The differential equation $\frac{dy}{dt} = ky$ models situations where the rate of change of a quantity is proportional to the quantity itself. This leads to exponential growth when $k > 0$ and exponential decay when $k < 0$. This model appears in population growth, radioactive decay, compound interest, and many other real-world phenomena.
🎯 In this section you will learn
- The exponential growth/decay model: $\frac{dy}{dt} = ky$
- The general solution: $y = y_0 e^{kt}$
- How to find $k$ using given data
- Half-life and doubling time formulas
- Real-world applications: population, radioactive decay, Newton's law of cooling
📌 The Exponential Model
$$ \frac{dy}{dt} = ky \quad \Longrightarrow \quad y = y_0 e^{kt} $$
where $y_0 = y(0)$ is the initial quantity
💡 Derivation
This is a separable differential equation: $\frac{dy}{y} = k\, dt$ → $\ln|y| = kt + C$ → $y = e^{kt + C} = y_0 e^{kt}$.
A bacteria culture starts with 500 bacteria and grows at a rate proportional to its size. After 3 hours, there are 8000 bacteria. Find the number after 5 hours.
①
General solution
$P(t) = P_0 e^{kt} = 500 e^{kt}$
②
Use $P(3)=8000$ to find $k$
$8000 = 500 e^{3k}$ → $e^{3k} = 16$ → $3k = \ln 16 = 4\ln 2$ → $k = \frac{4\ln 2}{3}$
③
Find $P(5)$
$P(5) = 500 e^{5k} = 500 e^{5 \cdot \frac{4\ln 2}{3}} = 500 e^{\frac{20}{3}\ln 2} = 500 \cdot 2^{20/3} \approx 500 \cdot 101.6 \approx 50,800$
A radioactive substance decays from 100 mg to 25 mg in 8 days. Find the half-life.
①
General solution
$A(t) = A_0 e^{kt} = 100 e^{kt}$ (note $k < 0$)
②
Use $A(8)=25$ to find $k$
$25 = 100 e^{8k}$ → $e^{8k} = 0.25 = \frac{1}{4}$ → $8k = \ln(1/4) = -2\ln 2$ → $k = -\frac{\ln 2}{4}$
③
Half-life: $A(t) = A_0/2$
$\frac{A_0}{2} = A_0 e^{kt}$ → $\frac{1}{2} = e^{kt}$ → $kt = \ln(1/2) = -\ln 2$
④
Solve for $t$
$-\frac{\ln 2}{4} \cdot t = -\ln 2$ → $t = 4$ days
An investment grows at a rate proportional to its value. If it doubles in 10 years, what is the growth rate $k$?
- $A(t) = A_0 e^{kt}$
- Doubling: $2A_0 = A_0 e^{10k}$ → $e^{10k} = 2$ → $10k = \ln 2$ → $k = \frac{\ln 2}{10} \approx 0.0693$ (about 6.93%)
📌 Key Formulas
| Concept | Formula |
|---|
| Growth/Decay Equation | $y = y_0 e^{kt}$ |
| Half-Life ($t_{1/2}$) | $t_{1/2} = \frac{\ln 2}{k}$ (for decay, $k$ is positive magnitude) |
| Doubling Time ($T_2$) | $T_2 = \frac{\ln 2}{k}$ (for growth, $k > 0$) |
| Find $k$ from data | $k = \frac{1}{t} \ln\left(\frac{y}{y_0}\right)$ |
A cup of coffee at 90°C is placed in a room at 20°C. The temperature $T$ satisfies $\frac{dT}{dt} = k(T - 20)$. After 5 minutes, the coffee has cooled to 80°C. Find $k$.
①
Let $u = T - 20$
Then $\frac{du}{dt} = k u$, so $u(t) = u_0 e^{kt}$
②
Initial condition
$u_0 = T(0) - 20 = 90 - 20 = 70$
③
Use $T(5)=80$
$u(5) = 80 - 20 = 60 = 70 e^{5k}$ → $e^{5k} = \frac{60}{70} = \frac{6}{7}$ → $k = \frac{1}{5} \ln\left(\frac{6}{7}\right) \approx -0.0308$
⚠️ Common Mistakes
- Confusing growth and decay: $k > 0$ for growth, $k < 0$ for decay.
- Half-life formula: $t_{1/2} = \frac{\ln 2}{|k|}$, not $\frac{\ln 2}{k}$ (since $k$ is negative for decay).
- Forgetting the initial condition: The solution is $y = y_0 e^{kt}$, not $y = Ce^{kt}$ without determining $C$.
🔍 Key Takeaways
- Exponential growth/decay: $\frac{dy}{dt} = ky$ → $y = y_0 e^{kt}$.
- Growth: $k > 0$; Decay: $k < 0$.
- Half-life: $t_{1/2} = \frac{\ln 2}{|k|}$.
- Doubling time: $T_2 = \frac{\ln 2}{k}$ (for growth).
- Real-world applications include population, radioactive decay, compound interest, and Newton's law of cooling.
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