Skip to content

Applications - Growth and Decay

You will learn how to model and solve real-world growth and decay problems using separable differential equations, including unlimited exponential growth, exponential decay, and limited logistic growth.

Many natural processes involve a rate of change that is proportional to the current amount. These lead to the differential equation:

dAdt=kA\frac{dA}{dt} = kA

This is the classic exponential growth/decay model. Its solution is:

A(t)=A0ektA(t) = A_0 e^{kt}

When resources are limited, growth slows down near a carrying capacity. This is modeled by the logistic equation:

dPdt=rP(1PK)\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)

where KK is the carrying capacity. This is still separable but requires partial fractions to solve.

The visual above compares pure exponential growth (red) with logistic growth (blue) approaching a carrying capacity. Notice how the logistic curve flattens out while the exponential keeps climbing.

Example 1: Exponential Decay

The half-life of Carbon-14 is 5730 years. How much of a 100-gram sample remains after 10,000 years?

Solution: The decay constant is k=ln(0.5)57300.000121k = \frac{\ln(0.5)}{5730} \approx -0.000121.

A(t)=100e0.0001211000029.8 gramsA(t) = 100 e^{-0.000121 \cdot 10000} \approx 29.8 \text{ grams}

Example 2: Unlimited Growth

A bacteria culture grows at a rate proportional to its size. It doubles every 3 hours. How long until it reaches 10 times the original amount?

Solution: k=ln23k = \frac{\ln 2}{3}. Set A(t)=10A0A(t) = 10A_0:

10=ekt    t=ln10k=3ln10ln29.97 hours10 = e^{kt} \implies t = \frac{\ln 10}{k} = \frac{3\ln 10}{\ln 2} \approx 9.97 \text{ hours}

Example 3: Logistic Growth

A population satisfies dPdt=0.4P(1P1000)\frac{dP}{dt} = 0.4P\left(1 - \frac{P}{1000}\right) with P(0)=100P(0) = 100. Find the population after 10 units of time.

Solution: The solution is:

P(t)=10001+9e0.4tP(t) = \frac{1000}{1 + 9e^{-0.4t}}

At t=10t = 10: P(10)731P(10) \approx 731.

These models are used everywhere. Radiocarbon dating relies on exponential decay of Carbon-14. Banks use continuous compounding (exponential growth) for interest. Biologists model fish populations in lakes with logistic growth to set sustainable fishing limits. Epidemiologists used logistic-style models during the COVID-19 pandemic to predict infection curves and the effect of interventions. Even your phone’s battery percentage decreasing over time follows a rough exponential decay pattern.

The differential equation for exponential growth or decay is:
The general solution to $\frac{dA}{dt} = kA$ is:
In the logistic model $\frac{dP}{dt} = rP(1 - P/K)$, $K$ represents:
Exponential decay is characterized by:
Logistic growth differs from pure exponential growth because: