# Differential equations for growth, decay and logistic models

> IB Mathematics Applications and Interpretation HL · IB Math AI HL
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u4-differential-equations-for-growth-decay/

This subtopic covers constructing and solving differential equations that model exponential growth, exponential decay, and constrained logistic growth with a limiting carrying capacity. These are common modelling questions in IB AI HL exams.

**Prerequisites:** [Separable differential equations](https://www.owlsprep.com/study/ib-math-ai-hl-u4-separable-differential-equations/); [Basic integration techniques](https://www.owlsprep.com/study/ib-math-ai-hl-u3-integration-basics/)

## Learning objectives

- Set up differential equations for exponential growth and decay contexts
- Solve separable differential equations for growth/decay problems
- Set up and interpret logistic differential equations for constrained growth
- Solve logistic DEs and interpret model parameters in context
- Apply these models to answer common IB exam questions

## Exponential Growth Models

**Exponential growth** — Growth of a quantity where the instantaneous rate of change is proportional to the current size of the quantity. This applies when there are unlimited resources for growth.

*Notation:* \frac{dP}{dt} = kP, \quad k>0

*Example:* Bacteria growth in an unconstrained lab environment

This is a separable differential equation that can be solved explicitly to get the general form $P(t) = P_0 e^{kt}$, where $P_0 = P(0)$ is the initial quantity.

**Worked example:** A population of bacteria grows exponentially. At $t=0$, there are 100 bacteria. After 2 hours, there are 500 bacteria. Find the growth constant $k$, and predict the population after 5 hours, to 3 significant figures.

1. Set up the differential equation with initial condition:
2. $$\frac{dP}{dt} = kP, \quad P(0) = 100$$
3. Separate variables and integrate both sides:
4. $$\int \frac{1}{P} dP = \int k dt \implies \ln|P| = kt + C$$
5. Exponentiate both sides to get the general solution, where $A = e^C$:
6. $$P(t) = Ae^{kt}$$
7. Apply the initial condition $P(0) = 100$ to find $A$:
8. $$100 = Ae^{0} \implies A = 100$$
9. Use the second condition $P(2) = 500$ to solve for $k$:
10. $$500 = 100e^{2k} \implies \ln 5 = 2k \implies k = \frac{\ln 5}{2} \approx 0.805$$
11. Calculate the population at $t=5$:
12. $$P(5) = 100e^{0.805 \times 5} \approx 100e^{4.025} \approx 5620$$

> **Exam tip:** Always round constants to 3 significant figures unless the question specifies otherwise, which is the IB standard.

*Calculator:* allowed

## Exponential Decay and Half-Life

**Half-life** — The time taken for an exponentially decaying quantity to decrease to half of its initial value. Half-life is constant for any exponential decay process, regardless of initial quantity.

*Example:* Radioactive decay of unstable isotopes

For exponential decay, the differential equation includes a negative sign to show the quantity is decreasing over time: $\frac{dN}{dt} = -kN$, where $k>0$ is the decay constant.

**Worked example:** The half-life of Carbon-14 is 5730 years. A sample initially contains 10 g of Carbon-14. How much Carbon-14 remains after 10000 years, to 3 significant figures?

1. Set up the decay model with initial condition:
2. $$\frac{dN}{dt} = -kN, \quad N(0) = 10 \implies N(t) = 10e^{-kt}$$
3. Use the half-life to find $k$:
4. $$N(5730) = 5 = 10e^{-k(5730)} \implies \ln(0.5) = -5730k \implies k = \frac{\ln 2}{5730} \approx 1.210 \times 10^{-4}$$
5. Evaluate at $t=10000$:
6. $$N(10000) = 10e^{-(1.210 \times 10^{-4})(10000)} = 10e^{-1.21} \approx 2.98$$
7. Final answer: 2.98 g of Carbon-14 remains.

> **warning**
>
> Always include the negative sign in the decay differential equation to avoid getting an increasing quantity, which is a common exam mistake.

*Calculator:* allowed

## Logistic Growth Models

**Logistic differential equation** — A model for population growth that accounts for limited resources, where $r$ is the intrinsic growth rate (growth when population is small) and $K$ is the carrying capacity (maximum sustainable population).

*Notation:* \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)

*Example:* Wild animal populations in a confined habitat

The general solution to the logistic differential equation for initial population $P_0$ is: $P(t) = \frac{K}{1 + Ae^{-rt}}$, where $A = \frac{K - P_0}{P_0}$.

**Worked example:** A deer population in a forest follows logistic growth with carrying capacity 1000 and intrinsic growth rate 0.1 per year. The initial population in 2020 was 200. What is the predicted population in 2030, to the nearest whole number?

1. Identify parameters: $K = 1000$, $r = 0.1$, $P_0 = 200$, $t = 10$ years
2. Calculate constant $A$:
3. $$A = \frac{K - P_0}{P_0} = \frac{1000 - 200}{200} = 4$$
4. Substitute into the general solution:
5. $$P(t) = \frac{1000}{1 + 4e^{-0.1t}}$$
6. Evaluate at $t=10$:
7. $$P(10) = \frac{1000}{1 + 4e^{-1}} \approx \frac{1000}{2.4716} \approx 405$$

**Check your understanding**

Test your understanding of logistic growth:

1. What is the value of $\frac{dP}{dt}$ when $P = K$?

   - A) $rK$
   - B) $0$
   - C) $r$
   - D) $\frac{rK}{2}$

   *Why:* When $P=K$, $1 - \frac{P}{K} = 0$, so the rate of growth is zero, and the population stabilizes at the carrying capacity.

*Calculator:* allowed

## Interpreting Model Parameters

IB exams very frequently ask you to interpret the meaning of parameters in context, not just calculate values. This is worth multiple marks in extended response questions, so it is important to phrase your answer correctly.

**Exam command terms**

Common command terms for these questions:

- **Interpret** — Explain what the parameter means in the given context, including units if appropriate *(Interpret the value $K=1000$ in this model)*

- **Justify** — Explain why a given model is (or is not) appropriate for the context *(Justify the use of a logistic model for this population)*

**Worked example:** A city's population is modelled by $\frac{dP}{dt} = 0.02P\left(1 - \frac{P}{5000000}\right)$, where $P$ is the number of people and $t$ is years. Interpret $0.02$ and $5000000$ in context.

1. Recall the standard logistic form $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$
2. Interpret $r=0.02$: This is the intrinsic growth rate. When the population is very small, it grows at an instantaneous rate of 2% per year.
3. Interpret $K=5000000$: This is the carrying capacity. The maximum sustainable population of the city predicted by this model is 5 million people.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting the negative sign in the exponential decay DE, leading to an increasing quantity instead of decreasing.
  - Why it fails: Mixing up the notation for growth and decay, resulting in opposite results that lose all marks.
  - Correct: Always write decay as $\frac{dN}{dt} = -kN$ with $k>0$, so the quantity decreases over time.
- **Wrong:** Solving for the growth/decay constant before finding the constant from the initial condition, leading to incorrect values.
  - Why it fails: Rushing through algebra and misordering solution steps.
  - Correct: Always find the constant $A$ from the initial condition $P(0) = P_0$ first, before using a second data point to find $k$ or $r$.
- **Wrong:** Interpreting $r$ in the logistic model as the actual growth rate for any population size.
  - Why it fails: Confusing the intrinsic growth rate (growth at small populations) with the actual current growth rate.
  - Correct: Remember $r$ is only the growth rate when the population is very small. The actual growth rate depends on the current population size.
- **Wrong:** Using an exponential growth model for a population that has a stated maximum size limit.
  - Why it fails: Overgeneralizing exponential models and forgetting that exponential growth assumes unlimited resources.
  - Correct: If the question mentions a maximum sustainable population or limited resources, use a logistic model instead of exponential.

## Cheatsheet

| Model | Differential Equation | General Solution | Key Parameters |
| --- | --- | --- | --- |
| Exponential Growth | $\frac{dP}{dt} = kP, k>0$ | $P(t) = P_0 e^{kt}$ | $k$ = growth constant, $P_0$ = initial quantity |
| Exponential Decay | $\frac{dN}{dt} = -kN, k>0$ | $N(t) = N_0 e^{-kt}$ | Half-life $t_{1/2} = \frac{\ln 2}{k}$ |
| Logistic Growth | $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$ | $P(t) = \frac{K}{1 + Ae^{-rt}}, A = \frac{K-P_0}{P_0}$ | $r$ = intrinsic growth, $K$ = carrying capacity |

## What's next

This subtopic builds on your knowledge of separable differential equations and basic integration, and forms the core of the frequently tested calculus modelling strand in IB AI HL Paper 2 extended response questions. Mastery of these models is also extremely useful for your internal assessment if you choose to complete a modelling-focused project. After mastering growth, decay and logistic models, you can progress to more advanced differential equation topics, including numerical solution methods for DEs that cannot be solved analytically, and further applications of modelling to real-world contexts.

- [Statistics and probability](https://www.owlsprep.com/study/ib-math-ai-hl-u5-overview/)
- [Data types, sampling, bias](https://www.owlsprep.com/study/ib-math-ai-hl-u5-data-types-sampling-bias/)
- [Descriptive statistics: measures of center and spread](https://www.owlsprep.com/study/ib-math-ai-hl-u5-descriptive-statistics-measures-of-center/)

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