Exponential Growth and Decay Models

By the end of this lesson, you will be able to derive and apply the exponential growth/decay model from the differential equation $dy/dt = ky$, solve related problems including half-life, radioactive decay, compound interest, and Newton's Law of Cooling.

The Exponential Differential Equation

The most important first-order ODE:

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

Solution: $y(t) = y_0 e^{kt}$, where $y_0 = y(0)$.

• $k > 0$: exponential growth (population, compound interest)
• $k < 0$: exponential decay (radioactive decay, cooling)

Key property: the rate of change is proportional to the current amount.

Deriving the Solution

From separation (Example 2, previous lesson):

$\dfrac{dy}{y} = k,dt \implies \ln|y| = kt + C \implies y = y_0 e^{kt}$

Example 1 — Population Growth

A population of bacteria doubles every 3 hours. Starting with 500 bacteria, find:

(a) The exponential model

$y(t) = 500e^{kt}$; doubling: $y(3) = 1000$: $e^{3k} = 2 \implies k = \dfrac{\ln 2}{3}$

$y(t) = 500e^{(\ln 2/3)t} = 500 \cdot 2^{t/3}$

(b) Population after 10 hours:

$y(10) = 500 \cdot 2^{10/3} \approx 500 \times 10.08 \approx 5040$

(c) When does the population reach 10,000?

$10000 = 500 \cdot 2^{t/3} \implies 2^{t/3} = 20 \implies t = 3\log_2 20 \approx 12.9$ hours

Half-Life

For $y(t) = y_0 e^{kt}$ with $k < 0$: the half-life $T_{1/2}$ satisfies:

$$y_0/2 = y_0 e^{kT_{1/2}} \implies T_{1/2} = -\frac{\ln 2}{k} = \frac{\ln 2}{|k|}$$

Example 2 — Radioactive Decay

Carbon-14 has half-life 5,730 years. What fraction remains after 10,000 years?

$k = -\dfrac{\ln 2}{5730}$

$\dfrac{y(10000)}{y_0} = e^{-10000\ln 2/5730} = 2^{-10000/5730} \approx 2^{-1.746} \approx 0.298$

About 29.8% remains.

Newton's Law of Cooling

Temperature approaches ambient temperature $T_s$ exponentially:

$$\frac{dT}{dt} = k(T - T_s)$$

Solution: $T(t) = T_s + (T_0 - T_s)e^{kt}$ where $T_0 = T(0)$.