A Friendly Introduction to Differential Equations

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A Friendly Introduction to Differential Equations

Copyright © 2015 Mohammed K A Kaabar

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About the Author A Friendly Introduction to Differential Equations

2 M. Kaabar

Mohammed Kaabar has a Bachelor of Science in Theoretical Mathematics from Washington State University, Pullman, WA. He is a graduate student in Applied Mathematics at Washington State University, Pullman, WA, and he is a math tutor at the Math Learning Center (MLC) at Washington State University, Pullman. He is the author of A First Course in Linear Algebra Book, and his research interests are applied optimization, numerical analysis, differential equations, linear algebra, and real analysis. He was invited to serve as a Technical Program Committee (TPC) member in many conferences such as ICECCS 14, ENCINS 15, eQeSS 15, SSCC 15, ICSoEB 15, CCA 14, WSMEAP 14, EECSI 14, JIEEEC 13 and WCEEENG 12. He is an online instructor of two free online courses in numerical analysis: Introduction to Numerical Analysis and Advanced Numerical Analysis at Udemy Inc, San Francisco, CA. He is a former member of Institute of Electrical and Electronics Engineers (IEEE), IEEE Antennas and Propagation Society, IEEE Consultants Network, IEEE Smart Grid Community, IEEE Technical Committee on RFID, IEEE Life Sciences Community, IEEE Green ICT Community, IEEE Cloud Computing Community, IEEE Internet of Things Community, IEEE Committee on Earth Observations, IEEE Electric Vehicles Community, IEEE Electron Devices Society, IEEE Communications Society, and IEEE Computer Society. He also received several educational awards and certificates from accredited institutions. For more information about the author and his free online courses, please visit his personal website: http://www.mohammed-kaabar.net.

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Table of Contents 6

1 The Laplace Transform

9

1.1

Introduction to Differential Equations..…........9

1.2

Introduction to the Laplace Transforms……..14

1.3

Inverse Laplace Transforms……….…….........24

1.4

Initial Value Problems……..……………..........27

1.5

Properties of Laplace Transforms……….……33

1.6

Systems of Linear Equations……….………....45

1.7

Exercises……………………………………..…...49

Systems

of

Homogeneous

Equations (HLDE)

Linear

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4 Extended Methods of First and Higher Orders

Introduction

2


Differential

Differential Equations

78

4.1

Bernoulli Method………………….….....………78

4.2

Separable Method…..…..................................85

4.3

Exact Method………..…..................................87

4.4

Reduced to Separable Method…..…...............90

4.5

Reduction of Order Method……...…...............92

4.6

Exercises……………………………………….....95

5 Applications of Differential Equations

96

5.1

Temperature Application……........................96

5.2

Growth and Decay Application…..………....100

5.3

Water Tank Application…………….....…….104

51 Appendices

109

2.1

HLDE with Constant Coefficients..................51

2.2

Method of Undetermined Coefficients…….…60

A

Determinants…………………….......................109

2.3

Exercises…………………………………….…...65

B

Vector Spaces…………….………..………….....116

C

Homogenous Systems…...…………….....…….135

3 Methods of First and Higher Orders Differential Equations

157

Index

159

Bibliography

163

67

3.1

Variation Method……………………………….67

3.2

Cauchy-Euler Method………..…………..……74

3.3

Exercises……………………………………..….76

4 M. Kaabar

Answers to Odd-Numbered Exercises

5


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Introduction

Table of Laplace Transform

In this book, I wrote five chapters: The Laplace Transform, Systems of Homogenous Linear Differential Equations (HLDE), Methods of First and Higher Orders Differential Equations, Extended Methods of First and Higher Orders Differential Equations, and Applications of Differential Equations. I also added exercises at the end of each chapter above to let students practice additional sets of problems other than examples, and they can also check their solutions to some of these exercises by looking at “Answers to Odd-Numbered Exercises” section at the end of this book. This book is a very useful for college students who studied Calculus II, and other students who want to review some concepts of differential equations before studying courses such as partial differential equations, applied mathematics, and electric circuits II. According to my experience as a math tutor, I have noticed that some students have difficulty to understand some concepts of laplace transforms because most authors of differential equations books did not start with laplace transforms as a first chapter, and they left it at the end of their books. Therefore, I decided to start with a different approach by choosing laplace transforms to be in the first chapter of this book. If you have any comments related to the contents of this book, please email your comments to [email protected]. I wish to express my gratitude and appreciation to my father, my mother, and my only lovely 13-year old brother who is sick, and I want to spend every dollar in his heath care. I would also like to give a special thanks to all administrators and professors of mathematics at WSU for their educational support. In conclusion, I would appreciate to consider this book as a milestone for developing more math books that can serve our mathematical society in the area of differential equations.

1

Mohammed K A Kaabar

6 M. Kaabar

λ

ࣦሼ݂ሺ‫ݐ‬ሻሽ ൌ න ݁ ି௦௧ ݂ሺ‫ݐ‬ሻ݀‫ݐ‬ Ͳ

2

ͳ ࣦሼͳሽ ൌ ‫ݏ‬

ࣦሼ‫ ݐ‬௠ ሽ ൌ

3

ࣦሼ ܿ‫ݐ‬ሽ ൌ

4

ࣦሼ݁ ௕௧ ሽ ൌ

‫ݏ‬ଶ

௠Ǩ

where m is a

௦ ೘శభ

positive integer (whole number) ‫ݏ‬ ࣦሼ ܿ‫ݐ‬ሽ ൌ ଶ ‫ ݏ‬൅ ܿଶ

ܿ ൅ ܿଶ

ͳ ‫ݏ‬െܾ

ࣦ ିଵ ሼ‫ܨ‬ሺ‫ݏ‬ሻ ȁ‫ ݏ‬՜ ‫ ݏ‬െ ܾ ሽ ൌ ܾ݁‫݂ ݐ‬ሺ‫ݐ‬ሻ

5

ࣦሼ݁ ௕௧ ݂ሺ‫ݐ‬ሻሽ ൌ ‫ܨ‬ሺ‫ݏ‬ሻ ȁ‫ ݏ‬՜ ‫ ݏ‬െ ܾ

6

ࣦሼ݄ሺ‫ݐ‬ሻܷሺ‫ ݐ‬െ ܾሻሽ ൌ ݁ ି௕௦ ࣦሼ݄ሺ‫ ݐ‬൅ ܾሻሽ ࣦ ିଵ ሼ݁ ି௕௦ ‫ܨ‬ሺ‫ݏ‬ሻሽ ൌ ݂ ሺ‫ ݐ‬െ ܾሻܷሺ‫ ݐ‬െ ܾሻ

7

ࣦሼܷሺ‫ ݐ‬െ ܾሻሽ ൌ

݁ ି௕௦ ‫ݏ‬

ࣦ ିଵ ቊ

݁ ି௕௦ ቋ ൌ ܷሺ‫ ݐ‬െ ܾሻ ‫ݏ‬

8

ࣦ൛݂ ሺ௠ሻ ሺ‫ݐ‬ሻൟ ൌ ‫ ݏ‬௠ ‫ܨ‬ሺ‫ݏ‬ሻ െ ‫ ݏ‬௠ିଵ ݂ሺͲሻ െ ‫ ݏ‬௠ିଶ ݂ ᇱ ሺ‫ݏ‬ሻ െ ‫ ڮ‬െ ݂ ሺ௠ିଵሻ ሺ‫ݏ‬ሻ

9

ࣦሼ‫ ݐ‬௠ ݂ሺ‫ݐ‬ሻሽሺ‫ݏ‬ሻ ൌ ሺെͳሻ௠

where m is a positive integer ௗ ೘ிሺ௦ሻ ௗ௦ ೘

ࣦ ିଵ ቄ

ௗ ೘ிሺ௦ሻ ௗ௦ ೘

ቅ ൌ ሺെͳሻ௠ ‫ ݐ‬௠ ݂ሺ‫ݐ‬ሻ

where m is a positive where m is a positive integer integer

10 11

௧

ࣦሼ݂ሺ‫ݐ‬ሻ ‫݄ כ‬ሺ‫ݐ‬ሻሽ ൌ ‫ ܨ‬ሺ‫ݏ‬ሻ ή ‫ܪ‬ሺ‫ݏ‬ሻ

݂ሺ‫ݐ‬ሻ ‫݄ כ‬ሺ‫ݐ‬ሻ ൌ න ݂ሺ߰ሻ݄ሺ‫ ݐ‬െ ߰ሻ݀߰ ଴

௧

ࣦ ቐන ݂ ሺ߰ሻ݀߰ቑ ൌ ଴

‫ ܨ‬ሺ‫ݏ‬ሻ ‫ݏ‬

ࣦ ିଵ ሼ‫ܨ‬ሺ‫ݏ‬ሻ ή ‫ ܪ‬ሺ‫ݏ‬ሻሽ ൌ ݂ሺ‫ݐ‬ሻ ‫݄ כ‬ሺ‫ݐ‬ሻ

ࣦሼߜሺ‫ ݐ‬െ ܾሻሽ ൌ ݁ ି௕௦ 12 ࣦሼߜሺ‫ݐ‬ሻሽ ൌ ͳ 13 Assume that ݂ሺ‫ݐ‬ሻ is periodic with period ܲ, then: ௉

ࣦሼ݂ሺ‫ݐ‬ሻሽ ൌ

ͳ න ݁ ି௦௧ ݂ሺ‫ݐ‬ሻ݀‫ݐ‬ ͳ െ ݁ ି௉௦ ଴

Table 1.1.1: Laplace Transform

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Chapter 1 This page intentionally left blank

The Laplace Transform In this chapter, we start with an introduction to Differential Equations (DEs) including linear DEs, nonlinear DEs, independent variables, dependent variables, and the order of DEs. Then, we define the laplace transforms, and we give some examples of Initial Value Problems (IVPs). In addition, we discuss the inverse laplace transforms. We cover in the remaining sections an important concept known as the laplace transforms of derivatives, and we mention some properties of laplace transforms. Finally, we learn how to solve systems of linear equations (LEs) using Cramer’s Rule.

1.1 Introductions to Differential Equations In this section, we are going to discuss how to determine whether the differential equation is linear or nonlinear, and we will find the order of differential equations. At the end of this section, we will show the purpose of differential equations.

8 M. Kaabar

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let’s start with the definition of differential equation and with a simple example about differential equation. Definition 1.1.1 A mathematical equation is called differential equation if it has two types of variables: dependent and independent variables where the dependent variable can be written in terms of independent variable. Example 1.1.1 Given that ‫ ݕ‬ᇱ ൌ ͳͷ‫ݔ‬. a) Find ‫ݕ‬. (Hint: Find the general solution of ‫ ݕ‬ᇱ ) b) Determine whether ‫ ݕ‬ᇱ ൌ ͳͷ‫ ݔ‬is a linear differential equation or nonlinear differential equation. Why? c) What is the order of this differential equation? Solution: Part a: To find ‫ݕ‬, we need to find the general solution of ‫ ݕ‬ᇱ by taking the integral of both sides as follows: න ‫ ݕ‬ᇱ ݀‫ ݕ‬ൌ න ͳͷ‫ݔ݀ ݔ‬ Since ‫ ݕ ׬‬ᇱ ݀‫ ݕ‬ൌ ‫ ݕ‬because the integral of derivative function is the original function itself (In general, ‫ ݂ ׬‬ᇱ ሺ‫ݔ‬ሻ ݀‫ ݔ‬ൌ ݂ሺ‫ݔ‬ሻ), then ‫ ݕ‬ൌ ‫ͳ ׬‬ͷ‫ ݔ݀ ݔ‬ൌ

ଵହ ଶ

‫ݔ‬ଶ ൅ ܿ ൌ

͹Ǥͷ‫ ݔ‬ଶ ൅ ܿ. Thus, the general solution is the following: ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ͹Ǥͷ‫ ݔ‬ଶ ൅ ܿ where ܿ is constant. This means that ‫ݕ‬ is called dependent variable because it depends on ‫ݔ‬, and ‫ ݔ‬is called independent variable because it is independent from ‫ݕ‬. Part b: To determine whether ‫ ݕ‬ᇱ ൌ ͳͷ‫ ݔ‬is a linear differential equation or nonlinear differential equation, we need to introduce the following definition:

10 M. Kaabar


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Definition 1.1.2 The differential equation is called linear if the dependent variable and all its derivatives are to the power 1. Otherwise, the differential equation is nonlinear. According to the above question, we have the following: ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ͹Ǥͷ‫ ݔ‬ଶ ൅ ܿ where ܿ is constant. Since the dependent variable ‫ ݕ‬and all its derivatives are to the power 1, then using definition 1.1.2, this differential equation is linear. Part c: To find the order of this differential equation, we need to introduce the following definition: Definition 1.1.3 The order of differential equation is the highest derivative in the equation (i.e. The order of ‫ ݕ‬ᇱᇱᇱ ൅ ͵‫ ݕ‬ᇱᇱ ൅ ʹ‫ ݕ‬ᇱ ൌ ͳʹ‫ ݔ‬ଶ ൅ ʹʹ is 3). Using definition 1.1.3, the order of ‫ ݕ‬ᇱ ൌ ͳͷ‫ ݔ‬is 1. Example 1.1.2 Given that ‫ ݖ‬ᇱᇱᇱ ൅ ʹ‫ ݖ‬ᇱᇱ ൅ ‫ ݖ‬ᇱ ൌ ʹ‫ ݔ‬ଷ ൅ ʹʹ. a) Determine whether ‫ ݖ‬ᇱᇱᇱ ൅ ʹ‫ ݖ‬ᇱᇱ ൅ ‫ ݖ‬ᇱ ൌ ʹ‫ ݔ‬ଷ ൅ ʹʹ is a linear differential equation or nonlinear differential equation. Why? b) What is the order of this differential equation? Solution: Part a: Since ‫ ݖ‬is called dependent variable because it depends on ‫ݔ‬, and ‫ ݔ‬is called independent variable because it is independent from ‫ݖ‬, then to determine whether ‫ ݖ‬ᇱᇱᇱ ൅ ʹ‫ ݖ‬ᇱᇱ ൅ ‫ ݖ‬ᇱ ൌ ʹ‫ ݔ‬ଷ ൅ ʹʹ is a linear differential equation or nonlinear differential equation, we need to use definition 1.1.2 as follows: Since the dependent variable ‫ ݖ‬and all its derivatives are to the power 1, then this differential equation is linear. Part b: To find the order of ‫ ݖ‬ᇱᇱᇱ ൅ ʹ‫ ݖ‬ᇱᇱ ൅ ‫ ݖ‬ᇱ ൌ ʹ‫ ݔ‬ଷ ൅ ʹʹ, we use definition 1.1.3 which implies that the order is 3 because the highest derivative is 3.

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Example 1.1.3 Given that ݉ሺସሻ ൅ ሺ͵݉ᇱᇱ ሻଷ െ ݉ ൌ ξ‫ ݔ‬൅ ͳ. (Hint: Do not confuse between ݉ሺସሻ and ݉ସ because ݉ሺସሻ means the fourth derivative of ݉, while ݉ସ means the fourth power of m). a) Determine whether ݉ሺସሻ ൅ ሺ͵݉ᇱᇱ ሻଷ െ ݉ ൌ ξ‫ ݔ‬൅ ͳ is a linear differential equation or nonlinear differential equation. Why? b) What is the order of this differential equation? Solution: Part a: Since ݉ is called dependent variable because it depends on ‫ݔ‬, and ‫ ݔ‬is called independent variable because it is independent from ݉, then to determine whether ݉ሺସሻ ൅ ሺ͵݉ᇱᇱ ሻଷ െ ݉ ൌ ξ‫ ݔ‬൅ ͳ is a linear differential equation or nonlinear differential equation, we need to use definition 1.1.2 as follows: Since the dependent variable ݉ and all its derivatives are not to the power 1, then this differential equation is nonlinear. Part b: To find the order of ݉ሺସሻ ൅ ሺ͵݉ᇱᇱ ሻଷ െ ݉ ൌ ξ‫ ݔ‬൅ ͳ, we use definition 1.1.3 which implies that the order is 4 because the highest derivative is 4. The following are some useful notations about differential equations: ‫ ݖ‬ሺ௠ሻ is the ݉th derivative of ‫ݖ‬. ‫ ݖ‬௠ is the ݉th power of ‫ݖ‬. The following two examples are a summary of this section: Example 1.1.4 Given that ʹ‫ܾݔ‬ሺଷሻ ൅ ሺ‫ ݔ‬൅ ͳሻܾሺଶሻ ൅ ͵ܾ ൌ ‫ ݔ‬ଶ ݁ ௫ . Determine whether it is a linear differential equation or nonlinear differential equation.

12 M. Kaabar


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Solution: To determine whether it is a linear differential equation or nonlinear differential equation, We need to apply what we have learned from the previous examples in the following five steps: Step 1: ܾ is a dependent variable, and ‫ ݔ‬is an independent variable. Step 2: Since ܾ and all its derivatives are to the power 1, then the above differential equation is linear. Step 3: Coefficients of ܾ and all its derivatives are in terms of the independent variable ‫ݔ‬. Step 4: Assume that ‫ ܥ‬ሺ‫ ݔ‬ሻ ൌ ‫ ݔ‬ଶ ݁ ௫ Ǥ Then, ‫ ܥ‬ሺ‫ ݔ‬ሻ must be in terms of ‫ݔ‬. Step 5: Our purpose from the above differential equation is to find a solution where ܾ can be written in term of ‫ݔ‬. Thus, the above differential equation is a linear differential equation of order 3. Example 1.1.5 Given that ሺ‫ ݓ‬ଶ ൅ ͳሻ݄ሺସሻ െ ͵‫݄ݓ‬ᇱ ൌ ‫ ݓ‬ଶ ൅ ͳǤ Determine whether it is a linear differential equation or nonlinear differential equation. Solution: To determine whether it is a linear differential equation or nonlinear differential equation, We need to apply what we have learned from the previous examples in the following five steps: Step 1: ݄ is a dependent variable, and ‫ ݓ‬is an independent variable. Step 2: Since ݄ and all its derivatives are to the power 1, then the above differential equation is linear. Step 3: Coefficients of ݄ and all its derivatives are in terms of the independent variable ‫ݓ‬.

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Step 4: Assume that ‫ ܥ‬ሺ‫ݓ‬ሻ ൌ ‫ ݓ‬ଶ ൅ ͳǤ Then, ‫ ܥ‬ሺ‫ݓ‬ሻ must be in terms of ‫ݓ‬. Step 5: Our purpose from the above differential equation is to find a solution where ݄ can be written in term of ‫ݓ‬. Thus, the above differential equation is a linear differential equation of order 4.


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Step 2: Here in this example, ݂ ሺ‫ݔ‬ሻ ൌ ͳ because ࣦሼ݂ሺ‫ݔ‬ሻሽ ൌ ࣦሼͳሽ.

Step 3: λ

ࣦሼͳሽ ൌ නሺͳሻ݁ ି௦௫ ݀‫ݔ‬ Ͳ

ஶ

By the definition of integral, we substitute ‫׬‬଴ ሺͳሻ݁ ି௦௫ ݀‫ݔ‬ ௕

1.2 Introductions to the

with ௕՜ஶ ‫׬‬଴ ሺͳሻ݁ ି௦௫ ݀‫ݔ‬.

Laplace Transforms

It is easier to find what it is inside the above box

In this section, we are going to introduce the definition

‫׬‬଴ ݁ ି௦௫ ݀‫ݔ‬.

of the laplace transforms in general, and how can we

ൌ െ ௦ ݁ ି௦௕ ൅ ௦ ݁ ି௦ሺ଴ሻ ൌ Thus, ‫׬‬଴ ݁ ି௦௫ ݀‫ ݔ‬ൌ െ ௦ ݁ ି௦௫ ቚ ௫ୀ௕ ௫ୀ଴

use this definition to find the laplace transform of any

െ ௦ ݁ ି௦௕ ൅ ௦ .

function. Then, we will give several examples about

Step 5: We need find the limit of െ ௦ ݁ ି௦௕ ൅ ௦ as follows:

the laplace transforms, and we will show how the table 1.1.1 is helpful to find laplace transforms. Definition 1.2.1 the laplace transform, denoted by ࣦ , is defined in general as follows: λ

ࣦሼ݂ሺ‫ݔ‬ሻሽ ൌ න ݂ ሺ‫ݔ‬ሻ݁ ି௦௫ ݀‫ݔ‬ Ͳ

௕

Step 4: We need to find ௕՜ஶ ‫׬‬଴ ݁ ି௦௫ ݀‫ ݔ‬as follows: ௕

ቀ‫׬‬଴ ݁ ି௦௫ ݀‫ ݔ‬ቁ, and after that we can find the limit of ௕

௕

ଵ

ଵ

ଵ

ଵ

ଵ

ଵ

ଵ

ଵ

ଵ

ଵ

ቀെ ௦ ݁ ି௦௕ ൅ ௦ ቁ ൌ ௦ ‫ ݏ‬൐ ͲǤ

௕՜ஶ

To check if our answer is right, we need to look at table 1.1.1 at the beginning of this book. According to table ଵ 1.1.1 section 2, we found ࣦሼͳሽ ൌ ௦ which is the same answer we got. Thus, we can conclude our example with the following fact:

Example 1.2.1 Using definition 1.2.1, find ࣦሼͳሽ.

Fact 1.2.1 ࣦሼܽ݊‫ݐ݊ܽݐݏ݊݋ܿ ݕ‬ǡ ‫݉ ݕܽݏ‬ሽ ൌ ௦ .

Solution: To find ࣦሼͳሽ using definition 1.2.1, we need to do the following steps: Step 1: We write the general definition of laplace transform as follows:

Example 1.2.2 Using definition 1.2.1, find ࣦሼ݁ ଻௫ ሽ.

λ


௠

Solution: To find ࣦሼ݁ ଻௫ ሽ using definition 1.2.1, we need to do the following steps: Step 1: We write the general definition of laplace transform as follows: λ


14 M. Kaabar

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Then, we need to keep deriving ‫ ݔ‬ଷ till we get zero, and we stop integrating when the corresponding row is zero. The following table shows the table method to find ‫ ݔ ׬‬ଷ ݁ ଶ௫ ݀‫ݔ‬:

Step 2: Here in this example, ݂ ሺ‫ݔ‬ሻ ൌ ݁ ଻௫ because ࣦሼ݂ሺ‫ݔ‬ሻሽ ൌ ࣦሼ݁ ଻௫ ሽ.

Step 3: λ

ࣦሼ݁

଻௫ ሽ

ൌ නሺ݁ ଻௫ ሻ݁ ି௦௫ ݀‫ݔ‬ Ͳ

By the definition of integral, we substitute ஶ ‫׬‬଴ ሺ݁͹‫ ݔ‬ሻ݁ ି௦௫ ݀‫ݔ‬

Step 4: We

௕ with ௕՜ஶ ‫׬‬଴ ሺ݁͹‫ ݔ‬ሻ݁ ି௦௫ ݀‫ݔ‬. ௕ need to find ௕՜ஶ ‫׬‬଴ ሺ݁͹‫ ݔ‬ሻ݁ ି௦௫ ݀‫ݔ‬

as follows:

It is easier to find what it is inside the above box ௕ ቀ‫׬‬଴ ሺ݁͹‫ ݔ‬ሻ݁ ି௦௫ ݀‫ ݔ‬ቁ, and after that we can find the limit of ௕ ‫׬‬଴ ሺ݁͹‫ ݔ‬ሻ݁ ି௦௫ ݀‫ݔ‬. ௕ ௕ ଵ ൌ Thus, ‫׬‬଴ ሺ݁͹‫ ݔ‬ሻ݁ ି௦௫ ݀‫ ݔ‬ൌ ‫׬‬଴ ݁ ሺ଻ି௦ሻ௫ ݀‫ ݔ‬ൌ ሺ଻ି௦ሻ ݁ ሺ଻ି௦ሻ௫ ቚ ௫ୀ௕ ௫ୀ଴ ଵ ଵ ଵ ଵ ሺ଻ି௦ሻሺ଴ሻ ሺ଻ି௦ሻ௕ ሺ଻ି௦ሻ௕ ሺ଻ି௦ሻ

݁

െ ሺ଻ି௦ሻ ݁

െ ሺ଻ି௦ሻ .

ൌ ሺ଻ି௦ሻ ݁

ଵ

ଵ

Step 5: We need find the limit of ሺ଻ି௦ሻ ݁ ሺ଻ି௦ሻ௕ െ ሺ଻ି௦ሻ as follows: ቀ

ଵ

௕՜ஶ ሺ଻ି௦ሻ

ଵ

ଵ

ଵ

݁ ሺ଻ି௦ሻ௕ െ ሺ଻ି௦ሻቁ ൌ െ ሺ଻ି௦ሻ ൌ ሺ௦ି଻ሻ ‫ ݏ‬൐ ͹Ǥ

To check if our answer is right, we need to look at table 1.1.1 at the beginning of this book. According to table ଵ 1.1.1 section 4, we found ࣦሼ݁ ଻௫ ሽ ൌ ௦ି଻ which is the same answer we got. The following examples are two examples about finding the integrals to review some concepts that will help us finding the laplace transforms.

Example 1.2.3 Find ‫ ݔ ׬‬ଷ ݁ ଶ௫ ݀‫ݔ‬. Solution: To find ‫ ݔ ׬‬ଷ ݁ ଶ௫ ݀‫ݔ‬, it is easier to use a method known as the table method than using integration by parts. In the table method, we need to create two columns: one for derivatives of ‫ ݔ‬ଷ , and the other one for integrations of ݁ ଶ௫ .

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Derivatives Part ‫ݔ‬ଷ

Integration Part ݁ ଶ௫

͵‫ ݔ‬ଶ

ଵ ଶ௫ ݁ ଶ

͸‫ݔ‬

ͳ ଶ௫ ݁ Ͷ ଵ ଶ௫ ݁

͸

଼

ଵ

Ͳ

ଵ଺

݁ ଶ௫

Table 1.2.1: Table Method for ‫ ݔ ׬‬ଷ ݁ ଶ௫ ݀‫ݔ‬ We always start with positive sign, followed by negative sign, and so on as we can see in the above table 1.2.1. Now, from the above table 1.2.1, we can find ‫ ݔ ׬‬ଷ ݁ ଶ௫ ݀‫ ݔ‬as follows: ͳ ͳ ͳ ͳ න ‫ ݔ‬ଷ ݁ ଶ௫ ݀‫ ݔ‬ൌ ‫ ݔ‬ଷ ݁ ଶ௫ െ ሺ͵ሻ‫ ݔ‬ଶ ݁ ଶ௫ ൅ ሺ͸ሻ‫ ݁ݔ‬ଶ௫ െ ሺ͸ሻ݁ ଶ௫ ൅ ‫ܥ‬ ʹ Ͷ ͺ ͳ͸ ଵ

ଷ

ଷ

ଷ

Thus, ‫ ݔ ׬‬ଷ ݁ ଶ௫ ݀‫ ݔ‬ൌ ଶ ‫ ݔ‬ଷ ݁ ଶ௫ െ ସ ‫ ݔ‬ଶ ݁ ଶ௫ ൅ ସ ‫ ݁ݔ‬ଶ௫ െ ଼ ݁ ଶ௫ ൅ ‫ܥ‬.

In conclusion, we can always use the table method to find integrals like ‫׬‬ሺ‫ ݈ܽ݅݉݋݊ݕ݈݋݌‬ሻ݁ ௔௫ ݀‫ ݔ‬and ‫׬‬ሺ‫ ݈ܽ݅݉݋݊ݕ݈݋݌‬ሻሺܽ‫ݔ‬ሻ݀‫ݔ‬. Example 1.2.4 Find ‫ ݔ͵ ׬‬ଶ ሺͶ‫ݔ‬ሻ݀‫ݔ‬.

Solution: To find ‫ ݔ͵ ׬‬ଶ ሺͶ‫ݔ‬ሻ݀‫ݔ‬, it is easier to use the table method than using integration by parts. In the table method, we need to create two columns: one for derivatives of ͵‫ ݔ‬ଶ , and the other one for integrations of ሺͶ‫ݔ‬ሻ. Then, we need to keep deriving ͵‫ ݔ‬ଶ till we get zero, and we stop integrating when the corresponding row is zero. The following table shows the table method to find ‫ ݔ͵ ׬‬ଶ ሺͶ‫ݔ‬ሻ݀‫ݔ‬:

17


Derivatives Part ͵‫ ݔ‬ଶ

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Integration Part ሺͶ‫ݔ‬ሻ ଵ

͸‫ݔ‬

െ ସ ሺͶ‫ݔ‬ሻ ͳ ሺͶ‫ݔ‬ሻ ͳ͸ ଵ ሺͶ‫ݔ‬ሻ

͸

െ

Ͳ

଺ସ

Table 1.2.2: Table Method for ‫ ݔ͵ ׬‬ଶ ሺͶ‫ݔ‬ሻ݀‫ݔ‬ We always start with positive sign, followed by negative sign, and so on as we can see in the above table 1.2.2. Now, from the above table 1.2.2, we can find ‫ ݔ͵ ׬‬ଶ ሺͶ‫ݔ‬ሻ݀‫ݔ‬ as follows: න ͵‫ ݔ‬ଶ ሺͶ‫ݔ‬ሻ݀‫ݔ‬ ͳ ͳ ൌ െ ሺ͵ሻ‫ ݔ‬ଶ ሺͶ‫ݔ‬ሻ െ ൬െ ൰ ͸‫ ݔ‬ሺͶ‫ ݔ‬ሻ Ͷ ͳ͸ ͳ ൅ ൬ ൰ ͸ ሺͶ‫ ݔ‬ሻ ൅ ‫ܥ‬ ͸Ͷ ଷ

ଷ

ସ

଼


ஶ

By the definition of integral, we substitute ‫׬‬଴ ሺ‫ ʹݔ‬ሻ݁െ‫ݔ݀ ݔݏ‬ ௕

with ௕՜ஶ ‫׬‬଴ ሺ‫ ʹݔ‬ሻ݁െ‫ݔ݀ ݔݏ‬.

ଷଶ

It is easier to find what it is inside the above box ௕

ቀ‫׬‬଴ ሺ‫ ʹݔ‬ሻ݁െ‫ ݔ݀ ݔݏ‬ቁ, and after that we can find the limit of ௕

‫׬‬଴ ሺ‫ ʹݔ‬ሻ݁െ‫ݔ݀ ݔݏ‬.

method. In the table method, we need to create two columns: one for derivatives of ‫ ݔ‬ଶ , and the other one for integrations of ݁ ି௦௫ . Then, we need to keep deriving ‫ ݔ‬ଶ till we get zero, and we stop integrating when the corresponding row is zero. The following table shows the table method to find ‫׬‬ሺ‫ ݔ‬ଶ ሻ݁ ି௦௫ ݀‫ݔ‬: Derivatives Part Integration Part ݁ ି௦௫ ‫ݔ‬ଶ


Step 2: Here in this example, ݂ ሺ‫ݔ‬ሻ ൌ ‫ ݔ‬because ଶ

ࣦሼ݂ሺ‫ݔ‬ሻሽ ൌ ࣦሼ‫ ݔ‬ଶ ሽ. λ Step 3: ࣦሼ‫ ݔ‬ଶ ሽ ൌ ‫ Ͳ׬‬ሺ‫ ݔ‬ଶ ሻ݁ ି௦௫ ݀‫ݔ‬

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ଵ

ʹ‫ݔ‬

െ ݁ ି௦௫

ʹ

ͳ ି௦௫ ݁ ‫ݏ‬ଶ ଵ ି௦௫ െ ௦య ݁

௦

Ͳ

Example 1.2.5 Using definition 1.2.1, find ࣦሼ‫ ݔ‬ଶ ሽ.

λ

௕

Now, we need to find ‫׬‬଴ ሺ‫ ʹݔ‬ሻ݁െ‫ ݔ݀ ݔݏ‬using the table

ሺͶ‫ ݔ‬ሻ ൅ ‫ܥ‬.

Solution: To find ࣦሼ‫ ݔ‬ଶ ሽ using definition 1.2.1, we need to do the following steps: Step 1: We write the general definition of laplace transform as follows:

௕

Step 4: We need to find ௕՜ஶ ‫׬‬଴ ሺ‫ ʹݔ‬ሻ݁െ‫ ݔ݀ ݔݏ‬as follows:

Thus, ‫ ݔ͵ ׬‬ଶ ሺͶ‫ݔ‬ሻ݀‫ ݔ‬ൌ െ ‫ ݔ‬ଶ ሺͶ‫ݔ‬ሻ ൅ ‫ ݔ‬ሺͶ‫ ݔ‬ሻ ൅ ଷ

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Table 1.2.3: Table Method for ‫׬‬ሺ‫ ݔ‬ଶ ሻ݁ ି௦௫ ݀‫ݔ‬ We always start with positive sign, followed by negative sign, and so on as we can see in the above table 1.2.3. Now, from the above table 1.2.3, we can find ‫ ݔ͵ ׬‬ଶ ሺͶ‫ݔ‬ሻ݀‫ݔ‬ as follows: ଵ

ଶ

௦

௦మ

Thus, ‫׬‬ሺ‫ ݔ‬ଶ ሻ݁ ି௦௫ ݀‫ ݔ‬ൌ െ ‫ ݔ‬ଶ ݁ ି௦௫ െ

‫ି ݁ݔ‬௦௫ െ

ଶ ௦య

݁ ି௦௫ ൅ ‫ܥ‬.

Now, we need to evaluate the above integral from 0 to ܾ as follows: ௕

ͳ

‫׬‬଴ ሺ‫ ʹݔ‬ሻ݁െ‫ ݔ݀ ݔݏ‬ൌ െ ‫݁ ʹݔ ݏ‬ ͳ

ቀ െ ܾʹ ݁ ‫ݏ‬

െ‫ܾݏ‬

െ

ʹ ‫ʹݏ‬

ܾ݁

െ‫ܾݏ‬

െ

െ‫ݔݏ‬

ʹ ‫͵ݏ‬

݁

െ

െ‫ܾݏ‬

ʹ ‫ʹݏ‬

‫݁ݔ‬

െ‫ݔݏ‬

െ

ʹ ‫͵ݏ‬

݁

െ‫ݔݏ‬

ቚ ௫ୀ௕ ൌ ௫ୀ଴

ʹ

൅ ͵ቁ . ‫ݏ‬

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ଵ

ଶ

ܾ݁ ି௦௕ െ

ଶ ௦య

ͳ

ଶ

݁ ି௦௕ ൅ య ቁ as follows: ௦

ቀെ ܾʹ ݁

௕՜ஶ ʹ

‫͵ݏ‬

െ‫ܾݏ‬

‫ݏ‬

െ

ʹ ‫ʹݏ‬

ܾ݁

െ‫ܾݏ‬

െ

ʹ ‫͵ݏ‬

݁

െ‫ܾݏ‬

ʹ

ʹ

൅ ͵ ቁ ൌ ቀͲ ൅ ͵ ቁ ൌ ‫ݏ‬

௕՜ஶ

‫ݏ‬

‫ ݏ‬൐ ͲǤ

To check if our answer is right, we need to look at table 1.1.1 at the beginning of this book. According to table ଶǨ 1.1.1 section 2 at the right side, we found ࣦሼ‫ ݔ‬ଶ ሽ ൌ ௦మశభ ൌ ଶ

௦య

which is the same answer we got.

Example 1.2.6 Using table 1.1.1, find ࣦሼሺͷ‫ݔ‬ሻሽ. Solution: To find ࣦሼሺͷ‫ݔ‬ሻሽ using table 1.1.1, we need to do the following steps: Step 1: We look at the transform table (table 1.1.1). Step 2: We look at which section in table 1.1.1 contains ‫ ݊݅ݏ‬function. Step 3: We write down what we get from table 1.1.1 (Section 3 at the left side) as follows: ࣦሼ ܿ‫ݐ‬ሽ ൌ

‫ݏ‬ଶ

ܿ ൅ ܿଶ

ହ ௦ మାହమ

ൌ


‫ݏ‬ଶ

‫ݏ‬ ൅ ܿଶ

Step 4: We change what we got from step 3 to make it look like ࣦሼሺെͺ‫ݔ‬ሻሽ as follows: ࣦሼሺെͺ‫ݔ‬ሻሽ ൌ

௦

௦ మ ାሺି଼ሻమ

ൌ

௦

.

௦ మ ା଺ସ ௦

௦

Thus, ࣦሼሺെͺ‫ݔ‬ሻሽ ൌ ௦మ ାሺି଼ሻమ ൌ ௦మ ା଺ସ. We will give some important mathematical results about laplace transforms. Result 1.2.1 Assume that ܿ is a constant, and ݂ሺ‫ݔ‬ሻ, ݃ሺ‫ݔ‬ሻ are functions. Then, we have the following: (Hint: ‫ ܨ‬ሺ‫ݏ‬ሻ ‫ܩ‬ሺ‫ݏ‬ሻ are the laplace transforms of ݂ሺ‫ݔ‬ሻ and ݃ሺ‫ ݔ‬ሻǡ respectively). ଼

a) ࣦሼ݃ሺ‫ݔ‬ሻሽ ൌ ‫ܩ‬ሺ‫ݏ‬ሻ (i.e. ࣦሼͺሽ ൌ ௦ ൌ ‫ܩ‬ሺ‫ݏ‬ሻ where ݃ሺ‫ݔ‬ሻ ൌ ͺ). b) ࣦ൛ܿ ή ݃ሺ‫ݔ‬ሻൟ ൌ ܿ ή ࣦሼ݃ሺ‫ݔ‬ሻሽ ൌ ܿ ή ‫ ܩ‬ሺ‫ݏ‬ሻǤ c) ࣦሼ݂ሺ‫ݔ‬ሻ ‫݃ ט‬ሺ‫ݔ‬ሻሽ ൌ ‫ ܨ‬ሺ‫ݏ‬ሻ ‫ ܩ ט‬ሺ‫ݏ‬ሻǤ d) ࣦሼ݂ሺ‫ݔ‬ሻ ή ݃ሺ‫ݔ‬ሻሽ is not necessary equal to ‫ܨ‬ሺ‫ݏ‬ሻ ή ‫ ܩ‬ሺ‫ݏ‬ሻǤ Example 1.2.8 Using definition 1.2.1, find ࣦሼ‫ ݕ‬ᇱ ሽ.

Step 4: We change what we got from step 3 to make it look like ࣦሼሺͷ‫ݔ‬ሻሽ as follows: ࣦሼ ͷ‫ݔ‬ሽ ൌ

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Step 3: We write down what we get from table 1.1.1 (Section 3 at the right side) as follows:

Step 5: We need find the limit of ቀെ ௦ ܾଶ ݁ ି௦௕ െ ௦మ


ହ ௦ మ ାଶହ ହ

. ହ

Thus, ࣦሼ ͷ‫ݔ‬ሽ ൌ ௦మ ାହమ ൌ ௦మ ାଶହ. Example 1.2.7 Using table 1.1.1, find ࣦሼሺെͺ‫ݔ‬ሻሽ. Solution: To find ࣦሼሺെͺ‫ݔ‬ሻሽ using table 1.1.1, we need to do the following steps: Step 1: We look at the transform table (table 1.1.1). Step 2: We look at which section in table 1.1.1 contains ܿ‫ ݏ݋‬function.

Solution: To find ࣦሼ‫ ݕ‬ᇱ ሽ using definition 1.2.1, we need to do the following steps: Step 1: We write the general definition of laplace transform as follows: λ


Step 2: Here in this example, ݂ ሺ‫ݔ‬ሻ ൌ ‫ ݕ‬ᇱ because ࣦሼ݂ሺ‫ݔ‬ሻሽ ൌ ࣦሼ‫ ݕ‬ᇱ ሽ. λ Step 3: ࣦሼ‫ ݕ‬ᇱ ሽ ൌ ‫ Ͳ׬‬ሺ‫ ݕ‬ᇱ ሻ݁ ି௦௫ ݀‫ݔ‬

ஶ

By the definition of integral, we substitute ‫׬‬଴ ሺ‫ݕ‬Ԣ ሻ݁െ‫ݔ݀ ݔݏ‬ ௕

with ௕՜ஶ ‫׬‬଴ ሺ‫ݕ‬Ԣ ሻ݁െ‫ݔ݀ ݔݏ‬.

20 M. Kaabar

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௕

Step 4: We need to find ௕՜ஶ ‫׬‬଴ ‫ݕ‬Ԣ ݁െ‫ ݔ݀ ݔݏ‬as follows: It is easier to find what it is inside the above box ௕ ቀ‫׬‬଴ ‫ݕ‬Ԣ ݁െ‫ ݔ݀ ݔݏ‬ቁ, and after that we can find the limit of ௕ ‫׬‬଴ ‫ݕ‬Ԣ ݁െ‫ݔ݀ ݔݏ‬. ௕ To find ‫׬‬଴ ‫ݕ‬Ԣ ݁െ‫ݔ݀ ݔݏ‬, we need to use integration by parts

݀‫ ݒ‬ൌ ‫ ݕ‬ᇱ ݀‫ݔ‬ ‫ ݒ‬ൌ ‫ ݒ݀ ׬‬ൌ ‫ ݕ ׬‬ᇱ ݀‫ ݔ‬ൌ ‫ݕ‬

න ‫ ݒ݀ݑ‬ൌ ‫ ݒݑ‬െ න ‫ݑ݀ݒ‬

௕ ௕՜ஶ

௕՜ஶ

Ͳ λ

න ‫݁ ݕ‬

݀‫ ݔ‬ൌ ‫݁ݕ‬ ௕՜ஶ

଴ λ

Because ‫׬‬଴ ‫ݕ‬ሺ‫݁ݏ‬

ି௦௫

െ‫ ݔݏ‬ȁ

ሻ ݀‫ ݔ‬ൌ

௕

‫ݔ‬ൌܾ ൅ න ‫ݕ‬ሺ‫݁ݏ‬െ‫ ݔݏ‬ሻ ݀‫ݔ‬ ‫ݔ‬ൌͲ

Ͳ ௕ ‫ݕ ׬‬ሺെ‫ି ݁ݏ‬௦௫ ሻ ݀‫ݔ‬. ܾ՜λ ଴

௕՜ஶ

଴ ௕

න ‫݁ ݕ‬

௕՜ஶ

Ͳ λ

Ԣ െ‫ݔݏ‬

଴

݀‫ ݔ‬ൌ ൫‫ݕ‬ሺܾሻ݁ ௕՜ஶ

െ‫ܾݏ‬

െ ‫ݕ‬ሺͲሻ݁

െ‫ݏ‬ሺͲሻ

൯ ൅ ‫ ݏ‬න ‫݁ݕ‬

െ‫ݔݏ‬

Ͳ

ஶ

Since we have‫׬‬଴ ‫ି ݁ݕ‬௦௫ ݀‫ݔ‬, then using result 1.2.1 λ

൫‫׬‬଴ ‫ି ݁ݕ‬௦௫ ݀‫ ݔ‬ൌ ܻሺ‫ݏ‬ሻ൯, we obtain the following: ௕

λ

න ‫ݕ‬Ԣ ݁െ‫ ݔ݀ ݔݏ‬ൌ ሺ‫ݕ‬ሺܾሻ݁െ‫ ܾݏ‬െ ‫ݕ‬ሺͲሻሻ ൅ ‫ ݏ‬න ‫݁ݕ‬െ‫ݔ݀ ݔݏ‬

௕՜ஶ

଴

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௕՜ஶ

න ‫ݕ‬Ԣ ݁െ‫ ݔ݀ ݔݏ‬ൌ ൫‫ݕ‬ሺܾሻ݁െ‫ݏ‬ሺஶሻ െ ‫ݕ‬ሺͲሻ൯ ൅ ‫ܻݏ‬ሺ‫ݏ‬ሻ ଴ ௕

଴ ௕

න ‫ݕ‬Ԣ ݁െ‫ ݔ݀ ݔݏ‬ൌ ‫ܻݏ‬ሺ‫ݏ‬ሻ െ ‫ݕ‬ሺͲሻ

݂ ᇱᇱ ሺͲሻ. d) ࣦ൛݂ ሺସሻ ሺ‫ݔ‬ሻൟ ൌ ‫ ݏ‬ସ ‫ ܨ‬ሺ‫ݏ‬ሻ െ ‫ ݏ‬ଷ ݂ ሺͲሻ െ ‫ ݏ‬ଶ ݂ ᇱ ሺͲሻ െ ‫ ݂ݏ‬ᇱᇱ ሺͲሻ െ ݂ ᇱᇱᇱ ሺͲሻ.

λ

න ‫ݕ‬Ԣ ݁െ‫ ݔ݀ ݔݏ‬ൌ ሺ‫݁ݕ‬െ‫ ܾݏ‬െ ‫݁ݕ‬െ‫ݏ‬ሺͲሻ ሻ ൅ න ‫ݕ‬ሺ‫݁ݏ‬െ‫ ݔݏ‬ሻ ݀‫ݔ‬

௕՜ஶ

଴ ௕

We conclude this example with the following results: Result 1.2.2 Assume that ݂ሺ‫ݔ‬ሻ is a function, and ‫ ܨ‬ሺ‫ݏ‬ሻ is the laplace transform of ݂ሺ‫ݔ‬ሻ. Then, we have the following: a) ࣦሼ݂ ᇱ ሺ‫ݔ‬ሻሽ ൌ ࣦ൛݂ ሺଵሻ ሺ‫ݔ‬ሻൟ ൌ ‫ ܨݏ‬ሺ‫ݏ‬ሻ െ ݂ሺͲሻ. b) ࣦሼ݂ ᇱᇱ ሺ‫ݔ‬ሻሽ ൌ ࣦ൛݂ ሺଶሻ ሺ‫ݔ‬ሻൟ ൌ ‫ ݏ‬ଶ ‫ ܨ‬ሺ‫ݏ‬ሻ െ ‫݂ݏ‬ሺͲሻ െ ݂ ᇱ ሺͲሻ. c) ࣦሼ݂ ᇱᇱᇱ ሺ‫ݔ‬ሻሽ ൌ ࣦ൛݂ ሺଷሻ ሺ‫ݔ‬ሻൟ ൌ ‫ ݏ‬ଷ ‫ ܨ‬ሺ‫ݏ‬ሻ െ ‫ ݏ‬ଶ ݂ ሺͲሻ െ ‫ ݂ݏ‬ᇱ ሺͲሻ െ

ܾ

න ‫ݕ‬Ԣ ݁െ‫ ݔ݀ ݔݏ‬ൌ ‫݁ݕ‬െ‫ ݔݏ‬െ න ‫ݕ‬ሺെ‫݁ݏ‬െ‫ ݔݏ‬ሻ ݀‫ݔ‬

௕՜ஶ

න ‫ݕ‬Ԣ ݁െ‫ ݔ݀ ݔݏ‬ൌ ሺ‫ݕ‬ሺܾሻ݁െ‫ ܾݏ‬െ ‫ݕ‬ሺͲሻሻ ൅ ‫ܻݏ‬ሺ‫ݏ‬ሻ

Thus, ࣦሼ‫ ݕ‬ᇱ ሽ ൌ ‫ܻݏ‬ሺ‫ݏ‬ሻ െ ‫ݕ‬ሺͲሻ.

Now, we can find the limit as follows:

Ԣ െ‫ݔݏ‬

௕

଴

න ‫ ݕ‬ᇱ ݁ ି௦௫ ݀‫ ݔ‬ൌ ‫ି ݁ݕ‬௦௫ െ න ‫ݕ‬ሺെ‫ି ݁ݏ‬௦௫ ሻ ݀‫ݔ‬

଴ ௕

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න ‫ݕ‬Ԣ ݁െ‫ ݔ݀ ݔݏ‬ൌ ሺͲ െ ‫ݕ‬ሺͲሻሻ ൅ ‫ܻݏ‬ሺ‫ݏ‬ሻ

as follows: ‫ ݑ‬ൌ ݁െ‫ݔݏ‬ ݀‫ ݑ‬ൌ െ‫݁ݏ‬െ‫ݔ݀ ݔݏ‬

௕՜ஶ


௕՜ஶ

݀‫ݔ‬

For more information about this result, it is recommended to look at section 8 in table 1.1.1. If you look at it, you will find the following: ࣦ൛݂ ሺ௠ሻ ሺ‫ݐ‬ሻൟ ൌ ‫ ݏ‬௠ ‫ܨ‬ሺ‫ݏ‬ሻ െ ‫ ݏ‬௠ିଵ ݂ሺͲሻ െ ‫ ݏ‬௠ିଶ ݂ ᇱ ሺ‫ݏ‬ሻ െ ‫ ڮ‬െ ݂ ሺ௠ିଵሻ ሺ‫ݏ‬ሻ.

Result 1.2.3 Assume that ܿ is a constant, and ݂ሺ‫ݔ‬ሻ is a function where ‫ ܨ‬ሺ‫ݏ‬ሻ is the laplace transform of ݂ሺ‫ݔ‬ሻ. Then, we have the following: ࣦሼ݂ܿሺ‫ݔ‬ሻሽ ൌ ࣦܿሼ݂ሺ‫ݔ‬ሻሽ ൌ ܿ‫ܨ‬ሺ‫ݏ‬ሻ.

Ͳ

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௦

Since ࣦ൛ ξʹ‫ ݐ‬ൟ ൌ ௦మ ାሺξଶሻమ , then this means that ௦

1.3 Inverse Laplace

ࣦ ିଵ ቄ

Transforms

Solution: Using definition 1.3.1 and the left side of section 3 in table 1.1.1, we find the following:

In this section, we will discuss how to find the inverse


௦ మାଶ

ቅ ൌ ξʹ‫ݐ‬. ିଷ

Example 1.3.3 Find ࣦ ିଵ ቄ௦మାଽቅ.

laplace transforms of different types of mathematical

ିଷ ௦ మ ାଽ

‫ݏ‬ଶ

ܿ ൅ ܿଶ

ିଷ

is an equivalent to ௦మ ାሺିଷሻమ

functions, and we will use table 1.1.1 to refer to the

Since ࣦሼሺെ͵ሻ‫ݐ‬ሽ ൌ ࣦሼሺെ͵‫ݐ‬ሻሽ ൌ ௦మ ାሺିଷሻమ , then this

laplace transforms.

means that ࣦ ିଵ ቄ௦మାଽቅ ൌ ሺെ͵‫ݐ‬ሻ.

ିଷ

Definition 1.3.1 The inverse laplace transform, denoted by ࣦ ିଵ ǡ is defined as a reverse laplace transform, and to find the inverse laplace transform, we need to think about which function has a laplace transform that equals to the function in the inverse laplace transform. For example, suppose that ݂ሺ‫ݔ‬ሻ is a function where ‫ ܨ‬ሺ‫ݏ‬ሻ is the laplace transform of ݂ሺ‫ݔ‬ሻ. Then, the inverse laplace transform is ࣦ ିଵ ሼ‫ܨ‬ሺ‫ݏ‬ሻሽ ൌ ݂ሺ‫ݔ‬ሻ. (i.e. ࣦ ିଵ ቄଵ௦ቅ we need to think which function has a laplace transform that ଵ

equals to ௦ ǡ in this case the answer is 1). ଷସ

Example 1.3.1 Find ࣦ ିଵ ቄ ௦ ቅ. ଷସ

ଵ

Solution: First of all, ࣦ ିଵ ቄ ௦ ቅ ൌ ࣦ ିଵ ቄ͵Ͷ ቀ௦ ቁቅ. Using definition 1.3.1 and table 1.1.1, the answer is 34. Example 1.3.2 Find

ିଷ

௦ ࣦ ିଵ ቄ మ ቅ. ௦ ାଶ

ଵ

Example 1.3.4 Find ࣦ ିଵ ቄ௦ା଼ቅ. Solution: Using definition 1.3.1 and section 4 in table 1.1.1, we find the following: ଵ

ࣦሼ݁ ௕௧ ሽ ൌ ௦ି௕ . ଵ

௦ା଼

ଵ

is an equivalent to ௦ିሺି଼ሻ ଵ

ଵ

Since ࣦሼ݁ ି଼௧ ሽ ൌ ௦ା଼ , then this means that ࣦ ିଵ ቄ௦ା଼ቅ ൌ ݁ ି଼௧ .

Result 1.3.1 Assume that ܿ is a constant, and ݂ሺ‫ݔ‬ሻ is a function where ‫ ܨ‬ሺ‫ݏ‬ሻ ‫ܩ‬ሺ‫ݏ‬ሻ are the laplace transforms of ݂ሺ‫ ݔ‬ሻǡ ݃ ሺ‫ ݔ‬ሻǡ . a) ࣦ ିଵ ሼܿ‫ܨ‬ሺ‫ݏ‬ሻሽ ൌ ࣦܿ ିଵ ሼ‫ܨ‬ሺ‫ݏ‬ሻሽ ൌ ݂ܿሺ‫ݔ‬ሻ. b) ࣦ ିଵ ሼ‫ܨ‬ሺ‫ݏ‬ሻ ‫ ܩ ט‬ሺ‫ݏ‬ሻሽ ൌ ࣦ ିଵ ሼ‫ܨ‬ሺ‫ݏ‬ሻሽ ‫ି ࣦ ט‬ଵ ሼ‫ܩ‬ሺ‫ݏ‬ሻሽ ൌ ݂ሺ‫ݔ‬ሻ ‫ט‬ ݃ሺ‫ݔ‬ሻǤ ହ

Example 1.3.5 Find ࣦ ିଵ ቄ௦మାସቅ. ହ

ଵ

Solution: Using definition 1.3.1 and the right side of section 3 in table 1.1.1, we find the following:

Solution: Using result 1.3.1, ࣦ ିଵ ቄ௦మାସቅ ൌ ͷࣦ ିଵ ቄ௦మାସቅ. Now,


to make it look like ௦మ ା௖ మ because ࣦሼ ܿ‫ݐ‬ሽ ൌ ௦మ ା௖ మ.

௦ ௦ మ ାଶ

‫ݏ‬ ଶ ‫ ݏ‬൅ ܿଶ

by using the left side of section 3 in table 1.1.1, we need ௖

௦

௖

is an equivalent to ௦మ ାሺξଶሻమ

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ହ

ଵ

ହ

ଶ

Therefore, we do the following: ଶ ࣦ ିଵ ቄሺʹሻ ௦మାସቅ ൌ ଶ ࣦ ିଵ ቄ௦మାସቅ, ଶ

ଶ

and ௦మାସ is an equivalent to ହ

.

௦ మ ାሺଶሻమ ହ

ହ

ଶ

ହ

Since ଶ ࣦሼሺʹሻ‫ݐ‬ሽ ൌ ࣦ ቄଶ ሺʹ‫ݐ‬ሻቅ ൌ ଶ ቀ௦మ ାሺଶሻమቁ ൌ ௦మ ାସ , then by ହ ࣦ ିଵ ቄ௦మାସ ቅ

using definition 1.3.1, this means that ହ ଶ

ൌ

଻

Now, by using section 4 in table 1.1.1, we need to make it ଵ

ଵ

ଵ

ଶ

ࣦ

ିଵ ቊ

ଵ య మ

ቀ௦ି ቁ

ଵ య మ

ଶቀ௦ି ቁ

ቋൌ

ቋ. ଻

య ௧ మ

଻

ଵ

Since ଶ ࣦ ቄ݁ ቅ ൌ ࣦ ቄଶ ݁ ቅ ൌ ଶ ቆ

଻

య మ

ቀ௦ି ቁ

ቇ ൌ ଶ௦ିଷ , then by using ଻

଻ య

definition 1.3.1, this means that ࣦ ିଵ ቄଶ௦ିଷ ቅ ൌ ଶ ݁ మ௧ . ଵାଷ௦

ଵ

௖

section 3 in table 1.1.1, we need to make it look like ௦మ ା௖ మ ௦ మା௖

మ because ࣦሼ ܿ‫ݐ‬ሽ ൌ

௖

௦ మା௖ మ


௦

௦ మ ା௖

మ .

Therefore, we do the following: ࣦ ିଵ ቄ

ଵ ௦ మାଽ ଵ

൅

ଷ௦ ௦ మ ାଽ

ଵ

ଵሺଷሻ

ଷ

௦ మାଽ

ቅ ൌ ࣦ ିଵ ቄ

ቅ ൅ ͵ࣦ ିଵ ቄ ଵ

௦

from sections 1.2 and 1.3 to find the largest interval on

ଷ௦

ଵାଷ௦

using definition 1.3.1, this means that ࣦ ିଵ ቄ௦మାଽ ቅ ൌ ଵ ଷ௦ ࣦ ିଵ ቄ మାଽቅ ൅ ࣦ ିଵ ቄ మାଽቅ ௦ ௦

ଵ

ൌ ଷ ሺ͵‫ݐ‬ሻ ൅ ͵ ሺ͵‫ݐ‬ሻ. ଺

Example 1.3.8 Find ࣦ ିଵ ቄ௦ర ቅ.

Definition 1.4.1 Given ܽ௡ ሺ‫ݔ‬ሻ‫ ݕ‬ሺ௡ሻ ൅ ܽ௡ିଵ ሺ‫ ݔ‬ሻ‫ ݕ‬ሺ௡ିଵሻ ൅ ‫ ڮ‬൅ ܽ଴ ሺ‫ ݔ‬ሻ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ‫ܭ‬ሺ‫ݔ‬ሻ. Assume that ܽ௡ ሺ݉ሻ ് Ͳ for every where

‫ܫ‬

is

some are

interval, continuous

and on

‫ܫ‬.

Suppose that ‫ݕ‬ሺ‫ݓ‬ሻǡ ‫ ݕ‬ᇱ ሺ‫ݓ‬ሻǡ ǥ ǡ ‫ ݕ‬ሺ௡ିଵሻ ሺ‫ݓ‬ሻ for some ‫ܫ א ݓ‬. Then, the solution to the differential equations is unique which means that there exists exactly one ‫ݕ‬ሺ‫ ݔ‬ሻ in terms of ‫ݔ‬, and this type of mathematical problems

ቅ.

௦ మାଽ

Since ଷ ࣦሼሺ͵‫ݐ‬ሻሽ ൅ ͵ࣦሼሺ͵‫ݐ‬ሻሽ ൌ ቀ௦మାଽ ൅ ௦మ ାଽቁ , then by

26 M. Kaabar

(IVP), and we will use it with what we have learned

ܽ௡ ሺ‫ݔ‬ሻǡ ܽ௡ିଵ ሺ‫ݔ‬ሻǡ ǥ ǡ ܽ଴ ሺ‫ ݔ‬ሻǡ ‫ܭ‬ሺ‫ݔ‬ሻ

ଷ௦

Solution: ࣦ ିଵ ቄ௦మାଽ ቅ ൌ ࣦ ିଵ ቄ௦మାଽ ൅ ௦మ ାଽቅ. Now, by using ௦

଺

In this section, we will introduce the main theorem of

݉‫ܫא‬

ଵାଷ௦

Example 1.3.7 Find ࣦ ିଵ ቄ௦మାଽ ቅ.

and

଺

the x-axis.

య ௧ మ

଻

ଷǨ

is an equivalent to ௦యశభ

differential equations known as Initial Value Problems

because ࣦሼ݁ ௕௧ ሽ ൌ ௦ି௕.

Therefore, we do the following: ͹ࣦ ିଵ ቄଶ௦ିଷ ቅ ൌ ͹ࣦ ିଵ ቊ ଻

଺

௦ర

1.4 Initial Value Problems

ଵ

Solution: Using result 1.3.1, ࣦ ିଵ ቄଶ௦ିଷቅ ൌ ͹ࣦ ିଵ ቄଶ௦ିଷቅ. ௦ି௕

Solution: Using definition 1.3.1 and the right side of section 2 in table 1.1.1, we find the following: ௠Ǩ ࣦሼ‫ ݐ‬௠ ሽ ൌ ೘శభ where ݉ is a positive integer. ௦ ଷǨ

Example 1.3.6 Find ࣦ ିଵ ቄଶ௦ିଷቅ.

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Since ࣦሼ‫ ݐ‬ଷ ሽ ൌ ௦యశభ ൌ ௦ర , then this means that ࣦ ିଵ ቄ௦ర ቅ ൌ ‫ ݐ‬ଷ .

ሺʹ‫ݐ‬ሻ.

଻


is called Initial Value Problems (IVP). Example 1.4.1 Find the largest interval on the ‫ ݔ‬െ ܽ‫ݏ݅ݔ‬ ௫ିଷ

so that ௫ାଶ ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൅ ʹ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ξ‫ ݔ‬൅ ͳ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ͷ‫ ݔ‬൅ ͹ has a solution. Given ‫ ݕ‬ᇱ ሺͷሻ ൌ ͳͲǡ ‫ݕ‬ሺͷሻ ൌ ʹǡ ‫ ݕ‬ሺଶሻ ሺͷሻ ൌ െͷ.

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Solution: Finding largest interval on the ‫ ݔ‬െ ܽ‫ݏ݅ݔ‬ means that we need to find the domain for the solution of the above differential equation in other words we need to find for what values of ‫ ݔ‬the solution of the above differential equation holds. Therefore, we do the following:

Solution: Finding largest interval on the ‫ ݔ‬െ ܽ‫ݏ݅ݔ‬ means that we need to find the domain for the solution of the above differential equation in other words we need to find for what values of ‫ ݔ‬the solution of the above differential equation holds. Therefore, we do the following:

௫ିଷ

ሺ‫ ݔ‬ଶ ൅ ʹ‫ ݔ‬െ ͵ሻ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅

௫ାଶ

‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൅ ʹ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ξ‫ ݔ‬൅ ͳ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ͷ‫ ݔ‬൅ ͹

Using definition 1.4.1, we also suppose the following: ‫ݔ‬െ͵ ܽଷ ሺ ‫ ݔ‬ሻ ൌ ‫ݔ‬൅ʹ ܽଶ ሺ ‫ ݔ‬ሻ ൌ ʹ ܽଵ ሺ‫ ݔ‬ሻ ൌ ξ‫ ݔ‬൅ ͳ

ଵ ௫ାଷ

‫ݕ‬ሺ‫ݔ‬ሻ ൌ ͳͲ

Using definition 1.4.1, we also suppose the following: ܽଶ ሺ‫ ݔ‬ሻ ൌ ሺ‫ ݔ‬ଶ ൅ ʹ‫ ݔ‬െ ͵ሻ ܽଵ ሺ‫ ݔ‬ሻ ൌ Ͳ ܽ଴ ሺ ‫ ݔ‬ሻ ൌ

ͳ ‫ݔ‬൅͵

‫ܭ‬ሺ‫ ݔ‬ሻ ൌ ͷ‫ ݔ‬൅ ͹

‫ܭ‬ሺ‫ ݔ‬ሻ ൌ ͳͲ

Now, we need to determine the interval of each

Now, we need to determine the interval of each

coefficient above as follows:

coefficient above as follows:

ܽଷ ሺ ‫ ݔ‬ሻ ൌ

௫ିଷ ௫ାଶ

has a solution which is continuous

everywhere (Ը) except ‫ ݔ‬ൌ െʹ ‫ ݔ‬ൌ ͵Ǥ ܽଶ ሺ ‫ ݔ‬ሻ ൌ ʹ


ܽଶ ሺ‫ ݔ‬ሻ ൌ ሺ‫ ݔ‬ଶ ൅ ʹ‫ ݔ‬െ ͵ሻ ൌ ሺ‫ ݔ‬െ ͳሻሺ‫ ݔ‬൅ ͵ሻ has a solution which is continuous everywhere (Ը) except ‫ ݔ‬ൌ ͳ and ‫ ݔ‬ൌ െ͵. ଵ

everywhere (Ը).

ܽ଴ ሺ ‫ ݔ‬ሻ ൌ

ܽଵ ሺ‫ ݔ‬ሻ ൌ ξ‫ ݔ‬൅ ͳ has a solution which is continuous on

everywhere (Ը) except ‫ ݔ‬ൌ െ͵.

the interval ሾെͳǡ λሻ.

‫ܭ‬ሺ‫ ݔ‬ሻ ൌ ͳͲ

‫ܭ‬ሺ‫ ݔ‬ሻ ൌ ͷ‫ ݔ‬൅ ͹ has a solution which is continuous

everywhere (Ը).

everywhere (Ը).

Thus, the largest interval on the ‫ ݔ‬െ ܽ‫ ݏ݅ݔ‬is ሺͳǡ λሻ.

Thus, the largest interval on the ‫ ݔ‬െ ܽ‫ ݏ݅ݔ‬is ሺ͵ǡ λሻ.

Example 1.4.3 Solve the following Initial Value Problem (IVP): ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ. Given ‫ݕ‬ሺͲሻ ൌ ͵.

Example 1.4.2 Find the largest interval on the ‫ ݔ‬െ ܽ‫ݏ݅ݔ‬ so that ሺ‫ ݔ‬ଶ ൅ ʹ‫ ݔ‬െ ͵ሻ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅

ଵ ௫ାଷ

‫ݕ‬ሺ‫ݔ‬ሻ ൌ ͳͲ has a

solution. Given ‫ ݕ‬ᇱ ሺʹሻ ൌ ͳͲǡ ‫ݕ‬ሺʹሻ ൌ Ͷ.

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௫ାଷ


has

a

solution

which

is

continuous

Solution: ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ is a linear differential equation of order 1. First, we need to find the domain for the solution of the above differential equation in

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other words we need to find for what values of ‫ ݔ‬the solution of the above differential equation holds. Therefore, we do the following: ͳ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ Using definition 1.4.1, we also suppose the following: ܽଵ ሺ‫ ݔ‬ሻ ൌ ͳ


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‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ͵݁ ିଷ௫ (It is written in terms of ‫ ݔ‬instead of ‫ݐ‬ because we need it in terms of ‫)ݔ‬. Then, we will find ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ by finding the derivative of what we got above (‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ͵݁ ିଷ௫ ሻ as follows: ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൌ ሺ͵݁ ିଷ௫ ሻᇱ ൌ ሺ͵ሻሺെ͵ሻ݁ ିଷ௫ ൌ െͻ݁ ିଷ௫ .

ܽ଴ ሺ ‫ ݔ‬ሻ ൌ ͵

Finally, to check our solution if it is right, we

‫ ܭ‬ሺ‫ ݔ‬ሻ ൌ Ͳ

substitute what we got from ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ͵݁ ିଷ௫ and ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൌ

The domain of solution is ሺെλǡ λሻ.

െͻ݁ ିଷ௫ in

Now, to find the solution of the above differential

‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ as follows:

equation, we need to take the laplace transform for

െͻ݁ ିଷ௫ ൅ ͵ሺ͵݁ ିଷ௫ ሻ ൌ െͻ݁ ିଷ௫ ൅ ሺͻ݁ ିଷ௫ ሻ ൌ Ͳ

both sides as follows

Thus, our solution is correct which is ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ͵݁ ିଷ௫ and

ᇱ

ࣦሼ‫ ݕ‬ሺ‫ݔ‬ሻሽ ൅ ࣦሼ͵‫ݕ‬ሺ‫ݔ‬ሻሽ ൌ ࣦሼͲሽ

‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൌ െͻ݁ ିଷ௫ .

ᇱ

ࣦሼ‫ ݕ‬ሺ‫ݔ‬ሻሽ ൅ ͵ࣦሼ‫ݕ‬ሺ‫ݔ‬ሻሽ ൌ Ͳ because (ࣦሼͲሽ ൌ Ͳ).

൫‫ܻݏ‬ሺ‫ݏ‬ሻ െ ‫ݕ‬ሺͲሻ൯ ൅ ͵ܻሺ‫ݏ‬ሻ ൌ Ͳ

from result 1.2.2.

‫ܻݏ‬ሺ‫ݏ‬ሻ െ ‫ݕ‬ሺͲሻ ൅ ͵ܻሺ‫ݏ‬ሻ ൌ Ͳ ܻሺ‫ݏ‬ሻሺ‫ ݏ‬൅ ͵ሻ ൌ ‫ݕ‬ሺͲሻ We substitute ‫ݕ‬ሺͲሻ ൌ ͵ because it is given in the question itself. ܻሺ‫ݏ‬ሻሺ‫ ݏ‬൅ ͵ሻ ൌ ͵ ଷ

ܻሺ‫ݏ‬ሻ ൌ ሺ௦ାଷሻ To find a solution, we need to find the inverse laplace transform as follows: ࣦ ିଵ ሼܻሺ‫ݏ‬ሻሽ ൌ ࣦ ିଵ ቄሺ

1.1.1 section 4.

͵ ቅ ‫ݏ‬൅͵ሻ

ͳ

ൌ ͵ࣦ ିଵ ቄሺ‫ݏ‬൅͵ሻቅ and we use table

Example 1.4.4 Solve the following Initial Value Problem (IVP): ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ. Given ‫ݕ‬ሺͲሻ ൌ Ͳ, and ‫ ݕ‬ᇱ ሺͲሻ ൌ ͳ. Solution: ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ is a linear differential equation of order 2. First, we need to find the domain for the solution of the above differential equation in other words we need to find for what values of ‫ ݔ‬the solution of the above differential equation holds. Therefore, we do the following: ͳ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ Using definition 1.4.1, we also suppose the following: ܽଶ ሺ ‫ ݔ‬ሻ ൌ ͳ ܽଵ ሺ‫ ݔ‬ሻ ൌ Ͳ ܽ଴ ሺ ‫ ݔ‬ሻ ൌ ͵ ‫ ܭ‬ሺ‫ ݔ‬ሻ ൌ Ͳ The domain of solution is ሺെλǡ λሻ.

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Now, to find the solution of the above differential equation, we need to take the laplace transform for


‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൌ ൬

ͳ ξ͵

ᇱ

ሺξ͵ ‫ݔ‬ሻ൰ ൌ

ͳ ξ͵

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൫ξ͵൯ ൫ξ͵ ‫ݔ‬൯ ൌ ൫ξ͵ ‫ݔ‬൯


Now, we will find ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ by finding the derivative of

ࣦሼ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻሽ ൅ ࣦሼ͵‫ݕ‬ሺ‫ݔ‬ሻሽ ൌ ࣦሼͲሽ

what we got above as follows:

ࣦሼ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻሽ ൅ ͵ࣦሼ‫ݕ‬ሺ‫ݔ‬ሻሽ ൌ Ͳ because (ࣦሼͲሽ ൌ Ͳ).

‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൌ ൫൫ξ͵ ‫ݔ‬൯൯ ൌ െξ͵ ൫ξ͵‫ݔ‬൯

ሺ‫ ݏ‬ଶ ܻሺ‫ݏ‬ሻ െ ‫ݕݏ‬ሺͲሻ െ ‫ ݕ‬ᇱ ሺͲሻሻ ൅ ͵ܻሺ‫ݏ‬ሻ ൌ Ͳ from result 1.2.2.

Finally, to check our solution if it is right, we

ଶ

ᇱ

ᇱ

‫ܻ ݏ‬ሺ‫ݏ‬ሻ െ ‫ݕݏ‬ሺͲሻ െ ‫ ݕ‬ሺͲሻ ൅ ͵ܻሺ‫ݏ‬ሻ ൌ Ͳ

substitute what we got from ‫ݕ‬ሺ‫ ݔ‬ሻ and ‫ ݕ‬ሺଶሻ ሺ‫ ݔ‬ሻ in

ܻሺ‫ݏ‬ሻሺ‫ ݏ‬ଶ ൅ ͵ሻ ൌ ‫ݕݏ‬ሺͲሻ ൅ ‫ ݕ‬ᇱ ሺͲሻ

‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ as follows:

We substitute ‫ݕ‬ሺͲሻ ൌ Ͳ, and ‫ ݕ‬ᇱ ሺͲሻ ൌ ͳ because it is

െξ͵ ൫ξ͵‫ݔ‬൯ ൅ ξ͵ ‫݊݅ݏ‬൫ξ͵ ‫ݔ‬൯ ൌ Ͳ

given in the question itself.

Thus, our solution is correct which is

ଶ

ܻሺ‫ݏ‬ሻሺ‫ ݏ‬൅ ͵ሻ ൌ ሺ‫ݏ‬ሻሺͲሻ ൅ ͳ

‫ ݕ‬ሺ‫ ݔ‬ሻ ൌ

ଶ

ܻሺ‫ݏ‬ሻሺ‫ ݏ‬൅ ͵ሻ ൌ Ͳ ൅ ͳ ܻሺ‫ݏ‬ሻሺ‫ ݏ‬ଶ ൅ ͵ሻ ൌ ͳ

ଵ ξଷ

ሺξ͵ ‫ݔ‬ሻ and ‫ ݕ‬ሺଶሻ ሺ‫ ݔ‬ሻ ൌ െξ͵ ൫ξ͵‫ݔ‬൯.

1.5 Properties of Laplace

ͳ ܻ ሺ‫ ݏ‬ሻ ൌ ଶ ሺ‫ ݏ‬൅ ͵ሻ To find a solution, we need to find the inverse laplace

Transforms

transform as follows:

In this section, we discuss several properties of laplace

ࣦ ିଵ ሼܻሺ‫ݏ‬ሻሽ ൌ ࣦ ିଵ ൜൫

ͳ ൠ ‫ ʹݏ‬൅͵൯

ͳ

ൌ ξ͵ ࣦ ିଵ ൝

ͳ ʹ

ൡ

൬‫ ʹݏ‬൅ሺξ͵ሻ ൰

and we use

table 1.1.1 at the left side of section 3. ‫ ݕ‬ሺ‫ ݔ‬ሻ ൌ

ଵ ξଷ

ሺξ͵ ‫ݔ‬ሻ (It is written in terms of ‫ ݔ‬instead of

transforms such as shifting, unit step function, periodic function, and convolution. We start with some examples of shifting property. Example 1.5.1 Find ࣦሼ݁ ଷ௫ ‫ ݔ‬ଷ ሽ.

‫ ݐ‬because we need it in terms of ‫)ݔ‬.

Solution: By using shifting property at the left side of

Then, we will find ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ by finding the derivative of

section 5 in table 1.1.1, we obtain:

what we got above ቀ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ

ଵ ξଷ

ሺξ͵ ‫ݔ‬ሻቁ as follows:

ࣦሼ݁ ௕௫ ݂ሺ‫ݔ‬ሻሽ ൌ ‫ ܨ‬ሺ‫ݏ‬ሻ ȁ‫ ݏ‬՜ ‫ ݏ‬െ ܾ

Let ܾ ൌ ͵, and ݂ሺ‫ ݔ‬ሻ ൌ ‫ ͵ݔ‬. ‫ ܨ‬ሺ‫ݏ‬ሻ ൌ ࣦሼ‫ ݔ‬ଷ ሽ.

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ଷǨ

Hence, ࣦሼ݁ ଷ௫ ‫ ݔ‬ଷ ሽ ൌ ࣦሼ‫ ݔ‬ଷ ሽ ȁ‫ ݏ‬՜ ‫ ݏ‬െ ͵ ൌ ሺ௦ሻయశభ ȁ‫ ݏ‬՜ ‫ ݏ‬െ ͵ ൌ ଷǨ ሺ௦ሻర

ȁ‫ ݏ‬՜ ‫ ݏ‬െ ͵.


‫ݏ‬

ࣦെͳ ൜

ሺ‫ ݏ‬െ ʹሻʹ ൅ Ͷ

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ʹ ‫ݏ‬െʹ ൠ ൌ ࣦ ିଵ ൜ ൠ ൅ ࣦെͳ ൜ ൠ ሺ‫ ݏ‬െ ʹሻ ଶ ൅ Ͷ ሺ‫ ݏ‬െ ʹሻʹ ൅ Ͷ

By using shifting property at the right side of section 5

Now, we need to substitute ‫ ݏ‬with ‫ ݏ‬െ ͵ as follows:

in table 1.1.1, we obtain:

͵Ǩ ͸ ൌ ሺ‫ ݏ‬െ ͵ሻସ ሺ‫ ݏ‬െ ͵ሻସ

ࣦ ିଵ ሼ‫ܨ‬ሺ‫ݏ‬ሻ ȁ‫ ݏ‬՜ ‫ ݏ‬െ ܾ ሽ ൌ ݁ ௕௫ ݂ሺ‫ݔ‬ሻ ‫ݏ‬െʹ

ଷǨ

଺

Thus, ࣦሼ݁ ଷ௫ ‫ ݔ‬ଷ ሽ ൌ ሺ௦ିଷሻర ൌ ሺ௦ିଷሻర.

ʹ

Let ܾ ൌ ʹ, ‫ܨ‬ଵ ሺ‫ݏ‬ሻ ൌ ࣦ ିଵ ቄሺ‫ݏ‬െʹሻʹ ൅Ͷ ቅ, and ‫ܨ‬ଶ ሺ‫ݏ‬ሻ ൌ ࣦ ିଵ ቄሺ‫ݏ‬െʹሻʹ ൅Ͷ ቅ. ௦

Example 1.5.2 Find ࣦሼ݁ ିଶ௫ ሺͶ‫ݔ‬ሻሽ.

ࣦ ିଵ ቄሺ௦ିଶሻమ ାସ ቅ ൌ ࣦ ିଵ ሼ‫ܨ‬ଵ ሺ‫ݏ‬ሻ ȁ‫ ݏ‬՜ ‫ ݏ‬െ ʹ ሽ ൅ ࣦ ିଵ ሼ‫ܨ‬ଶ ሺ‫ݏ‬ሻ ȁ‫ ݏ‬՜ ‫ ݏ‬െ ʹ ሽ

Solution: By using shifting property at the left side of

Thus, ࣦ ିଵ ቄሺ௦ିଶሻమ ାସ ቅ ൌ ݁ ଶ௫ ሺʹ‫ݔ‬ሻ ൅ ݁ ଶ௫ ሺʹ‫ݔ‬ሻǤ

section 5 in table 1.1.1, we obtain:

Example 1.5.4 Find ࣦ ିଵ ቄሺ௦ାଶሻయ ቅ.

ࣦሼ݁ ௕௫ ݂ሺ‫ݔ‬ሻሽ ൌ ‫ ܨ‬ሺ‫ݏ‬ሻ ȁ‫ ݏ‬՜ ‫ ݏ‬െ ܾ

Let ܾ ൌ െʹ, and ݂ሺ‫ݔ‬ሻ ൌ ሺͶ‫ݔ‬ሻ. ‫ ܨ‬ሺ‫ݏ‬ሻ ൌ ࣦሼሺͶ‫ݔ‬ሻሽ.

௦

௦

Solution: Since we have a shift such as ‫ ݏ‬൅ ʹ, we need to do the following: ʹ ‫ݏ‬ ‫ݏ‬൅ʹെʹ ‫ݏ‬൅ʹ ൠ ൌ ࣦ ିଵ ൜ ൠ ൌ ࣦ ିଵ ൜ ൠ െ ሺ‫ ݏ‬൅ ʹሻଷ ሺ‫ ݏ‬൅ ʹሻଷ ሺ‫ ݏ‬൅ ʹሻଷ ሺ‫ ݏ‬൅ ʹሻଷ

Hence, ࣦሼ݁ ିଶ௫ ሺͶ‫ݔ‬ሻሽ ൌ ࣦሼሺͶ‫ݔ‬ሻሽ ȁ‫ ݏ‬՜ ‫ ݏ‬൅ ʹ ൌ

ࣦ ିଵ ൜

Ͷ ȁ‫ ݏ‬՜ ‫ ݏ‬൅ ʹ ‫ ݏ‬ଶ ൅ ͳ͸

ࣦെͳ ቊ

Now, we need to substitute ‫ ݏ‬with ‫ ݏ‬൅ ʹ as follows: ࣦെͳ ቊ

Ͷ ሺ‫ ݏ‬൅ ʹሻଶ ൅ ͳ͸ ସ

Thus, ࣦሼ݁ ିଶ௫ ሺͶ‫ݔ‬ሻሽ ൌ ሺ௦ାଶሻమାଵ଺ . ௦

‫ݏ‬ ͵

ቋ ൌ ࣦ ିଵ ቊ

͵

ቋ ൌ ࣦ ିଵ ቊ

ሺ‫ ݏ‬൅ ʹሻ ‫ݏ‬ ሺ‫ ݏ‬൅ ʹሻ

‫ݏ‬൅ʹ ͵

ቋ ൅ ࣦെͳ ቊ

ʹ

ቋ ൅ ࣦെͳ ቊ

ሺ‫ ݏ‬൅ ʹሻ ͳ

ሺ‫ ݏ‬൅ ʹሻ

െʹ ሺ‫ ݏ‬൅ ʹሻ͵ െʹ ሺ‫ ݏ‬൅ ʹሻ͵

ቋ ቋ

By using shifting property at the right side of section 5 in table 1.1.1, we obtain:

Example 1.5.3 Find ࣦ ିଵ ቄሺ௦ିଶሻమ ାସ ቅ.

ࣦ ିଵ ሼ‫ܨ‬ሺ‫ݏ‬ሻ ȁ‫ ݏ‬՜ ‫ ݏ‬െ ܾ ሽ ൌ ݁ ௕௫ ݂ሺ‫ݔ‬ሻ

Solution: Since we have a shift such as ‫ ݏ‬െ ʹ, we need

Let ܾ ൌ െʹ, ‫ܨ‬ଵ ሺ‫ݏ‬ሻ ൌ ࣦ ିଵ ቄሺ௦ାଶሻమ ቅ, and ‫ܨ‬ଶ ሺ‫ݏ‬ሻ ൌ ࣦ ିଵ ቄሺ௦ାଶሻయ ቅ.

to do the following: ‫ݏ‬ ‫ݏ‬െʹ൅ʹ ൠ ൌ ࣦ ିଵ ൜ ൠ ࣦ ିଵ ൜ ଶ ሺ‫ ݏ‬െ ʹሻ ൅ Ͷ ሺ‫ ݏ‬െ ʹሻଶ ൅ Ͷ ൌ ࣦ ିଵ ൜

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‫ݏ‬െʹ ʹ ൠ ൅ ሺ‫ ݏ‬െ ʹሻଶ ൅ Ͷ ሺ‫ ݏ‬െ ʹሻଶ ൅ Ͷ

ଵ

ࣦ ିଵ ቄ

௦ ሺ௦ାଶሻయ

ିଶ

ቅ ൌ ࣦ ିଵ ሼ‫ܨ‬ଵ ሺ‫ݏ‬ሻ ȁ‫ ݏ‬՜ ‫ ݏ‬൅ ʹ ሽ ൅ ࣦ ିଵ ሼ‫ܨ‬ଶ ሺ‫ݏ‬ሻ ȁ‫ ݏ‬՜ ‫ ݏ‬൅ ʹ ሽ ௦

Thus, ࣦ ିଵ ቄሺ௦ାଶሻయ ቅ ൌ ݁ ିଶ௫ ‫ ݔ‬െ ݁ ିଶ௫ ‫ ݔ‬ଶ Ǥ ଵ

Example 1.5.5 Find ࣦ ିଵ ቄ௦మିସ ቅ.

35


ଵ

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ଵ


ଵ

ଵ

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ଵ

Solution: ࣦ ିଵ ቄ௦మିସ ቅ ൌ ࣦ ିଵ ቄሺ௦ିଶሻሺ௦ାଶሻ ቅ.

ǡ ࣦ ିଵ ቄ௦ మିସ ቅ ൌ ସ ݁ ଶ௫ െ ସ ݁ ିଶ௫ .

Since the numerator has a polynomial of degree 0

Now, we will introduce a new property from table 1.1.1

(‫ ݔ‬଴ ൌ ͳሻ, and the denominator a polynomial of degree

in the following two examples.

2, then this means the degree of numerator is less than

Example 1.5.6 Find ࣦሼ‫ ݁ݔ‬௫ ሽ.

the degree of denominator. Thus, in this case, we need

Solution: By using the left side of section 9 in table

to use the partial fraction as follows:

1.1.1, we obtain:

ͳ ܽ ܾ ൌ ൅ ሺ‫ ݏ‬െ ʹሻሺ‫ ݏ‬൅ ʹሻ ሺ‫ ݏ‬െ ʹሻ ሺ‫ ݏ‬൅ ʹሻ


It is easier to use a method known as cover method

where ࣦሼ݂ሺ‫ݔ‬ሻሽ ൌ ‫ ܨ‬ሺ‫ݏ‬ሻ, and ݂ ሺ‫ݔ‬ሻ ൌ ݁ ௫ .

than using the traditional method that takes long time

Hence, ࣦሼ‫ ݁ݔ‬௫ ሽ ൌ ሺെͳሻଵ ‫ ܨ‬ሺଵሻ ሺ‫ݏ‬ሻ ൌ െ‫ ܨ‬ሺଵሻ ሺ‫ݏ‬ሻ

to finish it. In the cover method, we cover the original,

Now, we need to find ‫ ܨ‬ሺଵሻ ሺ‫ݏ‬ሻ as follows: This means

ଵ

݀ ௠ ‫ܨ‬ሺ‫ݏ‬ሻ ൌ ሺെͳሻ௠ ‫ ܨ‬ሺ௠ሻ ሺ‫ݏ‬ሻ ݀‫ ݏ‬௠

say ሺ‫ ݏ‬െ ʹሻ, and substitute ‫ ݏ‬ൌ ʹ in ሺ௦ିଶሻሺ௦ାଶሻ to find the

that we first need to find the laplace transform of ݂ ሺ‫ ݔ‬ሻ,

value of ܽ. Then, we cover the original, say ሺ‫ ݏ‬൅ ʹሻ, and

and then we need to find the first derivative of the

substitute ‫ ݏ‬ൌ െʹ in

ଵ ሺ௦ିଶሻሺ௦ାଶሻ

ଵ

ଵ

ସ

ସ

to find the value of ܾ.

Thus, ܽ ൌ and ܾ ൌ െ . This implies that

result from the laplace transform. ‫ ܨ‬ሺଵሻ ሺ‫ݏ‬ሻ ൌ ‫ ܨ‬ᇱ ሺ‫ݏ‬ሻ ൌ ሺࣦሼ݂ሺ‫ݔ‬ሻሽሻᇱ ൌ ሺࣦሼ݁ ௫ ሽሻᇱ ൌ ቀ

ሺ௦ିଵሻమ

Now, we need to do the following:

Example 1.5.7 Find ࣦሼ‫ ݔ‬ଶ ሺ‫ݔ‬ሻሽ.

ଵ

ͳ ͳ െͶ ͳ Ͷ ିଵ ൠ ቐ ቑ ൌ ࣦ ൅ ‫ݏ‬ଶ െ Ͷ ሺ‫ ݏ‬െ ʹሻ ሺ‫ ݏ‬൅ ʹሻ ͳ Ͷ

.

Solution: By using the left side of section 9 in table 1.1.1, we obtain: ݀ ௠ ‫ܨ‬ሺ‫ݏ‬ሻ ൌ ሺെͳሻ௠ ‫ ܨ‬ሺ௠ሻ ሺ‫ݏ‬ሻ ݀‫ ݏ‬௠

ͳ Ͷ ቑ ൠ ൌ ࣦ ିଵ ቐ ቑ ൅ ࣦെͳ ቐ ࣦെͳ ൜ ʹ ‫ ݏ‬െͶ ሺ‫ ݏ‬െ ʹሻ ሺ‫ ݏ‬൅ ʹሻ


ͳ ͳ ͳ ͳ ͳ ൠ ൌ ࣦ ିଵ ൜ ൠ െ ࣦെͳ ൜ ൠ ࣦെͳ ൜ ʹ ‫ ݏ‬െͶ ሺ‫ ݏ‬െ ʹሻ ሺ‫ ݏ‬൅ ʹሻ Ͷ Ͷ

Hence, ࣦሼ‫ ݔ‬ଶ ሺ‫ݔ‬ሻሽ ൌ ሺെͳሻଶ ‫ ܨ‬ሺଶሻ ሺ‫ݏ‬ሻ ൌ ‫ ܨ‬ሺଶሻ ሺ‫ݏ‬ሻ

ͳ

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ଵ ሺ௦ିଵሻమ

ଵ

ͳ െ ͳ Ͷ ൌ ൅ ሺ‫ ݏ‬െ ʹሻሺ‫ ݏ‬൅ ʹሻ ሺ‫ ݏ‬െ ʹሻ ሺ‫ ݏ‬൅ ʹሻ

ࣦ

ᇱ

ቁ ൌെ

Thus, ࣦሼ‫ ݁ݔ‬௫ ሽ ൌ ሺെͳሻଵ ‫ ܨ‬ሺଵሻ ሺ‫ݏ‬ሻ ൌ െ‫ ܨ‬ሺଵሻ ሺ‫ݏ‬ሻ ൌ െ ቀെ ሺ௦ିଵሻమ ቁ ൌ

ͳ Ͷ

ିଵ ൜

ଵ

௦ିଵ

െ

where ࣦሼ݂ሺ‫ݔ‬ሻሽ ൌ ‫ ܨ‬ሺ‫ݏ‬ሻ, and ݂ ሺ‫ݔ‬ሻ ൌ ሺ‫ݔ‬ሻ.

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Now, we need to find ‫ ܨ‬ሺଶሻ ሺ‫ݏ‬ሻ as follows: This means that we first need to find the laplace transform of ݂ ሺ‫ ݔ‬ሻ, and then we need to find the second derivative of the result from the laplace transform. ‫ ܨ‬ሺଶሻ ሺ‫ݏ‬ሻ ൌ ‫ ܨ‬ᇱᇱ ሺ‫ݏ‬ሻ ൌ ሺࣦሼ݂ሺ‫ݔ‬ሻሽሻᇱᇱ ൌ ሺࣦሼሺ‫ݔ‬ሻሽሻᇱᇱ ൌ ቀ ቀ

ିଶ௦ ሺ௦ మାଵሻమ

ᇱ

ቁ ൌ

ଵ ௦ మାଵ

ᇱᇱ

ቁ ൌ

ሺ௦ మ ାଵሻర

ࣦሼ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻሽ ൅ ࣦሼͷ‫ ݕ‬ሺଵሻ ሺ‫ ݔ‬ሻሽ ൅ ࣦሼ͸‫ݕ‬ሺ‫ݔ‬ሻሽ ൌ ࣦሼͳሽ ࣦሼ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻሽ ൅ ͷࣦሼ‫ ݕ‬ሺଵሻ ሺ‫ ݔ‬ሻሽ ൅ ͸ࣦሼ‫ݕ‬ሺ‫ݔ‬ሻሽ ൌ

ͳ ‫ݏ‬

ͳ

because ቀࣦሼͳሽ ൌ ‫ ݏ‬ቁ. ሺ‫ ݏ‬ଶ ܻሺ‫ݏ‬ሻ െ ‫ݕݏ‬ሺͲሻ െ ‫ ݕ‬ᇱ ሺͲሻሻ ൅ ͷሺ‫ܻݏ‬ሺ‫ݏ‬ሻ െ ‫ݕ‬ሺͲሻሻ ൅ ͸ܻሺ‫ݏ‬ሻ ൌ

ଵ ௦

from result 1.2.2. We substitute ‫ݕ‬ሺͲሻ ൌ Ͳ, and ‫Ͳ ݕ‬ሻ ൌ Ͳ

ሺ௦ మାଵሻర ିଶሺ௦ మାଵሻమା଼௦ మሺ௦ మ ାଵሻ

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ᇱሺ

ିଶሺ௦ మ ାଵሻమା଼௦ మ ሺ௦ మ ାଵሻ

Thus, ࣦሼ‫ ݔ‬ଶ ሺ‫ݔ‬ሻሽ ൌ


because it is given in the question itself. .

Example 1.5.8 Solve the following Initial Value Problem (IVP): ‫ ݕ‬ሺଶሻ ሺ‫ ݔ‬ሻ ൅ ͷ‫ ݕ‬ሺଵሻ ሺ‫ݔ‬ሻ ൅ ͸‫ݕ‬ሺ‫ݔ‬ሻ ൌ ͳ. Given ‫ݕ‬ሺͲሻ ൌ ‫ ݕ‬ᇱ ሺͲሻ ൌ Ͳ. Solution: ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ͷ‫ ݕ‬ሺଵሻ ሺ‫ݔ‬ሻ ൅ ͸‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ͳ is a linear differential equation of order 2. First, we need to find the domain for the solution of the above differential equation in other words we need to find for what values of ‫ ݔ‬the solution of the above differential equation holds. Therefore, we do the following: ͳ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ͷ‫ ݕ‬ሺଵሻ ሺ‫ݔ‬ሻ ൅ ͸‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ͳ Using definition 1.4.1, we also suppose the following: ܽଶ ሺ ‫ ݔ‬ሻ ൌ ͳ

ሺ‫ ݏ‬ଶ ܻሺ‫ݏ‬ሻ െ Ͳ െ Ͳሻ ൅ ͷሺ‫ܻݏ‬ሺ‫ݏ‬ሻ െ Ͳሻ ൅ ͸ܻሺ‫ݏ‬ሻ ൌ ‫ ݏ‬ଶ ܻሺ‫ݏ‬ሻ ൅ ͷ‫ܻݏ‬ሺ‫ݏ‬ሻ ൅ ͸ܻሺ‫ݏ‬ሻ ൌ

ଵ ௦

ଵ ௦

ଵ

ܻሺ‫ݏ‬ሻሺ‫ ݏ‬ଶ ൅ ͷ‫ ݏ‬൅ ͸ሻ ൌ ௦ ܻ ሺ‫ ݏ‬ሻ ൌ

‫ ݏ‬ሺ‫ ݏ‬ଶ

ͳ ͳ ൌ ൅ ͷ‫ ݏ‬൅ ͸ሻ ‫ݏ‬ሺ‫ ݏ‬൅ ͵ሻሺ‫ ݏ‬൅ ʹሻ

To find a solution, we need to find the inverse laplace transform as follows: ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ࣦ ିଵ ሼܻሺ‫ݏ‬ሻሽ ൌ ࣦ ିଵ ቄ

ͳ ቅ ‫ݏ‬ሺ‫ݏ‬൅͵ሻሺ‫ݏ‬൅ʹሻ

Since the numerator has a polynomial of degree 0 (‫ ݔ‬଴ ൌ ͳሻ, and the denominator a polynomial of degree

ܽଵ ሺ‫ ݔ‬ሻ ൌ ͷ

3, then this means the degree of numerator is less than

ܽ଴ ሺ ‫ ݔ‬ሻ ൌ ͸

the degree of denominator. Thus, in this case, we need

‫ ܭ‬ሺ‫ ݔ‬ሻ ൌ ͳ

to use the partial fraction as follows:


ͳ ܽ ܾ ܿ ൌ ൅ ൅ ‫ݏ‬ሺ‫ ݏ‬൅ ͵ሻሺ‫ ݏ‬൅ ʹሻ ‫ ݏ‬ሺ‫ ݏ‬൅ ͵ሻ ሺ‫ ݏ‬൅ ʹሻ

Now, to find the solution of the above differential equation, we need to take the laplace transform for both sides as follows

Now, we use the cover method. In the cover method, we cover the original, say ‫ݏ‬, and substitute ‫ ݏ‬ൌ Ͳ in ଵ ௦ሺ௦ାଷሻሺ௦ାଶሻ

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to find the value of ܽ. We cover the original,

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say ሺ‫ ݏ‬൅ ͵ሻ, and substitute ‫ ݏ‬ൌ െ͵ in ௦ሺ௦ାଷሻሺ௦ାଶሻ to find

ଵ

Example 1.5.10 Find ܷሺ‫ ݔ‬െ ͵ሻȁ‫ ݔ‬ൌ ͳ.

the value of ܾ. Then, we cover the original, say ሺ‫ ݏ‬൅ ʹሻ,

Solution: Since ‫ ݔ‬ൌ ͳ is between 0 and ൌ ͵ , then by


and substitute ‫ ݏ‬ൌ െʹ in

ଵ ௦ሺ௦ାଷሻሺ௦ାଶሻ

ଵ

ଵ

ଵ

଺

ଷ

ଶ

to find the value of

ܿ. Thus, ൌ , ܾ ൌ and ܿ ൌ െ . This implies that ͳ ‫ݏ‬ሺ‫ ݏ‬൅ ͵ሻሺ‫ ݏ‬൅ ʹሻ

ͳ ൌ͸ ൅ ‫ݏ‬

ͳ ͵ ሺ‫ ݏ‬൅ ͵ሻ

൅

ͳ െʹ

ࣦ

െͳ

ࣦ

ሺ‫ ݏ‬൅ ʹሻ

ܷሺ‫ ݔ‬െ ͵ሻȁ‫ ݔ‬ൌ ͳ ൌ ܷ ሺͳ െ ͹ሻ ൌ Ͳ. Example 1.5.11 Find ࣦሼܷሺ‫ ݔ‬െ ͵ሻሽ.

1.1.1, we obtain:

ͳ ͳ ͳ െʹ ͳ ͸ ͵ ିଵ ቋൌࣦ ቐ ൅ ቑ ൅ ‫ݏ‬ሺ‫ ݏ‬൅ ͵ሻሺ‫ ݏ‬൅ ʹሻ ‫ ݏ‬ሺ‫ ݏ‬൅ ͵ሻ ሺ‫ ݏ‬൅ ʹሻ

ͳ ͳ ͳ ͳ ͳ ൜ ൠ ൌ ࣦ ିଵ ൜ ൠ ൅ ࣦെͳ ൜ ൠ ሺ‫ ݏ‬൅ ͵ሻ ‫ݏ‬ ‫ݏ‬ሺ‫ ݏ‬൅ ͵ሻሺ‫ ݏ‬൅ ʹሻ ͸ ͵ ͳ ͳ ൠ െ ࣦെͳ ൜ ሺ‫ ݏ‬൅ ʹሻ ʹ ଵ

using definition 1.5.1, we obtain:

Solution: By using the left side of section 7 in table

Now, we need to do the following: ିଵ ቊ

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ࣦሼܷሺ‫ ݐ‬െ ܾሻሽ ൌ

݁ ି௕௦ ‫ݏ‬

Thus, ࣦሼܷሺ‫ ݔ‬െ ͵ሻሽ ൌ

௘ షయೞ ௦

.

Example 1.5.12 Find ࣦ ିଵ ቄ

௘ షమೞ ௦

ቅ.

Solution: By using the right side of section 7 in table 1.1.1, we obtain:

ଵ

ଵ

ଵ

ǡ ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ࣦ ିଵ ቄ௦ሺ௦ାଷሻሺ௦ାଶሻቅ ൌ ଺ ൅ ଷ ݁ ିଷ௫ െ ଶ ݁ ିଶ௫ . Definition 1.5.1 Given ܽ ൐ Ͳ. Unit Step Function is

ࣦ ିଵ ቊ

݁ ି௕௦ ቋ ൌ ܷሺ‫ ݐ‬െ ܾሻ ‫ݏ‬

a) ܷሺ‫ ݔ‬െ Ͳሻ ൌ ܷሺ‫ݔ‬ሻ ൌ ͳ for every Ͳ ൑ ‫ ݔ‬൑ λ.

݂݅ Ͳ ൑ ‫ ݔ‬൏ ʹ ݂݅ ʹ ൑ ‫ ݔ‬൏ λ ͵ ݂݅ ͳ ൑ ‫ ݔ‬൏ Ͷ ݂݅ Ͷ ൑ ‫ ݔ‬൏ ͳͲ Example 1.5.13 Given ݂ሺ‫ݔ‬ሻ ൌ ቐ െ‫ݔ‬ ሺ‫ ݔ‬൅ ͳሻ ݂݅ ͳͲ ൑ ‫ ݔ‬൏ λ Rewrite ݂ሺ‫ ݔ‬ሻ in terms of ܷ݊݅‫݊݋݅ݐܿ݊ݑܨ ݌݁ݐܵ ݐ‬.

b) ܷሺ‫ ݔ‬െ λሻ ൌ Ͳ for every Ͳ ൑ ‫ ݔ‬൑ λ.

Solution: To re-write ݂ሺ‫ ݔ‬ሻ in terms of

Ͳ defined as follows: ܷሺ‫ ݔ‬െ ܽሻ ൌ ൜ ͳ

݂݅ Ͳ ൑ ‫ ݔ‬൏ ܽ ݂݅ ܽ ൑ ‫ ݔ‬൏ λ

Result 1.5.1 Given ܽ ൒ Ͳ. Then, we obtain:

Thus, ࣦ ିଵ ቄ

௘ షమೞ ௦

Ͳ ͳ

ቅ ൌ ܷሺ‫ ݔ‬െ ʹሻ ൌ ൜

Example 1.5.9 Find ܷሺ‫ ݔ‬െ ͹ሻȁ‫ ݔ‬ൌ ͺ.

ܷ݊݅‫݊݋݅ݐܿ݊ݑܨ ݌݁ݐܵ ݐ‬, we do the following:

Solution: Since ‫ ݔ‬ൌ ͺ is between ܽ ൌ ͹ and λ, then by

݂ሺ‫ ݔ‬ሻ ൌ ͵൫ܷሺ‫ ݔ‬െ ͳሻ െ ܷሺ‫ ݔ‬െ Ͷሻ൯

using definition 1.5.1, we obtain:

൅ ሺെ‫ ݔ‬ሻ൫ܷሺ‫ ݔ‬െ Ͷሻ െ ܷሺ‫ ݔ‬െ ͳͲሻ൯

ܷሺ‫ ݔ‬െ ͹ሻȁ‫ ݔ‬ൌ ͺ ൌ ܷ ሺͺ െ ͹ሻ ൌ ͳ.

൅ ሺ‫ ݔ‬൅ ͳሻሺܷሺ‫ ݔ‬െ ͳͲሻ െ ͲሻǤ

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Now, we need to check our unit step functions as

Result 1.5.3 ࣦሼ݂ሺ‫ݔ‬ሻ ή ݄ሺ‫ݔ‬ሻሽ ് ࣦሼ݄ሺ‫ݔ‬ሻሽ ή ࣦሼ݂ሺ‫ݔ‬ሻሽ.

follows: We choose ‫ ݔ‬ൌ ͻ.

Example 1.5.15 Use definition 1.5.2 to find

݂ሺͻሻ ൌ ͵൫ܷ ሺͻ െ ͳሻ െ ܷሺͻ െ Ͷሻ൯

ࣦ൛‫׬‬଴ ‫݊݅ݏ‬ሺ߰ሻ݀߰ൟ.

௫

൅ ሺെͻሻ൫ܷሺͻ െ Ͷሻ െ ܷሺͻ െ ͳͲሻ൯

Solution: By using definition 1.5.2 and section 10 in

൅ ሺͻ ൅ ͳሻሺܷሺͻ െ ͳͲሻ െ Ͳሻ

table 1.1.1, we obtain:

݂ሺͻሻ ൌ ͵ሺͳ െ ͳሻ ൅ ሺെͻሻሺͳ െ Ͳሻ ൅ ሺͳͲሻሺͲ െ Ͳሻ ݂ሺͻሻ ൌ ͵ሺͲሻ ൅ ሺെͻሻሺͳሻ ൅ ሺͳͲሻሺͲሻ

௫

ࣦ ൝න ‫݊݅ݏ‬ሺ߰ሻ݀߰ൡ ൌ ࣦሼͳ ‫ כ‬ሺ‫ݔ‬ሻሽ ൌ ࣦሼͳሽ ή ࣦሼሺ‫ݔ‬ሻሽ ൌ ଴

݂ሺͻሻ ൌ Ͳ ൅ ሺെͻሻሺͳሻ ൅ Ͳ ൌ െͻǤ Thus, our unit step functions are correct. Example 1.5.14 Find ࣦሼ‫ܷݔ‬ሺ‫ ݔ‬െ ʹሻሽ. Solution: By using the upper side of section 6 in table 1.1.1, we obtain: ࣦሼ݄ሺ‫ݔ‬ሻܷሺ‫ ݔ‬െ ܾሻሽ ൌ ݁ ି௕௦ ࣦሼ݄ሺ‫ ݔ‬൅ ܾሻሽ

where ܾ ൌ ʹ, and ݄ሺ‫ݔ‬ሻ ൌ ‫ݔ‬. Hence, ࣦሼ‫ܷݔ‬ሺ‫ ݔ‬െ ʹሻሽ ൌ ݁ ିଶ௦ ࣦሼ݄ሺ‫ ݔ‬൅ ʹሻሽ ൌ ݁ ିଶ௦ ࣦሼ‫ ݔ‬൅ ʹሽ ൌ ଵ

ଶ

௦

௦

݁ ିଶ௦ ቀ మ ൅ ቁ.

Definition 1.5.2 Convolution, denoted by ‫כ‬ǡ is defined as follows: ௫

݂ ሺ‫ݔ‬ሻ ‫݄ כ‬ሺ‫ݔ‬ሻ ൌ න ݂ሺ߰ሻ݄ሺ‫ ݔ‬െ ߰ሻ݀߰ ଴

where ݂ሺ‫ ݔ‬ሻ and ݄ሺ‫ݔ‬ሻ are functions. (Note: do not confuse between multiplication and convolution). Result 1.5.2 ࣦሼ݂ሺ‫ݔ‬ሻ ‫݄ כ‬ሺ‫ݔ‬ሻሽ ൌ ࣦሼ݄ሺ‫ݔ‬ሻ ‫݂ כ‬ሺ‫ݔ‬ሻሽ ൌ ‫ܨ‬ሺ‫ݏ‬ሻ ή ‫ܪ‬ሺ‫ݏ‬ሻ where ݂ ሺ‫ ݔ‬ሻ and ݄ሺ‫ݔ‬ሻ are functions. (The proof

ൌ

ͳ ͳ ή ‫ ݏ‬ሺ‫ ݏ‬ଶ ൅ ͳሻ

ͳ ൅ ͳሻ

‫ݏ‬ሺ‫ ݏ‬ଶ

௫

ଵ

Thus, ࣦ൛‫׬‬଴ ‫݊݅ݏ‬ሺ߰ሻ݀߰ൟ ൌ ௦ሺ௦మାଵሻ. Definition 1.5.3 ݂ሺ‫ݔ‬ሻ is a periodic function on ሾͲǡ λሻ if ݂ሺ‫ݔ‬ሻ has a period ܲ such that ݂ ሺܾሻ ൌ ݂ሺܾ െ ܲሻ for every ܾ ൒ ܲ. Example 1.5.16 Given ݂ሺ‫ݔ‬ሻ is periodic on ሾͲǡ λሻ such that the first period of ݂ሺ‫ݔ‬ሻ is given by the following piece-wise continuous function: ͵ ݂݅ Ͳ ൑ ‫ ݔ‬൏ ʹ ൜ െʹ ݂݅ ʹ ൑ ‫ ݔ‬൏ ͺ a) Find the 8th period of this function. b) Suppose ܲ ൌ ͺ. Find ݂ሺͳͲሻ. c) Suppose ܲ ൌ ͺ. Find ݂ሺ͵Ͳሻ. Solution: Part a: By using section 13 in table 1.1.1, we obtain: ௉

ͳ න ݁ ି௦௧ ݂ ሺ‫ݔ‬ሻ݀‫ݔ‬ ࣦሼ݂ሺ‫ݔ‬ሻሽ ൌ ͳ െ ݁ ି௉௦ ଴

Since we need find the

8th

period, then this means that

ܲ ൌ ͺ, and we can apply what we got above as follows:

of this result left as an exercise 16 in section 1.7).

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଼

଼


ͳ ͳ න ݁ ି௦௧ ݂ሺ‫ݔ‬ሻ݀‫ ݔ‬ൌ න ݂ሺ‫ݔ‬ሻ݁ ି௦௧ ݀‫ݔ‬ ࣦሼ݂ሺ‫ݔ‬ሻሽ ൌ ͳ െ ݁ ି଼௦ ͳ െ ݁ ି଼௦

Example 1.5.17 Find ࣦሼߜሺ‫ ݐ‬൅ ͳʹሻሽ.

Using the given first period function, we obtain:

1.1.1, we obtain: ࣦሼߜሺ‫ ݐ‬െ ܾሻሽ ൌ ݁ ି௕௦

଴

଴

଼

ࣦሼ݂ሺ‫ݔ‬ሻሽ ൌ

Solution: By using the right side of section 12 in table Thus, ࣦሼߜሺ‫ ݐ‬൅ ͳʹሻሽ ൌ ݁ଵଶ௦ .

ͳ න ݂ሺ‫ݔ‬ሻ݁ ି௦௧ ݀‫ݔ‬ ͳ െ ݁ ି଼௦ ଴

ଶ

଼

ͳ ቎͵ න ݁ ି௦௧ ݀‫ ݔ‬െ ʹ න ݁ ି௦௧ ݀‫ݔ‬቏ ൌ ͳ െ ݁ ି଼௦ ଴

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ଶ

ͳ ͵ ‫ ݔ‬ൌ ʹ ʹ ି௦௧ ‫ ݔ‬ൌ ͺ ൨ ൤െ ݁ ି௦௧ ฬ ൅ ݁ ฬ ͳ െ ݁ ି଼௦ ‫ݏ‬ ‫ݔ‬ൌͲ ‫ݏ‬ ‫ݔ‬ൌʹ ͳ ͵ ʹ ൤െ ሺ݁ ିଶ௧ െ ͳሻ ൅ ሺ݁ ି଼௧ െ ݁ ିଶ௧ ሻ൨ ൌ ି଼௦ ͳെ݁ ‫ݏ‬ ‫ݏ‬ ൌ

1.6 Systems of Linear Equations Most of the materials of this section are taken from section 1.8 in my published book titled A First Course

Part b: By using definition 1.5.3, we obtain:

in Linear Algebra: Study Guide for the Undergraduate

݂ሺͳͲሻ ൌ ݂ሺͳͲ െ ͺሻ ൌ ݂ ሺʹሻ ൌ െʹ from the given first

Linear Algebra Course, First Edition1, because it is

period function.

very important to give a review from linear algebra

Part c: By using definition 1.5.3, we obtain:

about Cramer’s rule, and how some concepts of linear

݂ሺ͵Ͳሻ ൌ ݂ሺ͵Ͳ െ ͺሻ ൌ ݂ ሺʹʹሻǤ

algebra can be used to solve some problems in

݂ሺʹʹሻ ൌ ݂ሺʹʹ െ ͺሻ ൌ ݂ሺͳͶሻ.

differential equations. In this section, we discuss how

݂ሺͳͶሻ ൌ ݂ሺͳͶ െ ͺሻ ൌ ݂ ሺ͸ሻ ൌ െʹ from the given first

to use what we have learned from previous sections

period function.

such as initial value problems (IVP), and how to use

Definition 1.5.4 Suppose that ݅ ൐ Ͳ is fixed, and

Cramer’s rule to solve systems of linear equations.

ߜ ൏ ݆ ൏ ݅ is chosen arbitrary. Then, we obtain:

Definition 1.6.1 Given ݊ ൈ ݊ system of linear equations.

ߜ௝ ሺ‫ ݐ‬െ ݅ ሻ ൌ

Ͳ ‫ͳۓ‬ ‫݆ʹ۔‬ ‫Ͳە‬

݂݅ Ͳ ൑ ‫ ݐ‬൏ ሺ݅ െ ݆ሻ ݂݅ ሺ݅ െ ݆ሻ ൑ ‫ ݐ‬൏ ሺ݅ ൅ ݆ሻ ‫ ݐ‬൒ ሺ݅ ൅ ݆ሻ

ߜ௝ ሺ‫ ݐ‬െ ݅ ሻ is called Dirac Delta Function.

Let ൌ be the matrix form of the given system: ‫ݔ‬ଵ ܽଵ ‫ ݔ ۍ‬ଶ ‫ܽ ۍ ې‬ଶ ‫ې‬ ‫ۑ ێ ۑ ێ‬ ‫ ݔ ێ‬ଷ ‫ ۑ‬ൌ ‫ܽ ێ‬ଷ ‫ۑ‬ ‫ۑڭێ ۑڭێ‬ ‫ ݔ ۏ‬௡ ‫ܽ ۏ ے‬௡ ‫ے‬

Result 1.5.4 ߜ ሺ‫ ݐ‬െ ݅ ሻ ൌ ௜՜଴శ ߜ௝ ሺ‫ ݐ‬െ ݅ ሻ.

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The system has a unique solution if and only if ሺሻ ് Ͳ. Cramer’s Rule tells us how to find ‫ݔ‬ଵ ǡ ‫ݔ‬ଶ ǡ ǥ ǡ ‫ݔ‬௡ as follows: ͳ ͵ Let W = ൥ͳ ʹ ͹ Ͷ

Ͷ ͳ൩ Then, the solutions for the system of ͵

linear equations are: ܽଵ ͵ Ͷ ൥ ‫ͳ ʹ ڭ‬൩ ܽ௡ Ͷ ͵ ‫ݔ‬ଵ ൌ ሺሻ ͳ ܽଵ Ͷ ൥ͳ ‫ͳ ڭ‬൩ ͹ ܽ௡ ͵ ‫ݔ‬ଶ ൌ ሺሻ ͳ ͵ ܽଵ ൥ͳ ʹ ‫ ڭ‬൩ ͹ Ͷ ܽ௡ ‫ݔ‬ଷ ൌ ሺሻ


ͳ͵ ͹ ͳ͵ ቃ ቂ ቂ െͶ ͵ െͶ ‫ݔ‬ଵ ൌ ൌ ሺሻ ͹͸

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͹ ቃ ͵ ൌ ͸͹ ͹͸

ʹ ͳ͵ ʹ ͳ͵ ቃ ቂ ቃ ͳʹʹ ቂ െͳͲ െͶ െͳͲ െͶ ‫ݔ‬ଶ ൌ ൌ ൌ ሺሻ ͹͸ ͹͸ ଺଻

Thus, the solutions are ‫ݔ‬ଵ ൌ ଻଺ ‫ݔ‬ଶ ൌ

ଵଶଶ ଻଺

Ǥ

Example 1.6.2 Solve for ݊ሺ‫ݐ‬ሻ and ݉ሺ‫ݐ‬ሻ: ݊ᇱᇱ ሺ‫ݐ‬ሻ െ Ͷ݉ሺ‫ݐ‬ሻ ൌ Ͳ ǥ ǥ ǥ ǥ ǥ ǥ ‫ͳ ݊݋݅ݐܽݑݍܧ‬ ൜ ݊ሺ‫ݐ‬ሻ ൅ ʹ݉ᇱ ሺ‫ݐ‬ሻ ൌ ͷ݁ ଶ௧ ǥ ǥ ǥ Ǥ Ǥ ǥ ‫ʹ ݊݋݅ݐܽݑݍܧ‬ Given that ݊ሺͲሻ ൌ ͳǡ ݊ᇱ ሺͲሻ ൌ ʹǡ ݉ሺͲሻ ൌ ͳǤ Solution: First, we need to take the laplace transform of both sides for each of the above two equations. For ‫ͳ ݊݋݅ݐܽݑݍܧ‬ǣ We take the laplace transform of both sides: ࣦሼ݊ԢԢ ሺ‫ݐ‬ሻሽ ൅ ࣦሼെͶ݉ሺ‫ݐ‬ሻሽ ൌ ࣦሼͲሽ ࣦሼ݊ԢԢ ሺ‫ݐ‬ሻሽ െ Ͷࣦሼ݉ሺ‫ݐ‬ሻሽ ൌ ࣦሼͲሽ

ሺ‫ ݏ‬ଶ ܰሺ‫ݏ‬ሻ െ ‫݊ݏ‬ሺͲሻ െ ݊ᇱ ሺͲሻሻ െ Ͷ‫ܯ‬ሺ‫ݏ‬ሻ ൌ Ͳ

Example 1.6.1 Solve the following system of linear equations using Cramer’s Rule: ʹ‫ݔ‬ଵ ൅ ͹‫ݔ‬ଶ ൌ ͳ͵ ൜ െͳͲ‫ݔ‬ଵ ൅ ͵‫ݔ‬ଶ ൌ െͶ

Now, we substitute what is given in this question to

Solution: First of all, we write ʹ ൈ ʹ system in the form ൌ according to definition 1.6.1. ʹ ͹ ‫ݔ‬ଵ ͳ͵ ቂ ቃቂ ቃ ൌ ቂ ቃ െͳͲ ͵ ‫ݔ‬ଶ െͶ ʹ ͹ ቃ, then Since W in this form is ቂ െͳͲ ͵

Thus, ‫ ݏ‬ଶ ܰ ሺ‫ݏ‬ሻ െ Ͷ‫ܯ‬ሺ‫ݏ‬ሻ ൌ ‫ ݏ‬൅ ʹ.

ሺሻ ൌ ሺʹ ή ͵ሻ െ ൫͹ ή ሺെͳͲሻ൯ ൌ ͸ െ ሺെ͹Ͳሻ ൌ ͹͸ ് ͲǤ The solutions for this system of linear equations are:

46 M. Kaabar

obtain the following: ሺ‫ ݏ‬ଶ ܰሺ‫ݏ‬ሻ െ ‫ ݏ‬െ ʹሻ െ Ͷ‫ܯ‬ሺ‫ݏ‬ሻ ൌ Ͳ For ‫ʹ ݊݋݅ݐܽݑݍܧ‬ǣ We take the laplace transform of both sides: ࣦሼ݊ሺ‫ݐ‬ሻሽ ൅ ࣦሼʹ݉Ԣ ሺ‫ݐ‬ሻሽ ൌ ࣦሼͷ݁ʹ‫ ݐ‬ሽ ࣦሼ݊ሺ‫ݐ‬ሻሽ ൅ ʹࣦሼ݉Ԣ ሺ‫ݐ‬ሻሽ ൌ ͷࣦሼ݁ʹ‫ ݐ‬ሽ

ܰሺ‫ݏ‬ሻ ൅ ʹ൫‫ ܯݏ‬ሺ‫ݏ‬ሻ െ ݉ሺͲሻ൯ ൌ

ͷ ‫ݏ‬െʹ

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Now, we substitute what is given in this question to obtain the following: ͷ ܰሺ‫ݏ‬ሻ ൅ ʹሺ‫ܯݏ‬ሺ‫ݏ‬ሻ െ ͳሻ ൌ ‫ݏ‬െʹ ͷ ʹ‫ ݏ‬൅ ͳ ܰሺ‫ݏ‬ሻ ൅ ʹ‫ ܯݏ‬ሺ‫ݏ‬ሻ ൌ ൅ʹ ൌ ‫ݏ‬െʹ ‫ݏ‬െʹ Thus, ܰሺ‫ݏ‬ሻ ൅ ʹ‫ܯݏ‬ሺ‫ݏ‬ሻ ൌ

ଶ௦ାଵ ௦ିଶ

.


݊ᇱᇱ ሺ‫ݐ‬ሻ െ Ͷ݉ሺ‫ݐ‬ሻ ൌ Ͳ ՜ ݉ሺ‫ݐ‬ሻ ൌ Thus, ݉ሺ‫ݐ‬ሻ ൌ ݁ ଶ௧ Ǥ

ଵ଴

1. Find ࣦ ିଵ ቄሺ௦ିସሻర ቅ. ௦ାହ

2. Find ࣦ ିଵ ቄሺ௦ିଵሻమାଵ଺ቅ.

need to find ܰሺ‫ݏ‬ሻ and ‫ܯ‬ሺ‫ݏ‬ሻ as follows:

3. Find ࣦ ିଵ ቄሺ௦ାଷሻర ቅ.

‫ ݏ‬ଶ ܰሺ‫ݏ‬ሻ െ Ͷ‫ܯ‬ሺ‫ݏ‬ሻ ൌ ‫ ݏ‬൅ ʹ ቐ ʹ‫ ݏ‬൅ ͳ ܰሺ‫ݏ‬ሻ ൅ ʹ‫ܯݏ‬ሺ‫ݏ‬ሻ ൌ ‫ݏ‬െʹ

4. Find ࣦ ିଵ ቄଷ௦ିଵ଴ቅ.

Now, we use Cramer’s rule as follows:

6. Find ࣦ ିଵ ቄ௦మ ି଻௦ି଼ቅ.

௦ାହ ହ

ଶ

5. Find ࣦ ିଵ ቄ௦మ ି଺௦ାଵଷቅ.

‫ ݏ‬൅ ʹ െͶ ቉ ʹ‫ ݏ‬ଶ ൅ Ͷ‫ ݏ‬ͺ‫ ݏ‬൅ Ͷ ቈʹ‫ ݏ‬൅ ͳ ʹ‫ݏ‬ ൅ ‫ݏ‬െʹ ‫ݏ‬െʹ ͳ ܰ ሺ‫ ݏ‬ሻ ൌ ൌ ଶ െͶቃ ʹ‫ ݏ‬ଷ ൅ Ͷ ቂ‫ݏ‬ ͳ ʹ‫ݏ‬ ሺ‫ ݏ‬െ ʹሻሺʹ‫ ݏ‬ଶ ൅ Ͷ‫ݏ‬ሻ ൅ ͺ‫ ݏ‬൅ Ͷ ሺ‫ ݏ‬െ ʹሻሺʹ‫ ݏ‬ଷ ൅ Ͷሻ

ʹ‫ ݏ‬ଷ ൅ Ͷ‫ ݏ‬ଶ െ Ͷ‫ ݏ‬ଶ െ ͺ‫ ݏ‬൅ ͺ‫ ݏ‬൅ Ͷ ͳ ൌ ൌ ሺ‫ ݏ‬െ ʹሻሺʹ‫ ݏ‬ଷ ൅ Ͷሻ ‫ݏ‬െʹ Hence, ݊ሺ‫ݐ‬ሻ ൌ ࣦെͳ ሼܰሺ‫ݏ‬ሻሽ ൌ ࣦെͳ ቄ

ଵ

ቅ ൌ ݁ ଶ௧ .

௦ିଶ

Further, we can use one of the given equations to find ݉ሺ‫ݐ‬ሻ as follows: ݊ᇱᇱ ሺ‫ݐ‬ሻ െ Ͷ݉ሺ‫ݐ‬ሻ ൌ Ͳ We need to find the second derivative of ݊ሺ‫ݐ‬ሻ. ݊ᇱ ሺ‫ݐ‬ሻ ൌ ʹ݁ ଶ௧ . ݊ᇱᇱ ሺ‫ݐ‬ሻ ൌ Ͷ݁ ଶ௧ Ǥ Now, we can find ݉ሺ‫ݐ‬ሻ as follows:

48 M. Kaabar

݊ᇱᇱ ሺ‫ݐ‬ሻ Ͷ݁ ଶ௧ ൌ ൌ ݁ ଶ௧ Ǥ Ͷ Ͷ

1.7 Exercises

From what we got from Equation 1 and Equation 2, we

ൌ

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ହ

7. Solve the following Initial Value Problem (IVP): ʹ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ ͸‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ. Given ‫ݕ‬ሺͲሻ ൌ െͶ. 8. Solve the following Initial Value Problem (IVP): ‫ ݕ‬ᇱᇱ ሺ‫ݔ‬ሻ ൅ Ͷ‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ. Given ‫ݕ‬ሺͲሻ ൌ ʹ, and ‫ ݕ‬ᇱ ሺͲሻ ൌ Ͳ. ସ

9. Find ࣦ ିଵ ቄሺ௦ିଵሻమሺ௦ାଷሻቅ. ߨ

10. Find ࣦሼܷ ቀ‫ ݔ‬െ ʹ ቁ ሺ‫ݔ‬ሻሽ. 11. Find ࣦሼܷሺ‫ ݔ‬െ ʹሻ݁͵‫ ݔ‬ሽ. Ͷ 12. Given ݂ሺ‫ݔ‬ሻ ൌ ൜ ଶ௫ ݁

݂݅ Ͳ ൑ ‫ ݔ‬൏ ͵ ݂݅ ͵ ൑ ‫ ݔ‬൏ λ

Rewrite ݂ሺ‫ ݔ‬ሻ in terms of ܷ݊݅‫݊݋݅ݐܿ݊ݑܨ ݌݁ݐܵ ݐ‬. ௦ ݁െͶ‫ݔ‬

13. Find ࣦ ିଵ ቄ ௦మ ାସ ቅ.

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14. Solve the following Initial Value Problem (IVP): ‫ ݕ‬ᇱᇱ ሺ‫ ݔ‬ሻ൅͵‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ െ Ͷ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݂ሺͲሻ. Given ͳ ݂ሺ‫ݔ‬ሻ ൌ ቄ Ͳ

݂݅ Ͳ ൑ ‫ ݔ‬൏ ͵ ܱ‫݁ݏ݅ݓݎ݄݁ݐ‬

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Chapter 2 Systems of Homogeneous

‫ݕ‬ሺͲሻ ൌ ‫ ݕ‬ᇱ ሺͲሻ ൌ Ͳ. 15. Solve the following Initial Value Problem (IVP): ‫ ݕ‬ᇱᇱ ሺ‫ ݔ‬ሻ൅͹‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ െ ͺ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݂ሺ‫ݔ‬ሻ where ͵ ݂ሺ‫ݔ‬ሻ ൌ ൜ െʹ


Linear Differential Equations (HLDE)

݂݅ Ͳ ൑ ‫ ݔ‬൏ ͷ ݂݅ ͷ ൑ ‫ ݔ‬൏ λ

‫ݕ‬ሺͲሻ ൌ ‫ ݕ‬ᇱ ሺͲሻ ൌ Ͳ.

In this chapter, we start introducing the homogeneous

16. Prove result 1.5.2.

linear differential equations (HLDE) with constant

17. Solve the following Initial Value Problem (IVP):

coefficients. In addition, we discuss how to find the

‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ൅‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ܷ ሺ‫ ݔ‬െ ͵ሻǤ Given ‫ݕ‬ሺͲሻ ൌ ‫ ݕ‬ᇱ ሺͲሻ ൌ ‫ ݕ‬ᇱᇱ ሺͲሻ ൌ

general solution of HLDE. At the end of this chapter,

Ͳ.

we introduce a new method called Undetermined Coefficient Method.

௫

18. Find ࣦ൛‫׬‬଴ ܿ‫ݏ݋‬ሺʹ߰ሻ݁ሺ͵‫ݔ‬൅ʹటሻ ݀߰ൟ. ଶ௦

19. Find ࣦ ିଵ ቄሺ௦మାସሻమቅ. 20. Solve for ‫ݎ‬ሺ‫ݐ‬ሻ and ݇ሺ‫ݐ‬ሻ: ‫ ݎ‬ᇱ ሺ‫ݐ‬ሻ െ ʹ݇ ሺ‫ݐ‬ሻ ൌ Ͳ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ Ǥ ‫ͳ ݊݋݅ݐܽݑݍܧ‬ ൜ ᇱᇱ ‫ ݎ‬ሺ‫ݐ‬ሻ െ ʹ݇ ᇱ ሺ‫ݐ‬ሻ ൌ Ͳ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ‫ʹ ݊݋݅ݐܽݑݍܧ‬ Given that ݇ሺͲሻ ൌ ͳǡ ‫ ݎ‬ᇱ ሺͲሻ ൌ ʹǡ ‫ݎ‬ሺͲሻ ൌ ͲǤ 21. Solve for ‫ݓ‬ሺ‫ݐ‬ሻ and ݄ሺ‫ݐ‬ሻ: ‫ݐ‬

2.1 HLDE with Constant Coefficients In this section, we discuss how to find the general solution

of

the

homogeneous

linear

differential

‫ ݓ‬ሺ‫ݐ‬ሻ െ න ݄ሺ߰ሻ݀߰ ൌ ͳ ǥ ǥ ǥ ǥ ǥ Ǥ ‫ͳ ݊݋݅ݐܽݑݍܧ‬ ൞

equations (HLDE) with constant coefficients.

‫ ݓ‬ᇱᇱ ሺ‫ݐ‬ሻ ൅ ݄ᇱ ሺ‫ݐ‬ሻ ൌ Ͷ ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ‫ʹ ݊݋݅ݐܽݑݍܧ‬ Given that ‫ݓ‬ሺͲሻ ൌ ͳǡ ‫ ݓ‬ᇱ ሺͲሻ ൌ ݄ሺͲሻ ൌ ͲǤ

To give an introduction about HLDE, it is important to

Ͳ

22. Find ݂ሺ‫ݐ‬ሻ such that ݂ሺ‫ݐ‬ሻ ൌ ݁

ିଷ௦

൅

௧ ‫׬‬଴

݂ሺ߰ሻ݀߰.

(Hint: Use laplace transform to solve this problem)

50 M. Kaabar

start with the definition of homogeneous system. *Definition 2.1.1 Homogeneous System is defined as a ݉ ൈ ݊ system of linear equations that has all zero constants. (i.e. the following is an example of

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ʹܽ ൅ ܾ െ ܿ ൅ ݀ ൌ Ͳ homogeneous system): ൝͵ܽ ൅ ͷܾ ൅ ͵ܿ ൅ Ͷ݀ ൌ Ͳ െܾ ൅ ܿ െ ݀ ൌ Ͳ *Definition 2.1.1 is taken from section 3.1 in my published book titled A First Course in Linear Algebra:


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Thus, ͵‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻെʹ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ ͹‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳ is a homogeneous linear differential equation of order 3. Example 2.1.3 Describe the following differential equation: ͵‫ ݕ‬ሺଶሻ ሺ‫ ݔ‬ሻെʹ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ͳʹ.

Study Guide for the Undergraduate Linear Algebra Course, First Edition1.

Solution: Since the above differential equation has a

Example 2.1.1 Describe the following differential

nonzero constant, then according to definition 2.1.1, it

ᇱᇱ ሺ

ᇱ

equation: ‫ ݔ ݕ‬ሻ൅͵‫ ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ.

is a non-homogeneous differential equation. In

Solution: Since the above differential equation has a

addition, it is linear because the dependent variable ‫ݕ‬

zero constant, then according to definition 2.1.1, it is a

and all its derivatives are to the power 1. For the order

homogeneous differential equation. In addition, it is

of this non-homogeneous differential equation, since

linear because the dependent variable ‫ ݕ‬and all its

the highest derivative is 2, then the order is 2.

derivatives are to the power 1. For the order of this

Thus, ͵‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻെʹ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ͳʹ is a non-homogeneous

homogeneous differential equation, since the highest

linear differential equation of order 2.

derivative is 2, then the order is 2. Thus,

Example 2.1.4 Find the general solution the following Initial Value Problem (IVP): ʹ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ Ͷ‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ. Given: ‫ ݕ‬ᇱ ሺͲሻ ൌ ͳǤ (Hint: Use the concepts of section 1.4)

‫ݕ‬

ᇱᇱ ሺ

ᇱ

‫ ݔ‬ሻ൅͵‫ ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ is a homogeneous linear differential

equation of order 2.

derivatives are to the power 1. For the order of this

Solution: ʹ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ Ͷ‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ is a homogeneous linear differential equation of order 1. First, we need to find the domain for the solution of the above differential equation in other words we need to find for what values of ‫ ݔ‬the solution of the above differential equation holds. Therefore, we do the following: ʹ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ Ͷ‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ Using definition 1.4.1, we also suppose the following: ܽଵ ሺ‫ ݔ‬ሻ ൌ ʹ

homogeneous differential equation, since the highest

ܽ଴ ሺ ‫ ݔ‬ሻ ൌ Ͷ

derivative is 3, then the order is 3.

‫ ܭ‬ሺ‫ ݔ‬ሻ ൌ Ͳ

Example 2.1.2 Describe the following differential equation: ͵‫ ݕ‬ሺଷሻ ሺ‫ ݔ‬ሻെʹ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ ͹‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ. Solution: Since the above differential equation has a zero constant, then according to definition 2.1.1, it is a homogeneous differential equation. In addition, it is linear because the dependent variable ‫ ݕ‬and all its


52 M. Kaabar

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Now, to find the solution of the above differential

Result 2.1.2 Assume that ‫ݕ‬ଵ ൌ ݁ ௞భ ௫ , ‫ݕ‬ଶ ൌ ݁ ௞మ ௫ , …,

equation, we need to take the laplace transform for

‫ݕ‬௡ ൌ ݁ ௞೙௫ are independent if and only if ݇ଵ ǡ ݇ଶ ǡ ǥ ǡ ݇௡ are


distinct real numbers.

ᇱ

ࣦሼʹ‫ ݕ‬ሺ‫ݔ‬ሻሽ ൅ ࣦሼͶ‫ݕ‬ሺ‫ݔ‬ሻሽ ൌ ࣦሼͲሽ

Example 2.1.5 Given ‫ ݕ‬ᇱᇱ ሺ‫ ݔ‬ሻ െ Ͷ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳǡ

ʹࣦሼ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻሽ ൅ Ͷࣦሼ‫ݕ‬ሺ‫ݔ‬ሻሽ ൌ Ͳ because (ࣦሼͲሽ ൌ Ͳ).

‫ݕ‬ሺͲሻ ൌ Ͷǡ ‫ ݕ‬ᇱ ሺͲሻ ൌ ͳͲǤ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ.

ʹሺ‫ܻݏ‬ሺ‫ݏ‬ሻ െ ‫ ݕ‬ᇱ ሺͲሻሻ ൅ Ͷܻሺ‫ݏ‬ሻ ൌ Ͳ from result 1.2.2.

(Hint: Use results 2.1.1 and 2.1.2)

ᇱ

ʹ‫ܻݏ‬ሺ‫ݏ‬ሻ െ ʹ‫ ݕ‬ሺͲሻ ൅ Ͷܻሺ‫ݏ‬ሻ ൌ Ͳ

Solution: ‫ ݕ‬ᇱᇱ ሺ‫ ݔ‬ሻ െ Ͷ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳ is a homogeneous linear

ܻሺ‫ݏ‬ሻሺʹ‫ ݏ‬൅ Ͷሻ ൌ ʹ‫ ݕ‬ᇱ ሺͲሻ

differential equation of order 2. In this example, we

ܻሺ‫ݏ‬ሻሺʹ‫ ݏ‬൅ Ͷሻ ൌ ʹሺͳሻ

will use a different approach from example 2.1.4

ʹ ͳ ʹ ൌ ൌ ܻ ሺ‫ ݏ‬ሻ ൌ ʹ‫ ݏ‬൅ Ͷ ʹሺ‫ ݏ‬൅ ʹሻ ‫ ݏ‬൅ ʹ

(laplace transform approach) to solve it. Since

To find a solution, we need to find the inverse laplace transform as follows: ͳ

‫ݕ‬ሺ‫ݔ‬ሻ ൌ ࣦ ିଵ ሼܻሺ‫ݏ‬ሻሽ ൌ ࣦ ିଵ ቄ‫ݏ‬൅ʹቅ ൌ ݁ ିଶ௫ Ǥ

Thus, the general solution ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ܿ݁ ିଶ௫ ǡ for some

‫ ݕ‬ᇱᇱ ሺ‫ ݔ‬ሻ െ Ͷ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳ is HLDE with constant coefficients, then we will do the following: Let ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ , we need to find ݇. First of all, we will find the first and second derivatives as follows:

constant ܿ. Here, ܿ ൌ ͳ.

‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫

Result 2.1.1 Assume that ݉ଵ ‫ ݕ‬ሺ௡ሻ ሺ‫ݔ‬ሻ ൅ ݉ଶ ‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ is a

‫ ݕ‬ᇱᇱ ሺ‫ݔ‬ሻ ൌ ݇ ଶ ݁ ௞௫

homogeneous linear differential equation of order ݊

Now, we substitute ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ and ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫ in

with constant coefficients ݉ଵ and ݉ଶ . Then,

‫ ݕ‬ᇱᇱ ሺ‫ ݔ‬ሻ െ Ͷ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳ as follows:

a) ݉ଵ ‫ ݕ‬ሺ௡ሻ ሺ‫ݔ‬ሻ ൅ ݉ଶ ‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ must have exactly ݊

݇ ଶ ݁ ௞௫ െ Ͷ݁ ௞௫ ൌ Ͳ

independent solutions, say ݂ଵ ሺ‫ ݔ‬ሻǡ ݂ଶ ሺ‫ ݔ‬ሻǡ ǥ ǡ ݂௡ ሺ‫ ݔ‬ሻǤ b) Every solution of ݉ଵ ‫ݕ‬

ሺ௡ሻ

ሺ‫ݔ‬ሻ ൅ ݉ଶ ‫ݕ‬ሺ‫ݔ‬ሻ ൌ Ͳ is of

݁ ௞௫ ሺ݇ ଶ െ Ͷሻ ൌ Ͳ ݁ ௞௫ ሺ݇ െ ʹሻሺ݇ ൅ ʹሻ ൌ Ͳ

the form: ܿଵ ݂ଵ ሺ‫ ݔ‬ሻ ൅ ܿଶ ݂ଶ ሺ‫ ݔ‬ሻ ൅ ‫ ڮ‬൅ ܿ௡ ݂௡ ሺ‫ ݔ‬ሻ, for

Thus, ݇ ൌ ʹ ‫ ݇ ݎ݋‬ൌ െʹ. Then, we use our values to

some constants ܿଵ ǡ ܿଶ ǡ ǥ ǡ ܿ௡ .

substitute ݇ in our assumption which is ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ : at ݇ ൌ ʹ, ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ଶ௫

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at ݇ ൌ െʹ, ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ିଶ௫

constant coefficients, then we will do the following: Let

Hence, using result 2.1.1, the general solution for ‫ݕ‬ሺ‫ݔ‬ሻ

‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ , we need to find ݇.

is: ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൌ ܿଵ ݁ ଶ௫ ൅ ܿଶ ݁ ିଶ௫ , for some ܿଵ ǡ ܿଶ ‫ א‬Ը. (Note:

First of all, we will find the first, second, and third

݄‫ ݋݉݋‬denotes to homogeneous). Now, we need to find

derivatives as follows: ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫

the values of ܿଵ ܿଶ as follows: at ‫ ݔ‬ൌ Ͳ,

‫ݕ‬௛௢௠௢ ሺͲሻ ൌ ܿଵ ݁ ଶሺ଴ሻ ൅ ܿଶ ݁ ିଶሺ଴ሻ

‫ ݕ‬ᇱᇱ ሺ‫ݔ‬ሻ ൌ ݇ ଶ ݁ ௞௫

‫ݕ‬௛௢௠௢ ሺͲሻ ൌ ܿଵ ݁ ଴ ൅ ܿଶ ݁ ଴

‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൌ ݇ ଷ ݁ ௞௫

‫ݕ‬௛௢௠௢ ሺͲሻ ൌ ܿଵ ൅ ܿଶ

Now, we substitute ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫ , ‫ ݕ‬ᇱᇱ ሺ‫ݔ‬ሻ ൌ ݇ ଶ ݁ ௞௫ , and

Since ‫ݕ‬ሺͲሻ ൌ Ͷ, then ܿଵ ൅ ܿଶ ൌ Ͷ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺͳሻ

‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൌ ݇ ଷ ݁ ௞௫ in ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ െ ͷ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ͸‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൌ Ͳ as

‫ݕ‬௛௢௠௢ ᇱ ሺ‫ ݔ‬ሻ ൌ ʹܿଵ ݁ ଶ௫ െ ʹܿଶ ݁ ିଶ௫ at ‫ ݔ‬ൌ Ͳ,

‫ݕ‬௛௢௠௢ ᇱ ሺͲሻ ൌ ʹܿଵ ݁ ଶሺ଴ሻ െ ʹܿଶ ݁ ିଶሺ଴ሻ

follows: ݇ ଷ ݁ ௞௫ െ ͷ݇ ଶ ݁ ௞௫ ൅ ͸݇݁ ௞௫ ൌ Ͳ

‫ݕ‬௛௢௠௢ ᇱ ሺͲሻ ൌ ʹܿଵ ݁ ଴ െ ʹܿଶ ݁ ଴

݁ ௞௫ ሺ݇ ଷ െ ͷ݇ ଶ ൅ ͸݇ሻ ൌ Ͳ

‫ݕ‬௛௢௠௢ ᇱ ሺͲሻ ൌ ʹܿଵ െ ʹܿଶ

݁ ௞௫ ሺ݇ሺ݇ ଶ െ ͷ݇ ൅ ͸ሻሻ ൌ Ͳ

Since ‫ ݕ‬ᇱ ሺͲሻ ൌ ͳͲ, then ʹܿଵ െ ʹܿଶ ൌ ͳͲ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺʹሻ

݁ ௞௫ ሺ݇ሺ݇ െ ʹሻሺ݇ െ ͵ሻሻ ൌ Ͳ

From ሺͳሻ and ሺʹሻ, ܿଵ ൌ ͶǤͷ and ܿଶ ൌ െͲǤͷ.

Thus, ݇ ൌ Ͳǡ ݇ ൌ ʹ ‫ ݇ ݎ݋‬ൌ ͵. Then, we use our values to

Thus, the general solution is:

substitute ݇ in our assumption which is ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ :

‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൌ ͶǤͷ݁ ଶ௫ െ ͲǤͷ݁ ିଶ௫ .

at ݇ ൌ Ͳ, ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ሺ଴ሻ௫ ൌ ͳ

Example 2.1.6 Given ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ െ ͷ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ͸‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ͲǤ

at ݇ ൌ ʹ, ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ଶ௫

Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: Use results

at ݇ ൌ ͵, ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ ൌ ݁ ଷ௫

2.1.1 and 2.1.2, and in this example, no need to find

Thus, using result 2.1.1, the general solution for ‫ݕ‬ሺ‫ݔ‬ሻ

the values of ܿଵ ǡ ܿଶ ǡ ܿଷ )

is: ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൌ ܿଵ ൅ ܿଶ ݁ ଶ௫ ൅ ܿଷ ݁ ଷ௫ , for some ܿଵ ǡ ܿଶ ǡ ܿଷ ‫ א‬Ը.

Solution: ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ െ ͷ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ͸‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൌ Ͳ is a

(Note: ݄‫ ݋݉݋‬denotes to homogeneous).

homogeneous linear differential equation of order 3.

Example 2.1.7 Given ‫ ݕ‬ሺହሻ ሺ‫ݔ‬ሻ െ ‫ ݕ‬ሺସሻ ሺ‫ݔ‬ሻ െ ʹ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൌ ͲǤ

Since ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ െ ͷ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ͸‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ Ͳ is HLDE with

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Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: Use results

at ݇ ൌ Ͳ, ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ሺ଴ሻ௫ ൌ ͳ

2.1.1 and 2.1.2, and in this example, no need to find

at ݇ ൌ Ͳ, ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ሺ଴ሻ௫ ൌ ͳ ή ‫ݔ‬

the values of ܿଵ ǡ ܿଶ ǡ ܿଷ ǡ ܿସ ܿହ)

at ݇ ൌ Ͳ, ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ ൌ ݁ ሺ଴ሻ௫ ൌ ͳ ή ‫ ݔ‬ଶ

Solution: ‫ ݕ‬ሺହሻ ሺ‫ݔ‬ሻ െ ‫ ݕ‬ሺସሻ ሺ‫ݔ‬ሻ െ ʹ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൌ Ͳ is a

because ݇ ଷ ൌ ܵ‫݊ܽ݌‬ሼͳǡ ‫ݔ‬ǡ ‫ ݔ‬ଶ ሽ ൌ Ͳ (Note: ܵ‫݊ܽ݌‬ሼͳǡ ‫ݔ‬ǡ ‫ ݔ‬ଶ ሽ ൌ Ͳ

homogeneous linear differential equation of order 5.

means ܽ ή ͳ ൅ ܾ ή ‫ ݔ‬൅ ܿ ή ‫ ݔ‬ଶ ൌ Ͳ)

Since ‫ ݕ‬ሺହሻ ሺ‫ݔ‬ሻ െ ‫ ݕ‬ሺସሻ ሺ‫ ݔ‬ሻ െ ʹ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൌ Ͳ is HLDE with

In other words ܵ‫݊ܽ݌‬ሼͳǡ ‫ݔ‬ǡ ‫ ݔ‬ଶ ሽ is the set of all linear

constant coefficients, then we will do the following: Let

combinations of ͳ, ‫ݔ‬, and ‫ ݔ‬ଶ .

‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ , we need to find ݇.

at ݇ ൌ ʹ, ‫ݕ‬ସ ሺ‫ ݔ‬ሻ ൌ ݁ ଶ௫

First of all, we will find the first, second, third, fourth,

at ݇ ൌ െͳ, ‫ݕ‬ହ ሺ‫ ݔ‬ሻ ൌ ݁ ି௫

and fifth derivatives as follows:


‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫

is: ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൌ ܿଵ ൅ ܿଶ ‫ ݔ‬൅ ܿଷ ‫ ݔ‬ଶ ൅ܿସ ݁ ଶ௫ ൅ ܿହ ݁ ି௫ , for some

‫ ݕ‬ᇱᇱ ሺ‫ݔ‬ሻ ൌ ݇ ଶ ݁ ௞௫

ܿଵ ǡ ܿଶ ǡ ܿଷ ǡ ܿସ ǡ ܿହ ‫ א‬Ը. (Note: ݄‫ ݋݉݋‬denotes to

‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൌ ݇ ଷ ݁ ௞௫

homogeneous).

‫ݕ‬

ሺସሻ

ସ ௞௫

ሺ‫ݔ‬ሻ ൌ ݇ ݁

‫ ݕ‬ሺହሻ ሺ‫ݔ‬ሻ ൌ ݇ ହ ݁ ௞௫

Example 2.1.8 Given ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ଷ௫ ǡ ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ିଷ௫ ǡ ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ ൌ ݁ ௫ Ǥ Are ‫ݕ‬ଵ ሺ‫ݔ‬ሻǡ ‫ݕ‬ଷ ሺ‫ ݔ‬ሻǡ ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ independent?

Now, we substitute ‫ ݕ‬ሺହሻ ሺ‫ݔ‬ሻ ൌ ݇ ହ ݁ ௞௫ , ‫ ݕ‬ሺସሻ ሺ‫ݔ‬ሻ ൌ ݇ ସ ݁ ௞௫ ,

Solution: We cannot write ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ as a linear

and ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൌ ݇ ଷ ݁ ௞௫ in ‫ ݕ‬ሺହሻ ሺ‫ݔ‬ሻ െ ‫ ݕ‬ሺସሻ ሺ‫ݔ‬ሻ െ ʹ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൌ Ͳ

combination of ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ and ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ as follows:

as follows:

݁ ௫ ് ሺ‫ݎܾ݁݉ݑܰ ݀݁ݔ݅ܨ‬ሻ ή ݁ ଷ௫ ൅ ሺ‫ݎܾ݁݉ݑܰ ݀݁ݔ݅ܨ‬ሻ ή ݁ ିଷ௫ ݇ ହ ݁ ௞௫ െ ݇ ସ ݁ ௞௫ െ ʹ݇ ଷ ݁ ௞௫ ൌ Ͳ

Thus, ‫ݕ‬ଵ ሺ‫ ݔ‬ሻǡ ‫ݕ‬ଷ ሺ‫ݔ‬ሻǡ ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ are independent.

݁ ௞௫ ሺ݇ ହ െ ݇ ସ െ ʹ݇ ଷ ሻ ൌ Ͳ

Example 2.1.9 Given ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ሺ௫ାଷሻ ǡ ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ଷ ǡ

݁ ௞௫ ሺ݇ ଷ ሺ݇ ଶ െ ݇ െ ʹሻሻ ൌ Ͳ

‫ݕ‬ଷ ሺ‫ ݔ‬ሻ ൌ ݁ ௫ Ǥ Are ‫ݕ‬ଵ ሺ‫ݔ‬ሻǡ ‫ݕ‬ଷ ሺ‫ ݔ‬ሻǡ ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ independent?

݁ ௞௫ ሺ݇ ଷ ሺ݇ െ ʹሻሺ݇ ൅ ͳሻሻ ൌ Ͳ

Solution: We can write ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ as a linear combination of

Thus, ݇ ൌ Ͳǡ ݇ ൌ Ͳǡ ݇ ൌ Ͳǡ ݇ ൌ ʹ ‫ ݇ ݎ݋‬ൌ െͳ. Then, we use

‫ݕ‬ଶ ሺ‫ ݔ‬ሻ and ‫ݕ‬ଷ ሺ‫ݔ‬ሻ as follows:

our values to substitute ݇ in our assumption which is

݁ ሺ௫ାଷሻ ൌ ݁ ௫ ή ݁ ଷ . Thus, ‫ݕ‬ଵ ሺ‫ ݔ‬ሻǡ ‫ݕ‬ଷ ሺ‫ ݔ‬ሻǡ ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ are

‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ :

dependent (not independent).

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Let ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ , we need to find ݇.

2.2 Method of Undetermined Coefficients In this section, we discuss how to use what we have learned from section 2.1 to combine it with what we

First of all, we will find the first, second, third, fourth, and fifth derivatives as follows: ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫ Now, we substitute ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫ in ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳ as follows: ݇݁ ௞௫ ൅ ͵݁ ௞௫ ൌ Ͳ

will learn from section 2.2 in order to find the general solution using a method known as undetermined coefficients method. In this method, we will find a general solution consisting of homogeneous solution and particular solution together. We give the following examples to introduce the undetermined coefficient method. Example 2.2.1 Given ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ‫ݔ‬ǡ ‫ݕ‬ሺͲሻ ൌ ͳǤ Find the general solution for ‫ݕ‬ሺ‫ݔ‬ሻ. (Hint: No need to find the value of ܿଵ ) Solution: Since ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ‫ ݔ‬does not have a constant coefficient, then we need to use the undetermined coefficients method as follows: Step 1: We need to find the homogeneous solution by letting ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ݔ‬ሻ equal to zero as follows: ᇱሺ

‫ ݔ ݕ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳ. Now, it is a homogeneous linear differential equation of order 1. Since ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳ is a HLDE with constant coefficients, then we will do the following:

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݁ ௞௫ ሺ݇ ൅ ͵ሻ ൌ Ͳ Thus, ݇ ൌ െ͵. Then, we use our value to substitute ݇ in our assumption which is ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ : at ݇ ൌ െ͵, ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ሺିଷሻ௫ ൌ ݁ ିଷ௫ Thus, using result 2.1.1, the general solution for ‫ݕ‬ሺ‫ݔ‬ሻ is: ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൌ ܿଵ ݁ ିଷ௫ , for some ܿଵ ‫ א‬Ը. (Note: ݄‫݋݉݋‬ denotes to homogeneous). Step 2: We need to find the particular solution as follows: Since ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ equals ‫ݔ‬, then the particular solution should be in the following form: ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ൌ ܽ ൅ ܾ‫ ݔ‬because ‫ ݔ‬is a polynomial of the first degree, and the general form for first degree polynomial is ܽ ൅ ܾ‫ݔ‬. Now, we need to find ܽ and ܾ as follows: ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ᇱ ሺ‫ ݔ‬ሻ ൌ ܾ We substitute ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ᇱ ሺ‫ ݔ‬ሻ ൌ ܾ in ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ‫ݔ‬. ܾ ൅ ͵ሺܽ ൅ ܾ‫ ݔ‬ሻ ൌ ‫ݔ‬ ܾ ൅ ͵ܽ ൅ ͵ܾ‫ ݔ‬ൌ ‫ݔ‬

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at ‫ ݔ‬ൌ Ͳ, we obtain: ܾ ൅ ͵ܽ ൅ ሺ͵ܾሻሺͲሻ ൌ Ͳ

Since ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳ is a HLDE with constant

ܾ ൅ ͵ܽ ൅ Ͳ ൌ Ͳ

coefficients, then we will do the following:

ܾ ൅ ͵ܽ ൌ Ͳ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺͳሻ


at ‫ ݔ‬ൌ ͳ, we obtain: ܾ ൅ ͵ܽ ൅ ሺ͵ܾሻሺͳሻ ൌ ͳ

First of all, we will find the first, second, third, fourth,

ܾ ൅ ͵ܽ ൅ ͵ܾ ൌ ͳ

and fifth derivatives as follows:

Ͷܾ ൅ ͵ܽ ൌ ͳ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺʹሻ ଵ

ଵ

From ሺͳሻ and ሺʹሻ, we get: ܽ ൌ െ ଽ and ܾ ൌ ଷ . ଵ

‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫ Now, we substitute ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫ in ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳ as follows:

ଵ

Thus, ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ൌ െ ൅ ‫ݔ‬. ଽ ଷ Step 3: We need to find the general solution as follows: ‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ ݔ‬ሻ ൌ ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൅ ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ଵ

ଵ

ଽ

ଷ

Thus, ‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ ݔ‬ሻ ൌ ܿଵ ݁ ିଷ௫ ൅ ቀെ ൅ ‫ݔ‬ቁ.

݇݁ ௞௫ ൅ ͵݁ ௞௫ ൌ Ͳ ݁ ௞௫ ሺ݇ ൅ ͵ሻ ൌ Ͳ Thus, ݇ ൌ െ͵. Then, we use our value to substitute ݇ in our assumption which is ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ :

Example 2.2.2 Given ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ݁ ିଷ௫ ǡ ‫ݕ‬ሺͲሻ ൌ ͳǤ

at ݇ ൌ െ͵, ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ሺିଷሻ௫ ൌ ݁ ିଷ௫

Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: No need to


find the value of ܿଵ )

is: ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൌ ܿଵ ݁ ିଷ௫ , for some ܿଵ ‫ א‬Ը. (Note: ݄‫݋݉݋‬

Solution: In this example, we will have the same

denotes to homogeneous).

homogeneous solution as we did in example 2.2.1 but

Step 2: We need to find the particular solution as

the only difference is the particular solution. We will

follows: Since ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ equals ݁ ିଷ௫ , then the

repeat some steps in case you did not read example

particular solution should be in the following form:

ᇱሺ

2.2.1. Since ‫ ݔ ݕ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ݁

ିଷ௫

does not have a

‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ൌ ሺܽ݁ ିଷ௫ ሻ‫ݔ‬

constant coefficient, then we need to use the

Now, we need to find ܽ as follows:

undetermined coefficients method as follows:

‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ᇱ ሺ‫ ݔ‬ሻ ൌ ሺܽ݁ ିଷ௫ ሻ െ ͵ሺܽ‫ି ݁ݔ‬ଷ௫ ሻ

Step 1: We need to find the homogeneous solution by letting ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ݔ‬ሻ equal to zero as follows: ᇱሺ

‫ ݔ ݕ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ Ͳ. Now, it is a homogeneous linear differential equation of order 1.

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We substitute ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ᇱ ሺ‫ ݔ‬ሻ ൌ ሺܽ݁ ିଷ௫ ሻ െ ͵ሺܽ‫ି ݁ݔ‬ଷ௫ ሻ in ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ݁ ିଷ௫ . ሾሺܽ݁ ିଷ௫ ሻ െ ͵ሺܽ‫ି ݁ݔ‬ଷ௫ ሻሿ ൅ ͵ሺܽ݁ ିଷ௫ ሻ‫ ݔ‬ൌ ݁ ିଷ௫

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ܽ݁ ିଷ௫ ൌ ݁ ିଷ௫


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Example 2.2.5 Given ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ െ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ‫ ݔ‬ଶ ݁ ௫ Ǥ Describe

ܽൌͳ

‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ but do not find it.

Thus, ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ൌ ሺͳ݁

ିଷ௫ ሻ

‫ ݔ‬ൌ ‫݁ݔ‬

ିଷ௫

Ǥ

Solution: To describe ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ݔ‬ሻ, we do the following:

Step 3: We need to find the general solution as follows:

In this example, we look at ‫ ݔ‬ଶ , and we write it as:

‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ ݔ‬ሻ ൌ ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൅ ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ

ܽ ൅ ܾ‫ ݔ‬൅ ܿ‫ ݔ‬ଶ , and then we multiply it by ݁ ௫ .

Thus, ‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ ݔ‬ሻ ൌ ܿଵ ݁ ିଷ௫ ൅ ‫ି ݁ݔ‬ଷ௫ Ǥ

‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ൌ ሾܽ ൅ ܾ‫ ݔ‬൅ ܿ‫ ݔ‬ଶ ሿ݁ ௫ .

Result 2.2.1 Suppose that you have a linear differential

Example 2.2.6 Given ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ െ ͵‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ሺ͵‫ݔ‬ሻ݁ ௫ Ǥ

equation with the least derivative, say ݉, and this

Describe ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ but do not find it.

differential equation equals to a polynomial of degree

Solution: To describe ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ݔ‬ሻ, we do the following:

‫ݓ‬. Then, we obtain the following:

In this example, we look at ሺ͵‫ݔ‬ሻ, and we write it as:

‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ൌ ሾܲ‫ݓ ݁݁ݎ݃݁ܦ ݂݋ ݈ܽ݅݉݋݊ݕ݈݋‬ሿ‫ ݔ‬௠ .

ሺܽ ሺ͵‫ ݔ‬ሻ ൅ ܾሺ͵‫ݔ‬ሻሻ, and then we multiply it by ݁ ௫ .

Result 2.2.2 Suppose that you have a linear differential

‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ൌ ሾܽ ሺ͵‫ ݔ‬ሻ ൅ ܾሺ͵‫ݔ‬ሻሿ݁ ௫ .

equation, then the general solution is always written as: ‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ݔ‬ሻ ൌ ‫ݕ‬௛௢௠௢௚௘௡௘௢௨௦ ሺ‫ ݔ‬ሻ ൅ ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ݔ‬ሻ. Example 2.2.3 Given ‫ݕ‬

ሺସሻ ሺ

‫ݔ‬ሻ െ ͹‫ݕ‬

ሺଷሻ ሺ

‫ݔ‬ሻ ൌ ‫ ݔ‬Ǥ Describe

‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ but do not find it. Solution: To describe ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ݔ‬ሻ, we do the following: By using result 2.2.1, we obtain the following: ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ൌ ሾܽ ൅ ܾ‫ ݔ‬൅ ܿ‫ ݔ‬ଶ ሿ‫ ݔ‬ଷ . Example 2.2.4 Given ‫ݕ‬

ሺସሻ ሺ

‫ݔ‬ሻ െ ͹‫ݕ‬

ሺଷሻ ሺ

‫ݔ‬ሻ ൌ ͵Ǥ Describe

‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ but do not find it. Solution: To describe ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ݔ‬ሻ, we do the following: By using result 2.2.1, we obtain the following: ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ൌ ሾܽሿ‫ ݔ‬ଷ ൌ ܽ‫ ݔ‬ଷ Ǥ

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2.3 Exercises

ଶ

1. Given ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ʹ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൅ ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ͲǤ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: Use results 2.1.1 and 2.1.2, and in this exercise, no need to find the values of ܿଵ ǡ ܿଶ ) 2. Given ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ െ ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൌ ͵‫ݔ‬Ǥ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: No need to find the value of ܿଵ ǡ ܿଶ ǡ ܿଷ ) 3. Given ‫ ݕ‬ሺସሻ ሺ‫ݔ‬ሻ െ ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ ൌ ͵‫ ݔ‬ଶ Ǥ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: No need to find the value of ܿଵ ǡ ܿଶ ǡ ܿଷ ǡ ܿସ )

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4. Given ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ െ ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൌ ݁ ௫ Ǥ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: No need to find the value of ܿଵ ǡ ܿଶ ǡ ܿଷ ) 5. Given ‫ ݕ‬ሺଷሻ ሺ‫ݔ‬ሻ െ ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൌ ሺʹ‫ݔ‬ሻǤ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: No need to find the value of ܿଵ ǡ ܿଶ ǡ ܿଷ ) 6. Given ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ െ ͵‫ݕ‬ሺ‫ݔ‬ሻ ൌ ͵‫ݔ‬ሺͷ‫ݔ‬ሻǤ Describe ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ but do not find it. మ

7. Given ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ െ ͵‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௫ Ǥ Describe ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ݔ‬ሻ but do not find it.


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Chapter 3 Methods of First and Higher Orders Differential Equations In this chapter, we introduce two new methods called Variation Method and Cauchy-Euler Method in order to solve first and higher orders differential equations. In addition, we give several examples about these methods, and the difference between them and the previous methods in chapter 2.

3.1 Variation Method In this section, we discuss how to find the particular solution using Variation Method. For the homogeneous solution, it will be similar to what we learned in chapter 2. Definition 3.1.1 Given ܽଶ ሺ‫ ݔ‬ሻ‫ ݕ‬ሺଶሻ ൅ ܽଵ ሺ‫ ݔ‬ሻ‫ ݕ‬ᇱ ൌ ‫ܭ‬ሺ‫ݔ‬ሻ is a linear differential equation of order 2. Assume that ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ and ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ are independent solution to the homogeneous solution. Then, the particular solution using Variation Method is written as:

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‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ൌ ݄ଵ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ሺ‫ݔ‬ሻ ൅ ݄ଶ ሺ‫ݔ‬ሻ‫ݕ‬ଶ ሺ‫ݔ‬ሻ. To find ݄ଵ ሺ‫ ݔ‬ሻ and ݄ଶ ሺ‫ ݔ‬ሻ, we need to solve the following two equations: ݄ଵ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൅ ݄ଶ ᇱ ሺ‫ݔ‬ሻ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ Ͳ ‫ ܭ‬ሺ‫ ݔ‬ሻ ݄ଵ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ᇱ ሺ‫ ݔ‬ሻ ൅ ݄ଶ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଶ ᇱ ሺ‫ ݔ‬ሻ ൌ ܽଶ ሺ‫ ݔ‬ሻ ሺଷሻ Definition 3.1.2 Given ܽଷ ሺ‫ ݔ‬ሻ‫ ݕ‬൅ ‫ ڮ‬൅ ܽଵ ሺ‫ݔ‬ሻ‫ ݕ‬ᇱ ൌ ‫ܭ‬ሺ‫ݔ‬ሻ is a linear differential equation of order 3. Assume that ‫ݕ‬ଵ ሺ‫ ݔ‬ሻǡ ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ and ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ are independent solution to the homogeneous solution. Then, the particular solution using Variation Method is written as: ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ൌ ݄ଵ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ሺ‫ݔ‬ሻ ൅ ݄ଶ ሺ‫ ݔ‬ሻ‫ݕ‬ଶ ሺ‫ݔ‬ሻ ൅ ݄ଷ ሺ‫ݔ‬ሻ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ. To find ݄ଵ ሺ‫ ݔ‬ሻǡ ݄ଶ ሺ‫ ݔ‬ሻ and ݄ଷ ሺ‫ ݔ‬ሻ, we need to solve the following three equations: ݄ଵ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ሺ‫ݔ‬ሻ ൅ ݄ଶ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൅ ݄ଷ ᇱ ሺ‫ݔ‬ሻ‫ݕ‬ଷ ሺ‫ ݔ‬ሻ ൌ Ͳ ݄ଵ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ᇱ ሺ‫ ݔ‬ሻ ൅ ݄ଶ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଶ ᇱ ሺ‫ ݔ‬ሻ ൅ ݄ଷ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଷ ᇱ ሺ‫ ݔ‬ሻ ൌ Ͳ ‫ ܭ‬ሺ‫ ݔ‬ሻ ݄ଵ ሺଶሻ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ሺଶሻ ሺ‫ ݔ‬ሻ ൅ ݄ଶ ሺଶሻ ሺ‫ ݔ‬ሻ‫ݕ‬ଶ ሺଶሻ ሺ‫ ݔ‬ሻ ൅ ݄ଷ ሺଶሻ ሺ‫ ݔ‬ሻ‫ݕ‬ଷ ሺଶሻ ሺ‫ ݔ‬ሻ ൌ ܽଷ ሺ ‫ ݔ‬ሻ ଵ


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Since ‫ ݕ‬ሺଶሻ ൅ ͵‫ ݕ‬ᇱ ൌ Ͳ is a HLDE with constant coefficients, then we will do the following: Let ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ , we need to find ݇. First of all, we will find the first and second derivatives as follows: ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫ ‫ ݕ‬ᇱᇱ ሺ‫ݔ‬ሻ ൌ ݇ ଶ ݁ ௞௫ Now, we substitute ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫ and ‫ ݕ‬ᇱᇱ ሺ‫ݔ‬ሻ ൌ ݇ ଶ ݁ ௞௫ in ‫ ݕ‬ሺଶሻ ൅ ͵‫ ݕ‬ᇱ ൌ Ͳ as follows: ݇ ଶ ݁ ௞௫ ൅ ͵݇݁ ௞௫ ൌ Ͳ ݁ ௞௫ ሺ݇ ଶ ൅ ͵݇ሻ ൌ Ͳ ݁ ௞௫ ሺ݇ሺ݇ ൅ ͵ሻሻ ൌ Ͳ Thus, ݇ ൌ Ͳ and ݇ ൌ െ͵. Then, we use our values to substitute ݇ in our assumption which is ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ :

Example 3.1.1 Given ‫ ݕ‬ሺଶሻ ൅ ͵‫ ݕ‬ᇱ ൌ ௫ Ǥ Find the general

at ݇ ൌ Ͳ, ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ሺ଴ሻ௫ ൌ ݁ ଴ ൌ ͳ

solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: No need to find the values of

at ݇ ൌ െ͵, ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ሺିଷሻ௫ ൌ ݁ ିଷ௫

ܿଵ and ܿଶ )

Notice that ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ and ‫ݕ‬ଶ ሺ‫ݔ‬ሻ are independent. ଵ

Solution: Since ‫ ݕ‬ሺଶሻ ൅ ͵‫ ݕ‬ᇱ ൌ does not have a constant ௫

coefficient, then we need to use the variation method as follows: Step 1: We need to find the homogeneous solution by letting ‫ݕ‬

ሺଶሻ

ᇱ

൅ ͵‫ ݕ‬equal to zero as follows:

‫ ݕ‬ሺଶሻ ൅ ͵‫ ݕ‬ᇱ ൌ Ͳ. Now, it is a homogeneous linear differential equation of order 2.

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Thus, using result 2.1.1, the general homogenous solution for ‫ݕ‬ሺ‫ݔ‬ሻ is: ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൌ ܿଵ ൅ ܿଶ ݁ ିଷ௫ , for some ܿଵ ܿଶ ‫ א‬Ը. (Note: ݄‫ ݋݉݋‬denotes to homogeneous). Step 2: We need to find the particular solution using ଵ

definition 3.1.1 as follows: Since ‫ ݕ‬ሺଶሻ ൅ ͵‫ ݕ‬ᇱ equals , ௫

then the particular solution should be in the following form: ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ൌ ݄ଵ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൅ ݄ଶ ሺ‫ݔ‬ሻ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ. To find ݄ଵ ሺ‫ ݔ‬ሻ and ݄ଶ ሺ‫ ݔ‬ሻ, we need to solve the following two equations:

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݄ଵ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൅ ݄ଶ ᇱ ሺ‫ݔ‬ሻ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ Ͳ ‫ ܭ‬ሺ‫ ݔ‬ሻ ݄ଵ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ᇱ ሺ‫ ݔ‬ሻ ൅ ݄ଶ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଶ ᇱ ሺ‫ ݔ‬ሻ ൌ ܽଶ ሺ‫ ݔ‬ሻ ᇱ ------Æ ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ Ͳ ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ͳ

ଵ

Thus, ‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ ݔ‬ሻ ൌ ሺܿଵ ൅ ܿଶ ݁ ିଷ௫ ሻ ൅ ሺȁ‫ ݔ‬ȁሻ ൅ ଷ

௫ ଵ ݁ ିଷ௫ ቀ‫׬‬଴ െ ଷ௧ ݁ ଷ௧

Now, we substitute what we got above in the particular solution form as follows:

ଵ

method as follows:

ିଷ௫ ሻ

݁ ଷ௫ ሺ݁ ିଷ௫ ሻ ൌ

ଵ ଷ௫

ଵ

Since it is impossible to integrate ݄ଶ ሺ‫ ݔ‬ሻ ൌ െ ଷ௫ ݁

ଷ௫

to

find ݄ଶ ሺ‫ ݔ‬ሻ, then it is enough to write as: ݄ଶ ሺ‫ ݔ‬ሻ ൌ

݁ ଷ௧ ݀‫ݐ‬.

then we do the following:

Since ‫ ݕ‬ሺଶሻ ൅ ͸‫ ݕ‬ᇱ ൅ ͺ‫ ݕ‬ൌ Ͳ is a HLDE with constant coefficients, then we will do the following:

as follows: ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫ ‫ ݕ‬ᇱᇱ ሺ‫ݔ‬ሻ ൌ ݇ ଶ ݁ ௞௫

ଵ

݀‫ ݔ‬ൌ ଷ ሺȁ‫ ݔ‬ȁሻ, ‫ ݔ‬൐ Ͳ. ଷ௫

Thus, we write the particular solution as follows: ௫

ͳ ͳ ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ൌ ሺȁ‫ ݔ‬ȁሻ ൅ ݁ ିଷ௫ ቌන െ ݁ ଷ௧ ݀‫ݐ‬ቍ ͵ ͵‫ݐ‬ ଴

Step 3: We need to find the general solution as follows:

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‫ ݕ‬ሺଶሻ ൅ ͸‫ ݕ‬ᇱ ൅ ͺ‫ ݕ‬ൌ Ͳ. Now, it is a homogeneous linear

First of all, we will find the first and second derivatives ଵ

ଵ

letting ‫ ݕ‬ሺଶሻ ൅ ͸‫ ݕ‬ᇱ ൅ ͺ‫ ݕ‬equal to zero as follows:


Since it is possible to integrate ݄ଵ ᇱ ሺ‫ ݔ‬ሻ ൌ ଷ௫ to find ݄ଵ ሺ‫ ݔ‬ሻ, ݄ଵ ሺ‫ ݔ‬ሻ ൌ ‫׬‬

Step 1: We need to find the homogeneous solution by

differential equation of order 2.

. ᇱ

௫ ଵ ‫׬‬଴ െ ଷ௧

general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: No need to find the

constant coefficient, then we need to use the variation

ൌ Ͳ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ ሺͳሻ ݄ଵ ሺ‫ ݔ‬ሻሺͳሻ ൅ ݄ଶ ሺ‫ ݔ‬ሻሺ݁ ͳ ݄ଶ ᇱ ሺ‫ ݔ‬ሻሺെ͵݁ ିଷ௫ ሻ ൌ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺʹሻ ‫ݔ‬ ଵ By solving ሺͳሻ and ሺʹሻ, ݄ଶ ᇱ ሺ‫ ݔ‬ሻ ൌ െ ଷ௫ ݁ ଷ௫ and ଷ௫

Example 3.1.2 Given ‫ ݕ‬ሺଶሻ ൅ ͸‫ ݕ‬ᇱ ൅ ͺ‫ ݕ‬ൌ ݁ ିସ௫ Ǥ Find the

Solution: Since ‫ ݕ‬ሺଶሻ ൅ ͸‫ ݕ‬ᇱ ൅ ͺ‫ ݕ‬ൌ ݁ ିସ௫ does not have a

ͳ ݄ଵ ᇱ ሺ‫ ݔ‬ሻሺͲሻ ൅ ݄ଶ ᇱ ሺ‫ ݔ‬ሻሺെ͵݁ ିଷ௫ ሻ ൌ ‫ݔ‬ ͳ

݄ଵ ᇱ ሺ‫ ݔ‬ሻ ൌ

݀‫ݐ‬ቁǡ for some ܿଵ ܿଶ ‫ א‬ԸǤ

values of ܿଵ and ܿଶ )

݄ଵ ᇱ ሺ‫ ݔ‬ሻሺͳሻ ൅ ݄ଶ ᇱ ሺ‫ ݔ‬ሻሺ݁ ିଷ௫ ሻ ൌ Ͳ

ᇱ

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‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ ݔ‬ሻ ൌ ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൅ ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ

‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ିଷ௫ ------Æ ‫ݕ‬ଶ ᇱ ሺ‫ ݔ‬ሻ ൌ െ͵݁ ିଷ௫

ᇱ


Now, we substitute‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ݁ ௞௫ ǡ ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ݇݁ ௞௫ and ‫ ݕ‬ᇱᇱ ሺ‫ݔ‬ሻ ൌ ݇ ଶ ݁ ௞௫ in ‫ ݕ‬ሺଶሻ ൅ ͸‫ ݕ‬ᇱ ൅ ͺ‫ ݕ‬ൌ Ͳ as follows: ݇ ଶ ݁ ௞௫ ൅ ͸݇݁ ௞௫ ൅ ͺ݁ ௞௫ ൌ Ͳ ݁ ௞௫ ሺ݇ ଶ ൅ ͸݇ ൅ ͺሻ ൌ Ͳ ݁ ௞௫ ሺሺ݇ ൅ ʹሻሺ݇ ൅ Ͷሻሻ ൌ Ͳ

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Thus, ݇ ൌ െʹ and ݇ ൌ െͶ. Then, we use our values to

By solving ሺͳሻ and ሺʹሻ, and using Cramer’s rule, we

substitute ݇ in our assumption which is ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ݁ ௞௫ :

obtain:

at ݇ ൌ െʹ, ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ሺିଶሻ௫ ൌ ݁ ିଶ௫

ᇱ

݄ଵ ሺ‫ ݔ‬ሻ ൌ ‫ݔ‬ଵ ൌ

at ݇ ൌ െͶ, ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ሺିସሻ௫ ൌ ݁ ିସ௫ Notice that ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ and ‫ݕ‬ଶ ሺ‫ݔ‬ሻ are independent.

ൌ

Thus, using result 2.1.1, the general homogenous solution for ‫ݕ‬ሺ‫ݔ‬ሻ is: ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൌ ܿଵ ݁ ିଶ௫ ൅ ܿଶ ݁ ିସ௫ , for some ܿଵ ܿଶ ‫ א‬Ը. (Note: ݄‫ ݋݉݋‬denotes to Step 2: We need to find the particular solution using definition 3.1.1 as follows: Since ‫ ݕ‬ሺଶሻ ൅ ͸‫ ݕ‬ᇱ ൅ ͺ‫ ݕ‬equals ݁ ିସ௫ , then the particular solution should be in the following form: ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ൌ ݄ଵ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൅ ݄ଶ ሺ‫ݔ‬ሻ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ. To find ݄ଵ ሺ‫ ݔ‬ሻ and ݄ଶ ሺ‫ ݔ‬ሻ, we need to solve the following two equations: ݄ଵ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൅ ݄ଶ ᇱ ሺ‫ݔ‬ሻ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ Ͳ ‫ ܭ‬ሺ‫ ݔ‬ሻ ݄ଵ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଵ ᇱ ሺ‫ ݔ‬ሻ ൅ ݄ଶ ᇱ ሺ‫ ݔ‬ሻ‫ݕ‬ଶ ᇱ ሺ‫ ݔ‬ሻ ൌ ܽଶ ሺ‫ ݔ‬ሻ ᇱ ିଶ௫ ିଶ௫ ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ------Æ ‫ݕ‬ଵ ሺ‫ݔ‬ሻ ൌ െʹ݁ ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ିସ௫ ------Æ ‫ݕ‬ଶ ᇱ ሺ‫ ݔ‬ሻ ൌ െͶ݁ ିସ௫ Now, we substitute what we got above in the particular solution form as follows:

݄ଵ ᇱ ሺ‫ ݔ‬ሻሺെʹ݁ ିଶ௫ ሻ ൅ ݄ଶ ᇱ ሺ‫ ݔ‬ሻሺെͶ݁ ିସ௫ ሻ ൌ ᇱ

ିଶ௫ ሻ

ᇱ

ିସ௫ ሻ

݁ ିସ௫ ͳ

൅ ݄ଶ ሺ‫ݔ‬ሻሺ݁ ൌ Ͳ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ሺͳሻ ݄ଵ ሺ‫ ݔ‬ሻሺ݁ ᇱ ᇱ ିଶ௫ ݄ଵ ሺ‫ ݔ‬ሻሺെʹ݁ ሻ ൅ ݄ଶ ሺ‫ ݔ‬ሻሺെͶ݁ ିସ௫ ሻ ൌ ݁ ିସ௫ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺʹሻ

ͳ ൬ ݁ ିଶ௫ ൰ ሺ݁ ିଶ௫ ሻ ൅ ݄ଶ ᇱ ሺ‫ ݔ‬ሻሺ݁ ିସ௫ ሻ ൌ Ͳ ʹ ͳ ݄ଶ ᇱ ሺ‫ ݔ‬ሻ ൌ െ ʹ ଵ

Since it is possible to integrate ݄ଵ ᇱ ሺ‫ ݔ‬ሻ ൌ ଶ ݁ ିଶ௫ to find ݄ଵ ሺ‫ ݔ‬ሻ, then we do the following: ଵ

ଵ

݄ଵ ሺ‫ ݔ‬ሻ ൌ ‫ି ݁ ׬‬ଶ௫ ݀‫ ݔ‬ൌ െ ݁ ିଶ௫ . ଶ ସ ଵ

Since it is possible to integrate ݄ଶ ᇱ ሺ‫ ݔ‬ሻ ൌ െ ଶ to find ݄ଶ ሺ‫ ݔ‬ሻ, then we do the following: ଵ

ଵ

ଶ

ଶ

݄ଶ ሺ‫ ݔ‬ሻ ൌ ‫ ׬‬െ ݀‫ ݔ‬ൌ െ ‫ݔ‬. Thus, we write the particular solution as follows: ͳ ͳ ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ൌ െ ݁ ିଶ௫ ሺ݁ ିଶ௫ ሻ െ ‫ݔ‬ሺ݁ ିସ௫ ሻ Ͷ ʹ Step 3: We need to find the general solution as follows: ‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ ݔ‬ሻ ൌ ‫ݕ‬௛௢௠௢ ሺ‫ ݔ‬ሻ ൅ ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ ଵ

Thus, ‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ ݔ‬ሻ ൌ ሺܿଵ ݁ ିଶ௫ ൅ ܿଶ ݁ ିସ௫ ሻ ൅ ቀെ ସ ݁ ିଶ௫ ሺ݁ ିଶ௫ ሻ െ ଵ ଶ

72 M. Kaabar

െ݁ ି଼௫ െ݁ ି଼௫ ൌ െͶ݁ ି଺௫ ൅ ʹ݁ ି଺௫ െʹ݁ ି଺௫ ͳ ൌ ݁ ିଶ௫ ʹ

By substituting ݄ଵ ᇱ ሺ‫ ݔ‬ሻ in ሺͳሻ to find ݄ଶ ᇱ ሺ‫ ݔ‬ሻ as follows:

homogeneous).

݄ଵ ᇱ ሺ‫ ݔ‬ሻሺ݁ ିଶ௫ ሻ ൅ ݄ଶ ᇱ ሺ‫ ݔ‬ሻሺ݁ ିସ௫ ሻ ൌ Ͳ

଴ ௘ షరೣ ൨ ୢୣ୲൤ షరೣ ௘ ିସ௘ షరೣ షమೣ ௘ షరೣ ቃ ୢୣ୲ቂ ௘ షమೣ ିଶ௘ ିସ௘ షరೣ

‫ݔ‬ሺ݁ ିସ௫ ሻቁǡ for some ܿଵ ܿଶ ‫ א‬ԸǤ

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3.2 Cauchy-Euler Method


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‫ ݔ‬௞ିଵ ሺ݇ ଶ െ ʹ݇ ൅ ͳሻ ൌ Ͳ ‫ ݔ‬௞ିଵ ሺሺ݇ െ ͳሻሺ݇ െ ͳሻሻ ൌ Ͳ

In this section, we will show how to use Cauchy-Euler

Thus, ݇ ൌ ͳ and ݇ ൌ ͳ. Then, we use our values to

Method to find the general solution for differential

substitute ݇ in our assumption which is ‫ ݕ‬ൌ ‫ ݔ‬௞ :

equations that do not have constant coefficients.

at ݇ ൌ ͳ, ‫ݕ‬ଵ ൌ ‫ ݔ‬ଵ ൌ ‫ݔ‬

To introduce this method, we start with some examples as follows:

at ݇ ൌ ͳ, ‫ݕ‬ଶ ൌ ‫ ݔ‬ଵ ൌ ‫ ݔ‬ή ሺ‫ݔ‬ሻ

ଵ

In the above case, we multiplied ‫ ݔ‬by ሺ‫ݔ‬ሻ because we

Example 3.2.1 Given ‫ ݕݔ‬ሺଶሻ െ ‫ ݕ‬ᇱ ൅ ௫ ‫ ݕ‬ൌ ͲǤ Find the

had a repeating for ‫ݔ‬, and in Cauchy-Euler Method, we

general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: No need to find the

should multiply any repeating by natural logarithm.

values of ܿଵ and ܿଶ )

Thus, the general solution for ‫ݕ‬ሺ‫ݔ‬ሻ is: ଵ

Solution: Since ‫ ݕݔ‬ሺଶሻ െ ‫ ݕ‬ᇱ ൅ ௫ ‫ ݕ‬ൌ Ͳ does not have constant coefficients, then we need to use the CauchyEuler method by letting ‫ ݕ‬ൌ ‫ ݔ‬௞ , and after substitution all terms must be of the same degree as follows: First of all, we will find the first and second derivatives

Example 3.2.2 Given ‫ ݔ‬ଷ ‫ ݕ‬ሺଶሻ െ ‫ ݔ‬ଶ ‫ ݕ‬ᇱ ൅ ‫ ݕݔ‬ൌ ͲǤ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: No need to find the values of ܿଵ and ܿଶ ) Solution: Since ‫ ݔ‬ଷ ‫ ݕ‬ሺଶሻ ൅ ‫ ݔ‬ଶ ‫ ݕ‬ᇱ ൅ ‫ ݕݔ‬ൌ Ͳ does not have constant coefficients, then we need to use the Cauchy-

as follows:

Euler method by letting ‫ ݕ‬ൌ ‫ ݔ‬௞ , and after substitution

‫ ݕ‬ᇱ ൌ ݇‫ ݔ‬௞ିଵ ‫ ݕ‬ᇱᇱ ൌ ݇ሺ݇ െ ͳሻ‫ ݔ‬௞ିଶ Now, we substitute ‫ ݕ‬ൌ ‫ ݔ‬௞ ǡ ‫ ݕ‬ᇱ ൌ ݇‫ ݔ‬௞ିଵ and ଵ

‫ ݕ‬ᇱᇱ ൌ ݇ሺ݇ െ ͳሻ‫ ݔ‬௞ିଶ in ‫ ݕݔ‬ሺଶሻ െ ‫ ݕ‬ᇱ ൅ ‫ ݕ‬ൌ Ͳ as follows: ௫

all terms must be of the same degree as follows: First of all, we will find the first, second and third derivatives as follows: ‫ ݕ‬ᇱ ൌ ݇‫ ݔ‬௞ିଵ

ͳ ‫݇ݔ‬ሺ݇ െ ͳሻ‫ ݔ‬௞ିଶ െ ݇‫ ݔ‬௞ିଵ ൅ ‫ ݔ‬௞ ൌ Ͳ ‫ݔ‬

‫ ݕ‬ᇱᇱ ൌ ݇ሺ݇ െ ͳሻ‫ ݔ‬௞ିଶ

݇ሺ݇ െ ͳሻ‫ ݔ‬௞ିଵ െ ݇‫ ݔ‬௞ିଵ ൅ ‫ ݔ‬௞ିଵ ൌ Ͳ

‫ ݕ‬ᇱᇱᇱ ൌ ݇ሺ݇ െ ͳሻሺ݇ െ ʹሻ‫ ݔ‬௞ିଷ

‫ ݔ‬௞ିଵ ሺ݇ሺ݇ െ ͳሻ െ ݇ ൅ ͳሻ ൌ Ͳ ‫ ݔ‬௞ିଵ ሺ݇ ଶ െ ݇ െ ݇ ൅ ͳሻ ൌ Ͳ

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‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ܿଵ ‫ ݔ‬൅ ܿଶ ‫݈݊ݔ‬ሺ‫ݔ‬ሻ, for some ܿଵ ܿଶ ‫ א‬Ը.

Now, we substitute ‫ ݕ‬ൌ ‫ ݔ‬௞ ǡ ‫ ݕ‬ᇱ ൌ ݇‫ ݔ‬௞ିଵ , and ‫ ݕ‬ᇱᇱ ൌ ݇ሺ݇ െ ͳሻ‫ ݔ‬௞ିଶ in ‫ ݔ‬ଷ ‫ ݕ‬ሺଶሻ ൅ ‫ ݔ‬ଶ ‫ ݕ‬ᇱ ൅ ‫ ݕݔ‬ൌ Ͳ as follows:

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‫ ݔ‬ଷ ሺ݇ሺ݇ െ ͳሻ‫ ݔ‬௞ିଶ ሻ ൅ ‫ ݔ‬ଶ ሺ݇‫ ݔ‬௞ିଵ ሻ ൅ ‫ݔ‬ሺ‫ ݔ‬௞ ሻ ൌ Ͳ ሺ݇ሺ݇ െ ͳሻ‫ ݔ‬௞ାଵ ሻ ൅ ሺ݇‫ ݔ‬௞ାଵ ሻ ൅ ሺ‫ ݔ‬௞ାଵ ሻ ൌ Ͳ ‫ ݔ‬௞ାଵ ሺ݇ ଶ െ ݇ ൅ ݇ ൅ ͳሻ ൌ Ͳ ‫ ݔ‬௞ାଵ ሺ݇ ଶ ൅ ͳሻ ൌ Ͳ Thus, ݇ ൌ േξͳ ൌ േ݅ ൌ Ͳ േ ሺͳሻሺ݅ሻ.


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4. Given ‫ ݔ‬ଷ ‫ ݕ‬ሺଷሻ െ ʹ‫ ݕݔ‬ᇱ ൌ ͲǤ Find the general solution for ‫ݕ‬ሺ‫ݔ‬ሻ. (Hint: No need to find the values of ܿଵ , ܿଶ and ܿଷ ) 5. Given ‫ ݔ‬ଶ ‫ ݕ‬ሺଶሻ ൅ ‫ ݕ‬ᇱ ൌ ʹ‫ ݔ‬ଶ Ǥ Is it possible to find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ using Cauchy-Euler Method? Why?

Then, we use our values to substitute ݇ in our assumption which is ‫ ݕ‬ൌ ‫ ݔ‬௞ : Since we have two parts (real and imaginary), then by using the Cauchy-Euler Method, we need to write our solution as follows: ‫ݕ‬ଵ ൌ ‫ ݔ‬ሺ௥௘௔௟ ௣௔௥௧ሻ ሺ‫ ݐݎܽ݌ ݕݎܽ݊݅݃ܽ݉ܫ‬ή ሺ‫ ݔ‬ሻ ൌ ‫ ݔ‬ሺ଴ሻ ሺͳ ή ሺ‫ ݔ‬ሻሻ ൌ ሺሺ‫ ݔ‬ሻሻ ‫ݕ‬ଶ ൌ ‫ ݔ‬ሺ௥௘௔௟ ௣௔௥௧ሻ ሺ‫ ݐݎܽ݌ ݕݎܽ݊݅݃ܽ݉ܫ‬ή ሺ‫ ݔ‬ሻ ൌ ‫ ݔ‬ሺ଴ሻ ሺͳ ή ሺ‫ ݔ‬ሻሻ ൌ ሺሺ‫ ݔ‬ሻሻ Thus, the general solution for ‫ݕ‬ሺ‫ݔ‬ሻ is: ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ܿଵ ሺሺ‫ ݔ‬ሻሻ ൅ ܿଶ ሺሺ‫ ݔ‬ሻሻ, for some ܿଵ ܿଶ ‫ א‬Ը.

3.3 Exercises 1. Given ‫ ݕ‬ሺଶሻ ൅ ‫ ݕ‬ᇱ ൅ Ͷ‫ ݕ‬ൌ ͲǤ Find the general solution for ‫ݕ‬ሺ‫ݔ‬ሻ. (Hint: No need to find the values of ܿଵ and ܿଶ ) 2. Given ‫ ݕ‬ሺଶሻ ൅ ͷ‫ ݕ‬ᇱ ൅ ͹‫ ݕ‬ൌ ͲǤ Find the general solution for ‫ݕ‬ሺ‫ݔ‬ሻ. (Hint: No need to find the values of ܿଵ and ܿଶ ) 3. Given ‫ ݔ‬ଷ ‫ ݕ‬ሺଷሻ ൅ ‫ ݕݔ‬ᇱ ൌ ͲǤ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: No need to find the values of ܿଵ , ܿଶ and ܿଷ )

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Chapter 4 Extended Methods of First and Higher Orders Differential Equations


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ܽଵ ሺ‫ ݔ‬ሻ ᇱ ܽଶ ሺ‫ ݔ‬ሻ ‫ ܭ‬ሺ‫ ݔ‬ሻ ‫ ݕ‬൅ ‫ݕ‬ൌ ሺ ሻ ሺ ሻ ܽଵ ‫ݔ‬ ܽଵ ‫ݔ‬ ܽଵ ሺ‫ ݔ‬ሻ ܽଶ ሺ ‫ ݔ‬ሻ ‫ ܭ‬ሺ‫ ݔ‬ሻ ‫ݕ‬ᇱ ൅ ‫ݕ‬ൌ ሺ ሻ ܽଵ ‫ݔ‬ ܽଵ ሺ‫ ݔ‬ሻ Assume that ݃ሺ‫ ݔ‬ሻ ൌ

௔మ ሺ௫ሻ ௔భ ሺ௫ሻ

and ‫ ܨ‬ሺ‫ ݔ‬ሻ ൌ

௄ሺ௫ሻ

, then:

௔భ ሺ௫ሻ

‫ ݕ‬ᇱ ൅ ݃ሺ‫ ݔ‬ሻ‫ ݕ‬ൌ ‫ ܨ‬ሺ‫ ݔ‬ሻ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺͳሻ Thus, the solution using Integral Factor Method is written in the following steps: Step 1: Multiply both sides of ሺͳሻ by letting

In this chapter, we discuss some new methods such as

‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ : ‫ ݕ‬ᇱ ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ൅ ݃ሺ‫ݔ‬ሻ‫ ׬ ݁ݕ‬௚ሺ௫ሻௗ௫ ൌ ‫ ܨ‬ሺ‫ݔ‬ሻ݁ ‫ ׬‬௚ሺ௫ሻௗ௫

Bernulli Method, Separable Method, Exact Method,

Step 2: ‫ ݕ‬ᇱ ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ൅ ݃ሺ‫ ݔ‬ሻ‫ ׬ ݁ݕ‬௚ሺ௫ሻௗ௫ ൌ ൣ‫ ݕ‬ή ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ൧ ǥ ሺʹሻ

Reduced to Separable Method and Reduction of Order

Step 3: ൣ‫ ݕ‬ή ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ൧ ൌ ‫ ܨ‬ሺ‫ ݔ‬ሻ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ሺ͵ሻ Step 4: Integrate both sides of ሺ͵ሻ, we obtain:

Method. We use higher

orders

these methods to solve first and

linear

and

non-linear

differential

equations. In addition, we give examples about these methods, and the differences between them and the previous methods in chapter 2 and chapter 3.

4.1 Bernoulli Method In this section, we start with two examples about using integral factor to solve first order linear differential equations. Then, we introduce Bernulli Method to solve some examples of first order non-linear differential equations. Definition 4.1.1 Given ܽଵ ሺ‫ ݔ‬ሻ‫ ݕ‬ᇱ ൅ ܽଶ ሺ‫ ݔ‬ሻ‫ ݕ‬ൌ ‫ ܭ‬ሺ‫ ݔ‬ሻǡ ܽଵ ሺ‫ ݔ‬ሻ ് Ͳ is a linear differential equation of order 1. Dividing both sides by ܽଵ ሺ‫ ݔ‬ሻ, we obtain:

78 M. Kaabar

ᇱ

ᇱ

ᇱ

නൣ‫ ݕ‬ή ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ൧ ݀‫ ݔ‬ൌ නሺ‫ܨ‬ሺ‫ݔ‬ሻ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ሻ ݀‫ݔ‬ ‫ ݕ‬ή ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ൌ නሺ‫ܨ‬ሺ‫ݔ‬ሻ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ሻ ݀‫ݔ‬ Step 5: By solving for ‫ݕ‬, and substituting ‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ we obtain: ‫ ݕ‬ή ‫ ܫ‬ൌ නሺ‫ܨ‬ሺ‫ݔ‬ሻሻሺ‫ܫ‬ሻ ݀‫ݔ‬ ‫׬‬ሺ‫ ܨ‬ሺ‫ ݔ‬ሻሻሺ‫ܫ‬ሻ ݀‫ݔ‬ ‫ܫ‬ ‫ ܫ ׬‬ή ‫ܨ‬ሺ‫ ݔ‬ሻ ݀‫ݔ‬ ‫ݕ‬ൌ ‫ܫ‬ Thus, the final solution is: ‫ ܫ ׬‬ή ‫ܨ‬ሺ‫ ݔ‬ሻ ݀‫ݔ‬ ‫ݕ‬ൌ ‫ܫ‬ Example 4.1.1 Given ‫ ݔ‬ଶ ‫ ݕ‬ᇱ െ ʹ‫ ݕݔ‬ൌ Ͷ‫ ݔ‬ଷ Ǥ Find the ‫ݕ‬ൌ

general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: Use integral factor method and no need to find the value of ܿ)

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Solution: Since ‫ ݔ‬ଶ ‫ ݕ‬ᇱ െ ʹ‫ ݕݔ‬ൌ Ͷ‫ ݔ‬ଷ does not have

Solution: Since ሺ‫ ݔ‬൅ ͳሻ‫ ݕ‬ᇱ ൅ ‫ ݕ‬ൌ ͷ does not have

constant coefficients, and it is a first order non-linear

constant coefficients, and it is a first order non-linear

differential equation, then by using definition 4.1.1, we

differential equation, then by using definition 4.1.1, we

need to use the integral factor method by letting

need to use the integral factor method by letting

‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ , where ݃ሺ‫ ݔ‬ሻ ൌ ௄ሺ௫ሻ

‫ ܨ‬ሺ‫ ݔ‬ሻ ൌ ௔

భ ሺ௫ሻ

ൌ

ସ௫ య ௫మ


ൌെ

ଶ௫ ௫మ

ଶ

‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ , where ݃ሺ‫ ݔ‬ሻ ൌ

ൌ െ and ௫

ൌ Ͷ‫ݔ‬

‫ ܨ‬ሺ‫ ݔ‬ሻ ൌ మ

Hence, ‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ൌ ݁ ‫ି ׬‬ೣௗ௫ ൌ ݁ ିଶ୪୬ሺ௫ሻ ൌ ݁ ୪୬ሺ௫

షమሻ

ൌ

The general solution is written as follows:

Hence, ‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ൌ ݁

‫ ܫ ׬‬ή ‫ܨ‬ሺ‫ ݔ‬ሻ ݀‫ݔ‬ ‫ܫ‬ ͳ ‫ ׬‬ଶ ή Ͷ‫ݔ݀ ݔ‬ ‫ݕ‬ൌ ‫ݔ‬ ͳ ‫ݔ‬ଶ Ͷ ‫ݔ݀ ݔ ׬‬ ‫ݕ‬ൌ ͳ ‫ݔ‬ଶ Ͷ ሺ‫ ݔ‬ሻ ൅ ܿ ‫ݕ‬ൌ ͳ ‫ݔ‬ଶ

Thus, the general solution is: ‫ ݕ‬ൌ Ͷ‫ ݔ‬ଶ ሺ‫ ݔ‬ሻ ൅ ܿ‫ ݔ‬ଶ for some ܿ ‫ א‬Ը. Example 4.1.2 Given ሺ‫ ݔ‬൅ ͳሻ‫ ݕ‬ᇱ ൅ ‫ ݕ‬ൌ ͷǤ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: Use integral factor method and no need to find the value of ܿ)

80 M. Kaabar

భ

‫׬‬ሺೣశభሻ ௗ௫

ൌ ݁ ୪୬ሺ௫ାଵሻ ൌ ሺ‫ ݔ‬൅ ͳሻ

The general solution is written as follows: ‫ ܫ ׬‬ή ‫ܨ‬ሺ‫ ݔ‬ሻ ݀‫ݔ‬ ‫ܫ‬ ͷ ‫׬‬ሺ‫ ݔ‬൅ ͳሻ ή ሺ‫ ݔ‬൅ ͳሻ ݀‫ݔ‬ ‫ݕ‬ൌ ሺ‫ ݔ‬൅ ͳሻ

‫ݕ‬ൌ

‫ ݕ‬ൌ Ͷ‫ ݔ‬ଶ ሺ‫ ݔ‬ሻ ൅ ܿ‫ ݔ‬ଶ

ଵ

ൌ ሺ௫ାଵሻ and

‫ ܭ‬ሺ‫ ݔ‬ሻ ͷ ൌ ܽଵ ሺ‫ݔ‬ሻ ሺ‫ ݔ‬൅ ͳሻ

ଵ ௫మ


‫ݕ‬ൌ

‫ݕ‬ൌ

‫ݕ‬ൌ

‫ ׬‬ͷ ݀‫ݔ‬ ሺ‫ ݔ‬൅ ͳሻ

‫ݕ‬ൌ

ͷ‫ ݔ‬൅ ܿ ሺ‫ ݔ‬൅ ͳሻ

ͷ‫ݔ‬ ܿ ൅ ሺ‫ ݔ‬൅ ͳሻ ሺ‫ ݔ‬൅ ͳሻ ହ௫

௖

Thus, the general solution is: ‫ ݕ‬ൌ ሺ௫ାଵሻ ൅ ሺ௫ାଵሻ for some ܿ ‫ א‬Ը. Definition 4.1.2 Given ‫ ݕ‬ᇱ ൅ ݃ሺ‫ ݔ‬ሻ‫ ݕ‬ൌ ݂ሺ‫ ݔ‬ሻ‫ ݕ‬௡ where ݊ ‫ א‬Ը and ݊ ് Ͳ and ݊ ് ͳ is a non-linear differential equation of order 1. Thus, the solution using Bernoulli Method is written in the following steps: Step 1: Change it to first order linear differential equation by letting ‫ ݓ‬ൌ ‫ ݕ‬ଵି௡ .

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Step 2: Find the derivative of both sides for ‫ ݓ‬ൌ ‫ ݕ‬ଵି௡ as follows: ݀‫ݓ‬ ݀‫ݕ‬ ൌ ሺͳ െ ݊ሻ‫ ݕ‬ଵି௡ିଵ ή ݀‫ݔ‬ ݀‫ݔ‬ ݀‫ݕ‬ ݀‫ݓ‬ ି௡ ൌ ሺͳ െ ݊ሻ‫ ݕ‬ή ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ Ǥ ሺͳሻ ݀‫ݔ‬ ݀‫ݔ‬ ௗ௬ Step 3: Solve ሺͳሻ for as follows: ௗ௫

݀‫ݕ‬ ͳ ݀‫ݓ‬ ൌ ‫ݕ‬௡ ή ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺʹሻ ݀‫ ͳ ݔ‬െ ݊ ݀‫ݔ‬ Step 4: Since we assumed that ‫ ݓ‬ൌ ‫ ݕ‬ଵି௡ , then ቀ

భ

ቁ

ቀ

೙

ቁ

‫ ݕ‬ൌ ‫ ݓ‬భష೙ , and hence ‫ ݕ‬௡ ൌ ‫ ݓ‬భష೙ . Step 5: Substitute what we got above in ௗ௬ ௗ௫

൅ ݃ሺ‫ ݔ‬ሻ‫ ݕ‬ൌ ݂ሺ‫ݔ‬ሻ‫ ݕ‬௡ as follows:

ଵ ௡ ͳ ݀‫ݓ‬ ቀ ቁ ቀ ቁ ‫ݕ‬௡ ή ൅ ݃ሺ‫ ݔ‬ሻ‫ ݓ‬ଵି௡ ൌ ݂ሺ‫ݔ‬ሻ‫ ݓ‬ଵି௡ ǥ ǥ Ǥ ǥ ǥ ǥ Ǥ Ǥ ሺ͵ሻ ͳെ݊ ݀‫ݔ‬ ଵ Step 6: Divide ሺ͵ሻ by ‫ ݕ‬௡ as follows:

݀‫ݓ‬ ൅ ݀‫ݔ‬

ଵ ቀ ቁ ݃ሺ‫ ݔ‬ሻ‫ ݓ‬ଵି௡

ͳ ௡ ͳെ݊‫ݕ‬

ଵି௡

ൌ

௡ ቀ ቁ ݂ ሺ‫ ݔ‬ሻ‫ ݓ‬ଵି௡

ͳ ௡ ͳെ݊‫ݕ‬

ଵ ௡ ݀‫ݓ‬ ቀ ቁ ቀ ቁ ൅ ݃ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ‫ ݓ‬ଵି௡ ‫ି ݕ‬௡ ൌ ݂ ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ‫ ݓ‬ଵି௡ ‫ି ݕ‬௡ ݀‫ݔ‬ Step 7: After substitution, we obtain: ݀‫ݓ‬ ൅ ݃ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ‫ି ݕݕ‬௡ ൌ ݂ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ‫ ݕ‬௡ ‫ି ݕ‬௡ ݀‫ݔ‬ ݀‫ݓ‬ ൅ ݃ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ‫ ݕ‬ଵି௡ ൌ ݂ ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ‫ ݕ‬௡ି௡ ݀‫ݔ‬ ݀‫ݓ‬ ൅ ݃ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ‫ ݕ‬ଵି௡ ൌ ݂ ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ‫ ݕ‬଴ ݀‫ݔ‬ ݀‫ݓ‬ ൅ ݃ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ‫ ݕ‬ଵି௡ ൌ ݂ ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ ݀‫ݔ‬ Step 8: We substitute ‫ ݓ‬ൌ ‫ ݕ‬ଵି௡ in the above equation as follows: ݀‫ݓ‬ ൅ ݃ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ‫ ݓ‬ൌ ݂ሺ‫ ݔ‬ሻሺͳ െ ݊ሻ ݀‫ݔ‬

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Thus, the final solution is: ݀‫ݓ‬ ൅ ሺͳ െ ݊ሻ݃ሺ‫ ݔ‬ሻ‫ ݓ‬ൌ ሺͳ െ ݊ሻ݂ ሺ‫ ݔ‬ሻ ݀‫ݔ‬ In the following example, we will show how to use Bernoulli Method, and we will explore the relationship between Bernoulli Method and Integral Factor Method. Example 4.1.3 Given ‫ ݕݔ‬ᇱ ൅ ͵‫ ݔ‬ଶ ‫ ݕ‬ൌ ሺ͸‫ ݔ‬ଶ ሻ‫ ݕ‬ଷ Ǥ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: Use Bernoulli method and no need to find the value of ܿ) Solution: Since ‫ ݕݔ‬ᇱ ൅ ͵‫ ݔ‬ଶ ‫ ݕ‬ൌ ሺ͸‫ ݔ‬ଶ ሻ‫ ݕ‬ଷ does not have constant coefficients, and it is a first order non-linear differential equation, then by using definition 4.1.2, we need to do the following by letting ‫ ݓ‬ൌ ‫ ݕ‬ଵି௡ , where in this example ݊ ൌ ͵, and ݃ሺ‫ ݔ‬ሻ ൌ ͵‫ ݔ‬ଶ and ݂ ሺ‫ ݔ‬ሻ ൌ ͸‫ ݔ‬ଶ. Since we assumed that ‫ ݓ‬ൌ ‫ ݕ‬ଵିଷ ൌ ‫ି ݕ‬ଶ , then ‫ݕ‬ൌ‫ݓ‬

భ ቁ భష೙

ቀ

ൌ‫ݓ‬

భ ቁ భషయ

ቀ

భ

ൌ ‫ି ݓ‬మ ൌ

ଵ ξ௪

ௗ௬

ଵ

ௗ௪

, and ௗ௫ ൌ െ ଶ ‫ ݕ‬ଷ ή ௗ௫

We substitute what we got above in ‫ ݕݔ‬ᇱ ൅ ͵‫ ݔ‬ଶ ‫ ݕ‬ൌ ሺ͸‫ ݔ‬ଶ ሻ‫ ݕ‬ଷ as follows: ͳ ݀‫ݓ‬ ͳ ͳ ൰ ൅ ͵‫ ݔ‬ଶ ൬ ൰ ൌ ሺ͸‫ ݔ‬ଶ ሻ ൬ ൰ ǥ ǥ ǥ ǥ Ǥ ǥ ሺͳሻ ‫ ݔ‬൬െ ‫ ݕ‬ଷ ή ݀‫ݔ‬ ʹ ‫ ݓ‬ξ‫ݓ‬ ξ‫ݓ‬ ଵ

Now, we divide ሺͳሻ by െ ଶ ‫ ݕݔ‬ଷ as follows: ͳ ͵‫ ݔ‬ଶ ൬ ൰ ሺ ଶ ሻ ଷ ͸‫ݕ ݔ‬ ݀‫ݓ‬ ‫ݓ‬ ξ ൅ ൌ ͳ ͳ ଷ ݀‫ݔ‬ െ ‫ ݕݔ‬ଷ െ ‫ݕݔ‬ ʹ ʹ ͳ ݀‫ݓ‬ ൅ ሺെʹሻ͵‫ ݔ‬൬ ൰ ‫ି ݕ‬ଷ ൌ ሺെʹሻ͸‫ ݔ‬ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺʹሻ ݀‫ݔ‬ ξ‫ݓ‬

83


Then, we substitute ‫ ݕ‬ൌ

ଵ ξ௪

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in ሺʹሻ as follows:

݀‫ݓ‬ ൅ ሺെʹሻ͵‫ݔ‬ሺ‫ݕ‬ሻ‫ି ݕ‬ଷ ൌ ሺെʹሻ͸‫ݔ‬ ݀‫ݔ‬ ݀‫ݓ‬ ൅ ሺെʹሻ͵‫ ݕݔ‬ଵିଷ ൌ ሺെʹሻ͸‫ݔ‬ ݀‫ݔ‬

‫ ݓ‬ൌʹ൅

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ܿ మ ݁ ିଷ௫

‫ ݓ‬ൌ ʹ ൅ ܿ݁ ଷ௫

మ మ

The general solution for ‫ݓ‬ሺ‫ݔ‬ሻ is: ‫ ݓ‬ሺ‫ ݔ‬ሻ ൌ ʹ ൅ ܿ݁ ଷ௫ . Thus, the general solution for ‫ݕ‬ሺ‫ݔ‬ሻ is:

݀‫ݓ‬ ൅ ሺെʹሻ͵‫ି ݕݔ‬ଶ ൌ ሺെʹሻ͸‫ ݔ‬ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺ͵ሻ ݀‫ݔ‬ Now, we substitute ‫ ݓ‬ൌ ‫ି ݕ‬ଶ in ሺ͵ሻ as follows: ݀‫ݓ‬ ൅ ሺെʹሻ͵‫ ݓݔ‬ൌ ሺെʹሻ͸‫ݔ‬ ݀‫ݔ‬ ݀‫ݓ‬ െ ͸‫ ݓݔ‬ൌ െͳʹ‫ ݔ‬ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ Ǥ ሺͶሻ ݀‫ݔ‬

‫ݕ‬ሺ‫ݔ‬ሻ ൌ

ͳ ඥ‫ݓ‬ሺ‫ݔ‬ሻ

ൌ

ͳ ඥʹ ൅ ܿ݁ ଷ௫ మ

for some ܿ ‫ א‬Ը.

4.2 Separable Method In this section, we will solve some differential

Then, we solve ሺͶሻ for ‫ݓ‬ሺ‫ݔ‬ሻ as follows:

equations using a method known as Separable Method.

To solve ሺͶሻ, we need to use the integral factor method:

This method is called separable because we separate

Hence, ‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௫ሻௗ௫ ൌ ݁ ‫଺ି ׬‬௫ ௗ௫ ൌ ݁

లೣమ ି మ

ൌ ݁ ିଷ௫

మ

The general solution for ‫ݓ‬ሺ‫ݔ‬ሻ is written as follows: ‫ݓ‬ൌ ‫ݓ‬ൌ

‫ ܫ ׬‬ή ‫ܨ‬ሺ‫ ݔ‬ሻ ݀‫ݔ‬ ‫ܫ‬

‫ ܫ ׬‬ή ሺെͳʹ‫ݔ‬ሻ ݀‫ݔ‬ ‫ܫ‬ మ

‫ି ݁ ׬‬ଷ௫ ή ሺെͳʹ‫ݔ‬ሻ ݀‫ݔ‬ ‫ݓ‬ൌ మ ݁ ିଷ௫ మ

ʹ ‫ି ݁ ׬‬ଷ௫ ή ሺെ͸‫ݔ‬ሻ ݀‫ݔ‬ ‫ݓ‬ൌ మ ݁ ିଷ௫ మ

ʹ݁ ିଷ௫ ൅ ܿ ‫ݓ‬ൌ మ ݁ ିଷ௫ మ

ʹ݁ ିଷ௫ ܿ ‫ ݓ‬ൌ ିଷ௫ మ ൅ ିଷ௫ మ ݁ ݁

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two different terms from each other. Definition 4.2.1 The standard form of Separable

Method is written as follows: ሺ‫ ݔ ݂݋ ݏ݉ݎ݁ݐ ݊݅ ݈݈ܣ‬ሻ݀‫ ݔ‬െ ሺ‫ݕ ݂݋ ݏ݉ݎ݁ݐ ݊݅ ݈݈ܣ‬ሻ݀‫ ݕ‬ൌ Ͳ Note: it does not matter whether it is the above form or in the following form: ሺ‫ݕ ݂݋ ݏ݉ݎ݁ݐ ݊݅ ݈݈ܣ‬ሻ݀‫ ݕ‬െ ሺ‫ ݔ ݂݋ ݏ݉ݎ݁ݐ ݊݅ ݈݈ܣ‬ሻ݀‫ ݔ‬ൌ Ͳ Example 4.2.1 Solve the following differential equation:

ௗ௬ ௗ௫

ൌ

௬య ሺ௫ାଷሻ

Solution: By using definition 4.2.1, we need to rewrite the above equation in a way that each term is separated from the other term as follows:

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ͳ ݀‫ݕ‬ ‫ݕ‬ଷ ሺ‫ ݔ‬൅ ͵ሻ ൌ ൌ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺͳሻ ͳ ݀‫ ݔ‬ሺ‫ ݔ‬൅ ͵ሻ ‫ݕ‬ଷ Now, we need to do a cross multiplication for ሺͳሻ as follows: ͳ ͳ ݀‫ ݕ‬ൌ ݀‫ݔ‬ ଷ ሺ‫ ݔ‬൅ ͵ሻ ‫ݕ‬ ͳ ͳ ݀‫ ݕ‬െ ݀‫ ݔ‬ൌ Ͳ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ ǥ ǥ ሺʹሻ ଷ ሺ‫ ݔ‬൅ ͵ሻ ‫ݕ‬


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݀‫ݕ‬ ݁ ଶ௫ ൌ ݁ ଷ௬ାଶ௫ ൌ ݁ ଷ௬ ή ݁ ଶ௫ ൌ ିଷ௬ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺͳሻ ݁ ݀‫ݔ‬ Now, we need to do a cross multiplication for ሺͳሻ as follows: ݁ ିଷ௬ ݀‫ ݕ‬ൌ ݁ ଶ௫ ݀‫ݔ‬ ݁ ିଷ௬ ݀‫ ݕ‬െ ݁ ଶ௫ ݀‫ ݔ‬ൌ Ͳ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ ǥ ሺʹሻ Then, we integrate both sides of ሺʹሻ as follows: නሺ݁ ିଷ௬ ݀‫ ݕ‬െ ݁ ଶ௫ ݀‫ ݔ‬ሻ ൌ න Ͳ

Then, we integrate both sides of ሺʹሻ as follows: ͳ ͳ න ൬ ଷ ݀‫ ݕ‬െ ݀‫ݔ‬൰ ൌ න Ͳ ሺ ‫ ݔ‬൅ ͵ሻ ‫ݕ‬ න൬

ͳ ͳ ൰ ݀‫ ݕ‬െ න ൬ ൰ ݀‫ ݔ‬ൌ ܿ ଷ ሺ ‫ ݔ‬൅ ͵ሻ ‫ݕ‬

නሺ‫ି ݕ‬ଷ ሻ݀‫ ݕ‬െ න ൬

ͳ ൰ ݀‫ ݔ‬ൌ ܿ ሺ‫ ݔ‬൅ ͵ሻ

ͳ െ ‫ି ݕ‬ଶ െ ሺȁሺ‫ ݔ‬൅ ͵ሻȁሻ ൌ ܿ ʹ

නሺ݁ ିଷ௬ ሻ݀‫ ݕ‬െ නሺ݁ ଶ௫ ሻ ݀‫ ݔ‬ൌ ܿ ͳ ͳ െ ݁ ିଷ௬ െ ݁ ଶ௫ ൌ ܿ ͵ ʹ Thus, the general solution is : ͳ ͳ െ ݁ ିଷ௬ െ ݁ ଶ௫ ൌ ܿ ͵ ʹ

4.3 Exact Method

Thus, the general solution is : ͳ െ ‫ି ݕ‬ଶ െ ሺȁሺ‫ ݔ‬൅ ͵ሻȁሻ ൌ ܿ ʹ

In this section, we will solve some differential

Example 4.2.2 Solve the following differential

Derivative Method.

equation:

ௗ௬ ௗ௫

ൌ ݁ ଷ௬ାଶ௫

Solution: By using definition 4.2.1, we need to rewrite the above equation in a way that each term is separated from the other term as follows:

86 M. Kaabar

equations using a method known as Exact Method. In other words, this method is called the Anti-Implicit Definition 4.3.1 The standard form of Exact Method is written as follows: ‫ܨ‬௫ ݀‫ݕ‬ ൌ െ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ ሺͳሻ ‫ܨ‬௬ ݀‫ݔ‬

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Then, we solve ሺͳሻ to find ‫ܨ‬ሺ‫ݔ‬ǡ ‫ݕ‬ሻ, and our general

Example 4.3.3 Solve the following differential

solution will be as follows: ‫ܨ‬ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ܿ for some constant

equation: ሺെ͵‫ ݔ‬൅ ‫ݕ‬ሻ݀‫ ݕ‬െ ሺͷ‫ ݔ‬൅ ͵‫ݕ‬ሻ݀‫ ݔ‬ൌ Ͳ

ܿ ‫ א‬Ը. In other words, the standard form for exact first

Solution: First of all, we need to check for the exact method as follows: We rewrite the above differential equation according to definition 4.3.1: ሺെ͵‫ ݔ‬൅ ‫ݕ‬ሻ݀‫ ݕ‬൅ െሺͷ‫ ݔ‬൅ ͵‫ݕ‬ሻ݀‫ ݔ‬ൌ Ͳ ሺെ͵‫ ݔ‬൅ ‫ݕ‬ሻ݀‫ ݕ‬൅ ሺെͷ‫ ݔ‬െ ͵‫ݕ‬ሻ݀‫ ݔ‬ൌ Ͳ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ Ǥ ሺͳሻ Thus, from the above differential equation, we obtain: ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ሺെͷ‫ ݔ‬െ ͵‫ݕ‬ሻ and ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ሺെ͵‫ ݔ‬൅ ‫ݕ‬ሻ

order differential equation is: ‫ܨ‬௬ ݀‫ ݕ‬൅ ‫ܨ‬௫ ݀‫ ݔ‬ൌ Ͳ, and it is considered exact if ‫ܨ‬௫௬ ൌ ‫ܨ‬௬௫ . Note: ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ is defined as the first derivative with respect to ‫ ݔ‬and considering ‫ ݕ‬as a constant, while ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ is defined as the first derivative with respect to ‫ ݕ‬and considering ‫ ݔ‬as a constant.

‫ܨ‬௫௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ‫ܨ‬௬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ as follows: ௗ௬

Example 4.3.1 Given ‫ ݔ‬ଶ ൅ ‫ ݕ‬ଶ െ Ͷ ൌ ͲǤ Find ௗ௫ . Solution: By using definition 4.3.1, we first find ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ by finding the first derivative with respect to ‫ ݔ‬and considering ‫ ݕ‬as a constant as follows: ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ʹ‫ݔ‬. Then, we find ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ by finding the first derivative with respect to ‫ ݕ‬and considering ‫ ݔ‬as a constant as follows: ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ʹ‫ݕ‬Ǥ ௗ௬

ிೣ

ଶ௫

௫

Thus, ௗ௫ ൌ െ ி ൌ െ ଶ௬ ൌ െ ௬ . ೤

Example 4.3.2 Given ‫ ݕ‬ଷ ݁ ௫ ൅ ͵‫ ݕݔ‬ଶ െ ‫ ݔ‬ଷ ൅ ‫ ݕݔ‬െ ͳ͵ ൌ ͲǤ Find

ௗ௬ ௗ௫

.

Solution: By using definition 4.3.1, we first find ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ by finding the first derivative with respect to ‫ ݔ‬and considering ‫ ݕ‬as a constant as follows: ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ‫ ݕ‬ଷ ݁ ௫ ൅ ͵‫ ݕ‬ଶ െ ͵‫ ݔ‬ଶ ൅ ‫ݕ‬. Then, we find ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ by finding the first derivative with respect to ‫ ݕ‬and considering ‫ ݔ‬as a constant as follows: ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ͵‫ ݕ‬ଶ ݁ ௫ ൅ ͸‫ ݕݔ‬൅ ‫ݔ‬Ǥ ௗ௬

ி

Thus, ௗ௫ ൌ െ ிೣ ൌ െ ೤

88 M. Kaabar

ሺ௬ య௘ ೣ ାଷ௬ మ ିଷ௫ మ ା௬ሻ ሺଷ௬ మ ௘ ೣ ା଺௫௬ା௫ሻ

Now, we need to check for the exact method by finding We first find ‫ܨ‬௫௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ by finding the first derivative of ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ with respect to ‫ ݕ‬and considering ‫ ݔ‬as a constant as follows: ‫ܨ‬௫௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ െ͵ Then, we find ‫ܨ‬௬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ by finding the first derivative of ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ with respect to ‫ ݔ‬and considering ‫ ݕ‬as a constant as follows: ‫ܨ‬௬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ െ͵ Since ‫ܨ‬௫௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ‫ܨ‬௬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ െ͵, then we can use the exact method. Now, we choose either ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ሺെͷ‫ ݔ‬െ ͵‫ݕ‬ሻ or ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ሺെ͵‫ ݔ‬൅ ‫ݕ‬ሻǡ and then we integrate. We will choose ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ሺെ͵‫ ݔ‬൅ ‫ݕ‬ሻ and we will integrate it as follows: ͳ න ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ݀‫ ݕ‬ൌ නሺെ͵‫ ݔ‬൅ ‫ݕ‬ሻ݀‫ ݕ‬ൌ െ͵‫ ݕݔ‬൅ ‫ ݕ‬ଶ ൅ ‫ܦ‬ሺ‫ ݔ‬ሻ ǥ ሺʹሻ ʹ ͳ ଶ ‫ܨ‬ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ െ͵‫ ݕݔ‬൅ ‫ ݕ‬൅ ‫ܦ‬ሺ‫ ݔ‬ሻ ʹ We need to find ‫ܦ‬ሺ‫ݔ‬ሻ as follows:

.

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Since we selected ‫ܨ‬௬ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ሺെ͵‫ ݔ‬൅ ‫ݕ‬ሻ previously for integration, then we need to find ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ for ሺʹሻ as follows: ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ െ͵‫ ݕ‬൅ ‫ܦ‬ᇱ ሺ‫ ݔ‬ሻ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺ͵ሻ Now, we substitute ‫ܨ‬௫ ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ሺെͷ‫ ݔ‬െ ͵‫ݕ‬ሻ in ሺ͵ሻ as follows: ሺെͷ‫ ݔ‬െ ͵‫ݕ‬ሻ ൌ െ͵‫ ݕ‬൅ ‫ܦ‬ᇱ ሺ‫ ݔ‬ሻ ‫ܦ‬ᇱ ሺ‫ ݔ‬ሻ ൌ െͷ‫ ݔ‬െ ͵‫ ݕ‬൅ ͵‫ ݕ‬ൌ െͷ‫ ݔ‬ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺͶሻ Then, we integrate both sides of ሺͶሻ as follows: න ‫ܦ‬ᇱ ሺ‫ ݔ‬ሻ݀‫ ݔ‬ൌ න െͷ‫ݔ݀ݔ‬ ͷ ‫ܦ‬ሺ‫ݔ‬ሻ ൌ න െͷ‫ ݔ݀ݔ‬ൌ െ ‫ ݔ‬ଶ ʹ Thus, the general solution of the exact method is : ‫ܨ‬ሺ‫ݔ‬ǡ ‫ݕ‬ሻ ൌ ܿ ͳ ͷ െ͵‫ ݕݔ‬൅ ‫ ݕ‬ଶ ൅ െ ‫ ݔ‬ଶ ൌ ܿ ʹ ʹ

4.4 Reduced to Separable Method

In this section, we will solve some differential equations using a method known as Reduced to Separable Method. Definition 4.4.1 The standard form of Reduced to

Separable Method is written as follows: ௗ௬ ௗ௫

ൌ ݂ሺܽ‫ ݔ‬൅ ܾ‫ ݕ‬൅ ܿሻ where ܽǡ ܾ ് Ͳ.

Example 4.4.1 Solve the following differential equation:

ௗ௬ ௗ௫

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ୱ୧୬ሺହ௫ା௬ሻ

ൌ ௖௢௦ሺହ௫ା௬ሻିଶୱ୧୬ሺହ௫ା௬ሻ െ ͷǤ


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Solution: By using definition 4.4.1, we first let ‫ ݑ‬ൌ ͷ‫ ݔ‬൅ ‫ݕ‬, and then, we need to find the first derivative of both sides of ‫ ݑ‬ൌ ͷ‫ ݔ‬൅ ‫ݕ‬. ݀‫ݑ‬ ݀‫ݕ‬ ൌͷ൅ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺͳሻ ݀‫ݔ‬ ݀‫ݔ‬ ௗ௬ Now, we solve ሺͳሻ for ௗ௫ as follows:

݀‫ݑ݀ ݕ‬ ൌ െ ͷ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺʹሻ ݀‫ݔ݀ ݔ‬ Then, we substitute ‫ ݑ‬ൌ ͷ‫ ݔ‬൅ ‫ ݕ‬and ሺʹሻ in ௗ௬ ௗ௫

ୱ୧୬ሺହ௫ା௬ሻ

ൌ ௖௢௦ሺହ௫ା௬ሻିଶୱ୧୬ሺହ௫ା௬ሻ െ ͷ as follows:

݀‫ݑ‬ ሺ‫ݑ‬ሻ െͷൌ െͷ ݀‫ݔ‬ ܿ‫ݏ݋‬ሺ‫ݑ‬ሻ െ ʹሺ‫ݑ‬ሻ ݀‫ݑ‬ ሺ‫ݑ‬ሻ ൌ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺ͵ሻ ሺ ݀‫ݑ ݏ݋ܿ ݔ‬ሻ െ ʹሺ‫ݑ‬ሻ Now, we can use the separable method to solve ሺ͵ሻ as follows: By using definition 4.2.1, we need to rewrite ሺ͵ሻ in a way that each term is separated from the other term as follows: ݀‫ݑ‬ ሺ‫ݑ‬ሻ ͳ ൌ ൌ ǥ ǥ ǥ ǥ Ǥ ǥ Ǥ ሺͶሻ ሺ ሻ ሺ ሻ ݀‫ ݑ ݏ݋ܿ ݔ‬െ ʹሺ‫ݑ‬ሻ ܿ‫ ݑ ݏ݋‬െ ʹሺ‫ݑ‬ሻ ሺ‫ݑ‬ሻ Now, we need to do a cross multiplication for ሺͶሻ as follows: ܿ‫ݏ݋‬ሺ‫ݑ‬ሻ െ ʹሺ‫ݑ‬ሻ ݀‫ ݑ‬ൌ ͳ݀‫ݔ‬ ሺ‫ݑ‬ሻ ܿ‫ݏ݋‬ሺ‫ݑ‬ሻ െ ʹሺ‫ݑ‬ሻ ݀‫ ݑ‬െ ͳ݀‫ ݔ‬ൌ Ͳ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺͷሻ ሺ‫ݑ‬ሻ Then, we integrate both sides of ሺͷሻ as follows: ܿ‫ݏ݋‬ሺ‫ݑ‬ሻ െ ʹሺ‫ݑ‬ሻ නቆ ݀‫ ݑ‬െ ͳ݀‫ݔ‬ቇ ൌ න Ͳ ሺ‫ݑ‬ሻ

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ሺ‫ ݔ‬൅ ͳሻ ሺଶሻ ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ Ͳ ‫ ݕ‬ሺ‫ ݔ‬ሻ െ ൌ ሺ‫ ݔ‬൅ ͳሻ ሺ‫ ݔ‬൅ ͳሻ ሺ‫ ݔ‬൅ ͳሻ

ܿ‫ݏ݋‬ሺ‫ݑ‬ሻ െ ʹሺ‫ݑ‬ሻ ቇ ݀‫ ݑ‬െ නሺͳሻ ݀‫ ݔ‬ൌ ܿ ሺ‫ݑ‬ሻ

ܿ‫ݏ݋‬ሺ‫ݑ‬ሻ ʹ ሺ‫ݑ‬ሻ නቆ െ ቇ ݀‫ ݑ‬െ ‫ ݔ‬ൌ ܿ ሺ‫ݑ‬ሻ ሺ‫ݑ‬ሻ ܿ‫ݏ݋‬ሺ‫ݑ‬ሻ නቆ െ ʹቇ ݀‫ ݑ‬െ ‫ ݔ‬ൌ ܿ ሺ‫ݑ‬ሻ ሺȁሺ‫ݑ‬ሻȁሻ െ ʹ‫ ݑ‬െ ‫ ݔ‬ൌ ܿ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ Ǥ ሺ͸ሻ

ͳ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ െ ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅

‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൌͲ ሺ‫ ݔ‬൅ ͳሻ

െͳ ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൌ Ͳ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺͳሻ ሺ‫ ݔ‬൅ ͳሻ ିଵ

Now, let ‫ܯ‬ሺ‫ ݔ‬ሻ ൌ ሺ௫ାଵሻ , and substitute it in ሺͳሻ as follows:

Now, we substitute ‫ ݑ‬ൌ ͷ‫ ݔ‬൅ ‫ ݕ‬in ሺ͸ሻ as follows: ሺȁሺͷ‫ ݔ‬൅ ‫ݕ‬ሻȁሻ െ ʹሺͷ‫ ݔ‬൅ ‫ݕ‬ሻ െ ‫ ݔ‬ൌ ܿ

‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ‫ܯ‬ሺ‫ ݔ‬ሻ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൌ Ͳ Hence, ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ is written as follows:

Thus, the general solution is :

‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ή න

ሺȁሺͷ‫ ݔ‬൅ ‫ݕ‬ሻȁሻ െ ʹሺͷ‫ ݔ‬൅ ‫ݕ‬ሻ െ ‫ ݔ‬ൌ ܿ

4.5 Reduction of Order

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݁ ‫ି ׬‬ெሺ௫ሻௗ௫ ݀‫ݔ‬ ‫ݕ‬ଵ ଶ ሺ‫ ݔ‬ሻ భ

In our example, ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ͳ ή ‫׬‬ ͳή‫׬‬

௘ ౢ౤ሺೣశభሻ ଵ

೏ೣ ‫׬‬ ௘ ሺೣశభሻ

ሺଵሻమ

݀‫ ݔ‬ൌ ଵ

݀‫ ݔ‬ൌ ‫ ݁ ׬‬୪୬ሺ௫ାଵሻ ݀‫ ݔ‬ൌ ‫׬‬ሺ‫ ݔ‬൅ ͳሻ ݀‫ ݔ‬ൌ ‫ ݔ‬ଶ ൅ ‫ݔ‬. ଶ

Method

Thus, the homogenous solution is written as follows:

In this section, we will solve differential equations

‫ݕ‬௛௢௠௢௚௘௡௢௨௦ ሺ‫ ݔ‬ሻ ൌ ܿଵ ൅ ܿଶ ቀଶ ‫ ݔ‬ଶ ൅ ‫ݔ‬ቁǡ for some ܿଵ ǡ ܿଶ ‫ א‬Ը.

using a method called Reduction of Order Method.

Example 4.5.1 Given the following differential

Definition 4.5.1 Reduction of Order Method is valid

equation: ‫ ݕݔ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ሺ‫ ݔ‬൅ ͳሻ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ െ ሺʹ‫ ݔ‬൅ ͳሻ‫ ݕ‬ൌ ‫଻ ݁ݔ‬௫ ,

method only for second order differential equations,

and ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ݁ ௫ is a solution to the associated

and one solution to the homogenous part must be

homogenous part. Find ‫ݕ‬௛௢௠௢௚௘௡௢௨௦ ሺ‫ ݔ‬ሻ? (Hint: Find first

given. For example, given ሺ‫ ݔ‬൅ ͳሻ‫ ݕ‬ሺଶሻ ሺ‫ ݔ‬ሻ െ ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ ൌ Ͳǡ

‫ݕ‬ଶ ሺ‫ ݔ‬ሻ, and then write ‫ݕ‬௛௢௠௢௚௘௡௢௨௦ ሺ‫ ݔ‬ሻ)

and ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ͳ. To find ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ, the differential equation

Solution: By using definition 4.5.1, To find ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ, the

must be written in the standard form (Coefficient of

differential equation must be equal to zero and must

ଵ

‫ ݕ‬ሺଶሻ must be 1) as follows:

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also be written in the standard form (Coefficient of ‫ ݕ‬ሺଶሻ must be 1) as follows: ‫ݕݔ‬

‫ݔ‬ሻ ൅ ሺ‫ ݔ‬൅ ͳሻ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ െ ሺʹ‫ ݔ‬൅ ͳሻ‫ ݕ‬ൌ Ͳ by ‫ ݔ‬as follows:

ሺ‫ ݔ‬൅ ͳሻ ᇱ ‫ ݔ‬ሺଶሻ ʹ‫ ݔ‬൅ ͳ Ͳ ‫ ݕ‬ሺ‫ ݔ‬ሻ ൅ ‫ ݕ‬ሺ‫ ݔ‬ሻ െ ‫ݕ‬ൌ ‫ݔ‬ ‫ݔ‬ ‫ݔ‬ ‫ݔ‬ ሺ ሻ ‫ݔ‬ ൅ ͳ ʹ‫ݔ‬ ൅ ͳ ͳ‫ ݕ‬ሺଶሻ ሺ‫ ݔ‬ሻ ൅ ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ െ ‫ ݕ‬ൌ Ͳ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ ǥ ሺͳሻ ‫ݔ‬ ‫ݔ‬ Now, let ሺ‫ݔ‬ሻ ൌ

ሺ௫ାଵሻ ௫

ଵ

ൌ ቀͳ ൅ ௫ቁ , and substitute it in ሺͳሻ

as follows: ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ ൅ ‫ܯ‬ሺ‫ݔ‬ሻ‫ ݕ‬ᇱ ሺ‫ ݔ‬ሻ െ

ʹ‫ ݔ‬൅ ͳ ‫ݕ‬ൌͲ ‫ݔ‬

Hence, ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ is written as follows: ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ή න ௫

݁ ‫ି ׬‬ெሺ௫ሻௗ௫ ݀‫ݔ‬ ‫ݕ‬ଵ ଶ ሺ‫ ݔ‬ሻ

In example 4.5.1, ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ή ‫׬‬ ଵ ୪୬ቀ ቁ ௫

݁ ି௫ ή ݁ ݁ ήන ݁ ଶ௫ ௫

௘

భ ‫׬‬ቀభశೣቁ೏ೣ

ሺ௘ ೣ ሻమ

݀‫ ݔ‬ൌ ݁ ௫ ή න

݀‫ ݔ‬ൌ

݁ ିଷ௫ ݀‫ݔ‬ ‫ݔ‬

Since it is impossible to integrate ݁ ௫ ή ‫׬‬ ௫ ௘ షయ೟

is enough to write it as: ݁ ௫ ή ‫׬‬଴ ௫ ௘ షయ೟

Therefore, ‫ݕ‬ଶ ሺ‫ ݔ‬ሻ ൌ ݁ ௫ ή ‫׬‬଴

௧

௧

௘ షయೣ ௫

݀‫ݔ‬, then it

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4.6 Exercises 1. Given ሺ‫ ݔ‬൅ ͳሻ‫ ݕ‬ᇱ ൅ ‫ ݕݔ‬ൌ

We divide both sides of ሺଶሻ ሺ


ሺ௫ାଵሻర ௬మ

Ǥ Find the general

solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: Use Bernoulli method and no need to find the value of ܿ) 2. Given ‫ ݔ‬ᇱ ൅ ͵‫ ݔݕ‬ൌ ͵‫ ݕ‬ଷ Ǥ Find the general solution for ‫ݕ‬ሺ‫ ݔ‬ሻ. (Hint: Use Bernoulli method and no need to find the value of ܿ) 3. Solve the following differential equation:

ௗ௬ ௗ௫ ௗ௬

ൌ

ଵା௬ మ ଵା௫ మ ଵ

4. Solve the following differential equation: ௗ௫ ൌ ଷ௫ା௫ మ௬ 5. Solve the following differential equation: ݀‫ݕ‬ ൌ ͵‫ ݁ݔ‬ሺ௫ାହ௬ሻ ݀‫ݔ‬ 6. Solve the following differential equation: ͳ ͵ ሺ݁ ௫ ‫ ݕ‬൅ ͵‫ ݔݕ‬െ ʹሻ݀‫ ݕ‬൅ ൬ ݁ ௫ ‫ ݕ‬ଶ ൅ ‫ ݕ‬ଶ ൅ ‫ ݔ‬ଶ ൰ ݀‫ ݔ‬ൌ Ͳ ʹ ʹ 7. Solve the following differential equation: ݀‫ݕ‬ ሺͷ‫ ݔ‬൅ ‫ݕ‬ሻ ൌ െͷ ݀‫ݏ݋ܿ ݔ‬ሺͷ‫ ݔ‬൅ ‫ݕ‬ሻ െ ʹሺͷ‫ ݔ‬൅ ‫ݕ‬ሻ 8. Given the following differential equation: ሺ‫ ݔ‬൅ ͳሻ‫ ݕ‬ሺଶሻ ሺ‫ݔ‬ሻ െ ‫ ݕ‬ᇱ ሺ‫ݔ‬ሻ ൌ ͳͲǡ and ‫ݕ‬ଵ ሺ‫ ݔ‬ሻ ൌ ͳ is a solution to the associated homogenous part. Find ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ?

݀‫ݐ‬.

݀‫ݐ‬.

Thus, the homogenous solution is written as follows: ௫ ௘ షయ೟

‫ݕ‬௛௢௠௢௚௘௡௢௨௦ ሺ‫ ݔ‬ሻ ൌ ܿଵ ݁ ௫ ൅ ܿଶ ቀ݁ ௫ ή ‫׬‬଴

௧

݀‫ݐ‬ቁǡ for some

ܿଵ ǡ ܿଶ ‫ א‬Ը.

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Chapter 5

a) How long will it take for the temperature of the

Applications of Differential

b) What will be the temperature of the engine 30

engine to cool to ͳͳ͹Ԭ? minutes from now?

Equations

Solution: Part a: To determine how long will it take for

In this chapter, we give examples of three different

to do the following:

applications of differential equations: temperature,

Assume that ܶሺ‫ݐ‬ሻ is the temperature of engine at the

growth and decay, and water tank. In each section, we

time ‫ݐ‬, and ܶ଴ is the constant outside air temperature.

give one example of each of the above applications, and

Now, we need to write the differential equation for this

we discuss how to use what we have learned previously

example as follows:

in this book to solve each problem.

݀ܶ ൌ ߚ ሺܶ െ ܶ଴ ሻ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺͳሻ ݀‫ݐ‬

5.1 Temperature Application

where ߚ is a constant.

In this section, we give an example of temperature application, and we introduce how to use one of the

the temperature of the engine to cool to ͳͳ͹Ԭ, we need

From ሺͳሻ, we can write as follows: ܶ ᇱ ൌ ߚܶ െ ߚܶ଴ ܶ ᇱ െ ߚܶ ൌ െߚܶ଴ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺʹሻ

differential equations methods to solve it.

From this example, it is given the following:

Example 5.1.1 Thomas drove his car from Pullman,

ܶሺͲሻ ൌ ͳͶͶԬ, ܶሺͳͲሻ ൌ ͳ͵͸Ԭ, and ܶ଴ ൌ ͳͲͶԬ

WA to Olympia, WA, and the outside air temperature

From ሺʹሻ, െߚܶ଴ is constant, and the dependent variable

was constant ͳͲͶԬ. During his trip, he took a break at

is ܶ, while the independent variable is the time ‫ݐ‬.

Othello, WA gas station, and then he switched off the

By substituting ܶ଴ ൌ ͳͲͶԬ in ሺʹሻ, we obtain:

engine of his car, and checked his car temperature

ܶ ᇱ െ ߚܶ ൌ െͳͲͶߚ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺ͵ሻ

gauge, and it was ͳͶͶԬ. After ten minutes, Thomas

Since ሺ͵ሻ is a first order linear differential equation,

checked his car temperature gauge, and it was ͳ͵͸Ԭ.

then by using definition 4.1.1, we need to use the

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integral factor method by letting ‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௧ሻௗ௧ , where


By substituting ܶሺͳͲሻ ൌ ͳ͵͸Ԭ in ሺͷሻ, we obtain: ܶሺͳͲሻ ൌ ͳͲͶ ൅ ͶͲ݁ ఉሺଵ଴ሻ

݃ሺ‫ݐ‬ሻ ൌ െߚ and ‫ ܨ‬ሺ‫ݐ‬ሻ ൌ െͳͲͶߚ. Hence, ‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௧ሻௗ௧ ൌ ݁ ‫ି ׬‬ఉ ௗ௧ ൌ ݁ ିఉ௧ .

ͳ͵͸ ൌ ͳͲͶ ൅ ͶͲ݁ ఉሺଵ଴ሻ


ͳ͵͸ െ ͳͲͶ ൌ ͶͲ݁ ఉሺଵ଴ሻ

ܶሺ‫ݐ‬ሻ ൌ ܶሺ‫ݐ‬ሻ ൌ

‫݁׬‬

ܶሺ‫ݐ‬ሻ ൌ

͵ʹ ൌ ͶͲ݁ ఉሺଵ଴ሻ

‫ ܫ ׬‬ή ‫ ܨ‬ሺ‫ݐ‬ሻ ݀‫ݐ‬ ‫ܫ‬

ିఉ௧

ή ሺെͳͲͶߚሻ ݀‫ݐ‬ ݁ ିఉ௧

‫׬‬ሺെͳͲͶߚሻ݁ ݁ ିఉ௧

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ିఉ௧

݀‫ݐ‬

݁ ఉሺଵ଴ሻ ൌ

By taking the natural logarithm for both sides of ሺ͸ሻ, we obtain: ݈݊൫ ݁ ఉሺଵ଴ሻ ൯ ൌ ݈݊ሺͲǤͺሻ

ͳͲͶ݁ ିఉ௧ ൅ ܿ ܶሺ‫ݐ‬ሻ ൌ ݁ ିఉ௧ ͳͲͶ݁ ିఉ௧ ܿ ൅ ିఉ௧ ିఉ௧ ݁ ݁ ܿ ܶሺ‫ݐ‬ሻ ൌ ͳͲͶ ൅ ିఉ௧ ݁

ܶሺ‫ ݐ‬ሻ ൌ

͵ʹ ൌ ͲǤͺ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺ͸ሻ ͶͲ

ߚ ሺͳͲሻ ൌ ݈݊ሺͲǤͺሻ ߚൌ

݈ ݊ሺͲǤͺሻ ൌ െͲǤͲʹʹ͵ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ ሺ͹ሻ ͳͲ

Now, we substitute ሺ͹ሻ in ሺͷሻ as follows: ܶሺ‫ݐ‬ሻ ൌ ͳͲͶ ൅ ͶͲ݁ ି଴Ǥ଴ଶଶଷ௧ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ ǥ Ǥ ሺͺሻ

ܶሺ‫ݐ‬ሻ ൌ ͳͲͶ ൅ ܿ݁ ఉ௧ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺͶሻ

Then, we need to find the time ‫ ݐ‬when ܶሺ‫ݐ‬ሻ ൌ ͳͳ͹Ԭ by

The general solution is: ܶሺ‫ݐ‬ሻ ൌ ͳͲͶ ൅ ܿ݁ ఉ௧ for some

substituting it in ሺͺሻ as follows:

ܿ ‫ א‬Ը.

ͳͳ͹ ൌ ͳͲͶ ൅ ͶͲ݁ ି଴Ǥ଴ଶଶଷሺ୲ሻ

Now, we need to find ܿ by substituting ܶሺͲሻ ൌ ͳͶͶԬ in

ͳͳ͹ െ ͳͲͶ ൌ ͶͲ݁ ି଴Ǥ଴ଶଶଷሺ୲ሻ

ሺͶሻ as follows:

͵ ൌ ͶͲ݁ ି଴Ǥ଴ଶଶଷሺ୲ሻ ܶሺͲሻ ൌ ͳͲͶ ൅ ܿ݁

ఉሺ଴ሻ

ͳͶͶ ൌ ͳͲͶ ൅ ܿ݁

଴

ͳͶͶ ൌ ͳͲͶ ൅ ܿሺͳሻ ͳͶͶ ൌ ͳͲͶ ൅ ܿ

݁ ି଴Ǥ଴ଶଶଷሺ୲ሻ ൌ

͵ ൌ ͲǤͲ͹ͷ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ሺͻሻ ͶͲ

By taking the natural logarithm for both sides of ሺͻሻ, we obtain:

ܿ ൌ ͳͶͶ െ ͳͲͶ ൌ ͶͲ

݈݊൫݁ ି଴Ǥ଴ଶଶଷሺ௧ሻ ൯ ൌ ݈݊ሺͲǤͲ͹ͷሻ

Thus, ܶሺ‫ݐ‬ሻ ൌ ͳͲͶ ൅ ͶͲ݁ ఉ௧ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ሺͷሻ

െͲǤͲʹʹ͵ሺ‫ݐ‬ሻ ൌ ݈݊ሺͲǤͲ͹ͷሻ

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݈ ݊ሺͲǤͲ͹ͷሻ ൎ ͳͳ͸Ǥͳ͸ െͲǤͲʹʹ͵

Thus, the temperature of the engine will take approximately ͳͳ͸Ǥͳ͸ minutes to cool to ͳͳ͹Ԭ Part b: To determine what will be the temperature of the engine 30 minutes from now, we need to do the following: We assume that ‫ ݐ‬ൌ ͵Ͳ, and then we substitute it in ሺͺሻ as follows: ܶሺ͵Ͳሻ ൌ ͳͲͶ ൅ ͶͲ݁ ି଴Ǥ଴ଶଶଷሺଷ଴ሻ ܶሺ͵Ͳሻ ൎ ͳʹͶǤͶͻԬ Thus, the temperature of the engine 30 minutes from now will be approximately ͳʹͶǤͶͻԬ.


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a) How long will it take to double number of WSU students in 2013? b) What will be the number of WSU students in 2018? Solution: Part a: To determine how long will it take to double number of WSU students in 2013, we need to do the following: Assume that ܹሺ‫ݐ‬ሻ is the number of WSU students at any time ‫ݐ‬. Now, we need to write the differential equation for this example as follows: ܹ݀ ൌ ߚඥܹሺ‫ݐ‬ሻ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺͳሻ ݀‫ݐ‬

5.2 Growth and Decay

where ߚ is a constant.

Application

ܹ ᇱ ൌ ߚඥܹሺ‫ݐ‬ሻ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺʹሻ

In this section, we give an example of growth and decay application, and we introduce how to use one of the differential equations methods to solve it. Example 5.2.1 The rate change of number of students at Washington State University (WSU) is proportional to the square root of the number of students at any time ‫ݐ‬. If the number of WSU students in 2013 was 28,686 students2, and suppose that the number of

From ሺͳሻ, we can write as follows:

From this example, it is given the following: ܹሺͲሻ ൌ ʹͺǡ͸ͺ͸, and ܹ ሺͳሻ ൌ ͵ʹǡͲͲͲ. From ሺʹሻ, the dependent variable is ܹ, while the independent variable is the time ‫ݐ‬. To solve ሺͳሻ, we need to use separable method as follows: By using definition 4.2.1, we need to rewrite ሺͳሻ in a way that each term is separated from the other term as follows:

students at WSU after one year was 32,000 students.

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ܹ݀ ߚ ൌ ߚඥܹሺ‫ݐ‬ሻ ൌ ଵ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺ͵ሻ ݀‫ݐ‬ ܹ ିଶ


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Now, we need to find ܿ by substituting ܹ ሺͲሻ ൌ ʹͺǡ͸ͺ͸ in ሺ͹ሻ as follows:

Now, we need to do a cross multiplication for ሺ͵ሻ as follows:

ܹ ሺͲሻ ൌ ൬

ܿ ൅ ߚሺͲሻ ଶ ൰ ʹ

ܿ൅Ͳ ଶ ൰ ʹ ܿ ଶ ʹͺǡ͸ͺ͸ ൌ ቀ ቁ ʹ

ଵ

൬ܹ ିଶ ൰ ܹ݀ ൌ ߚ݀‫ݐ‬

ʹͺǡ͸ͺ͸ ൌ ൬

ଵ

൬ܹ ିଶ ൰ ܹ݀ െ ߚ݀‫ ݐ‬ൌ Ͳ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ ሺͶሻ Then, we integrate both sides of ሺͶሻ as follows:

ʹͺǡ͸ͺ͸ ൌ

ܿଶ Ͷ

ܿ ଶ ൌ Ͷሺʹͺǡ͸ͺ͸ሻ

ଵ

න ቆ൬ܹ ିଶ ൰ ܹ݀ െ ߚ݀‫ݐ‬ቇ ൌ න Ͳ

ܿ ൌ ඥͶሺʹͺǡ͸ͺ͸ሻ ଵ න ൬ܹ ିଶ ൰ ܹ݀ ଵ ʹܹ ଶ

ܿ ൎ ͵͵ͺǤ͹Ͷ

െ නሺߚሻ ݀‫ ݐ‬ൌ ܿ Thus, ܹ ሺ‫ݐ‬ሻ ൌ ቀ

െ ߚ‫ ݐ‬ൌ ܿ

ଶ

ቁ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ሺͺሻ

By substituting ܹ ሺͳሻ ൌ ͵ʹǡͲͲͲ in ሺͺሻ, we obtain:

Thus, the general solution is : ଵ ʹܹ ଶ

ଷଷ଼Ǥ଻ସାఉ௧ ଶ

ܹ ሺ‫ ݐ‬ሻ ൌ ൬

െ ߚ‫ ݐ‬ൌ ܿ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺͷሻ

͵͵ͺǤ͹Ͷ ൅ ߚ ଶ ൰ ʹ

ሺ͵͵ͺǤ͹Ͷሻଶ ൅ ʹሺ͵͵ͺǤ͹Ͷሻߚ ൅ ߚ ଶ ቇ Ͷ

for some ܿ ‫ א‬Ը.

͵ʹǡͲͲͲ ൌ ቆ

Then, we rewrite ሺͷሻ as follows:

ሺ͵͵ͺǤ͹Ͷሻଶ ൅ ʹሺ͵͵ͺǤ͹Ͷሻߚ ൅ ߚ ଶ ൌ Ͷሺ͵ʹǡͲͲͲሻ

ଵ

ʹܹ ଶ ൌ ܿ ൅ ߚ‫ݐ‬ ଵ ܹଶ

ܿ ൅ ߚ‫ݐ‬ ൌ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ Ǥ ሺ͸ሻ ʹ

We square both sides of ሺ͸ሻ as follows: ଶ

ܹ ሺ‫ ݐ‬ሻ ൌ ൬

ܿ ൅ ߚ‫ݐ‬ ൰ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ ǥ Ǥ ǥ ሺ͹ሻ ʹ

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ʹሺ͵͵ͺǤ͹Ͷሻߚ ൅ ߚ ଶ ൌ Ͷሺ͵ʹǡͲͲͲሻ െ ሺ͵͵ͺǤ͹Ͷሻଶ ߚ ଶ ൅ ʹሺ͵͵ͺǤ͹Ͷሻߚ െ ͳ͵ǡʹͷͷǤʹͳʹͶ ൌ Ͳ Thus, ߚ ൎ ͸͵ǡ͸͸ͳǤʹ͸ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ ሺͻሻ Now, we substitute ሺͻሻ in ሺͺሻ as follows: ܹ ሺ‫ ݐ‬ሻ ൌ ൬

͵͵ͺǤ͹Ͷ ൅ ሺ͸͵ǡ͸͸ͳǤʹ͸ሻ‫ ݐ‬ଶ ൰ ǥ ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ Ǥ ሺͳͲሻ ʹ

Then, we need to find the time ‫ ݐ‬when

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ܹሺ‫ݐ‬ሻ ൌ ʹሺʹͺǡ͸ͺ͸ሻ ൌ ͷ͹ǡ͵͹ʹ by substituting it in ሺͳͲሻ as

has a tank that contains initially 350 gallons of

follows:

purified water, given that when ‫ ݐ‬ൌ Ͳ, the amount of ͷ͹ǡ͵͹ʹ ൌ ൬

͵͵ͺǤ͹Ͷ ൅ ሺ͸͵ǡ͸͸ͳǤʹ͸ሻ‫ ݐ‬ଶ ൰ ʹ

minerals is ͷ bounds. Suppose that there is a mixture of minerals containing 0.2 bound of minerals per gallon

ሺ͵͵ͺǤ͹Ͷሻଶ ൅ ʹሺ͵͵ͺǤ͹Ͷሻሺ͸͵ǡ͸͸ͳǤʹ͸ሻ‫ ݐ‬൅ ‫ ݐ‬ଶ ൌ Ͷሺͷ͹ǡ͵͹ʹሻ ‫ ݐ‬ଶ ൅ ʹሺ͵͵ͺǤ͹Ͷሻሺ͸͵ǡ͸͸ͳǤʹ͸ሻ‫ ݐ‬െ ͳͳͶǡ͹Ͷ͵ǤʹͳʹͶ ൌ Ͳ ‫ ݐ‬ൎ ͲǤͲͲʹ͸͸ years

is poured into the tank at rate of 5 gallons per minute, while the mixture of minerals goes out of the tank at rate of 2 gallons per minute.

Thus, it will take approximately ͲǤͲͲʹ͸͸ years to

a) What is the amount of minerals in the tank of

double the number of WSU students in 2013.

WSU Water Tower at any time ‫?ݐ‬

Part b: To determine what will be number of WSU

b) What is the concentration of minerals in the

students in 2018, we need to do the following:

tank of WSU Water Tower at ‫ ݐ‬ൌ ͵Ͷ minutes?

We assume that ‫ ݐ‬ൌ ʹͲͳͺ, and then we substitute it in ሺͳͲሻ as follows: ଶ

ܹሺʹͲͳͺሻ ൌ ቆ

͵͵ͺǤ͹Ͷ ൅ ሺ͸͵ǡ͸͸ͳǤʹ͸ሻሺʹͲͳͺሻ ቇ ʹ

ܹሺʹͲͳͺሻ ൎ ͶǤͳʹ͸ ൈ ͳͲଵହ students Thus, the number of WSU students will be approximately ͶǤͳʹ͸ ൈ ͳͲଵହ students in 2018.

5.3 Water Tank Application In this section, we give an example of water tank application, and we introduce how to use one of the

Solution: Part a: To determine the amount of minerals in the tank of WSU Water Tower at any time ‫ݐ‬, we need to do the following: Assume that ܹሺ‫ݐ‬ሻ is the amount of minerals at any time ‫ݐ‬, and ‫ܯ‬ሺ‫ݐ‬ሻ is the concentration of minerals in the tank at any time ‫ݐ‬. ‫ܯ‬ሺ‫ݐ‬ሻ is written in the following form: ‫ ܯ‬ሺ‫ ݐ‬ሻ ൌ ൌ

݄ܶ݁ ܽ݉‫ݏ݈ܽݎ݁݊݅ܯ ݂݋ ݐ݊ݑ݋‬ ݄ܶ݁ ܸ‫ݎ݁ݐܹܽ ݂݀݁݅݅ݎݑܲ ݂݋ ݁݉ݑ݈݋‬

ܹሺ‫ݐ‬ሻ ǥ ሺͳሻ ܲ‫ ݎ݁ݐܹܽ ݂݀݁݅݅ݎݑ‬൅ ሺሺ‫ ݁ݐܴܽ ݎ݁݊݊ܫ‬െ ܱ‫݁ݐܴܽ ݎ݁ݐݑ‬ሻ‫ݐ‬ሻ

differential equations methods to solve it.

From this example, it is given the following:

Example 5.3.1 One of the most beautiful places at

ܹሺͲሻ ൌ ͷ ܾ‫ݏ݀݊ݑ݋‬, ‫ ݁ݐܴܽ ݎ݁݊݊ܫ‬ൌ ͷ Ȁ, and

Washington State University campus is known as

ܱ‫ ݁ݐܴܽ ݎ݁ݐݑ‬ൌ ʹ Ȁ.

WSU Water Tower. Assume thatWSU Water Tower

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Now, we need to rewrite our previous equation for this

ܹሺ‫ݐ‬ሻ ൌ

example by substituting what is given in the example itself in ሺͳሻ as follows: ‫ ܯ‬ሺ‫ ݐ‬ሻ ൌ


ܹሺ‫ݐ‬ሻ ܹ ሺ‫ ݐ‬ሻ ൌ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǡ ǥ ǥ ሺʹሻ ሺ ሻ ͵ͷͲ ൅ ሺ ͷ െ ͵ ‫ݐ‬ሻ ͵ͷͲ ൅ ʹ‫ݐ‬

ܹ݀ ൌ ͲǤʹ ή ‫ ݁ݐܴܽ ݎ݁݊݊ܫ‬െ ‫ܯ‬ሺ‫ݐ‬ሻ ή ܱ‫݁ݐܴܽ ݎ݁ݐݑ‬ ݀‫ݐ‬ ܹ݀ ܹ ሺ‫ ݐ‬ሻ ൌ ͲǤʹ ή ሺͷሻ െ ቆ ቇ ή ሺʹሻ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ሺ͵ሻ ݀‫ݐ‬ ͵ͷͲ ൅ ʹ‫ݐ‬

ܹ ሺ‫ ݐ‬ሻ ൌ ቆ

‫ ܫ ׬‬ή ‫ܨ‬ሺ‫ݐ‬ሻ ݀‫ݐ‬ ‫ܫ‬

‫׬‬ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ ή ሺͳሻ ݀‫ݐ‬ ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ


‫׬‬ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ ݀‫ݐ‬ ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ


͵ͷͲ‫ ݐ‬൅ ‫ ݐ‬ଶ ൅ ܿ ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ

From ሺʹሻ, we can write the differential equation as follows:

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͵ͷͲ‫ݐ‬ ‫ݐ‬ଶ ܿ ൅ ൅ ቇ ǥ Ǥ Ǥ Ǥ Ǥ Ǥ ሺͷሻ ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ

The general solution is: ௧మ

ଷହ଴௧

௖

From ሺ͵ሻ, the dependent variable is ܹ, while the

ܹሺ‫ݐ‬ሻ ൌ ቀሺଷହ଴ାଶ௧ሻ ൅ ሺଷହ଴ାଶ௧ሻ ൅ ሺଷହ଴ାଶ௧ሻቁ for some ܿ ‫ א‬Ը.

independent variable is the time ‫ݐ‬. Then, we rewrite

Now, we need to find ܿ by substituting

ሺ͵ሻ as follows:

ܹሺͲሻ ൌ ͷ bounds in ሺͷሻ as follows:

ܹ ᇱ ሺ‫ݐ‬ሻ ൌ ͲǤʹ ή ሺͷሻ െ ቆ

ܹ ሺ‫ ݐ‬ሻ ቇ ή ሺʹሻ ͵ͷͲ ൅ ʹ‫ݐ‬

ܹ ᇱ ሺ‫ݐ‬ሻ ൌ ͳ െ ሺʹሻ ቆ ܹ ᇱ ሺ‫ ݐ‬ሻ ൅ ൬

ܹ ሺͲሻ ൌ ቆ

ሺͲሻଶ ܿ ͵ͷͲሺͲሻ ൅ ൅ ቇ ሺ͵ͷͲ ൅ ʹሺͲሻሻ ሺ͵ͷͲ ൅ ʹሺͲሻሻ ሺ͵ͷͲ ൅ ʹሺͲሻሻ

ܹ ሺ‫ ݐ‬ሻ ቇ ͵ͷͲ ൅ ʹ‫ݐ‬

ͷ ൌ ൬Ͳ ൅ Ͳ ൅

ʹ ൰ ܹ ሺ‫ݐ‬ሻ ൌ ͳ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ ሺͶሻ ͵ͷͲ ൅ ʹ‫ݐ‬

ܿ ൰ ሺ͵ͷͲ ൅ ʹሺͲሻሻ

ͷൌቀ

ܿ ቁ ͵ͷͲ

ܿ ൌ ሺͷሻሺ͵ͷͲሻ ൌ ͳ͹ͷͲ

Since ሺͶሻ is a first order linear differential equation,

ଷହ଴௧

௧మ

ଵ଻ହ଴

then by using definition 4.1.1, we need to use the

Thus, ܹ ሺ‫ݐ‬ሻ ൌ ቀሺଷହ଴ାଶ௧ሻ ൅ ሺଷହ଴ାଶ௧ሻ ൅ ሺଷହ଴ାଶ௧ሻቁ Ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ሺ͸ሻ

integral factor method by letting ‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௧ሻௗ௧ , where

The amount of minerals in the tank of WSU Water

ଶ

݃ሺ‫ݐ‬ሻ ൌ ቀଷହ଴ାଶ௧ ቁ and ‫ ܨ‬ሺ‫ݐ‬ሻ ൌ ͳ. మ

‫ ܫ‬ൌ ݁ ‫ ׬‬௚ሺ௧ሻௗ௧ ൌ ݁ ‫׬‬ቀయఱబశమ೟

ቁ ௗ௧

ൌ ݁ ୪୬ሺଷହ଴ାଶ௧ሻ ൌ ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ.

Tower at any time ‫ ݐ‬is: ܹ ሺ‫ ݐ‬ሻ ൌ ቆ

‫ݐ‬ଶ ͳ͹ͷͲ ͵ͷͲ‫ݐ‬ ൅ ൅ ቇ ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ ሺ͵ͷͲ ൅ ʹ‫ݐ‬ሻ


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Part b: To determine the concentration of minerals in the tank of WSU Water Tower at ‫ ݐ‬ൌ ͵Ͷ minutes, we need to do the following: We substitute ‫ ݐ‬ൌ ͵Ͷ minutes in ሺ͸ሻ as follows: ܹ ሺ͵Ͷሻ ൌ ቆ

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Appendices Review of Linear Algebra

ଶ

͵ͷͲሺ͵Ͷሻ ൅ ሺ͵Ͷሻ ൅ ͳ͹ͷͲ ቇ ൎ ͵ͷǤͶʹ ሺ͵ͷͲ ൅ ʹሺ͵Ͷሻሻ

Thus, the concentration of minerals in the tank of WSU Water Tower at ‫ ݐ‬ൌ ͵Ͷ minutes is approximately ͵ͷǤͶʹ minutes.


Appendix A: Determinants* *The materials of appendix A are taken from section 1.7 in my published book titled A First Course in

Linear Algebra: Study Guide for the Undergraduate Linear Algebra Course, First Edition1. In this section, we introduce step by step for finding determinant of a certain matrix. In addition, we discuss some important properties such as invertible and non-invertible. In addition, we talk about the effect of row-operations on determinants. Definition A.1 Determinant is a square matrix. Given ଶ ሺԹሻ ൌ Թ૛ൈ૛ ൌ Թ૛ൈ૛ , let ‫ א‬ଶ ሺԹሻ where A is ʹ ൈ ʹ ܽଵଵ ܽଵଶ matrix, ൌ ቂܽ ܽଶଶ ቃǤ The determinant of A is ଶଵ represented by ሺሻ ȁȁ. Hence, ሺሻ ൌ ȁȁ ൌ ܽଵଵ ܽଶଶ െ ܽଵଶ ܽଶଵ ‫ א‬Թ. (Warning: this definition works only for ʹ ൈ ʹ matrices). Example A.1 Given the following matrix: ͵ ʹ ቃ ൌቂ ͷ ͹ Find the determinant of A. Solution: Using definition A.1, we do the following:

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ሺሻ ൌ ȁȁ ൌ ሺ͵ሻሺ͹ሻ െ ሺʹሻሺͷሻ ൌ ʹͳ െ ͳͲ ൌ ͳͳǤ Thus, the determinant of A is 11. Example A.2 Given the following matrix: ͳ Ͳ ʹ ൌ ൥͵ ͳ െͳ൩ ͳ ʹ Ͷ Find the determinant of A. Solution: Since A is ͵ ൈ ͵ matrix such that ‫ א‬ଷ ሺԹሻ ൌ Թ૜ൈ૜ , then we cannot use definition A.1 because it is valid only for ʹ ൈ ʹ matrices. Thus, we need to use the following method to find the determinant of A. Step 1: Choose any row or any column. It is recommended to choose the one that has more zeros. In this example, we prefer to choose the second column or the first row. Let’s choose the second column as follows: ͳ Ͳ ʹ ൌ ൥͵ ͳ െͳ൩ ͳ ʹ Ͷ ܽଵଶ ൌ Ͳǡ ܽଶଶ ൌ ͳ ܽଷଶ ൌ ʹǤ Step 2: To find the determinant of A, we do the following: For ܽଵଶ , since ܽଵଶ is in the first row and second column, then we virtually remove the first row and second column. ͳ Ͳ ʹ ൌ ൥͵ ͳ െͳ൩ ͳ ʹ Ͷ ͵ െͳ ଵାଶ ቃ ሺെͳሻ ܽଵଶ ቂ ͳ Ͷ For ܽଶଶ , since ܽଶଶ is in the second row and second column, then we virtually remove the second row and second column.

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ͳ ൌ ൥͵ ͳ

Ͳ ͳ ʹ

ʹ െͳ൩ Ͷ ͳ ଶାଶ ሺെͳሻ ܽଶଶ ቂ ͳ

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ʹ ቃ Ͷ

For ܽଷଶ , since ܽଷଶ is in the third row and second column, then we virtually remove the third row and second column. ͳ Ͳ ʹ ൌ ൥͵ ͳ െͳ൩ ͳ ʹ Ͷ ͳ ʹ ଷାଶ ቃ ሺെͳሻ ܽଷଶ ቂ ͵ െͳ Step 3: Add all of them together as follows: ͵ െͳ ͳ ʹ ቃ ൅ ሺെͳሻଶାଶ ܽଶଶ ቂ ቃ ͳ Ͷ ͳ Ͷ ͳ ʹ ቃ ൅ ሺെͳሻଷାଶ ܽଷଶ ቂ ͵ െͳ ͵ െͳ ͳ ʹ ቃ ൅ ሺെͳሻସ ሺͳሻ ቂ ቃ ሺሻ ൌ ሺെͳሻଷ ሺͲሻ ቂ ͳ Ͷ ͳ Ͷ ͳ ʹ ቃ ൅ ሺെͳሻହ ሺʹሻ ቂ ͵ െͳ ͵ െͳ ͳ ʹ ቃ ൅ ሺͳሻሺͳሻ ቂ ቃ ሺሻ ൌ ሺെͳሻሺͲሻ ቂ ͳ Ͷ ͳ Ͷ ͳ ʹ ቃ ൅ ሺെͳሻሺʹሻ ቂ ͵ െͳ ሺሻ ൌ ሺെͳሻሺͲሻሺͳʹ െ െͳሻ ൅ ሺͳሻሺͳሻሺͶ െ ʹሻ ൅ ሺെͳሻሺʹሻሺെͳ

ሺሻ ൌ ሺെͳሻଵାଶ ܽଵଶ ቂ

െ ͸ሻ ሺሻ ൌ Ͳ ൅ ʹ ൅ ͳͶ ൌ ͳ͸Ǥ Thus, the determinant of A is 16. Result A.1 Let ‫ א‬௡ ሺԹሻ. Then, A is invertible (non-singular) if and only if ሺሻ ് ͲǤ

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The above result means that if ሺሻ ് Ͳ, then A is invertible (non-singular), and if A is invertible (nonsingular), then ሺሻ ് Ͳ. Example A.3 Given the following matrix: ʹ ͵ ቃ ൌቂ Ͷ ͸ Is A invertible (non-singular)? Solution: Using result A.1, we do the following: ሺሻ ൌ ȁȁ ൌ ሺʹሻሺ͸ሻ െ ሺ͵ሻሺͶሻ ൌ ͳʹ െ ͳʹ ൌ ͲǤ Since the determinant of A is 0, then A is noninvertible (singular). Thus, the answer is No because A is non-invertible (singular). ܽଵଵ ܽଵଶ Definition A.2 Given ൌ ቂܽ ܽଶଶ ቃ. Assume that ଶଵ ሺሻ ് Ͳ such that ሺሻ ൌ ܽଵଵ ܽଶଶ െ ܽଵଶ ܽଶଵ . To find ିଵ (the inverse of A), we use the following format that applies only for ʹ ൈ ʹ matrices: ିଵ

ିଵ ൌ

ͳ ܽଶଶ ቂെܽ ൌ ሺ ሻ ଶଵ ‫ܣ‬

െܽଵଶ ܽଵଵ ቃ

ͳ ܽଶଶ ቂ ܽଵଵ ܽଶଶ െ ܽଵଶ ܽଶଵ െܽଶଵ

െܽଵଶ ܽଵଵ ቃ

Example A.4 Given the following matrix: ͵ ʹ ቃ ൌቂ െͶ ͷ Is A invertible (non-singular)? If Yes, Find ିଵ . Solution: Using result A.1, we do the following: ሺሻ ൌ ȁȁ ൌ ሺ͵ሻሺͷሻ െ ሺʹሻሺെͶሻ ൌ ͳͷ ൅ ͺ ൌ ʹ͵ ് ͲǤ Since the determinant of A is not 0, then A is invertible (non-singular).

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Thus, the answer is Yes, there exists ିଵ according to definition 1.7.2 as follows: ͷ ʹ െ ͳ ͳ ͷ െʹ ͷ െʹ ʹ͵൪ ቃൌ ቃ ൌ ൦ʹ͵ ቂ ቂ ିଵ ൌ Ͷ ͵ ʹ͵ Ͷ ͵ ሺ‫ܣ‬ሻ Ͷ ͵ ʹ͵ ʹ͵ Result A.2 Let ‫ א‬௡ ሺԹሻ be a triangular matrix. Then, ሺሻ = multiplication of the numbers on the main diagonal of A. There are three types of triangular matrix: a) Upper Triangular Matrix: it has all zeros on the left side of the diagonal of ݊ ൈ ݊ matrix. ͳ ͹ (i.e. ൌ ൥Ͳ ʹ Ͳ Ͳ

͵ ͷ൩ is an Upper Triangular Matrix). Ͷ

b) Diagonal Matrix: it has all zeros on both left and right sides of the diagonal of ݊ ൈ ݊ matrix. ͳ (i.e. ൌ ൥Ͳ Ͳ

Ͳ Ͳ ʹ Ͳ൩ is a Diagonal Matrix). Ͳ Ͷ

c) Lower Triangular Matrix: it has all zeros on the right side of the diagonal of ݊ ൈ ݊ matrix. ͳ Ͳ (i.e. ൌ ൥ͷ ʹ ͳ ͻ

Ͳ Ͳ൩ is a Diagonal Matrix). Ͷ

Fact A.1 Let ‫ א‬௡ ሺԹሻ. Then, ሺሻ ൌ ሺ୘ ሻ. Fact A.2 Let ‫ א‬௡ ሺԹሻ. If A is an invertible (nonsingular) matrix, then ୘ is also an invertible (nonsingular) matrix. (i.e. ሺ୘ ሻିଵ ൌ ሺିଵ ሻ୘ ). Proof of Fact A.2 We will show that ሺ୘ ሻିଵ ൌ ሺିଵ ሻ୘ .

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We know from previous results that ିଵ ൌ ௡ .

* Ri՞Rk (Interchange two rows). It has no effect on

By taking the transpose of both sides, we obtain:

the determinants.

ሺିଵ ሻ୘ ൌ ሺ ௡ ሻ୘

In general, the effect of Column-Operations on

ିଵ ୘ ୘

Then, ሺ ሻ ൌ ሺ ௡ ሻ

୘

determinants is the same as for Row-Operations.

Since ሺ ௡ ሻ୘ ൌ ௡ , then ሺିଵ ሻ୘ ୘ ൌ ௡ . Similarly, ሺ୘ ሻିଵ ୘ ൌ ሺ ௡ ሻ୘ ൌ ௡ . Thus, ሺ୘ ሻିଵ ൌ ሺିଵ ሻ୘ . The effect of Row-Operations on determinants: Suppose ‫ ן‬is a non-zero constant, and ݅ ܽ݊݀ ݇ are row numbers in the augmented matrix. * ‫ן‬Ri

, ‫( Ͳ ്ן‬Multiply a row with a non-zero

constant ‫)ן‬. ͳ ʹ i.e. ൌ ൥Ͳ Ͷ ʹ Ͳ

͵ ͳ ͳ൩ 3R2 --Æ ൥Ͳ ͳ ʹ

ʹ ͵ ͳʹ ͵൩ ൌ Ͳ ͳ

Assume that ሺሻ ൌ ɀ where ɀ is known, then ሺሻ ൌ ͵ɀ. Similarly, if ሺሻ ൌ Ⱦ Ⱦ ǡ then ଵ

ሺሻ ൌ Ⱦ. ଷ * ‫ן‬Ri +Rk --Æ Rk (Multiply a row with a non-zero constant ‫ן‬ǡ ). ͳ ʹ ͵ i.e. ൌ ൥Ͳ Ͷ ͳ൩ ‫ן‬Ri +Rk --Æ Rk ʹ Ͳ ͳ ͳ ʹ ͵ ൥Ͳ ͳʹ ͵൩ ൌ ʹ Ͳ ͳ Then, ሺሻ ൌ ሺሻ.

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Example A.5 Given the following Ͷ ൈ Ͷ matrix A with some Row-Operations: 2R1 --Æ A1 3R3 --Æ A2 -2R4 --Æ A4 If ሺሻ ൌ Ͷ, then find ሺଷ ሻ Solution: Using what we have learned from the effect of determinants on Row-Operations: ሺଵ ሻ ൌ ʹ ‫ כ‬ሺሻ ൌ ʹ ‫ כ‬Ͷ ൌ ͺ because ଵ has the first row of A multiplied by 2. ሺଶ ሻ ൌ ͵ ‫ כ‬ሺଵ ሻ ൌ ͵ ‫ כ‬ͺ ൌ ʹͶ because ଶ has the third row of ଵ multiplied by 3. Similarly, ሺଷ ሻ ൌ െʹ ‫ כ‬ሺଶ ሻ ൌ െʹ ‫ʹ כ‬Ͷ ൌ െͶͺ because ଷ has the fourth row of ଶ multiplied by -2. Result A.3 Assume ݊ ൈ ݊ with a given ሺሻ ൌ ߛ . Let ߙ be a number. Then, ሺߙሻ ൌ ߙ ௡ ‫ߛ כ‬. Result A.4 Assume ݊ ൈ ݊ Ǥ Then: a) ሺሻ ൌ ሺሻ ‫ כ‬ሺሻǤ b) Assume ିଵ exists and ିଵ exists. Then, ሺሻିଵ ൌ ିଵ ିଵ Ǥ c) ሺሻ ൌ ሺሻǤ d) ሺሻ ൌ ሺ୘ ሻǤ ଵ

e) If ିଵ exists, then ሺିଵ ሻ ൌ ୢୣ୲ሺ୅ሻǤ Proof of Result A.4 (b) We will show that ሺሻିଵ ൌ ିଵ ିଵ .

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If we multiply ( ିଵ ିଵ ) by (AB), we obtain:

Before reviewing the concepts of set theory, it is

ିଵ ሺିଵ ሻ ൌ ିଵ ሺ ௡ ሻB = ିଵ ൌ ௡ Ǥ

recommended to revisit section 1.4, and read the

Thus, ሺሻିଵ ൌ ିଵ ିଵ Ǥ

notations of numbers and the representation of the

Proof of Result A.4 (e) We will show that

three sets of numbers in figure 1.4.1.

ଵ

Let’s explain some symbols and notations of set theory:

ሺିଵ ሻ ൌ ୢୣ୲ሺ୅ሻ. Since ିଵ ൌ ௡ , then ሺିଵ ሻ ൌ ሺ‫ܫ‬௡ ሻ ൌ ͳǤ ሺିଵ ሻ ൌ ሺሻ ‫ כ‬ሺିଵ ሻ ൌ ͳǤ Thus, ሺିଵ ሻ ൌ

ଵ ୢୣ୲ሺ୅ሻ

Ǥ

Appendix B: Vector Spaces*

͵ ‫ א‬Ժ means that 3 is an element of ԺǤ ଵ ଶ

ଵ

‫ ב‬Ժ means that ଶ is not an element of ԺǤ

{ } means that it is a set. {5} means that 5 is a subset of Ժ, and the set consists of exactly one element which is 5.

*The materials of appendix B are taken from chapter 2 in my published book titled A First Course in Linear

Definition B.1.1 The span of a certain set is the set of

Algebra: Study Guide for the Undergraduate Linear Algebra Course, First Edition1.

set.

We start this chapter reviewing some concepts of set

Solution: According to definition B.1.1, then the span

theory, and we discuss some important concepts of

of the set {1} is the set of all possible linear

vector spaces including span and dimension. In the

combinations of the subset of {1} which is 1.

remaining sections we introduce the concept of linear

Hence, Span{1} = Թ.

independence. At the end of this chapter we discuss

Example B.1.2 Find Span{(1,2),(2,3)}.

other concepts such as subspace and basis.

Solution: According to definition B.1.1, then the span

B.1 Span and Vector Spaces

of the set {(1,2),(2,3)} is the set of all possible linear

In this section, we review some concepts of set theory,

(1,2) and (2,3). Thus, the following is some possible

and we give an introduction to span and vector spaces

linear combinations:

including some examples related to these concepts.

ሺͳǡʹሻ ൌ ͳ ‫ כ‬ሺͳǡʹሻ ൅ Ͳ ‫ כ‬ሺʹǡ͵ሻ

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all possible linear combinations of the subset of that Example B.1.1 Find Span{1}.

combinations of the subsets of {(1,2),(2,3)} which are

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ሺʹǡ͵ሻ ൌ Ͳ ‫ כ‬ሺͳǡʹሻ ൅ ͳ ‫ כ‬ሺʹǡ͵ሻ ሺͷǡͺሻ ൌ ͳ ‫ כ‬ሺͳǡʹሻ ൅ ʹ ‫ כ‬ሺʹǡ͵ሻ Hence, ሼሺͳǡʹሻǡ ሺʹǡ͵ሻǡ ሺͷǡͺሻሽ ‫ א‬ሼሺͳǡʹሻǡ ሺʹǡ͵ሻሽ. Example B.1.3 Find Span{0}. Solution: According to definition B.1.1, then the span of the set {0} is the set of all possible linear combinations of the subset of {0} which is 0. Hence, Span{0} = 0.


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B.2 The Dimension of Vector Space In this section, we discuss how to find the dimension of vector space, and how it is related to what we have learned in section B.1. Definition B.2.1 Given a vector space ܸ, the dimension

Example B.1.4 Find Span{c} where c is a non-zero

of ܸ is the number of minimum elements needed in ܸ

integer.

so that their ܵ‫ ݊ܽ݌‬is equal to ܸ, and it is denoted by

Solution: Using definition B.1.1, the span of the set {c}

ሺܸሻ. (i.e. ሺԹሻ ൌ ͳ ሺԹଶ ሻ ൌ ʹ).

is the set of all possible linear combinations of the

Result B.2.1 ሺԹ௡ ሻ ൌ ݊.

subset of {c} which is ܿ ് Ͳ.

Proof of Result B.2.1 We will show that ሺԹ௡ ሻ ൌ ݊Ǥ

Thus, Span{c} = Թ.

Claim: ‫ ܦ‬ൌ ܵ‫݊ܽ݌‬ሼሺͳǡͲሻǡ ሺͲǤͳሻሽ ൌ Թଶ

Definition B.1.2 Թ௡ ൌ ሼሺܽଵ ǡ ܽଶ ǡ ܽଷ ǡ ǥ ǡ ܽ௡ ሻȁܽଵ ǡ ܽଶ ǡ ܽଷ ǡ ǥ ǡ ܽ௡ ‫ א‬Թሽ

ߙଵ ሺͳǡͲሻ ൅ ߙଶ ሺͲǡͳሻ ൌ ሺߙଵ ǡ ߙଶ ሻ ‫ א‬Թଶ

is a set of all points where each point has exactly ݊ coordinates. Definition B.1.3 ሺܸǡ ൅ǡήሻ is a vector space if satisfies the following: a. For every ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ‫ܸ א‬, ‫ݒ‬ଵ ൅ ‫ݒ‬ଶ ‫ܸ א‬Ǥ b. For every ߙ ‫ א‬Թ ‫ܸ א ݒ‬, ߙ‫ܸ א ݒ‬Ǥ (i.e. Given ܵ‫݊ܽ݌‬ሼ‫ݔ‬ǡ ‫ݕ‬ሽ ‫ ݐ݁ݏ‬ሼ‫ݔ‬ǡ ‫ݕ‬ሽǡ then ξͳͲ‫ ݔ‬൅ ʹ‫݊ܽ݌ܵ א ݕ‬ሼ‫ݔ‬ǡ ‫ݕ‬ሽ. Let’s assume that ‫݊ܽ݌ܵ א ݒ‬ሼ‫ݔ‬ǡ ‫ݕ‬ሽ, then ‫ ݒ‬ൌ ܿଵ ‫ ݔ‬൅ ܿଶ ‫ ݕ‬for some numbers ܿଵ ܿଶ ).

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Thus, ‫ ܦ‬is a subset of Թଶ (‫ ك ܦ‬Թଶ ). For every ‫ݔ‬ଵ ǡ ‫ݕ‬ଵ ‫ א‬Թ, ሺ‫ݔ‬ଵ ǡ ‫ݕ‬ଵ ሻ ‫ א‬Թଶ Ǥ Therefore, ሺ‫ݔ‬ଵ ǡ ‫ݕ‬ଵ ሻ ൌ ‫ݔ‬ଵ ሺͳǡͲሻ ൅ ‫ݕ‬ଵ ሺͲǡͳሻ ‫ܦ א‬Ǥ We prove the above claim, and hence ሺԹ௡ ሻ ൌ ݊. Fact 2B.2.1 ܵ‫݊ܽ݌‬ሼሺ͵ǡͶሻሽ ് Թଶ . Proof of Fact B.2.1 We will show that ܵ‫݊ܽ݌‬ሼሺ͵ǡͶሻሽ ് Թଶ Ǥ Claim: ‫ ܨ‬ൌ ܵ‫݊ܽ݌‬ሼሺ͸ǡͷሻሽ ് Թଶ where ሺ͸ǡͷሻ ‫ א‬Թଶ Ǥ

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We cannot find a number ߙ such that ሺ͸ǡͷሻ ൌ ߙሺ͵ǡͶሻ We prove the above claim, and hence ܵ‫݊ܽ݌‬ሼሺ͵ǡͶሻሽ ് Թଶ . Fact B.2.2 ܵ‫݊ܽ݌‬ሼሺͳǡͲሻǡ ሺͲǡͳሻሽ ൌ Թଶ . Fact B.2.3 ܵ‫݊ܽ݌‬ሼሺʹǡͳሻǡ ሺͳǡͲǤͷሻሽ ് Թଶ .


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Using the distribution property and algebra, we obtain: ‫ݒ‬ଵ ൌ ͵‫ݒ‬ଵ ܿଵ ൅ ‫ݒ‬ଶ ܿଵ ‫ݒ‬ଵ െ ͵‫ݒ‬ଵ ܿଵ ൌ ‫ݒ‬ଶ ܿଵ ‫ݒ‬ଵ ሺͳ െ ͵ܿଵ ሻ ൌ ‫ݒ‬ଶ ܿଵ ሺͳ െ ͵ܿଵ ሻ ‫ݒ‬ଵ ൌ ‫ݒ‬ଶ ܿଵ

B.3 Linear Independence

Thus, none of ‫ݒ‬ଵ and ͵‫ݒ‬ଵ ൅ ‫ݒ‬ଶ is a linear combination of

In this section, we learn how to determine whether

the others which means that ‫ݒ‬ଵ and ͵‫ݒ‬ଵ ൅ ‫ݒ‬ଶ are

vector spaces are linearly independent or not.

linearly independent. This is a contradiction.

Definition B.3.1 Given a vector space ሺܸǡ ൅ǡήሻ, we say

Therefore, our assumption that ‫ݒ‬ଵ and ͵‫ݒ‬ଵ ൅ ‫ݒ‬ଶ were

‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ǡ ǥ ǡ ‫ݒ‬௡ ‫ ܸ א‬are linearly independent if none of them is a linear combination of the remaining ‫ݒ‬௜ Ԣ‫ݏ‬.

linearly dependent is false. Hence, ‫ݒ‬ଵ and ͵‫ݒ‬ଵ ൅ ‫ݒ‬ଶ are

(i.e. ሺ͵ǡͶሻǡ ሺʹǡͲሻ ‫ א‬Թ are linearly independent because we cannot write them as a linear combination of each other, in other words, we cannot find a number ߙଵ ǡ ߙଶ such that ሺ͵ǡͶሻ ൌ ߙଵ ሺʹǡͲሻ and ሺʹǡͲሻ ൌ ߙଶ ሺ͵ǡͶሻ).

Example B.3.2 Given the following vectors:

Definition B.3.2 Given a vector space ሺܸǡ ൅ǡήሻ, we say ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ǡ ǥ ǡ ‫ݒ‬௡ ‫ ܸ א‬are linearly dependent if at least one of ‫ݒ‬௜ Ԣ‫ ݏ‬is a linear combination of the others. Example B.3.1 Assume ‫ݒ‬ଵ ‫ݒ‬ଶ are linearly independent. Show that ‫ݒ‬ଵ and ͵‫ݒ‬ଵ ൅ ‫ݒ‬ଶ are linearly independent. Solution: We will show that ‫ݒ‬ଵ and ͵‫ݒ‬ଵ ൅ ‫ݒ‬ଶ are linearly independent. Using proof by contradiction, we assume that ‫ݒ‬ଵ and ͵‫ݒ‬ଵ ൅ ‫ݒ‬ଶ are linearly dependent. For some non-zero number ܿଵ , ‫ݒ‬ଵ ൌ ܿଵ ሺ͵‫ݒ‬ଵ ൅ ‫ݒ‬ଶ ሻ.

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linearly independent. ‫ݒ‬ଵ ൌ ሺͳǡͲǡ െʹሻ ‫ݒ‬ଶ ൌ ሺെʹǡʹǡͳሻ ‫ݒ‬ଷ ൌ ሺെͳǡͲǡͷሻ Are these vectors independent elements? Solution: First of all, to determine whether these vectors are independent elements or not, we need to write these vectors as a matrix. ͳ Ͳ െʹ ൥െʹ ʹ ͳ ൩ Each point is a row-operation. We need to െͳ Ͳ ͷ reduce this matrix to Semi-Reduced Matrix. Definition B.3.3 Semi-Reduced Matrix is a reducedmatrix but the leader numbers can be any non-zero number.

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Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: ͳ Ͳ െʹ ʹܴ ൅ ܴ ՜ ܴ ͳ Ͳ ଵ ଶ ଶ ൥െʹ ʹ ͳ ൩ ൥ ܴଵ ൅ ܴଷ ՜ ܴଷ Ͳ ʹ െͳ Ͳ ͷ Ͳ Ͳ Reduced Matrix.

െʹ െ͵൩ This is a Semi͵

Since none of the rows in the Semi-Reduced Matrix become zero-row, then the elements are independent because we cannot write at least one of them as a linear combination of the others. Example 2.3.3 Given the following vectors:


ͳ െʹ Ͷ െܴଶ ൅ ܴଷ ՜ ܴଷ ൥Ͳ Ͳ Ͷ Ͳ Ͳ Ͳ Matrix.

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͸ ͺ൩ This is a Semi-Reduced Ͳ

Since there is a zero-row in the Semi-Reduced Matrix, then the elements are dependent because we can write at least one of them as a linear combination of the others.

B.4 Subspace and Basis In this section, we discuss one of the most important

‫ݒ‬ଵ ൌ ሺͳǡ െʹǡͶǡ͸ሻ

concepts in linear algebra that is known as subspace.

‫ݒ‬ଶ ൌ ሺെͳǡʹǡͲǡʹሻ

In addition, we give some examples explaining how to find the basis for subspace.

‫ݒ‬ଷ ൌ ሺͳǡ െʹǡͺǡͳͶሻ Are these vectors independent elements? Solution: First of all, to determine whether these vectors are independent elements or not, we need to write these vectors as a matrix.

Definition B.4.1 Subspace is a vector space but we call it a subspace because it lives inside a bigger vector space. (i.e. Given vector spaces ܸ and ‫ܦ‬, then according to the figure 2.4.1, ‫ ܦ‬is called a subspace of ܸ).

ͳ െʹ Ͷ ͸ ൥െͳ ʹ Ͳ ʹ ൩ Each point is a row-operation. We ͳ െʹ ͺ ͳͶ need to reduce this matrix to Semi-Reduced Matrix. Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: ͳ ൥െͳ ͳ

െʹ ʹ െʹ

Ͷ Ͳ ͺ

͸ ܴ ൅ܴ ՜ܴ ͳ ଵ ଶ ଶ ʹ ൩ െܴ ൅ ܴ ՜ ܴ ൥Ͳ ଵ ଷ ଷ Ͳ ͳͶ

െʹ Ͳ Ͳ

V

D

Ͷ ͸ Ͷ ͺ൩ Ͷ ͺ Figure B.4.1: Subspace of ܸ

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Fact B.4.1 Every vector space is a subspace of itself. Example B.4.1 Given a vector space ‫ ܮ‬ൌ ሼሺܿǡ ͵ܿሻȁܿ ‫ א‬Թሽ. a. Does ‫ ܮ‬live in Թଶ ? b. Does ‫ ܮ‬equal to Թଶ ? c. Is ‫ ܮ‬a subspace of Թଶ ? d. Does ‫ ܮ‬equal to ܵ‫݊ܽ݌‬ሼሺͲǡ͵ሻሽǫ e. Does ‫ ܮ‬equal to ܵ‫݊ܽ݌‬ሼሺͳǡ͵ሻǡ ሺʹǡ͸ሻሽǫ Solution: To answer all these questions, we need first to draw an equation from this vector space, say ‫ ݕ‬ൌ ͵‫ݔ‬. The following figure represents the graph of the above equation, and it passes through a point ሺͳǡ͵ሻ.


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Part b: No; ‫ ܮ‬does not equal to Թଶ . To show that we prove the following claim: Claim: ‫ ܮ‬ൌ ܵ‫݊ܽ݌‬ሼሺͷǡͳͷሻሽ ് Թଶ where ሺͷǡͳͷሻ ‫ א‬Թଶ Ǥ It is impossible to find a number ߙ ൌ ͵ such that ሺʹͲǡ͸Ͳሻ ൌ ߙሺͷǡͳͷሻ because in this case ߙ ൌ Ͷ where ሺʹͲǡ͸Ͳሻ ൌ Ͷሺͷǡͳͷሻ. We prove the above claim, and ܵ‫݊ܽ݌‬ሼሺͷǡͳͷሻሽ ് Թଶ . Thus, ‫ ܮ‬does not equal to Թଶ Part c: Yes; ‫ ܮ‬is a subspace of Թଶ because ‫ ܮ‬lives inside a bigger vector space which is Թଶ . Part d: No; according to the graph in figure 2.4.2, ሺͲǡ͵ሻ does not belong to ‫ܮ‬. Part e: Yes; because we can write ሺͳǡ͵ሻ and ሺʹǡ͸ሻ as a linear combination of each other. ߙଵ ሺͳǡ͵ሻ ൅ ߙଶ ሺʹǡ͸ሻ ൌ ሼሺߙଵ ൅ ʹߙଶ ሻǡ ሺ͵ߙଵ ൅ ͸ߙଶ ሻሽ ߙଵ ሺͳǡ͵ሻ ൅ ߙଶ ሺʹǡ͸ሻ ൌ ሼሺߙଵ ൅ ʹߙଶ ሻǡ ͵ሺߙଵ ൅ ʹߙଶ ሻሽ Assume ܿ ൌ ሺߙଵ ൅ ʹߙଶ ሻ, then we obtain: ߙଵ ሺͳǡ͵ሻ ൅ ߙଶ ሺʹǡ͸ሻ ൌ ሼሺܿǡ ͵ܿ ሻȁܿ ‫ א‬Թሽ ൌ ‫ܮ‬.

Figure B.4.2: Graph of ‫ ݕ‬ൌ ͵‫ݔ‬ Now, we can answer the given questions as follows: ଶ

Part a: Yes; ‫ ܮ‬lives in Թ .

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Thus, ‫ ܮ‬ൌ ܵ‫݊ܽ݌‬ሼሺͳǡ͵ሻǡ ሺʹǡ͸ሻሽ. Result B.4.1 ‫ ܮ‬is a subspace of Թଶ if satisfies the following: a. ‫ ܮ‬lives inside Թଶ . b. ‫ ܮ‬has only lines through the origin ሺͲǡͲሻ.

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Example B.4.2 Given a vector space ‫ ܦ‬ൌ ሼሺܽǡ ܾǡ ͳሻȁܽǡ ܾ ‫ א‬Թሽ. a. Does ‫ ܦ‬live in Թଷ ? b. Is ‫ ܦ‬a subspace of Թଷ ǫ


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Results B.4.3 and B.4.4 tell us the following: In order to get all Թ௡ , we need exactly ݊ independent points. Result B.4.5 Assume Թ௡ ൌ ܵ‫݊ܽ݌‬ሼܳଵ ǡ ܳଶ ǡ ǥ ǡ ܳ௞ ሽ, then ݇ ൒ ݊ (݊ points of the ܳ௞ Ԣ‫ ݏ‬are independents).

is a three-dimensional equation, there is no need to

Definition B.4.2 Basis is the set of points that is needed to ܵ‫ ݊ܽ݌‬the vector space.

draw it because it is difficult to draw it exactly. Thus,

Example B.4.3 Let ‫ ܦ‬ൌ ܵ‫݊ܽ݌‬ሼሺͳǡ െͳǡͲሻǡ ሺʹǡʹǡͳሻǡ ሺͲǡͶǡͳሻሽ.

Solution: Since the equation of the above vector space

we can answer the above questions immediately.

a. Find ሺ‫ܦ‬ሻ.

Part a: Yes; ‫ ܦ‬lives inside Թଷ .

b. Find a basis for ‫ܦ‬.

Part b: No; since ሺͲǡͲǡͲሻ ‫ܦ ב‬, then ‫ ܦ‬is not a subspace of Թଷ .

Solution: First of all, we have infinite set of points, and ‫ ܦ‬lives inside Թଷ . Let’s assume the following:

Fact B.4.2 Assume ‫ ܦ‬lives inside Թ௡ . If we can write ‫ܦ‬ as a ܵ‫݊ܽ݌‬, then it is a subspace of Թ௡ .

‫ݒ‬ଵ ൌ ሺͳǡ െͳǡͲሻ

Fact B.4.3 Assume ‫ ܦ‬lives inside Թ௡ . If we cannot write ‫ ܦ‬as a ܵ‫݊ܽ݌‬, then it is not a subspace of Թ௡ . Fact B.4.4 Assume ‫ ܦ‬lives inside Թ௡ . If ሺͲǡͲǡͲǡ ǥ ǡͲሻ is in ‫ܦ‬, then ‫ ܦ‬is a subspace of Թ௡ . Fact B.4.5 Assume ‫ ܦ‬lives inside Թ௡ . If ሺͲǡͲǡͲǡ ǥ ǡͲሻ is not in ‫ܦ‬, then ‫ ܦ‬is not a subspace of Թ௡ .

‫ݒ‬ଶ ൌ ሺʹǡʹǡͳሻ ‫ݒ‬ଷ ൌ ሺͲǡͶǡͳሻ Part a: To find ሺ‫ܦ‬ሻ, we check whether ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ and ‫ݒ‬ଷ are dependent elements or not. Using what we have learned so far from section 2.3: We need to write these vectors as a matrix.

Now, we list the main results on Թ௡ :

ͳ െͳ Ͳ ൥ʹ ʹ ͳ൩ Each point is a row-operation. We need to Ͳ Ͷ ͳ reduce this matrix to Semi-Reduced Matrix.

Result B.4.2 Maximum number of independent points is ݊.

Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows:

Result B.4.3 Choosing any ݊ independent points in Թ௡ , say ܳଵ ǡ ܳଶ ǡ ǥ ǡ ܳ௡ , then Թ௡ ൌ ܵ‫݊ܽ݌‬ሼܳଵ ǡ ܳଶ ǡ ǥ ǡ ܳ௡ ሽ.

ͳ ൥ʹ Ͳ

െͳ ʹ Ͷ

ͳ െͳ Ͳ ͳ൩ െʹܴଵ ൅ ܴଶ ՜ ܴଶ ൥Ͳ Ͷ Ͳ Ͷ ͳ

Ͳ ͳ൩ െܴଶ ൅ ܴଷ ՜ ܴଷ ͳ

Result B.4.4 ሺԹ௡ ሻ ൌ ݊.

126 M. Kaabar

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ͳ ൥Ͳ Ͳ

െͳ Ͷ Ͳ

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Ͳ ͳ൩ This is a Semi-Reduced Matrix. Ͳ

Since there is a zero-row in the Semi-Reduced Matrix, then these elements are dependent because we can write at least one of them as a linear combination of the others. Only two points survived in the SemiReduced Matrix. Thus, ሺ‫ܦ‬ሻ ൌ ʹ. Part b: ‫ ܦ‬is a plane that passes through the origin ሺͲǡͲǡͲሻ. Since ሺ‫ܦ‬ሻ ൌ ʹ, then any two independent points in ‫ ܦ‬will form a basis for ‫ܦ‬. Hence, the following are some possible bases for ‫ܦ‬: Basis for ‫ ܦ‬is ሼሺͳǡ െͳǡͲሻǡ ሺʹǡʹǡͳሻሽ. Another basis for ‫ ܦ‬is ሼሺͳǡ െͳǡͲሻǡ ሺͲǡͶǡͳሻሽ. Result B.4.6 It is always true that ȁ‫ݏ݅ݏܽܤ‬ȁ ൌ ݀݅݉ሺ‫ܦ‬ሻ. Example B.4.4 Given the following: ‫ ܯ‬ൌ ܵ‫݊ܽ݌‬ሼሺെͳǡʹǡͲǡͲሻǡ ሺͳǡ െʹǡ͵ǡͲሻǡ ሺെʹǡͲǡ͵ǡͲሻሽ. Find a basis for ‫ܯ‬. Solution: We have infinite set of points, and ‫ ܯ‬lives inside Թସ . Let’s assume the following:

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െͳ ʹ Ͳ Ͳ ൥ ͳ െʹ ͵ Ͳ൩ Each point is a row-operation. We െʹ Ͳ ͵ Ͳ need to reduce this matrix to Semi-Reduced Matrix. Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: െͳ ൥ͳ െʹ

ʹ െʹ Ͳ

Ͳ ͵ ͵

െͳ Ͳ ܴ ൅ܴ ՜ܴ ଵ ଶ ଶ Ͳ൩ െʹܴ ൅ ܴ ՜ ܴ ൥ Ͳ ଵ ଷ ଷ Ͳ Ͳ

െͳ െܴଶ ൅ ܴଷ ՜ ܴଷ ൥ Ͳ Ͳ Matrix.

ʹ Ͳ െͶ

ʹ Ͳ െͶ

Ͳ ͵ ͵

Ͳ Ͳ൩ Ͳ

Ͳ Ͳ ͵ Ͳ൩ This is a Semi-Reduced Ͳ Ͳ

Since there is no zero-row in the Semi-Reduced Matrix, then these elements are independent. All the three points survived in the Semi-Reduced Matrix. Thus, ሺ‫ܯ‬ሻ ൌ ͵. Since ሺ‫ܯ‬ሻ ൌ ͵, then any three independent points in ‫ ܯ‬from the above matrices will form a basis for ‫ܯ‬. Hence, the following are some possible bases for ‫ܯ‬: Basis for ‫ ܯ‬is ሼሺെͳǡʹǡͲǡͲሻǡ ሺͲǡͲǡ͵ǡͲሻǡ ሺͲǡ െͶǡͲǡͲሻሽ. Another basis for ‫ ܯ‬is ሼሺെͳǡʹǡͲǡͲሻǡ ሺͲǡͲǡ͵ǡͲሻǡ ሺͲǡ െͶǡ͵ǡͲሻሽ.

‫ݒ‬ଵ ൌ ሺെͳǡʹǡͲǡͲሻ

Another basis for ‫ ܯ‬is ሼሺെͳǡʹǡͲǡͲሻǡ ሺͳǡ െʹǡ͵ǡͲሻǡ ሺെʹǡͲǡ͵ǡͲሻሽ.

‫ݒ‬ଶ ൌ ሺͳǡ െʹǡ͵ǡͲሻ

Example B.4.5 Given the following:

‫ݒ‬ଷ ൌ ሺെʹǡͲǡ͵ǡͲሻ

ܹ ൌ ܵ‫݊ܽ݌‬ሼሺܽǡ െʹܽ ൅ ܾǡ െܽሻȁܽǡ ܾ ‫ א‬Թሽ.

We check if ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ and ‫ݒ‬ଷ are dependent elements. Using what we have learned so far from section 2.3 and example 2.4.3: We need to write these vectors as a matrix.

128 M. Kaabar


a. Show that ܹ is a subspace of Թଷ . b. Find a basis for ܹ. c. Rewrite ܹ as a ܵ‫݊ܽ݌‬.

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Solution: We have infinite set of points, and ܹ lives inside Թଷ .

‫ ܪ‬ൌ ܵ‫݊ܽ݌‬ሼሺܽଶ ǡ ͵ܾ ൅ ܽǡ െʹܿǡ ܽ ൅ ܾ ൅ ܿሻȁܽǡ ܾǡ ܿ ‫ א‬Թሽ.

Part a: We write each coordinate of ܹ as a linear combination of the free variables ܽ and ܾ.

Solution: We have infinite set of points, and ‫ ܪ‬lives inside Թସ . We try write each coordinate of ‫ ܪ‬as a linear combination of the free variables ܽǡ ܾ and ܿ.

ܽ ൌͳήܽ൅Ͳήܾ

ܽଶ ൌ ‫ ݎܾ݁݉ݑܰ ݀݁ݔ݅ܨ‬ή ܽ ൅ ‫ ݎܾ݁݉ݑܰ ݀݁ݔ݅ܨ‬ή ܾ ൅ ‫ ݎܾ݁݉ݑܰ ݀݁ݔ݅ܨ‬ή ܿ

െʹܽ ൅ ܾ ൌ െʹ ή ܽ ൅ ͳ ή ܾ

ܽଶ is not a linear combination of ܽǡ ܾ and ܿ.

െܽ ൌ െͳ ή ܽ ൅ Ͳ ή ܾ Since it is possible to write each coordinate of ܹ as a linear combination of the free variables ܽ and ܾ, then we conclude that ܹ is a subspace of Թଷ . Part b: To find a basis for ܹ, we first need to find ሺܹሻ. To find ሺܹሻ, let’s play a game called (ONOFF GAME) with the free variables ܽ ܾǤ ܽ ͳ Ͳ

ܾ Ͳ ͳ

ܲ‫ݐ݊݅݋‬ ሺͳǡ െʹǡ െͳሻ ሺͲǡͳǡͲሻ

Now, we check for independency: We already have the ͳ െʹ െͳ ቃǤ Thus, ሺܹሻ ൌ ʹ. Semi-Reduced Matrix: ቂ Ͳ ͳ Ͳ Hence, the basis for ܹ is ሼሺͳǡ െʹǡ െͳሻǡ ሺͲǡͳǡͲሻሽ. Part b: Since we found the basis for ܹ, then it is easy to rewrite ܹ as a ܵ‫ ݊ܽ݌‬as follows: ܹ ൌ ܵ‫݊ܽ݌‬ሼሺͳǡ െʹǡ െͳሻǡ ሺͲǡͳǡͲሻሽǤ Fact B.4.6 ሺܹሻ ൑ ܰ‫ ݁݁ݎܨ ݂݋ݎܾ݁݉ݑ‬െ ܸܽ‫ݏ݈ܾ݁ܽ݅ݎ‬Ǥ Example B.4.6 Given the following:

130 M. Kaabar

Is ‫ ܪ‬a subspace of Թସ ?

We assume that ‫ ݓ‬ൌ ሺͳǡͳǡͲǡͳሻ ‫ܪ א‬, and ܽ ൌ ͳǡ ܾ ൌ ܿ ൌ Ͳ. If ߙ ൌ െʹ, then െʹ ή ‫ ݓ‬ൌ െʹ ή ሺͳǡͳǡͲǡͳሻ ൌ ሺെʹǡ െʹǡͲǡ െʹሻ ‫ܪ ב‬. Since it is impossible to write each coordinate of ‫ ܪ‬as a linear combination of the free variables ܽǡ ܾ and ܿ, then we conclude that ‫ ܪ‬is not a subspace of Թସ . Example B.4.7 Form a basis for Թସ . Solution: We just need to select any random four independent points, and then we form a Ͷ ൈ Ͷ matrix with four independent rows as follows: ʹ ቎Ͳ Ͳ Ͳ

͵ ͷ Ͳ Ͳ

Ͳ Ͷ ͳ ͳ ቏ Note: ߨ ௘ is a number. ʹ ͵ Ͳ ߨ௘

Let’s assume the following: ‫ݒ‬ଵ ൌ ሺʹǡ͵ǡͲǡͶሻ ‫ݒ‬ଶ ൌ ሺͲǡͷǡͳǡͳሻ ‫ݒ‬ଷ ൌ ሺͲǡͲǡʹǡ͵ሻ ‫ݒ‬ସ ൌ ሺͲǡͲǡͲǡ ߨ ௘ ሻ Thus, the basis for Թସ ൌ ሼ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ǡ ‫ݒ‬ଷ ǡ ‫ݒ‬ସ ሽ, and

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ܵ‫݊ܽ݌‬ሼ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ǡ ‫ݒ‬ଷ ǡ ‫ݒ‬ସ ሽ ൌ Թସ .

‫ ܦ‬ൌ ܵ‫݊ܽ݌‬ሼሺͳǡͳǡͳǡͳሻǡ ሺെͳǡ െͳǡͲǡͲሻǡ ሺͲǡͲǡͳǡͳሻሽ

Example B.4.8 Form a basis for Թସ that contains the

Is ሺͳǡͳǡʹǡʹሻ ‫?ܦ א‬

following two independent points:

Solution: We have infinite set of points, and ‫ ܦ‬lives inside Թସ . There are two different to solve this example:

ሺͲǡʹǡͳǡͶሻ ሺͲǡ െʹǡ͵ǡ െͳͲሻ. Solution: We need to add two more points to the given one so that all four points are independent. Let’s

‫ݒ‬ଵ ൌ ሺͳǡͳǡͳǡͳሻ

assume the following:

‫ݒ‬ଶ ൌ ሺെͳǡ െͳǡͲǡͲሻ

‫ݒ‬ଵ ൌ ሺͲǡʹǡͳǡͶሻ

‫ݒ‬ଷ ൌ ሺͲǡͲǡͳǡͳሻ

‫ݒ‬ଶ ൌ ሺͲǡ െʹǡ͵ǡ െͳͲሻ ‫ݒ‬ଷ ൌ ሺͲǡͲǡͶǡ െ͸ሻ This is a random point. ‫ݒ‬ସ ൌ ሺͲǡͲǡͲǡͳͲͲͲሻ This is a random point. Then, we need to write these vectors as a matrix. Ͳ ʹ ͳ Ͷ Ͳ െʹ ͵ െͳͲ ቎ ቏ Each point is a row-operation. We Ͳ Ͳ Ͷ െ͸ Ͳ Ͳ Ͳ ͳͲͲͲ need to reduce this matrix to Semi-Reduced Matrix. Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows: Ͳ ʹ Ͳ െʹ ቎ Ͳ Ͳ Ͳ Ͳ

ͳ Ͷ Ͳ ͵ െͳͲ ቏ ܴ ൅ ܴ ՜ ܴ ቎͵ ଵ ଶ ଶ Ͳ Ͷ െ͸ Ͳ ͳͲͲͲ Ͳ

ʹ Ͳ Ͳ Ͳ

ͳ ͷ Ͷ Ͳ

Ͷ ͵Ͳ቏ െ͸ ͳͲͲͲ

This is a Semi-Reduced Matrix. Thus, the basis for Թସ is ሼሺͲǡʹǡͳǡͶሻ ǡ ሺͲǡ െʹǡ͵ǡ െͳͲሻǡ ሺ͵ǡͲǡͷǡ͵Ͳሻǡ ሺͲǡͲǡͲǡͳͲͲͲሻሽ. Example B.4.9 Given the following:

132 M. Kaabar

The First Way: Let’s assume the following:

We start asking ourselves the following question: Question: Can we find ߙଵ ǡ ߙଶ and ߙଷ such that ሺͳǡͳǡʹǡʹሻ ൌ ߙଵ ή ‫ݒ‬ଵ ൅ ߙଶ ή ‫ݒ‬ଶ ൅ ߙଷ ή ‫ݒ‬ଷ ? Answer: Yes but we need to solve the following system of linear equations: ͳ ൌ ߙଵ െ ߙଶ ൅ Ͳ ή ߙଷ ͳ ൌ ߙଵ െ ߙଶ ൅ Ͳ ή ߙଷ ʹ ൌ ߙଵ ൅ ߙଷ ʹ ൌ ߙଵ ൅ ߙଷ Using what we have learned from chapter 1 to solve the above system of linear equations, we obtain: ߙଵ ൌ ߙଶ ൌ ߙଷ ൌ ͳ Hence, Yes: ሺͳǡͳǡʹǡʹሻ ‫ܦ א‬Ǥ The Second Way (Recommended): We first need to find ݀݅݉ሺ‫ܦ‬ሻ, and then a basis for ‫ܦ‬. We have to write ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ‫ݒ‬ଷ as a matrix.

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ͳ ͳ ͳ ͳ ൥െͳ െͳ Ͳ Ͳ൩ Each point is a row-operation. We Ͳ Ͳ ͳ ͳ need to reduce this matrix to Semi-Reduced Matrix.

Appendix C: Homogenous

Now, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows:

Systems*

ͳ ൥െͳ Ͳ

ͳ െͳ Ͳ

ͳ Ͳ ͳ

ͳ ͳ ͳ Ͳ൩ ܴଵ ൅ ܴଶ ՜ ܴଶ ൥Ͳ Ͳ Ͳ Ͳ ͳ

ͳ ͳ െܴଶ ൅ ܴଷ ՜ ܴଷ ൥Ͳ Ͳ Ͳ Ͳ Matrix.

ͳ ͳ Ͳ

ͳ ͳ ͳ

ͳ ͳ൩ ͳ

ͳ ͳ൩ This is a Semi-Reduced Ͳ

*The materials of appendix C are taken from chapter 3 in my published book titled A First Course in Linear

Algebra: Study Guide for the Undergraduate Linear Algebra Course, First Edition1. In this chapter, we introduce the homogeneous systems, and we discuss how they are related to what

Since there is a zero-row in the Semi-Reduced Matrix, then these elements are dependent. Thus, ሺ‫ܦ‬ሻ ൌ ʹ.

we have learned in chapter B. We start with an

Thus, Basis for ‫ ܦ‬is ሼሺͳǡͳǡͳǡͳሻǡ ሺͲǡͲǡͳǡͳሻሽ, and

one of the most important topics in linear algebra

‫ ܦ‬ൌ ܵ‫݊ܽ݌‬ሼሺͳǡͳǡͳǡͳሻǡ ሺͲǡͲǡͳǡͳሻሽ.

which is linear transformation. At the end of this

Now, we ask ourselves the following question:

chapter we discuss how to find range and kernel, and

Question: Can we find ߙଵ ǡ ߙଶ and ߙଷ such that ሺͳǡͳǡʹǡʹሻ ൌ ߙଵ ή ሺͳǡͳǡͳǡͳሻ ൅ ߙଶ ή ሺͲǡͲǡͳǡͳሻ? Answer: Yes: ͳ ൌ ߙଵ ͳ ൌ ߙଵ

their relation to sections C.1 and C.2.

C.1 Null Space and Rank In this section, we first give an introduction to homogeneous systems, and we discuss how to find the null space and rank of homogeneous systems. In

ʹ ൌ ߙଵ ൅ ߙଶ

addition, we explain how to find row space and column

ʹ ൌ ߙଵ ൅ ߙଶ

space.

Thus, ߙଵ ൌ ߙଶ ൌ ߙଷ ൌ ͳ. Hence, Yes: ሺͳǡͳǡʹǡʹሻ ‫ܦ א‬Ǥ

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introduction to null space and rank. Then, we study

Definition C.1.1 Homogeneous System is a ݉ ൈ ݊ system of linear equations that has all zero constants.

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(i.e. the following is an example of homogeneous ʹ‫ݔ‬ଵ ൅ ‫ݔ‬ଶ െ ‫ݔ‬ଷ ൅ ‫ݔ‬ସ ൌ Ͳ ͵‫ݔ‬ system): ൝ ଵ ൅ ͷ‫ݔ‬ଶ ൅ ͵‫ݔ‬ଷ ൅ Ͷ‫ݔ‬ସ ൌ Ͳ െ‫ݔ‬ଶ ൅ ‫ݔ‬ଷ െ ‫ݔ‬ସ ൌ Ͳ Imagine we have the following solution to the homogeneous system: ‫ݔ‬ଵ ൌ ‫ݔ‬ଶ ൌ ‫ݔ‬ଷ ൌ ‫ݔ‬ସ ൌ Ͳ. Then, this solution can be viewed as a point of Թ௡ (here is Թସ ) : ሺͲǡͲǡͲǡͲሻ Result C.1.1 The solution of a homogeneous system ݉ ൈ ݊ can be written as ሼሺܽଵ ǡ ܽଶ ǡ ܽଷ ǡ ܽସ ǡ ǥ ǡ ܽ௡ ȁܽଵ ǡ ܽଶ ǡ ܽଷ ǡ ܽସ ǡ ǥ ǡ ܽ௡ ‫ א‬Թሽ. Result C.1.2 All solutions of a homogeneous system ݉ ൈ ݊ form a subset of Թ௡ , and it is equal to the number of variables. Result C.1.3 Given a homogeneous system ݉ ൈ ݊. We ‫ݔ‬ଵ Ͳ ‫ݔ ۍ‬ଶ ‫ېͲۍ ې‬ ‫ۑ ێ ۑ ێ‬ write it in the matrix-form: ‫ݔ ێ ܥ‬ଷ ‫ ۑ‬ൌ ‫ ۑͲێ‬where ‫ ܥ‬is a ‫ۑڭێ ۑ ڭ ێ‬ ‫ݔۏ‬௡ ‫ےͲۏ ے‬


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‫ ܯ‬൅ ܹ is a solution. We write them in the matrix-form: ݉ଵ ‫ݓ‬ଵ Ͳ Ͳ ‫݉ ۍ‬ଶ ‫ې Ͳ ۍ ې‬ ‫ݓۍ‬ଶ ‫ېͲۍ ې‬ ‫ۑ ێ ۑ ێ‬ ‫ۑ ێ ۑ ێ‬ ‫݉ ێ ܥ‬ଷ ‫ ۑ‬ൌ ‫ ۑͲێ‬and ‫ݓێ‬ଷ ‫ ۑ‬ൌ ‫ۑͲێ‬ ‫ۑڭێ ۑ ڭ ێ‬ ‫ۑڭێ ۑ ڭ ێ‬ ‫ݓۏ‬௡ ‫ےͲۏ ے‬ ‫݉ ۏ‬௡ ‫ے Ͳ ۏ ے‬ ݉ଵ ‫ݓ‬ଵ Ͳ ‫݉ ۍ‬ଶ ‫ې‬ ‫ݓۍ‬ଶ ‫ېͲۍ ې‬ ‫ۑ ێ ۑ ێ‬ ‫ۑ ێ‬ Now, using algebra: ‫ ܯ‬൅ ܹ ൌ ‫݉ ێ ܥ‬ଷ ‫ ۑ‬൅ ‫ݓێ ܥ‬ଷ ‫ ۑ‬ൌ ‫ۑͲێ‬ ‫ۑ ڭ ێ‬ ‫ۑڭێ ۑ ڭ ێ‬ ‫ݓۏ‬௡ ‫ےͲۏ ے‬ ‫݉ۏ‬௡ ‫ے‬ By taking ‫ ܥ‬as a common factor, we obtain: ݉ଵ ‫ݓ‬ଵ Ͳ ‫݉ ۍ‬ଶ ‫ݓ ۍ ې‬ଶ ‫ې‬ ‫ېͲۍ‬ ‫ۑ ێ ۊ ۑ ێ ۑ ێۇ‬ ‫݉ ێۈ ܥ‬ଷ ‫ ۑ‬൅ ‫ݓ ێ‬ଷ ‫ ۋۑ‬ൌ ‫ۑͲێ‬ ‫ۑ ڭ ێ ۑ ڭ ێ‬ ‫ۑڭێ‬ ‫݉ۏۉ‬௡ ‫ݓۏ ے‬௡ ‫ےͲۏ یے‬ ݉ଵ ൅ ‫ݓ‬ଵ Ͳ ‫ ݉ۍ‬൅ ‫ې ۍ ې ݓ‬ ଶ ଶ ‫ێ‬ ‫ۑͲێ ۑ‬ ‫݉ ێ ܥ‬ଷ ൅ ‫ݓ‬ଷ ‫ ۑ‬ൌ ‫ۑͲێ‬ ‫ڭ‬ ‫ێ‬ ‫ۑڭێ ۑ‬ ‫݉ۏ‬௡ ൅ ‫ݓ‬௡ ‫ےͲۏ ے‬ Thus, ‫ ܯ‬൅ ܹ is a solution. Fact C.1.1 If ‫ܯ‬ଵ ൌ ሺ݉ଵ ǡ ݉ଶ ǡ ǥ ǡ ݉௡ ሻ is a solution, and

coefficient. Then, the set of all solutions in this system

ߙ ‫ א‬Թ, then ߙ‫ ܯ‬ൌ ሺߙ݉ଵ ǡ ߙ݉ଶ ǡ ǥ ǡ ߙ݉௡ ሻ is a solution.

is a subspace of Թ௡ .

Fact C.1.2 The only system where the solutions form a

Proof of Result C.1.3 We assume that

vector space is the homogeneous system.

‫ܯ‬ଵ ൌ ሺ݉ଵ ǡ ݉ଶ ǡ ǥ ǡ ݉௡ ሻ and ܹଵ ൌ ሺ‫ݓ‬ǡ ‫ݓ‬ଶ ǡ ǥ ǡ ‫ݓ‬௡ ሻ are two

Definition C.1.2 Null Space of a matrix, say ‫ ܣ‬is a set

solutions to the above system. We will show that

of all solutions to the homogeneous system, and it is denoted by ܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻ or ܰሺ‫ܣ‬ሻ.

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Definition C.1.3 Rank of a matrix, say ‫ ܣ‬is the number of independent rows or columns of ‫ܣ‬, and it is denoted by ܴܽ݊݇ሺ‫ܣ‬ሻ. Definition C.1.4 Row Space of a matrix, say ‫ ܣ‬is the ܵ‫ ݊ܽ݌‬of independent rows of ‫ܣ‬, and it is denoted by ܴ‫ݓ݋‬ሺ‫ܣ‬ሻ. Definition C.1.5 Column Space of a matrix, say ‫ ܣ‬is the ܵ‫ ݊ܽ݌‬of independent columns of ‫ܣ‬, and it is denoted by ‫݊݉ݑ݈݋ܥ‬ሺ‫ܣ‬ሻ. Example C.1.1 Given the following ͵ ൈ ͷ matrix: ͳ ‫ ܣ‬ൌ ൥Ͳ Ͳ

െͳ ͳ Ͳ

ʹ ʹ Ͳ

Ͳ െͳ Ͳ ʹ ൩. ͳ Ͳ

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Step 2: Apply what we have learned from chapter 1 to solve systems of linear equations use Row-Operation Method. ͳ ൭Ͳ Ͳ

െͳ ʹ ͳ ʹ Ͳ Ͳ

ͳ Ͳ Ͷ ൭Ͳ ͳ ʹ Ͳ Ͳ Ͳ Matrix.

Ͳ Ͳ ͳ

െͳ Ͳ ʹ อͲ൱ ܴଶ ൅ ܴଵ ՜ ܴଵ Ͳ Ͳ

Ͳ ͳͲ Ͳ ʹอͲ൱ This is a Completely-Reduced ͳ ͲͲ

Step 3: Read the solution for the above system of linear equations after using Row-Operation. ‫ݔ‬ଵ ൅ Ͷ‫ݔ‬ଷ ൅ ‫ݔ‬ହ ൌ Ͳ ‫ݔ‬ଶ ൅ ʹ‫ݔ‬ଷ ൅ ʹ‫ݔ‬ହ ൌ Ͳ ‫ݔ‬ସ ൌ Ͳ

a. Find ܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻ.

Free variables are ‫ݔ‬ଷ and ‫ݔ‬ହ .

b. Find ݀݅݉ሺܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻሻ.

Assuming that ‫ݔ‬ଷ , ‫ݔ‬ହ ‫ א‬Թ. Then, the solution of the above homogeneous system is as follows:

c. Rewrite ܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻ as ܵ‫݊ܽ݌‬. d. Find ܴܽ݊݇ሺ‫ܣ‬ሻ. e. Find ܴ‫ݓ݋‬ሺ‫ܣ‬ሻ. Solution: Part a: To find the null space of ‫ܣ‬, we need to find the solution of ‫ ܣ‬as follows: Step 1: Write the above matrix as an AugmentedMatrix, and make all constants’ terms zeros. ͳ െͳ ʹ ൭Ͳ ͳ ʹ Ͳ Ͳ Ͳ


Ͳ Ͳ ͳ

െͳ Ͳ ʹ อͲ൱ Ͳ Ͳ

‫ݔ‬ଵ ൌ െͶ‫ݔ‬ଷ െ ‫ݔ‬ହ ‫ݔ‬ଶ ൌ െʹ‫ݔ‬ଷ െ ʹ‫ݔ‬ହ ‫ݔ‬ସ ൌ Ͳ Thus, according to definition 3.1.2, ܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻ ൌ ሼሺെͶ‫ݔ‬ଷ െ ‫ݔ‬ହ ǡ െʹ‫ݔ‬ଷ െ ʹ‫ݔ‬ହ ǡ ‫ݔ‬ଷ ǡ Ͳǡ ‫ݔ‬ହ ሻȁ‫ݔ‬ଷ , ‫ݔ‬ହ ‫ א‬Թሽ. Part b: It is always true that ݀݅݉൫ܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻ൯ ൌ ݀݅݉൫ܰሺ‫ܣ‬ሻ൯ ൌ ݄ܶ݁ ܰ‫ݏ݈ܾ݁ܽ݅ݎܸܽ ݁݁ݎܨ ݂݋ ݎܾ݁݉ݑ‬

Here, ݀݅݉൫ܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻ൯ ൌ ʹ.

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Definition C.1.6 The nullity of a matrix, say ‫ ܣ‬is the

Result C.1.6 Let ‫ ܣ‬be ݉ ൈ ݊ matrix. The geometric

dimension of the null space of ‫ܣ‬, and it is denoted by

meaning of ‫݊݉ݑ݈݋ܥ‬ሺ‫ܣ‬ሻ ൌ ܵ‫݊ܽ݌‬ሼ‫ݏ݊݉ݑ݈݋ܥ ݐ݊݁݀݊݁݌݁݀݊ܫ‬ሽ

݀݅݉ሺܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻሻ or ݀݅݉ሺܰሺ‫ܣ‬ሻሻ.

“lives” inside Թ௠ .

Part c: We first need to find a basis for ܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻ as follows: To find a basis for ܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻ, we play a game called (ON-OFF GAME) with the free variables ‫ݔ‬ଷ and ‫ݔ‬ହ Ǥ

Result C.1.7 Let ‫ ܣ‬be ݉ ൈ ݊ matrix. Then,

‫ݔ‬ଷ ͳ Ͳ

‫ݔ‬ହ Ͳ ͳ

ܲ‫ݐ݊݅݋‬ ሺെͶǡ െʹǡͳǡͲǡͲሻ ሺെͳǡ െʹǡͲǡͲǡͳሻ

ܴܽ݊݇ሺ‫ܣ‬ሻ ൌ ݀݅݉൫ܴ‫ݓ݋‬ሺ‫ܣ‬ሻ൯ ൌ ݀݅݉ሺ‫݊݉ݑ݈݋ܥ‬ሺ‫ܣ‬ሻሻ. Example C.1.2 Given the following ͵ ൈ ͷ matrix: ͳ ͳ ͳ ͳ ‫ ܤ‬ൌ ൥െͳ െͳ െͳ Ͳ Ͳ Ͳ Ͳ Ͳ

ͳ ʹ൩. Ͳ

a. Find ܴ‫ݓ݋‬ሺ‫ܤ‬ሻ.

The basis for ܰ‫݈݈ݑ‬ሺ‫ܣ‬ሻ ൌ ሼሺെͶǡ െʹǡͳǡͲǡͲሻǡ ሺെͳǡ െʹǡͲǡͲǡͳሻሽ.

b. Find ‫݊݉ݑ݈݋ܥ‬ሺ‫ܤ‬ሻ.

Thus, ܰ‫ ݈݈ݑ‬ሺ‫ܣ‬ሻ ൌ ܵ‫݊ܽ݌‬ሼሺെͶǡ െʹǡͳǡͲǡͲሻǡ ሺെͳǡ െʹǡͲǡͲǡͳሻሽ.

c. Find ܴܽ݊݇ሺ‫ܤ‬ሻ.

Part d: To find the rank of matrix ‫ܣ‬, we just need to change matrix ‫ ܣ‬to the Semi-Reduced Matrix. We already did that in part a. Thus, ܴܽ݊݇ ሺ‫ܣ‬ሻ ൌ ͵Ǥ Part e: To find the row space of matrix ‫ܣ‬, we just need to write the ܵ‫ ݊ܽ݌‬of independent rows. Thus, ܴ‫ݓ݋‬ሺ‫ܣ‬ሻ ൌ ܵ‫݊ܽ݌‬ሼሺͳǡ െͳǡʹǡͲǡ െͳሻǡ ሺͲǡͳǡʹǡͲǡʹሻǡ ሺͲǡͲǡͲǡͳǡͲሻሽǤ It is also a subspace of Թହ . Result C.1.4 Let ‫ ܣ‬be ݉ ൈ ݊ matrix. Then, ܴܽ݊݇ሺ‫ܣ‬ሻ ൅ ݀݅݉൫ܰ ሺ‫ܣ‬ሻ൯ ൌ ݊ ൌ ܰ‫ܣ ݂݋ ݏ݊݉ݑ݈݋ܥ ݂݋ ݎܾ݁݉ݑ‬. Result C.1.5 Let ‫ ܣ‬be ݉ ൈ ݊ matrix. The geometric meaning of ܴ‫ݓ݋‬ሺ‫ܣ‬ሻ ൌ ܵ‫݊ܽ݌‬ሼ‫ݏݓ݋ܴ ݐ݊݁݀݊݁݌݁݀݊ܫ‬ሽ “lives” inside Թ௡ .

Solution: Part a: To find the row space of ‫ܤ‬, we need to change matrix ‫ ܤ‬to the Semi-Reduced Matrix as follows: ͳ ൥െͳ Ͳ

ͳ െͳ Ͳ

ͳ ͳ െͳ Ͳ Ͳ Ͳ

ͳ ܴ ൅ܴ ՜ܴ ͳ ଵ ଶ ଶ ൥ ʹ൩ ܴଵ ൅ ܴଷ ՜ ܴଷ Ͳ Ͳ Ͳ

ͳ ͳ ͳ Ͳ Ͳ ͳ Ͳ Ͳ Ͳ

ͳ ͵൩ Ͳ

This is a Semi-Reduced Matrix. To find the row space of matrix ‫ܤ‬, we just need to write the ܵ‫ ݊ܽ݌‬of independent rows. Thus, ܴ‫ݓ݋‬ሺ‫ܤ‬ሻ ൌ ܵ‫݊ܽ݌‬ሼሺͳǡͳǡͳǡͳǡͳሻǡ ሺͲǡͲǡͲǡͳǡ͵ሻሽǤ Part b: To find the column space of ‫ܤ‬, we need to change matrix ‫ ܤ‬to the Semi-Reduced Matrix. We already did that in part a. Now, we need to locate the columns in the Semi-Reduced Matrix of ‫ ܤ‬that contain the leaders, and then we should locate them to the original matrix ‫ܤ‬.

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ͳ ൥Ͳ Ͳ

ͳ ൥െͳ Ͳ

ͳ ͳ ͳ Ͳ Ͳ ͳ Ͳ Ͳ Ͳ

ͳ െͳ Ͳ

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ͳ (i.e. ቂ Ͳ

ͳ ͵൩ Semi-Reduced Matrix Ͳ

ͳ ͳ െͳ Ͳ Ͳ Ͳ


ʹ ͳ

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͵ ቃ ሺͳǡʹǡ͵ǡͲǡͳǡͳሻ ). ͳ

Fact C.2.3 Թଷൈଶ is equivalent to Թ଺ as a vector space. ͳ (i.e. ൥͵ ͳ

ͳ ʹ൩ Matrix ‫ܤ‬ Ͳ

ʹ Ͳ൩ ሺͳǡʹǡ͵ǡͲǡͳǡͳሻ ). ͳ

After knowing the above polynomials as follows:

Each remaining columns is a linear combination of the first and fourth columns.

facts,

we

introduce

ܲ௡ ൌ ܵ݁‫ ݁݁ݎ݃݁݀ ݂݋ ݏ݈ܽ݅݉݋݊ݕ݈݋݌ ݈݈ܽ ݂݋ ݐ‬൏ ݊Ǥ

Thus, ‫݊݉ݑ݈݋ܥ‬ሺ‫ܤ‬ሻ ൌ ܵ‫݊ܽ݌‬ሼሺͳǡ െͳǡͲሻǡ ሺͳǡͲǡͲሻሽ.

The algebraic expression of polynomials is in the following from: ܽ௡ ‫ ݔ‬௡ ൅ ܽ௡ିଵ ‫ ݔ‬௡ିଵ ൅ ‫ ڮ‬൅ ܽଵ ‫ ݔ‬ଵ ൅ ܽ଴

Part c: To find the rank of matrix ‫ܤ‬, we just need to

ܽ௡ ǡ ܽ௡ିଵ ܽଵ are coefficients.

change matrix ‫ ܣ‬to the Semi-Reduced Matrix. We

݊ ݊ െ ͳ are exponents that must be positive integers whole numbers.

already did that in part a. Thus, ܴܽ݊݇ሺ‫ܣ‬ሻ ൌ ݀݅݉൫ܴ‫ݓ݋‬ሺ‫ܤ‬ሻ൯ ൌ ݀݅݉ሺ‫݊݉ݑ݈݋ܥ‬ሺ‫ܤ‬ሻሻ ൌ ʹǤ

ܽ଴ is a constant term.

C.2 Linear Transformation We start

this section

with

an

introduction

to

polynomials, and we explain how they are similar to

The degree of polynomial is determined by the highest power (exponent). We list the following examples of polynomials: x

ܲଶ ൌ ܵ݁‫ ݁݁ݎ݃݁݀ ݂݋ ݏ݈ܽ݅݉݋݊ݕ݈݋݌ ݈݈ܽ ݂݋ ݐ‬൏ ʹ (i.e.

Before discussing polynomials, we need to know the following mathematical facts:

x

Fact C.2.1 Թ௡ൈ௠ ൌ Թ௡ൈ௠ ൌ ‫ܯ‬௡ൈ௠ ሺԹሻ is a vector space.

x

͵‫ ݔ‬൅ ʹ ‫ܲ א‬ଶ , Ͳ ‫ܲ א‬ଶ , ͳͲ ‫ܲ א‬ଶ , ξ͵ ‫ܲ א‬ଶ but ξ͵ξ‫ܲ ב ݔ‬ଶ ). ܲସ ൌ ܵ݁‫ ݁݁ݎ݃݁݀ ݂݋ ݏ݈ܽ݅݉݋݊ݕ݈݋݌ ݈݈ܽ ݂݋ ݐ‬൏ Ͷ (i.e. ͵ͳ‫ ݔ‬ଶ ൅ Ͷ ‫ܲ א‬ସ ). If ܲሺ‫ ݔ‬ሻ ൌ ͵, then ݀݁݃൫ܲ ሺ‫ ݔ‬ሻ൯ ൌ Ͳ.

x

ξ‫ ݔ‬൅ ͵ is not a polynomial.

Թ௡ as vector spaces. At the end of this section we discuss a new concept called linear transformation.

Fact C.2.2 Թଶൈଷ is equivalent to Թ଺ as a vector space.

Result C.2.1 ܲ௡ is a vector space.

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Fact C.2.4 Թଶൈଷ ൌ ‫ܯ‬ଶൈଷ ሺԹሻ as a vector space same as Թ଺ .

Then, we apply the Row-Reduction Method to get the Semi-Reduced Matrix as follows:

Result C.2.2 ܲ௡ is a vector space, and it is the same as Թ௡ . (i.e. ܽ଴ ൅ ܽଵ ‫ ݔ‬ଵ ൅ ‫ ڮ‬൅ ܽ௡ିଵ ‫ ݔ‬௡ିଵ ՞ ሺܽ଴ ǡ ܽଵ ǡ ǥ ǡ ܽ௡ିଵ ሻ. Note: The above form is in an ascending order.

െʹ ൥Ͳ െͶ

Result C.2.3 ݀݅݉ሺܲ௡ ሻ ൌ ݊Ǥ Fact C.2.5 ܲଷ ൌ ܵ‫݊ܽ݌‬ሼ͵ ‫ݏ݈ܽ݅݉݋݊ݕ݈݋ܲ ݐ݊݁݀݊݁݌݁݀݊ܫ‬ǡ ܽ݊݀ ‫ ݁݁ݎ݃݁ܦ ݂݋ ݄ܿܽܧ‬൏ ͵ሽ. (i.e. ܲଷ ൌ ܵ‫݊ܽ݌‬ሼͳǡ ‫ݔ‬ǡ ‫ ʹݔ‬ሽ). Example C.2.1 Given the following polynomials: ͵‫ ݔ‬ଶ െ ʹǡ െͷ‫ݔ‬ǡ ͸‫ ݔ‬ଶ െ ͳͲ‫ ݔ‬െ Ͷ. a. Are these polynomials independent? b. Let ‫ ܦ‬ൌ ܵ‫݊ܽ݌‬ሼ͵‫ ݔ‬ଶ െ ʹǡ െͷ‫ݔ‬ǡ ͸‫ ݔ‬ଶ െ ͳͲ‫ ݔ‬െ Ͷሽ. Find a basis for ‫ܦ‬.

Ͳ െͷ െͳͲ

͵ െʹ Ͳ൩ െʹܴଵ ൅ ܴଷ ՜ ܴଷ ൥ Ͳ ͸ Ͳ

Ͳ െͷ െͳͲ

͵ Ͳ൩ Ͳ

െʹ Ͳ ͵ െʹܴଶ ൅ ܴଷ ՜ ܴଷ ൥ Ͳ െͷ Ͳ൩ This is a Semi-Reduced Ͳ Ͳ Ͳ Matrix. Since there is a zero-row in the Semi-Reduced Matrix, then these elements are dependent. Thus, the answer to this question is NO. Part b: Since there are only 2 vectors survived after checking for dependency in part a, then the basis for ሺͲǡ െͷǡͲሻ ՞ െͷ‫ݔ‬.

Solution: Part a: We know that these polynomial live in ܲଷ , and as a vector space ܲଷ is the same as Թଷ . According to result 3.2.2, we need to make each polynomial equivalent to Թ௡ as follows:

Result C.2.4 Given ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ǡ ǥ ǡ ‫ݒ‬௞ points in Թ௡ where ݇ ൏ ݊. Choose one particular point, say ܳ, such that ܳ ൌ ܿଵ ‫ݒ‬ଵ ൅ ܿଶ ‫ݒ‬ଶ ൅ ‫ ڮ‬൅ ܿ௞ ‫ݒ‬௞ where ܿଵ ǡ ܿଶ ǡ ǥ ǡ ܿ௞ are

͵‫ ݔ‬ଶ െ ʹ ൌ െʹ ൅ Ͳ‫ ݔ‬൅ ͵‫ ݔ‬ଶ ՞ ሺെʹǡͲǡ͵ሻ

constants. If ܿଵ ǡ ܿଶ ǡ ǥ ǡ ܿ௞ are unique, then ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ǡ ǥ ǡ ‫ݒ‬௞ are independent.

െͷ‫ ݔ‬ൌ Ͳ െ ͷ‫ ݔ‬൅ Ͳ‫ ݔ‬ଶ ՞ ሺͲǡ െͷǡͲሻ ͸‫ ݔ‬ଶ െ ͳͲ‫ ݔ‬െ Ͷ ൌ െͶ െ ͳͲ‫ ݔ‬൅ ͸‫ ݔ‬ଶ ՞ ሺെͶǡ െͳͲǡ͸ሻ Now, we need to write these vectors as a matrix. െʹ Ͳ ͵ ൥Ͳ െͷ Ͳ൩ Each point is a row-operation. We need െͶ െͳͲ ͸ to reduce this matrix to Semi-Reduced Matrix.

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Note: The word “unique” in result 3.2.4 means that there is only one value for each of ܿଵ ǡ ܿଶ ǡ ǥ ǡ ܿ௞ . Proof of Result C.2.4 By using proof by contradiction, we assume that ‫ݒ‬ଵ ൌ ߙଶ ‫ݒ‬ଶ ൅ ߙଷ ‫ݒ‬ଷ ൅ ‫ ڮ‬൅ ߙ௞ ‫ݒ‬௞ where ߙଶ ǡ ߙଷ ǡ ǥ ǡ ߙ௞ are constants. Our assumption means that it is dependent. Using algebra, we obtain:

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ܳ ൌ ܿଵ ߙଶ ‫ݒ‬ଶ ൅ ܿଵ ߙଷ ‫ݒ‬ଷ ൅ ‫ ڮ‬൅ ܿଵ ߙ௞ ‫ݒ‬௞ ൅ ܿଶ ‫ݒ‬ଶ ൅ ‫ ڮ‬൅ ܿ௞ ‫ݒ‬௞ .

Part c: Proof: We assume that ‫ݒ‬ଵ ൌ ሺܽଵ ǡ ܽଶ ሻ,

ܳ ൌ ሺܿଵ ߙଶ ൅ܿଶ ሻ‫ݒ‬ଶ ൅ ሺܿଵ ߙଷ ൅ ܿଷ ሻ‫ݒ‬ଷ ൅ ‫ ڮ‬൅ ሺܿଵ ߙ௞ ൅ ܿ௞ ሻ‫ݒ‬௞ ൅

‫ݒ‬ଶ ൌ ሺܾଵ ǡ ܾଶ ሻ, and ߙ ‫ א‬Թ. We will show that ܶ is a linear

Ͳ‫ݒ‬ଵ Ǥ Thus, none of them is a linear combination of the

transformation. Using algebra, we start from the Left-

others which means that they are linearly

Hand-Side (LHS):

independent. This is a contradiction. Therefore, our

ߙ‫ݒ‬ଵ ൅ ‫ݒ‬ଶ ൌ ሺߙܽଵ ൅ ܾଵ ǡ ߙܽଶ ൅ ܾଶ ሻ

assumption that ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ǡ ǥ ǡ ‫ݒ‬௞ were linearly

ܶሺߙ‫ݒ‬ଵ ൅ ‫ݒ‬ଶ ሻ ൌ ܶሺሺߙܽଵ ൅ ܾଵ ǡ ߙܽଶ ൅ ܾଶ ሻሻ

dependent is false. Hence, ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ǡ ǥ ǡ ‫ݒ‬௞ are linearly

ܶሺߙ‫ݒ‬ଵ ൅ ‫ݒ‬ଶ ሻ ൌ ሺ͵ߙܽଵ ൅ ͵ܾଵ ൅ ߙܽଶ ൅ ܾଶ ǡ ߙܽଶ ൅ ܾଶ ǡ െߙܽଵ െ ܾଵ ሻ

independent.

Now, we start from the Right-Hand-Side (RHS):

Result C.2.5 Assume ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ǡ ǥ ǡ ‫ݒ‬௞ are independent and ܳ ‫݊ܽ݌ܵ א‬ሼ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ǡ ǥ ǡ ‫ݒ‬௞ ሽ. Then, there exists unique number ܿଵ ǡ ܿଶ ǡ ǥ ǡ ܿ௞ such that ܳ ൌ ܿଵ ‫ݒ‬ଵ ൅ ܿଶ ‫ݒ‬ଶ ൅ ‫ ڮ‬൅ ܿ௞ ‫ݒ‬௞ .

ߙܶሺ‫ݒ‬ଵ ሻ ൅ ܶሺ‫ݒ‬ଶ ሻ ൌ ߙܶሺܽଵ ǡ ܽଶ ሻ ൅ ܶሺܾଵ ǡ ܾଶ ሻ

Linear Transformation: Definition C.2.1 ܶǣ ܸ ՜ ܹ where ܸ is a domain and ܹ is a co-domain. ܶ is a linear transformation if for every ‫ݒ‬ଵ ǡ ‫ݒ‬ଶ ‫ ܸ א‬and ߙ ‫ א‬Թ, we have the following: ܶሺߙ‫ݒ‬ଵ ൅ ‫ݒ‬ଶ ሻ ൌ ߙܶሺ‫ݒ‬ଵ ሻ ൅ ܶሺ‫ݒ‬ଶ ሻ. Example C.2.2 Given ܶǣ Թଶ ՜ Թଷ where Թଶ is a domain and Թଷ is a co-domain. ܶ൫ሺܽଵ ǡ ܽଶ ሻ൯ ൌ ሺ͵ܽଵ ൅ ܽଶ ǡ ܽଶ ǡ െܽଵ ሻ.

Thus, ܶ is a linear transformation.

a. Find ܶሺሺͳǡͳሻሻ. b. Find ܶሺሺͳǡͲሻሻ. c. Show that ܶ is a linear transformation. Solution: Part a: Since ܶ൫ሺܽଵ ǡ ܽଶ ሻ൯ ൌ ሺ͵ܽଵ ൅ ܽଶ ǡ ܽଶ ǡ െܽଵ ሻ, then ܽଵ ൌ ܽଶ ൌ ͳ. Thus, ܶ൫ሺͳǡͳሻ൯ ൌ ሺ͵ሺͳሻ ൅ ͳǡͳǡ െͳሻ ൌ ሺͶǡͳǡ െͳሻ. Part b: Since ܶ൫ሺܽଵ ǡ ܽଶ ሻ൯ ൌ ሺ͵ܽଵ ൅ ܽଶ ǡ ܽଶ ǡ െܽଵ ሻ, then ܽଵ ൌ ͳ ܽଶ ൌ Ͳ. Thus, ܶ൫ሺͳǡͲሻ൯ ൌ ሺ͵ሺͳሻ ൅ ͲǡͲǡ െͳሻ ൌ ሺ͵ǡͲǡ െͳሻ.

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ߙܶሺ‫ݒ‬ଵ ሻ ൅ ܶሺ‫ݒ‬ଶ ሻ ൌ ߙሺ͵ܽଵ ൅ ܽଶ ǡ ܽଶ ǡ െܽଵ ሻ ൅ ሺ͵ܾଵ ൅ ܾଶ ǡ ܾଶ ǡ െܾଵ ሻ

ൌ ሺ͵ߙܽଵ ൅ ߙܽଶ ǡ ߙܽଶ ǡ െߙܽଵ ሻ ൅ ሺ͵ܾଵ ൅ ܾଶ ǡ ܾଶ ǡ െܾଵ ሻ ൌ ሺ͵ߙܽଵ ൅ ߙܽଶ ൅ ͵ܾଵ ൅ ܾଶ ǡ ߙܽଶ ൅ ܾଶ ǡ െߙܽଵ െ ܾଵ ሻ

Result C.2.6 Given ܶǣ Թ௡ ՜ Թ௠ . Then, ܶሺሺܽଵ ǡ ܽଶ ǡ ܽଷ ǡ ǥ ǡ ܽ௡ ሻሻ ൌ Each coordinate is a linear combination of the ܽ௜ Ԣ‫ݏ‬. Example C.2.3 Given ܶǣ Թଷ ՜ Թସ where Թଷ is a domain and Թସ is a co-domain. a. If ܶ൫ሺ‫ݔ‬ଵ ǡ ‫ݔ‬ଶ ǡ ‫ݔ‬ଷ ሻ൯ ൌ ሺെ͵‫ݔ‬ଷ ൅ ͸‫ݔ‬ଵ ǡ െͳͲ‫ݔ‬ଶ ǡ ͳ͵ǡ െ‫ݔ‬ଷ ሻ, is ܶ a linear transformation? b. If ܶ൫ሺ‫ݔ‬ଵ ǡ ‫ݔ‬ଶ ǡ ‫ݔ‬ଷ ሻ൯ ൌ ሺെ͵‫ݔ‬ଷ ൅ ͸‫ݔ‬ଵ ǡ െͳͲ‫ݔ‬ଶ ǡ Ͳǡ െ‫ݔ‬ଷ ሻ, is ܶ a linear transformation? Solution: Part a: Since 13 is not a linear combination of ‫ݔ‬ଵ ǡ ‫ݔ‬ଶ ‫ݔ‬ଷ . Thus, ܶ is not a linear transformation. Part b: Since 0 is a linear combination of ‫ݔ‬ଵ ǡ ‫ݔ‬ଶ ‫ݔ‬ଷ . Thus, ܶ is a linear transformation.

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Example C.2.4 Given ܶǣ Թଶ ՜ Թଷ where Թଶ is a domain and Թଷ is a co-domain. If ܶ൫ሺܽଵ ǡ ܽଶ ሻ൯ ൌ ሺܽଵ ଶ ൅ ܽଶ ǡ െܽଶ ሻ, is ܶ a linear transformation? Solution: Since ܽଵ ଶ ൅ ܽଶ is not a linear combination of ܽଵ ܽଶ . Hence, ܶ is not a linear transformation. Example C.2.5 Given ܶǣ Թ ՜ Թ. If ܶሺ‫ ݔ‬ሻ ൌ ͳͲ‫ݔ‬, is ܶ a linear transformation? Solution: Since it is a linear combination of ܽଵ such that ߙܽଵ ൌ ͳͲ‫ݔ‬. Hence, ܶ is a linear transformation. ଶ

Example C.2.6 Find the standard basis for Թ . Solution: The standard basis for Թଶ is the rows of ‫ܫ‬ଶ . ͳ Since ‫ܫ‬ଶ ൌ ቂ Ͳ ሼሺͳǡͲሻǡ ሺͲǡͳሻሽ.

Ͳ ቃ, then the standard basis for Թଶ is ͳ

Example C.2.7 Find the standard basis for Թଷ . Solution: The standard basis for Թଷ is the rows of ‫ܫ‬ଷ . ͳ Ͳ Ͳ Since ‫ܫ‬ଷ ൌ ൥Ͳ ͳ Ͳ൩, then the standard basis for Թଷ is Ͳ Ͳ ͳ ሼሺͳǡͲǡͲሻǡ ሺͲǡͳǡͲሻǡ ሺͲǡͲǡͳሻሽ. Example C.2.8 Find the standard basis for ܲଷ . Solution: The standard basis for ܲଷ is ሼͳǡ ‫ݔ‬ǡ ‫ ݔ‬ଶ ሽ. Example C.2.9 Find the standard basis for ܲସ . Solution: The standard basis for ܲସ is ሼͳǡ ‫ݔ‬ǡ ‫ ݔ‬ଶ ǡ ‫ ݔ‬ଷ ሽ.

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Example C.2.10 Find the standard basis for Թଶൈଶ ൌ ‫ܯ‬ଶൈଶ ሺԹሻ. Solution: The standard basis for Թଶൈଶ ൌ ‫ܯ‬ଶൈଶ ሺԹሻ is ͳ Ͳ Ͳ ͳ Ͳ Ͳ Ͳ Ͳ ቃǡቂ ቃǡቂ ቃǡቂ ቃሽ because Թଶൈଶ ൌ ሼቂ Ͳ Ͳ Ͳ Ͳ ͳ Ͳ Ͳ ͳ ‫ܯ‬ଶൈଶ ሺԹሻ ൌ Թସ as a vector space where standard basis ͳ Ͳ Ͳ Ͳ for Թଶൈଶ ൌ ‫ܯ‬ଶൈଶ ሺԹሻ is the rows of ‫ܫ‬ସ ൌ ቎Ͳ ͳ Ͳ Ͳ቏that Ͳ Ͳ ͳ Ͳ Ͳ Ͳ Ͳ ͳ are represented by ʹ ൈ ʹ matrices. Example C.2.11 Let ܶǣ Թଶ ՜ Թଷ be a linear transformation such that ܶሺʹǡͲሻ ൌ ሺͲǡͳǡͶሻ ܶሺെͳǡͳሻ ൌ ሺʹǡͳǡͷሻ Find ܶሺ͵ǡͷሻ. Solution: The given points are ሺʹǡͲሻ and ሺെͳǡͳሻ. These two points are independent because of the following: ʹ Ͳ ͳ ʹ ቂ ቃ ܴ ൅ ܴଶ ՜ ܴଶ ቂ െͳ ͳ ʹ ଵ Ͳ

Ͳ ቃ ͳ

Every point in Թଶ is a linear combination of ሺʹǡͲሻ and ሺെͳǡͳሻ. There exists unique numbers ܿଵ and ܿଶ such that ሺ͵ǡͷሻ ൌ ܿଵ ሺʹǡͲሻ ൅ ܿଶ ሺെͳǡͳሻ. ͵ ൌ ʹܿଵ െ ܿଶ ͷ ൌ ܿଶ Now, we substitute ܿଶ ൌ ͷ in ͵ ൌ ʹܿଵ െ ܿଶ , we obtain: ͵ ൌ ʹܿଵ െ ͷ ܿଵ ൌ Ͷ Hence, ሺ͵ǡͷሻ ൌ ͶሺʹǡͲሻ ൅ ͷሺെͳǡͳሻ. ܶሺ͵ǡͷሻ ൌ ܶሺͶሺʹǡͲሻ ൅ ͷሺെͳǡͳሻሻ

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ܶሺ͵ǡͷሻ ൌ ͶܶሺʹǡͲሻ ൅ ͷܶሺെͳǡͳሻ ܶሺ͵ǡͷሻ ൌ ͶሺͲǡͳǡͶሻ ൅ ͷሺʹǡͳǡͷሻ ൌ ሺͳͲǡͻǡͶͳሻ Thus, ܶሺ͵ǡͷሻ ൌ ሺͳͲǡͻǡͶͳሻ. Example C.2.12 Let ܶǣ Թ ՜ Թ be a linear transformation such that ܶሺͳሻ ൌ ͵. Find ܶሺͷሻ. Solution: Since it is a linear transformation, then ܶሺͷሻ ൌ ܶሺͷ ή ͳሻ ൌ ͷܶሺͳሻ ൌ ͷሺ͵ሻ ൌ ͳͷ. If it is not a linear transformation, then it is impossible to find ܶሺͷሻ.

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ܶ൫ሺ‫ݔ‬ଵ ǡ ‫ݔ‬ଶ ǡ ‫ݔ‬ଷ ሻ൯ ൌ ሺെͷ‫ݔ‬ଵ ǡ ʹ‫ݔ‬ଶ ൅ ‫ݔ‬ଷ ǡ െ‫ݔ‬ଵ ǡ Ͳሻ a. b. c. d.

Find the Standard Matrix Representation. Find ܶሺሺ͵ǡʹǡͳሻሻ. Find ‫ݎ݁ܭ‬ሺܶሻ. Find ܴܽ݊݃݁ሺܶሻ.

Solution: Part a: According to definition 3.3.1, the Standard Matrix Representation, let’s call it ‫ܯ‬, here is Ͷ ൈ ͵. We know from section 3.2 that the standard basis for domain (here is Թଷ ) is ሼሺͳǡͲǡͲሻǡ ሺͲǡͳǡͲሻǡ ሺͲǡͲǡͳሻሽ. We assume the following: ‫ݒ‬ଵ ൌ ሺͳǡͲǡͲሻ

matrix representation, and we give examples of how to

‫ݒ‬ଶ ൌ ሺͲǡͳǡͲሻ

find kernel and range.

‫ݒ‬ଷ ൌ ሺͲǡͲǡͳሻ

݀݅݉ሺ‫ ݋ܥ‬െ ‫ ݊݅ܽ݉݋ܦ‬ሻ ൈ ݀݅݉ሺ‫݊݅ܽ݉݋ܦ‬ሻ matrix.

Now, we substitute each point of the standard basis for domain in ܶ൫ሺ‫ݔ‬ଵ ǡ ‫ݔ‬ଶ ǡ ‫ݔ‬ଷ ሻ൯ ൌ ሺെͷ‫ݔ‬ଵ ǡ ʹ‫ݔ‬ଶ ൅ ‫ݔ‬ଷ ǡ െ‫ݔ‬ଵ ǡ Ͳሻ as follows: ܶ൫ሺͳǡͲǡͲሻ൯ ൌ ሺെͷǡͲǡ െͳǡͲሻ ܶ൫ሺͲǡͳǡͲሻ൯ ൌ ሺͲǡʹǡͲǡͲሻ

Definition C.3.2 Given ܶǣ Թ௡ ՜ Թ௠ where Թ௡ is a

ܶ൫ሺͲǡͲǡͳሻ൯ ൌ ሺͲǡͳǡͲǡͲሻ

domain and Թ௠ is a co-domain. Kernel is a set of all

‫ݔ‬ଵ ‫ݔ‬ Our goal is to find ‫ ܯ‬so that ܶ൫ሺ‫ݔ‬ଵ ǡ ‫ݔ‬ଶ ǡ ‫ݔ‬ଷ ሻ൯ ൌ ‫ ܯ‬൥ ଶ ൩. ‫ݔ‬ଷ

Definition C.3.1 Given ܶǣ Թ௡ ՜ Թ௠ where Թ௡ is a domain and Թ௠ is a co-domain. Then, Standard Matrix Representation is a ݉ ൈ ݊ matrix. This means that it is

points in the domain that have image which equals to the origin point, and it is denoted by ‫ݎ݁ܭ‬ሺܶሻ. This means that ‫ݎ݁ܭ‬ሺܶሻ ൌ ܰ‫ܶ ݂݋ ݁ܿܽ݌ܵ ݈݈ݑ‬. Definition C.3.3 Range is the column space of standard matrix representation, and it is denoted by ܴܽ݊݃݁ሺܶሻ. Example C.3.1 Given ܶǣ Թଷ ՜ Թସ where Թଷ is a domain and Թସ is a co-domain.

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െͷ Ͳ Ͳ ‫ ܯ‬ൌ ቎ Ͳ ʹ ͳ቏ This is the Standard Matrix െͳ Ͳ Ͳ Ͳ Ͳ Ͳ Representation. The first, second and third columns represent ܶሺ‫ݒ‬ଵ ሻǡ ܶሺ‫ݒ‬ଶ ሻ ܶሺ‫ݒ‬ଷ ሻ.

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‫ݔ‬ଵ ‫ݔ‬ ሻ൯ Part b: Since ൫ሺ‫ݔ‬ଵ ǡ ‫ݔ‬ଶ ǡ ‫ݔ‬ଷ ൌ ‫ ܯ‬൥ ଶ ൩ , then ‫ݔ‬ଷ െͷ Ͳ Ͳ ͵ ܶ൫ሺ͵ǡʹǡͳሻ൯ ൌ ቎ Ͳ ʹ ͳ቏ ൥ʹ൩ െͳ Ͳ Ͳ ͳ Ͳ Ͳ Ͳ െͷ Ͳ Ͳ െͳͷ Ͳ ʹ ͳ ܶ൫ሺ͵ǡʹǡͳሻ൯ ൌ ͵ ή ቎ ቏ ൅ ʹ ή ቎ ቏ ൅ ͳ ή ቎ ቏ ൌ ቎ ͷ ቏ Ͳ Ͳ െͳ െ͵ Ͳ Ͳ Ͳ Ͳ െͳͷ ቎ ͷ ቏ is equivalent to ሺെͳͷǡͷǡ െ͵ǡͲሻ. This lives in the െ͵ Ͳ co-domain. Thus, ܶ൫ሺ͵ǡʹǡͳሻ൯ ൌ ሺെͳͷǡͷǡ െ͵ǡͲሻ.

ͳ Ͳ ͲͲ ͳ Ͳ Ͳ Ͳ ଵ Ͳ Ͳ ʹ ͳ ቌ ቮ ቍ ܴ ቌͲ ͳ ͲǤͷቮͲቍ This is a CompletelyͲ Ͳ ͲͲ ଶ ଶ Ͳ Ͳ Ͳ Ͳ Ͳ Ͳ ͲͲ Ͳ Ͳ Ͳ Ͳ Reduced Matrix. Now, we need to read the above matrix as follows: ‫ݔ‬ଵ ൌ Ͳ ͳ ‫ݔ‬ଶ ൅ ‫ݔ‬ଷ ൌ Ͳ ʹ ͲൌͲ ͲൌͲ To write the solution, we need to assume that ‫ݔ‬ଷ ‫ א‬Թ ሺ‫݈ܾ݁ܽ݅ݎܸܽ ݁݁ݎܨ‬ሻ.

Part c: According to definition 3.3.2, ‫ݎ݁ܭ‬ሺܶሻ is a set of all points in the domain that have imageൌ ሺͲǡͲǡͲǡͲሻ. Hence, ܶ൫ሺ‫ݔ‬ଵ ǡ ‫ݔ‬ଶ ǡ ‫ݔ‬ଷ ሻ൯ ൌ ሺͲǡͲǡͲǡͲሻ. This means the Ͳ ‫ݔ‬ଵ ‫ݔ‬ following: ‫ ܯ‬൥ ଶ ൩ ൌ ቎Ͳ቏ Ͳ ‫ݔ‬ଷ Ͳ Ͳ െͷ Ͳ Ͳ ‫ݔ‬ଵ ቎ Ͳ ʹ ͳ ቏ ൥‫ ݔ‬ଶ ൩ ൌ ቎ Ͳ ቏ Ͳ െͳ Ͳ Ͳ ‫ݔ‬ ଷ Ͳ Ͳ Ͳ Ͳ

ܰሺ‫ܯ‬ሻ ൌ ሼሺͲǡ െ ଶ ‫ݔ‬ଷ ǡ ‫ݔ‬ଷ ሻȁ‫ݔ‬ଷ ‫ א‬Թሽ.

Since ‫ݎ݁ܭ‬ሺܶሻ ൌ ܰ‫݈݈ݑ‬ሺ‫ܯ‬ሻ, then we need to find ܰሺ‫ܯ‬ሻ as follows: ͳ Ͳ ͲͲ െͷ Ͳ Ͳ Ͳ ͳ Ͳ ቌ Ͳ ʹ ͳቮ ቍ െ ܴଵ ቌ Ͳ ʹ ͳቮͲቍ ܴଵ ൅ ܴଷ ՜ ܴଷ െͳ Ͳ Ͳ Ͳ െͳ Ͳ Ͳ Ͳ ͷ Ͳ Ͳ ͲͲ Ͳ Ͳ ͲͲ

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ଵ

Hence, ‫ݔ‬ଵ ൌ Ͳ and ‫ݔ‬ଶ ൌ െ ଶ ‫ݔ‬ଷ . ଵ

By letting ‫ݔ‬ଷ ൌ ͳ, we obtain: ܰ‫ݕݐ݈݈݅ݑ‬ሺ‫ܯ‬ሻ ൌ ܰ‫ ݏ݈ܾ݁ܽ݅ݎܸܽ ݁݁ݎܨ ݂݋ ݎܾ݁݉ݑ‬ൌ ͳ, and ଵ

‫ ݏ݅ݏܽܤ‬ൌ ሼሺͲǡ െ ǡ ͳሻሽ ଶ

ଵ

Thus, ‫ݎ݁ܭ‬ሺܶሻ ൌ ܰሺ‫ ܯ‬ሻ ൌ ܵ‫݊ܽ݌‬ሼሺͲǡ െ ǡ ͳሻሽ. ଶ

Part d: According to definition 3.3.3, ܴܽ݊݃݁ሺܶሻ is the column space of ‫ܯ‬. Now, we need to locate the columns in the Completely-Reduced Matrix in part c that contain the leaders, and then we should locate them to the original matrix as follows:

ͳ ቌͲ Ͳ Ͳ

Ͳ Ͳ Ͳ ͳ ͲǤͷቮͲቍ Completely-Reduced Matrix Ͳ Ͳ Ͳ Ͳ Ͳ Ͳ

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Solution: െͷ ቌͲ െͳ Ͳ

Ͳ ʹ Ͳ Ͳ

ͲͲ ͳቮͲቍ Orignial Matrix ͲͲ ͲͲ

Thus, ܴܽ݊݃݁ሺܶሻ ൌ ܵ‫݊ܽ݌‬ሼሺെͷǡͲǡ െͳǡͲሻǡ ሺͲǡʹǡͲǡͲሻሽ. Result C.3.1 Given ܶǣ Թ௡ ՜ Թ௠ where Թ௡ is a domain and Թ௠ is a co-domain. Let ‫ ܯ‬be a standard matrix representation. Then, ܴܽ݊݃݁ሺܶሻ ൌ ܵ‫݊ܽ݌‬ሼ‫ܯ ݂݋ݏ݊݉ݑ݈݋ܥ ݐ݊݁݀݊݁݌݁݀݊ܫ‬ሽ. Result C.3.2 Given ܶǣ Թ௡ ՜ Թ௠ where Թ௡ is a domain and Թ௠ is a co-domain. Let ‫ ܯ‬be a standard matrix representation. Then, ݀݅݉൫ܴܽ݊݃݁ሺܶሻ൯ ൌ ܴܽ݊݇ሺ‫ܯ‬ሻ = Number of Independent Columns. Result C.3.3 Given ܶǣ Թ௡ ՜ Թ௠ where Թ௡ is a domain and Թ௠ is a co-domain. Let ‫ ܯ‬be a standard matrix representation. Then, ݀݅݉൫‫ݎ݁ܭ‬ሺܶሻ൯ ൌ ܰ‫ݕݐ݈݈݅ݑ‬ሺ‫ܯ‬ሻ. Result C.3.4 Given ܶǣ Թ௡ ՜ Թ௠ where Թ௡ is a domain and Թ௠ is a co-domain. Let ‫ ܯ‬be a standard matrix representation. Then, ݀݅݉൫ܴܽ݊݃݁ሺܶሻ൯ ൅ ݀݅݉ሺ‫ݎ݁ܭ‬ሺܶሻሻ ൌ ݀݅݉ሺ‫݊݅ܽ݉݋ܦ‬ሻ.

ଵ ‫ݔ‬ൌͳ ൌ ͲǤ Part a: ܶሺʹ‫ ݔ‬െ ͳሻ ൌ ‫׬‬଴ ሺʹ‫ ݔ‬െ ͳሻ݀‫ ݔ‬ൌ ‫ ݔ‬ଶ െ ‫ ݔ‬ቚ ‫ݔ‬ൌͲ Part b: To find ‫ݎ݁ܭ‬ሺܶሻ, we set equation of ܶ ൌ Ͳ, and ݂ሺ‫ ݔ‬ሻ ൌ ܽ଴ ൅ ܽଵ ‫ܲ א ݔ‬ଶ . ଵ ௔ ‫ݔ‬ൌͳ Thus, ܶሺ݂ሺ‫ݔ‬ሻሻ ൌ ‫׬‬଴ ሺܽ଴ ൅ ܽଵ ‫ ݔ‬ሻ݀‫ ݔ‬ൌ ܽ଴ ‫ ݔ‬൅ ଶభ ‫ ݔ‬ଶ ቚ ൌͲ ‫ݔ‬ൌͲ ܽଵ ܽ଴ ൅ െ Ͳ ൌ Ͳ ʹ ௔ ܽ଴ ൌ െ ଶభ

Hence, ‫ݎ݁ܭ‬ሺܶሻ ൌ ሼെ

௔భ ଶ

൅ ܽଵ ‫ݔ‬ȁܽଵ ‫ א‬Թሽ. We also know that

݀݅݉ሺ‫ݎ݁ܭ‬ሺܶሻ ൌ ͳ because there is one free variable. In addition, we can also find basis by letting ܽଵ be any real number not equal to zero, say ܽଵ ൌ ͳ, as follows: ͳ ‫ ݏ݅ݏܽܤ‬ൌ ሼെ ൅ ‫ݔ‬ሽ ʹ ଵ ሺ Thus, ‫ܶ ݎ݁ܭ‬ሻ ൌ ܵ‫݊ܽ݌‬ሼെ ଶ ൅ ‫ݔ‬ሽ. Part c: It is very easy to find range here. ܴܽ݊݃݁ሺܶሻ ൌ Թ because we linearly transform from a second degree polynomial to a real number. For example, if we linearly transform from a third degree polynomial to a second degree polynomial, then the range will be ܲଶ .

ଵ

Example C.3.2 Given ܶǣ ܲଶ ՜ Թ. ܶሺ݂ ሺ‫ ݔ‬ሻሻ ൌ ‫׬‬଴ ݂ሺ‫ ݔ‬ሻ݀‫ ݔ‬is a linear transformation. a. Find ܶሺʹ‫ ݔ‬െ ͳሻ. b. Find ‫ݎ݁ܭ‬ሺܶሻ. c. Find ܴܽ݊݃݁ሺܶሻ.

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Answers to Odd-Numbered Exercises This page intentionally left blank

1.7 Exercises ଵ଴

1. ࣦ ିଵ ቄሺ௦ିସሻర ቅ ൌ ௦ାହ

ଵ଴ ସ௫ ଷ ݁ ‫ݔ‬ ଺ ଵ

3. ࣦ ିଵ ቄሺ௦ାଷሻర ቅ ൌ ଶ ‫ ݔ‬ଶ ݁ ିଷ௫ ൅ ͸‫ ݔ‬ଷ ݁ ିଷ௫ ଶ

5. ࣦ ିଵ ቄ௦మି଺௦ାଵଷቅ ൌ ݁ ଷ௫ ሺʹ‫ݔ‬ሻ 7. ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ െͶ݁െ͵‫ݔ‬ ସ

ଵ

ଵ

9. ࣦ ିଵ ቄሺ௦ିଵሻమሺ௦ାଷሻቅ ൌ ସ ݁ ௫ ൅ ‫ ݁ݔ‬௫ ൅ ସ ݁ ିଷ௫ 11. ࣦ൛ܷሺ‫ ݔ‬െ ʹሻ݁͵‫ ݔ‬ൟ ൌ 13.

௦݁െͶ‫ݔ‬ ࣦ ିଵ ቄ మ ቅ ௦ ାସ

௘ లషమೞ

‫ݏ‬െ͵

ൌ ܷሺ‫ ݔ‬െ Ͷሻሺʹ‫ ݔ‬െ ͺሻ ଷ

ଵ

ͳ

15. Assume ܹሺ‫ ݔ‬ሻ ൌ െ ଼ ൅ ଷ ݁‫ ݔ‬൅ ʹͶ ݁െͺ‫ ݔ‬. Then, we obtain: ‫ݕ‬ሺ‫ݔ‬ሻ ൌ ܹሺ‫ݔ‬ሻ െ ͵ܷሺ‫ ݔ‬െ ͷሻܹ ሺ‫ ݔ‬െ ͷሻ െ ʹܷሺ‫ ݔ‬െ ͷሻܹሺ‫ ݔ‬െ ͷሻ ௘ షయೣ

17. ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ࣦെͳ ቄ‫ ʹݏ‬ሺ‫ ʹݏ‬൅ͳሻቅ ଶ௦

ସ௦

19. ࣦ ିଵ ቄሺ௦మାସሻమቅ ൌ ሺ௦మ ାସሻమ 21. ‫ ݓ‬ሺ‫ݐ‬ሻ ൌ ͳ ൅ ‫ ʹݐ‬and ݄ሺ‫ݐ‬ሻ ൌ ʹ‫ݐ‬

2.3 Exercises 1. ‫ݕ‬௛௢௠௢௚௘௡௢௨௦ ሺ‫ ݔ‬ሻ ൌ ܿଵ ݁ ି௫ ൅ ܿଶ ‫ି ݁ݔ‬௫ for some ܿଵ ǡ ܿଶ ‫ א‬Ը 3. ‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ ݔ‬ሻ ൌ ሺܿଵ ൅ ܿଶ ‫ ݔ‬൅ ܿଷ ‫ ݔ‬ଶ ൅ ܿସ ݁ ௫ ሻ ൅ ሺܽ଴ ൅ ܽଵ ‫ ݔ‬൅ ܽଶ ‫ ݔ‬ଶ ሻ‫ ݔ‬ଷ for some ܿଵ ǡ ܿଶ ǡ ܿଷ ǡ ܿସ ǡ ܽ଴ ǡ ܽଵ ǡ ܽଶ ‫ א‬Ը

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5. ‫ݕ‬௚௘௡௘௥௔௟ ሺ‫ ݔ‬ሻ ൌ ሺܿଵ ൅ ܿଶ ‫ ݔ‬൅ ܿଷ ݁ ௫ ሻ ൅ ͲǤͲͷ ሺʹ‫ݔ‬ሻ ൅ ͲǤͳሺʹ‫ݔ‬ሻ for some ܿଵ ǡ ܿଶ ǡ ܿଷ ‫ א‬Ը


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Index

7. It is impossible to describe ‫ݕ‬௣௔௥௧௜௖௨௟௔௥ ሺ‫ ݔ‬ሻ

A

3.3 Exercises

Applications of Differential Equations, 96

భ

1. ‫ݕ‬௛௢௠௢௚௘௡௢௨௦ ሺ‫ ݔ‬ሻ ൌ ݁ ିమ௫ ቂܿଵ ቀ

ξଵହ ‫ݔ‬ቁ ൅ ܿଶ ଶ

ቀ

ξଵହ ‫ݔ‬ቁቃfor ଶ

some ܿଵ ǡ ܿଶ ‫ א‬Ը య మ

ξଷ

3. ‫ݕ‬௛௢௠௢௚௘௡௢௨௦ ሺ‫ ݔ‬ሻ ൌ ܿଵ ൅ ܿଶ ‫ ݔ‬ቀ ଶ ሺ‫ݔ‬ሻቁ ൅ య

ξଷ

ܿଷ ‫ ݔ‬మ ቀ ଶ ሺ‫ݔ‬ሻቁfor some ܿଵ ǡ ܿଶ ǡ ܿଷ ‫ א‬Ը 5. It is impossible to use Cauchy-Euler Method because the degrees of ‫ ݕ‬ᇱ and ‫ ݕ‬ᇱᇱ are not equal to each other when you substitute them in the given differential equation.

4.6 Exercises య 1. ‫ݕ‬ሺ‫ ݔ‬ሻ ൌ ඥሺ‫ ݔ‬൅ ͳሻଷ ൅ ሺ‫ ݔ‬൅ ͳሻܿ݁ ିଷ௫ 3. ‫ି݊ܽݐ‬ଵ ሺ‫ݕ‬ሻ െ ‫ି݊ܽݐ‬ଵ ሺ‫ ݔ‬ሻ ൌ ܿ for some ܿ ‫ א‬Ը

ଵ

5. െ ହ ݁ ିହ௬ െ ͵‫ ݁ݔ‬௫ ൅ ͵݁ ௫ ൌ ܿ for some ܿ ‫ א‬Ը 7. ȁሺͷ‫ ݔ‬൅ ‫ݕ‬ሻȁ െ ʹሺͷ‫ ݔ‬൅ ‫ݕ‬ሻ െ ‫ ݔ‬ൌ ܿ for some ܿ ‫ א‬Ը

B Basis, 123 Bernoulli Method, 78

C Cauchy-Euler Method, 74 Constant Coefficients, 51 Cramer’s Rule, 46

D Determinants, 109 Differential Equations, 9 Dimension of Vector Spaces, 119

E Exact Method, 87

G Growth and Decay Applications, 100

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H

S

Homogenous Linear Differential Equations (HLDE), 51

Separable Method, 85 Subspace, 123

I Initial Value Problems (IVP), 27 Inverse Laplace Transforms, 24

T Temperature Application, 96

K

U

Kernel and Range, 150

Undetermined Coefficients Method, 60

L

V

Laplace Transforms, 9 Linear Equations, 45 Linear Independence, 120 Linear Transformations, 142

Variation Method, 67

N

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W Water Tank Application, 104

Null Space and Rank, 135

P Properties of Laplace Transforms, 33

R Reduced to Separable Method, 90 Reduction of Order Method, 92

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Bibliography

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[1] Kaabar, M.K.: A First Course in Linear Algebra: Study Guide for the Undergraduate Linear Algebra Course. CreateSpace, Charleston, SC (2014) [2] Quick Facts. http://about.wsu.edu/about/facts.aspx (2015). Accessed 01 Jan 2015

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An introduction to ordinary differential equations

An introduction to nonlinear partial differential equations

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An introduction to ordinary differential equations

Introduction to Numerical Methods in Differential Equations

An Introduction to Partial Differential Equations

An Introduction to Nonlinear Partial Differential Equations

Introduction to linear systems of differential equations

An Introduction to Partial Differential Equations

An introduction to nonlinear partial differential equations

An introduction to partial differential equations

A Friendly Introduction to Differential Equations

Introduction to Partial Differential Equations

A Modern Introduction To Differential Equations

A Modern introduction to differential equations

Introduction to Partial Differential Equations

Introduction to partial differential equations

A modern introduction to differential equations

A Modern Introduction to Differential Equations

Difference equations to differential equations

An Introduction to Partial Differential Equations

An Introduction to Partial Differential Equations

Introduction to Numerical Methods in Differential Equations

An introduction to ordinary differential equations

An Introduction to Partial Differential Equations

An Introduction to Ordinary Differential Equations (Universitext)

An Introduction to Partial Differential Equations

An Introduction to Ordinary Differential Equations (Universitext)

Introduction to Numerical Methods in Differential Equations

Introduction to Partial Differential Equations. Second Edition

An Introduction to Ordinary Differential Equations

An introduction to ordinary differential equations

An introduction to nonlinear partial differential equations

Introduction to partial differential equations with applications

An introduction to ordinary differential equations

Introduction to Numerical Methods in Differential Equations

An Introduction to Partial Differential Equations

An Introduction to Nonlinear Partial Differential Equations

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An Introduction to Partial Differential Equations

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