#Differential Equations for Exponential Models

#Table of Contents

Introduction to Exponential Models
Differential Equation for Exponential Models
Solutions to Exponential Growth & Decay Models
- Using Separation of Variables
Worked Example
Practice Questions
Glossary
Summary and Key Takeaways

#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.

Exam Tip

Always give your answers in the context of the question, especially in applied problems.

#Differential Equations for Exponential Models

#Table of Contents

Introduction to Exponential Models
Differential Equation for Exponential Models
Solutions to Exponential Growth & Decay Models
- Using Separation of Variables
Worked Example
Practice Questions
Glossary
Summary and Key Takeaways

#Introduction to Exponential 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.

Exam Tip

Always give your answers in the context of the question, especially in applied problems.

First-Order Differential Equations

Emily Davis

#Differential Equations for Exponential Models

#Table of Contents

#Introduction to Exponential Models

#Differential Equation for Exponential Models

#What Type of Differential Equation Corresponds to an Exponential Model?

#Solutions to Exponential Growth & Decay Models

#How to Find Solutions for Exponential Growth and Decay Models?

#Solving the Exponential Growth and Decay Model Using Separation of Variables

#Worked Example

#Practice Questions

#Glossary

#Summary and Key Takeaways

#Summary

#Key Takeaways

Continue your learning journey

How are we doing?

First-Order Differential Equations

Emily Davis

#Differential Equations for Exponential Models

#Table of Contents

#Introduction to Exponential Models

#Differential Equation for Exponential Models

#What Type of Differential Equation Corresponds to an Exponential Model?

#Solutions to Exponential Growth & Decay Models

#How to Find Solutions for Exponential Growth and Decay Models?

#Solving the Exponential Growth and Decay Model Using Separation of Variables

#Worked Example

#Practice Questions

#Glossary

#Summary and Key Takeaways

#Summary

#Key Takeaways

Continue your learning journey

How are we doing?