find time to halve
Finding $k$	$k = \dfrac{\ln(y/y_0)}{t}$	Given two data points

Question 1 of 5

If $\dfrac{dy}{dt} = -0.03y$ and $y(0) = 200$, what is $y(10)$?

The solution is $y = 200e^{-0.03 \cdot 10} = 200e^{-0.3}$. Option C uses discrete decay $(1-r)^t$ which is incorrect for a differential equation. Option D treats the rate as linear.

Question 2 of 5

A population doubles every 5 years. What is the continuous growth rate $k$?

From $e^{5k} = 2$, we get $5k = \ln 2$, so $k = \dfrac{\ln 2}{5} \approx 0.1386$. Option B inverts the fraction. Options A and D are common guesses that don't use the exponential model.

Question 3 of 5

A substance decays according to $\dfrac{dN}{dt} = -0.0866N$. What is the half-life?

Half-life $= \dfrac{\ln 2}{|k|} = \dfrac{\ln 2}{0.0866} \approx 8$ years. Option D gives a negative time because it uses $\ln 0.5 = -\ln 2$ without taking the absolute value. Time must be positive!

Question 4 of 5

Which differential equation does NOT have an exponential solution?

Options A, B, and C all have the form $\dfrac{dy}{dt} = ky$, which gives $y = Ce^{kt}$. Option D has the form $\dfrac{dy}{dt} = ky + b$ (linear but non-homogeneous), whose solution involves an exponential plus a constant: not a pure exponential model.

Question 5 of 5

An investment of \$5000 grows continuously at rate $k$. After 6 years it is worth \$7500. Find $k$.

From $7500 = 5000e^{6k}$, we get $e^{6k} = 1.5$, so $6k = \ln 1.5$ and $k = \dfrac{\ln 1.5}{6} \approx 0.0676$. Option A uses $t=5$ instead of $t=6$. Option C divides the ratio by time without using $\ln$. Option D inverts the formula.

	Growth Rate \(k\)	Growth Factor \(e^k\)
Definition	Continuous rate of change	Multiplier per unit time
Appears in	\(\dfrac{dy}{dt} = ky\)	\(y = y_0 \cdot (e^k)^t\)
Sign meaning	\(k > 0\): growth; \(k < 0\): decay	\(e^k > 1\): growth; \(e^k < 1\): decay
Example	\(k = 0.06\) (6% per year)	\(e^{0.06} \approx 1.0618\)

Exponential Modelswith Differential Equations

What Is a Differential Equation?

The Exponential Growth and Decay Model

Visualizing Growth vs Decay

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

Cooling coffee with temperature and rate shown separately

Excess temperature decays toward zero

The Exponential Growth/Decay Theorem

Growth Rate vs Growth Factor

Continuous vs Discrete Models

How to Solve an Exponential Model Problem

Example 1: Population Growth

Example 2: Radioactive Decay

Example 3: Continuously Compounded Interest

Particular Solutions and Initial Conditions

Doubling Time and Half-Life

Common Mistakes to Avoid

Summary of Key Formulas

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Exponential Models
with Differential Equations