Exponential Growth and Decay: The ODE Model

5 min read

Reference: Stewart §6.5

When Change Is Proportional to Size

Why do populations explode? Why do radioactive materials decay? Why does your coffee cool quickly at first, then slower? In each case, the rate of change is proportional to the current amount. This single principle, expressed as $\frac{dy}{dt} = ky$, governs an astonishing range of natural phenomena.

This differential equation is the foundation for modeling growth and decay. Master it and you can predict populations, date ancient artifacts, and understand how investments compound.

Before You Start

Quick Self-Check

Can you answer these? If not, review the linked prerequisite first.

What is $\frac{d}{dx}[e^{5x}]$? → Derivative of Exponential
Solve for $t$: $e^{2t} = 7$ → Natural Logarithm

Prerequisite Map

Quick Reference

Property	Value
Concept	Differential Equations
Chapter	6, Section 5
Difficulty	Beginner
Time	~15 minutes

Key Concepts

The Differential Equation

The law of natural growth (or decay) states:

$$\boxed{\frac{dy}{dt} = ky}$$

where:

$y(t)$ = quantity at time $t$
$k$ = proportionality constant (growth/decay rate)
$\frac{dy}{dt}$ = rate of change

Interpretation:

If $k > 0$: exponential growth (population, investment)
If $k < 0$: exponential decay (radioactive decay, cooling)

The Solution

Theorem: The only solutions of $\frac{dy}{dt} = ky$ are:

$$\boxed{y(t) = y(0)e^{kt} = Ce^{kt}}$$

where $C = y(0)$ is the initial value.

Verification

Why is $y = Ce^{kt}$ the solution? Differentiate:

$$\frac{dy}{dt} = \frac{d}{dt}[Ce^{kt}] = C \cdot ke^{kt} = k(Ce^{kt}) = ky \quad \checkmark$$

Physical Meaning of $k$

Context	Name for $k$	Units
Population	Relative growth rate	per year, per hour
Radioactive decay	Decay constant	per year, per second
Finance	Interest rate	per year
Temperature	Cooling constant	per minute

The Relative Growth Rate

The quantity $\frac{dy/dt}{y} = k$ is called the relative growth rate, the rate of change per unit of current value.

Example: If $k = 0.03$ per year, the quantity grows at 3% per year.

Continuous Compounding (Special Case)

For an investment with principal $A_0$ and interest rate $r$ compounded continuously:

$$A(t) = A_0 e^{rt}$$

This follows from $\frac{dA}{dt} = rA$ (the rate of increase equals the interest rate times the current amount).

Common Pitfalls

Mistake	Correction
Writing $y(t) = e^{kt}$ without initial value	Always write $y(t) = y_0 e^{kt}$ or $y(t) = Ce^{kt}$
Confusing $k$ with the percentage change	$k = 0.05$ means 5% relative rate, but actual growth after 1 unit is $e^{0.05} - 1 \approx 5.13\%$
Thinking decay reaches zero	$e^{kt} \to 0$ as $t \to \infty$, but never equals 0
Using $\ln$ incorrectly: $\ln(a/b) \neq \ln a / \ln b$	Correct: $\ln(a/b) = \ln a - \ln b$
Mixing time units	If $k$ is per hour, $t$ must be in hours

Worked Example

Problem: A quantity satisfies $\frac{dy}{dt} = 0.05y$ with $y(0) = 200$. Find $y(t)$ and determine when the quantity doubles.

Solution:

Step 1: Apply the general solution. $$y(t) = y(0)e^{kt} = 200e^{0.05t}$$

Step 2: Find the doubling time by setting $y(t) = 400$. $$200e^{0.05t} = 400$$ $$e^{0.05t} = 2$$ $$0.05t = \ln 2$$ $$t = \frac{\ln 2}{0.05} = 20\ln 2 \approx 13.86$$

The quantity doubles in approximately 13.86 time units.

Practice Problems

Level 1 Verifying a Solution

Verify that $y(t) = 5e^{-2t}$ is a solution to the differential equation $\frac{dy}{dt} = -2y$.

Thought Process

To verify a solution, substitute it into the differential equation. Compute $\frac{dy}{dt}$ and check that it equals $-2y$.

Show Answer

Compute the left side: $$\frac{dy}{dt} = \frac{d}{dt}[5e^{-2t}] = 5 \cdot (-2)e^{-2t} = -10e^{-2t}$$

Compute the right side: $$-2y = -2(5e^{-2t}) = -10e^{-2t}$$

Since $\frac{dy}{dt} = -2y$, the function is a solution. ✓

Level 2 Solving an Initial Value Problem

Solve the initial value problem: $\frac{dy}{dt} = 3y$, $y(0) = 7$.

Thought Process

This is a direct application of the formula. Here $k = 3$ and the initial condition gives $C = y(0) = 7$.

Show Answer

The general solution of $\frac{dy}{dt} = ky$ is $y(t) = Ce^{kt}$.

With $k = 3$ and $y(0) = 7$: $$y(t) = 7e^{3t}$$

Level 3 Finding the Growth Constant

A quantity grows exponentially. At $t = 0$ it is 50, and at $t = 4$ it is 150. Find the growth constant $k$ and write the formula for $y(t)$.

Thought Process

Use the two data points to set up an equation. We know $y(t) = 50e^{kt}$, and $y(4) = 150$. Solve for $k$ using logarithms.

Show Answer

From $y(0) = 50$, we have $y(t) = 50e^{kt}$.

Using $y(4) = 150$: $$50e^{4k} = 150$$ $$e^{4k} = 3$$ $$4k = \ln 3$$ $$k = \frac{\ln 3}{4} \approx 0.275$$

Therefore: $$y(t) = 50e^{(\ln 3/4)t} = 50 \cdot 3^{t/4}$$

Level 4 Continuous Compounding

An investment of \$8000 earns interest compounded continuously.

If the investment doubles in 12 years, what is the annual interest rate?
How long will it take for the investment to reach \$25,000?

Thought Process

For continuous compounding, $A(t) = A_0 e^{rt}$.

(a) Doubling means $A = 2A_0$. Set up $2 = e^{12r}$ and solve for $r$.

(b) Use the rate from (a) to solve $25000 = 8000e^{rt}$ for $t$.

Show Answer

(a) Doubling in 12 years means: $$16000 = 8000e^{12r}$$ $$2 = e^{12r}$$ $$12r = \ln 2$$ $$r = \frac{\ln 2}{12} \approx 0.0578$$

The annual interest rate is approximately 5.78%.

(b) Using $r = \frac{\ln 2}{12}$: $$25000 = 8000e^{rt}$$ $$\frac{25}{8} = e^{rt}$$ $$t = \frac{\ln(25/8)}{r} = \frac{\ln(25/8)}{\ln 2/12} = \frac{12\ln(25/8)}{\ln 2}$$ $$t = \frac{12 \cdot 1.139}{0.693} \approx 19.7 \text{ years}$$

Level 5 Uniqueness of Exponential Solutions

Prove that if $y(t)$ is any solution to $\frac{dy}{dt} = ky$ with $y(0) = C$, then $y(t) = Ce^{kt}$.

Hint: Consider the function $g(t) = y(t)e^{-kt}$ and show that $g'(t) = 0$.

Thought Process

If we can show $g(t)$ is constant, then $y(t)e^{-kt} = C$ for some constant, which means $y(t) = Ce^{kt}$.

Use the product rule to differentiate $g(t)$, then substitute $y' = ky$.

Show Answer

Proof:

Let $g(t) = y(t)e^{-kt}$.

Differentiate using the product rule: $$g'(t) = y'(t)e^{-kt} + y(t) \cdot (-k)e^{-kt}$$ $$= y'(t)e^{-kt} - ky(t)e^{-kt}$$ $$= e^{-kt}[y'(t) - ky(t)]$$

Since $y$ satisfies $\frac{dy}{dt} = ky$, we have $y'(t) - ky(t) = 0$.

Therefore $g'(t) = 0$, so $g(t)$ is constant.

Since $g(0) = y(0)e^0 = y(0) = C$, we have $g(t) = C$ for all $t$.

Thus $y(t)e^{-kt} = C$, which gives $y(t) = Ce^{kt}$. $\square$

CCI-Style Conceptual Questions

Question 1: If a quantity satisfies $\frac{dy}{dt} = -0.1y$, is the quantity growing or decaying? What is the relative rate of change?

Answer

Decaying, because $k = -0.1 < 0$. The relative rate of change is $-0.1$ or $-10\%$ per unit time.

Question 2: Two quantities both satisfy exponential growth equations. Quantity A has $k = 0.02$ and quantity B has $k = 0.05$. If both start at the same value, which grows faster? By what factor will B exceed A after 10 time units?

Answer

B grows faster (larger $k$). After 10 units:

$A = Ce^{0.2}$
$B = Ce^{0.5}$
Ratio: $\frac{B}{A} = e^{0.3} \approx 1.35$

B exceeds A by a factor of about 1.35.

Question 3: The doubling time for quantity X is 5 years. The doubling time for quantity Y is 10 years. What is the ratio of their growth constants?

Answer

Doubling time $T_d$ satisfies $e^{kT_d} = 2$, so $k = \frac{\ln 2}{T_d}$.

$k_X = \frac{\ln 2}{5}$
$k_Y = \frac{\ln 2}{10}$

Ratio: $\frac{k_X}{k_Y} = \frac{10}{5} = 2$

Common Misconceptions

Common misconception

the differential equation $\frac{dy}{dt} = ky$ has solution $y(t) = y_0 + kt$, a linear function.

This is the action-view-of-function error: the equation is treated as a statement about arithmetic addition rather than a rate law. The equation $\frac{dy}{dt} = ky$ says the rate of change equals $k$ times the current value, not that $y$ increases by $k$ at each step. The solution $y(t) = y_0 e^{kt}$ can be verified by differentiating: $\frac{d}{dt}[y_0 e^{kt}] = ky_0 e^{kt} = ky(t)$. A linear function $y_0 + kt$ has constant derivative $k$, which matches $ky$ only if $y$ itself is constant.

Mastery Checklist

I can write the differential equation for exponential growth/decay
I can verify that $y = Ce^{kt}$ satisfies $\frac{dy}{dt} = ky$
I can solve initial value problems of the form $y' = ky$, $y(0) = C$
I can find $k$ given two data points
I understand that $k > 0$ means growth and $k < 0$ means decay
I can find doubling time (or half-life) from the growth constant
I can apply continuous compounding formula $A = A_0 e^{rt}$

Mental Model

The “Interest on Interest” Analogy:

Think of exponential growth like compound interest that compounds every instant. Each tiny moment, you earn interest not just on your principal, but on all the interest you’ve already earned. The more you have, the more you gain; this feedback loop creates the exponential curve.

Still Confused?

If you’re struggling with...	Review this
The derivative of $e^{kt}$	Derivative of Exponential Functions
Solving equations with $\ln$	Natural Logarithm
What a differential equation is	This is a first-order ODE; see Introduction to Differential Equations

Connections

Looking back:

Derivative of $e^x$ is why $e^{kt}$ appears in the solution
The equation $y' = ky$ is the simplest first-order differential equation

Looking ahead:

Population growth applies this to biology
Half-life applies this to physics
Newton’s cooling extends this with a shift
Separable differential equations generalize this technique

Real-world connections:

Compound interest and investment growth
Bacterial population dynamics
Radioactive dating in archaeology
Drug metabolism in pharmacology

Previous	Up	Next
§6.4 Skills	Chapter 6 Skills	Population Growth

Last updated: 2026-01-22