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Differential Equations for Exponential Models

Introduction to Exponential Models

Exponential models are used in various real-world scenarios where the rate of change of a quantity is proportional to the quantity itself. This section will cover the key concepts and mathematical formulations of these models.

Differential Equation for Exponential Models

What Type of Differential Equation Corresponds to an Exponential Model?

Key Concept

Key Concept: In many situations, assuming that the rate of change of a quantity is proportional to the size of the quantity provides a good model.

For example, consider a population of bacteria: - The more bacteria there are, the more new bacteria will be produced.

Or consider a radioactive sample:

The more radioactive atoms there are, the greater the number of atoms that will undergo decay.

This is known as the exponential growth and decay model. Mathematically, for a quantity $y$ :

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

Where:

$y$ is the quantity.
$k$ is the constant of proportionality.

If the quantity

y

is **increasing**, then

k

is **positive**. If

y

is **decreasing**, then

k

is **negative**. Alternatively, you can assume

k

is positive and write

\frac{dy}{dt} = -ky

for decay.

Solutions to Exponential Growth & Decay Models

How to Find Solutions for Exponential Growth and Decay Models?

The solution to the exponential growth and decay model $\frac{dy}{dt} = ky$ , with the initial condition $y = y_0$ when $t = 0$ , is:

$y = y_0 e^{kt}$

Exam Tip

It's 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 $\frac{dy}{dt} = ky$ .
Separate the variables:

$\frac{1}{y} \frac{dy}{dt} = k$

Integrate both sides with respect to $t$ :

$\int \frac{1}{y} , dy = \int k , dt$

Integrate, including a constant of integration:

$\ln|y| = kt + C$

If $y$ represents a population, then $y$ can never be negative, so we can ignore the modulus sign:

$\ln y = kt + C$

Given the initial condition $y = y_0$ when $t = 0$ :

$\ln y_0 = k(0) + C \Rightarrow C = \ln y_0$

Rearrange:

$\begin{aligned} \ln y &= kt + \ln y_0 \\ e^{\ln y} &= e^{kt + \ln y_0} \\ y &= e^{kt} \cdot e^{\ln y_0} \\ y &= y_0 e^{kt} \end{aligned}$

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.

Answer: This is a description of an exponential growth model.

$\frac{dP}{dt} = kP$

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

Answer: For $\frac{dy}{dt} = ky$ , with the initial condition $y = y_0$ when $t = 0$ , the particular solution is $y = y_0 e^{kt}$ .

$P = 5000 e^{kt}$

(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: The question does not specify the units of time, but it will be easiest to use hours based on the information given.

Substitute the values to find $k$ :

$7000 = 5000 e^{k(1)}$

$e^k = \frac{7000}{5000} = 1.4$

$k = \ln(1.4)$

Using the value of $k$ , substitute $t$ to find the duration:

$100000 = 5000 e^{\ln(1.4)t}$

$e^{\ln(1.4)t} = \frac{100000}{5000} = 20$

$\ln(1.4)t = \ln(20)$

$t = \frac{\ln(20)}{\ln(1.4)} = 8.903356... \approx 8.903 \text{ hours (to 3 decimal places)}$

Practice Questions

Practice Question

A sample of a radioactive substance decays at a rate proportional to its current amount. If the initial amount of the substance is 100 grams and the half-life is 5 hours, write down the differential equation for the decay and solve for the amount remaining after 10 hours.

Practice Question

The population of a small town grows at a rate proportional to its current population. If the initial population is 2,000 and it doubles in 3 years, how long will it take for the population to reach 10,000?

Glossary

Exponential Growth and Decay Model: A model where the rate of change of a quantity is proportional to the quantity itself.
Constant of Proportionality ( $k$ ): A constant that determines the rate of growth or decay.
Separation of Variables: A method used to solve differential equations by separating the variables on each side of the equation.
Initial Condition: The value of the function at $t = 0$ , used to find the constant of integration.

Summary and Key Takeaways

Summary

The exponential growth and decay model is represented by the differential equation $\frac{dy}{dt} = ky$ .
The solution to this differential equation with an initial condition $y = y_0$ when $t = 0$ is $y = y_0 e^{kt}$ .
The method of separation of variables can be used to derive this solution.

Key Takeaways

Understand the differential equation $\frac{dy}{dt} = ky$ and its significance.
Remember the solution $y = y_0 e^{kt}$ for exponential growth and decay models.
Apply the separation of variables method to solve similar differential equations.
Be able to interpret and solve real-world problems using exponential models.