AP®MathsCollege BoardCalculus BCStudy GuidesUnit 7: Differential EquationsFirst-Order Differential EquationsExponential Models

Exponential Models (College Board AP® Calculus BC): Study Guide

Written by: Roger B

Reviewed by: Dan Finlay

Updated on 14 May 2026

Differential equations for exponential models

What type of differential equation corresponds to an exponential model?

There are many situations where assuming that the rate of change of a quantity is proportional to the size of the quantity provides a good model
- For example a population of bacteria
  - The more bacteria there are, the more new bacteria will be being produced
- Or a radioactive sample
  - The more radioactive atoms there are, the greater the number of atoms that will be undergoing decay
This is known as the exponential growth and decay model
The mathematical way of expressing this for a quantity $y$ is
- $\frac{d y}{d t} = k y$
  - $\frac{d y}{d t}$ is the rate of change of $y$
  - $k$ is the constant of proportionality
- If the quantity $y$ is increasing
  - then $k$ is positive
- If the quantity $y$ is decreasing
  - then $k$ is negative
  - Or alternatively, assume that $k$ is positive and write $\frac{d y}{d t} = - k y$

Solutions to exponential growth & decay models

How do I find the solutions for exponential growth and decay models?

The solution to the exponential growth and decay model $\frac{d y}{d t} = k y$ , with the initial condition $y = y_{0}$ when $t = 0$ , is
- $y = y_{0} e^{k t}$
It is a good idea to remember this result
- but it can also be derived using separation of variables

Solving the exponential growth and decay model using separation of variables

Start with $\frac{d y}{d t} = k y$
- Separate the variables
  - $\frac{1}{y} \frac{d y}{d t} = k$
- Integrate both sides with respect to $t$
  - $\int \frac{1}{y} d y = \int k d t$
- Integrate, including a constant of integration
  - $\ln |y| = k t + C$
- If $y$ represents a population then $y$ can never be negative, so we can ignore the modulus sign
  - $\ln y = k t + C$
- $y = y_{0}$ when $t = 0$ , so
  - $\ln (y_{0}) = k (0) + C \Rightarrow C = \ln (y_{0})$
- Rearrange
  - $\begin{array}{rcl} \ln y & = & k t + \ln (y_{0}) \\ e^{\ln y} & = & e^{k t + \ln (y_{0})} \\ y & = & e^{k t} \cdot e^{\ln (y_{0})} \\ y & = & y_{0} e^{k t} \end{array}$

Examiner Tips and Tricks

If an exam question is based on a real world example, be sure that your answers are given in the context of the question.

Worked Example

At any point in time, the rate of growth of a colony of bacteria is proportional to the current population size, $P$ .

(a) Write a differential equation to model the size of the population of bacteria.

At time $t = 0$ hours, the population size is 5000.

(b) Write down the particular solution of the differential equation from part (a).

After 1 hour, the population has grown to 7000.

(c) Determine how long it will take from time $t = 0$ , according to the model, for the population of bacteria to grow to 100 000.

Answer:

(a)

This is a description of an exponential growth model

$\frac{d P}{d t} = k P$

(b)

For $\frac{d y}{d t} = k y$ , with the initial condition $y = y_{0}$ when $t = 0$ , the particular solution is $y = y_{0} e^{k t}$

$P = 5000 e^{k t}$

(c)

The question doesn't say what units of time to use, but looking at the information given it will be easiest to use hours

Substitute the values for $t = 1$ hour into the particular solution, and solve for $k$

$\begin{array}{rcl} 7000 & = & 5000 e^{k (1)} \\ e^{k} & = & \frac{7000}{5000} = \frac{7}{5} = 1.4 \\ k & = & \ln (1.4) \end{array}$

Using that value of $k$ , substitute $P = 100000$ into the particular solution and solve for $t$

$\begin{array}{rcl} 100000 & = & 5000 e^{\ln (1.4) t} \\ e^{\ln (1.4) t} & = & \frac{100000}{5000} = 20 \\ \ln (1.4) t & = & \ln (20) \\ t & = & \frac{\ln (20)}{\ln (1.4)} = 8.903356 . . . \end{array}$

8.903 hours (to 3 decimal places)

What is doubling-time and half-life?

For an exponential growth model, the doubling-time is the time taken for the initial value to double
- If $y = y_{0} e^{k t}$ , then solve $2 y_{0} = y_{0} e^{k t}$
- You get $t = \frac{\ln 2}{k}$
For an exponential decay model, the half-life is the time taken for the initial value to halve
- If $y = y_{0} e^{- k t}$ , then solve $\frac{y_{0}}{2} = y_{0} e^{- k t}$
- You get $t = \frac{\ln 2}{k}$

Examiner Tips and Tricks

Notice that the doubling-time and half-life are independent of the initial value.

Worked Example

The mass of a radioactive isotope decays at a rate proportional to the mass present, modeled by the differential equation $\frac{d M}{d t} = k M$ , where $M$ is the mass in grams and $t$ is the time in years.

If the half-life of the isotope is 25 years, which of the following is the particular solution to the differential equation given an initial mass of 100 grams?

(A) $M (t) = 100 e^{- 25 \ln (2) t}$

(B) $M (t) = 100 e^{- \frac{\ln 2}{25} t}$

(D) $M (t) = 100 e^{- 0.5 t}$

Answer:

Identify the general solution

Note here that the value of $k$ is negative

$M (t) = 100 e^{k t}$

Apply the half-life condition and solve for $k$

$\begin{array}{rcl} M (25) & = & 50 \\ 100 e^{25 k} & = & 50 \\ e^{25 k} & = & \frac{1}{2} \\ 25 k & = & \ln (\frac{1}{2}) \\ 25 k & = & - \ln 2 \\ k & = & - \frac{\ln 2}{25} \end{array}$

Substitute the value of $k$ into the general solution

$M (t) = 100 e^{- \frac{\ln 2}{25} t}$

Option (B)

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Exponential Models (College Board AP® Calculus BC): Study Guide

Differential equations for exponential models

What type of differential equation corresponds to an exponential model?

Solutions to exponential growth & decay models

How do I find the solutions for exponential growth and decay models?

Solving the exponential growth and decay model using separation of variables

Examiner Tips and Tricks

Worked Example

What is doubling-time and half-life?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Dan Finlay