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A Friendly Introduction to Differential Equations
Copyright © 2015 Mohammed K A Kaabar
All Rights Reserved
Copyright © 2015 Mohammed K A Kaabar
All Rights Reserved
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.
3
Copyright © 2015 Mohammed K A Kaabar
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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
Copyright © 2015 Mohammed K A Kaabar
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
Copyright © 2015 Mohammed K A Kaabar
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Copyright © 2015 Mohammed K A Kaabar
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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 ሺሻ ሺݐሻൟ ൌ ݏ ܨሺݏሻ െ ݏିଵ ݂ሺͲሻ െ ݏିଶ ݂ ᇱ ሺݏሻ െ ڮെ ݂ ሺିଵሻ 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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Copyright © 2015 Mohammed K A Kaabar
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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: ݖሺሻ 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.
Copyright © 2015 Mohammed K A Kaabar
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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 ݁ ି௦ሺሻ ൌ 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 ࣦሼܽ݊ݐ݊ܽݐݏ݊ܿ 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
ࣦሼ݂ሺݔሻሽ ൌ න ݂ ሺݔሻ݁ ି௦௫ ݀ݔ Ͳ
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
ࣦሼ݂ሺݔሻሽ ൌ න ݂ ሺݔሻ݁ ି௦௫ ݀ݔ Ͳ
14 M. Kaabar
15
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Copyright © 2015 Mohammed K A 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 ݁ ሺି௦ሻ௫ ݀ ݔൌ ሺି௦ሻ ݁ ሺି௦ሻ௫ ቚ ௫ୀ ௫ୀ ଵ ଵ ଵ ଵ ሺି௦ሻሺሻ ሺି௦ሻ ሺି௦ሻ ሺି௦ሻ
݁
െ ሺି௦ሻ ݁
െ ሺି௦ሻ .
ൌ ሺି௦ሻ ݁
ଵ
ଵ
Step 5: We need find the limit of ሺି௦ሻ ݁ ሺି௦ሻ െ ሺି௦ሻ as follows: ቀ
ଵ
՜ஶ ሺି௦ሻ
ଵ
ଵ
ଵ
݁ ሺି௦ሻ െ ሺି௦ሻቁ ൌ െ ሺି௦ሻ ൌ ሺ௦ିሻ 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 .
16 M. Kaabar
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 ሺሻ ݁ݔଶ௫ െ ሺሻ݁ 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 ݈ܽ݅݉݊ݕ݈ሻ݁ ௫ ݀ ݔand ሺ ݈ܽ݅݉݊ݕ݈ሻሺܽݔሻ݀ݔ. 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
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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
଼
Copyright © 2015 Mohammed K A Kaabar
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
ࣦሼ݂ሺݔሻሽ ൌ න ݂ ሺݔሻ݁ ି௦௫ ݀ݔ Ͳ
Step 2: Here in this example, ݂ ሺݔሻ ൌ ݔbecause ଶ
ࣦሼ݂ሺݔሻሽ ൌ ࣦሼ ݔଶ ሽ. λ Step 3: ࣦሼ ݔଶ ሽ ൌ Ͳሺ 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
19
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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
ൌ
ࣦሼ
ܿݐሽ ൌ
ݏଶ
ݏ ܿଶ
Step 4: We change what we got from step 3 to make it look like ࣦሼ
ሺെͺݔሻሽ as follows: ࣦሼ
ሺെͺݔሻሽ ൌ
௦
௦ మ ାሺି଼ሻమ
ൌ
௦
.
௦ మ ାସ ௦
௦
Thus, ࣦሼ
ሺെͺݔሻሽ ൌ 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
Copyright © 2015 Mohammed K A Kaabar
ହ ௦ మ ାଶହ ହ
. ହ
Thus, ࣦሼ ͷݔሽ ൌ ௦మ ାହమ ൌ ௦మ ାଶହ. Example 1.2.7 Using table 1.1.1, find ࣦሼ
ሺെͺݔሻሽ. Solution: To find ࣦሼ
ሺെͺݔሻሽ 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.
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
ࣦሼ݂ሺݔሻሽ ൌ න ݂ ሺݔሻ݁ ି௦௫ ݀ݔ Ͳ
Step 2: Here in this example, ݂ ሺݔሻ ൌ ݕᇱ because ࣦሼ݂ሺݔሻሽ ൌ ࣦሼ ݕᇱ ሽ. λ Step 3: ࣦሼ ݕᇱ ሽ ൌ Ͳሺ ݕᇱ ሻ݁ ି௦௫ ݀ݔ
ஶ
By the definition of integral, we substitute ሺݕԢ ሻ݁െݔ݀ ݔݏ
with ՜ஶ ሺݕԢ ሻ݁െݔ݀ ݔݏ.
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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
Copyright © 2015 Mohammed K A Kaabar
՜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 ሺሻ ሺݐሻൟ ൌ ݏ ܨሺݏሻ െ ݏିଵ ݂ሺͲሻ െ ݏିଶ ݂ ᇱ ሺݏሻ െ ڮെ ݂ ሺିଵሻ 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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Copyright © 2015 Mohammed K A Kaabar
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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
ࣦሼ ܿݐሽ ൌ
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,
ࣦሼ
ܿݐሽ ൌ
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
ࣦሼ
ܿݐሽ ൌ
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 ݕሺሻ ܽିଵ ሺ ݔሻ ݕሺିଵሻ 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 ݕሺିଵሻ 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 ͵ࣦሼ
ሺ͵ݐሻሽ ൌ ቀ௦మାଽ ௦మ ାଽቁ , then by
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(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.
look like
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Since ࣦሼ ݐଷ ሽ ൌ ௦యశభ ൌ ௦ర , then this means that ࣦ ିଵ ቄ௦ర ቅ ൌ ݐଷ .
ሺʹݐሻ.
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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.
27
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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 ൌ ݔʹ ܽ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
has a solution which is continuous
ܽ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.
28 M. Kaabar
௫ାଷ
has a solution which is continuous
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
29
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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
Copyright © 2015 Mohammed K A Kaabar
ݕᇱ ሺ ݔሻ ൌ ൬
ͳ ξ͵
ᇱ
ሺξ͵ ݔሻ൰ ൌ
ͳ ξ͵
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൫ξ͵൯
൫ξ͵ ݔ൯ ൌ
൫ξ͵ ݔ൯
both sides as follows
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 ൌ ൫
൫ξ͵ 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 ʹݏሺξ͵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 ͵.
Copyright © 2015 Mohammed K A Kaabar
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 ݏʹെʹ ݏʹ 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
ݏʹ ͵
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 ݏെʹʹ 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.
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ଵ
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ଵ
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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
ࣦሼ ݐ ݂ሺݐሻሽሺݏሻ ൌ ሺെͳሻ
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 ܨሺሻ 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 ܨሺሻ ሺݏሻ ݀ ݏ
ͳ Ͷ ቑ ൠ ൌ ࣦ ିଵ ቐ ቑ ࣦെͳ ቐ ࣦെͳ ൜ ʹ ݏെͶ ሺ ݏെ ʹሻ ሺ ݏ ʹሻ
ࣦሼ ݐ ݂ሺݐሻሽሺݏሻ ൌ ሺെͳሻ
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 ࣦሼݕሺݔሻሽ ൌ ࣦ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 ൌ
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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͵ሻሺݏʹሻ
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:
The domain of solution is ሺെλǡ λሻ.
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
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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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଼
଼
Copyright © 2015 Mohammed K A Kaabar
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 ൌ ݂ ሺሻ ൌ െ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.
44 M. Kaabar
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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
Copyright © 2015 Mohammed K A Kaabar
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 ൌ െ ሺെͲሻ ൌ ് 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
.
Copyright © 2015 Mohammed K A Kaabar
݊ᇱᇱ ሺݐሻ െ Ͷ݉ሺݐሻ ൌ Ͳ ՜ ݉ሺݐሻ ൌ Thus, ݉ሺݐሻ ൌ ݁ ଶ௧ Ǥ
ଵ
1. Find ࣦ ିଵ ቄሺ௦ିସሻర ቅ. ௦ାହ
2. Find ࣦ ିଵ ቄሺ௦ିଵሻమାଵቅ.
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 ቄଷ௦ିଵቅ.
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
Copyright © 2015 Mohammed K A Kaabar
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 ܿݏሺʹ߰ሻ݁ሺ͵ݔʹటሻ ݀߰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:
Copyright © 2015 Mohammed K A Kaabar
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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
The domain of solution is ሺെλǡ λሻ.
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
both sides as follows
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 ቄݏʹቅ ൌ ݁ ି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 ݕሺሻ 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 ݕሺሻ 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
ሺሻ
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 ݇ ݎൌ െ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 ݁ ଶሺሻ ܿଶ ݁ ିଶሺሻ
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 ݁ ଶሺሻ െ ʹܿଶ ݁ ିଶሺሻ
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 ݇ ݎൌ ͵. 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 ൌ ݁ ሺሻ௫ ൌ 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
56 M. Kaabar
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Find the general solution for ݕሺ ݔሻ. (Hint: Use results
at ݇ ൌ Ͳ, ݕଵ ሺ ݔሻ ൌ ݁ ሺሻ௫ ൌ ͳ
2.1.1 and 2.1.2, and in this example, no need to find
at ݇ ൌ Ͳ, ݕଶ ሺ ݔሻ ൌ ݁ ሺሻ௫ ൌ ͳ ή ݔ
the values of ܿଵ ǡ ܿଶ ǡ ܿଷ ǡ ܿସ ܿହ)
at ݇ ൌ Ͳ, ݕଷ ሺ ݔሻ ൌ ݁ ሺሻ௫ ൌ 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:
Thus, using result 2.1.1, the general solution for ݕሺݔሻ
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 ݇ ݎൌ െ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: ݕ௧௨ 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: ݕ௧௨ ᇱ ሺ ݔሻ ൌ ܾ We substitute ݕ௧௨ ᇱ ሺ ݔሻ ൌ ܾ in ݕᇱ ሺ ݔሻ ͵ݕሺ ݔሻ ൌ ݔ. ܾ ͵ሺܽ ܾ ݔሻ ൌ ݔ ܾ ͵ܽ ͵ܾ ݔൌ ݔ
61
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Copyright © 2015 Mohammed K A Kaabar
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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
Let ݕሺݔሻ ൌ ݁ ௫ , we need to find ݇.
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, ݕ௧௨ ሺ ݔሻ ൌ െ ݔ. ଽ ଷ Step 3: We need to find the general solution as follows: ݕ ሺ ݔሻ ൌ ݕ ሺ ݔሻ ݕ௧௨ 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
Thus, using result 2.1.1, the general solution for ݕሺݔሻ
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
ݕ௧௨ 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:
ݕ௧௨ 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 ݕ௧௨ 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
ݕ௧௨ ሺ ݔሻ but do not find it.
Thus, ݕ௧௨ ሺ ݔሻ ൌ ሺͳ݁
ିଷ௫ ሻ
ݔൌ ݁ݔ
ିଷ௫
Ǥ
Solution: To describe ݕ௧௨ 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 ݕ௧௨ ሺ ݔሻ
ܽ ܾ ݔ ܿ ݔଶ , and then we multiply it by ݁ ௫ .
Thus, ݕ ሺ ݔሻ ൌ ܿଵ ݁ ିଷ௫ ି ݁ݔଷ௫ Ǥ
ݕ௧௨ ൌ 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 ݕ௧௨ ሺ ݔሻ but do not find it.
differential equation equals to a polynomial of degree
Solution: To describe ݕ௧௨ ሺݔሻ, we do the following:
ݓ. Then, we obtain the following:
In this example, we look at ሺ͵ݔሻ, and we write it as:
ݕ௧௨ ൌ ሾܲݓ ݁݁ݎ݃݁ܦ ݂ ݈ܽ݅݉݊ݕ݈ሿ ݔ .
ሺܽ ሺ͵ ݔሻ ܾሺ͵ݔሻሻ, and then we multiply it by ݁ ௫ .
Result 2.2.2 Suppose that you have a linear differential
ݕ௧௨ ൌ ሾܽ ሺ͵ ݔሻ ܾሺ͵ݔሻሿ݁ ௫ .
equation, then the general solution is always written as: ݕ ሺݔሻ ൌ ݕ௨௦ ሺ ݔሻ ݕ௧௨ ሺݔሻ. Example 2.2.3 Given ݕ
ሺସሻ ሺ
ݔሻ െ ݕ
ሺଷሻ ሺ
ݔሻ ൌ ݔǤ Describe
ݕ௧௨ ሺ ݔሻ but do not find it. Solution: To describe ݕ௧௨ ሺݔሻ, we do the following: By using result 2.2.1, we obtain the following: ݕ௧௨ ൌ ሾܽ ܾ ݔ ܿ ݔଶ ሿ ݔଷ . Example 2.2.4 Given ݕ
ሺସሻ ሺ
ݔሻ െ ݕ
ሺଷሻ ሺ
ݔሻ ൌ ͵Ǥ Describe
ݕ௧௨ ሺ ݔሻ but do not find it. Solution: To describe ݕ௧௨ ሺݔሻ, we do the following: By using result 2.2.1, we obtain the following: ݕ௧௨ ൌ 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 ݕ௧௨ ሺ ݔሻ but do not find it. మ
7. Given ݕሺଶሻ ሺݔሻ െ ͵ݕሺݔሻ ൌ ݁ ௫ Ǥ Describe ݕ௧௨ 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:
66 M. Kaabar
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ݕ௧௨ 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: ݕ௧௨ 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 ൌ ݁ ሺሻ௫ ൌ ݁ ൌ 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.
68 M. Kaabar
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: ݕ௧௨ 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 ݕ௧௨ ሺ ݔሻ ൌ ሺȁ ݔȁሻ ݁ ିଷ௫ ቌන െ ݁ ଷ௧ ݀ݐቍ ͵ ͵ݐ
Step 3: We need to find the general solution as follows:
70 M. Kaabar
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:
Let ݕሺݔሻ ൌ ݁ ௫ , we need to find ݇.
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 ݕ௧௨ ሺ ݔሻ
ݕଶ ሺ ݔሻ ൌ ݁ ିଷ௫ ------Æ ݕଶ ᇱ ሺ ݔሻ ൌ െ͵݁ ିଷ௫
ᇱ
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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: ݕ௧௨ 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 ݕ௧௨ 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 ݕ௧௨ ሺ ݔሻ ଵ
Thus, ݕ ሺ ݔሻ ൌ ሺܿଵ ݁ ିଶ௫ ܿଶ ݁ ିସ௫ ሻ ቀെ ସ ݁ ିଶ௫ ሺ݁ ିଶ௫ ሻ െ ଵ ଶ
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െ݁ ି଼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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ݔିଵ ሺ݇ ଶ െ ʹ݇ ͳሻ ൌ Ͳ ݔିଵ 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 ൌ ݇ ݔିଵ ݕᇱᇱ ൌ ݇ሺ݇ െ ͳሻ ݔିଶ Now, we substitute ݕൌ ݔ ǡ ݕᇱ ൌ ݇ ݔିଵ and ଵ
ݕᇱᇱ ൌ ݇ሺ݇ െ ͳሻ ݔିଶ 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 ൌ ݇ ݔିଵ
ͳ ݇ݔሺ݇ െ ͳሻ ݔିଶ െ ݇ ݔିଵ ݔ ൌ Ͳ ݔ
ݕᇱᇱ ൌ ݇ሺ݇ െ ͳሻ ݔିଶ
݇ሺ݇ െ ͳሻ ݔିଵ െ ݇ ݔିଵ ݔିଵ ൌ Ͳ
ݕᇱᇱᇱ ൌ ݇ሺ݇ െ ͳሻሺ݇ െ ʹሻ ݔିଷ
ݔିଵ ሺ݇ሺ݇ െ ͳሻ െ ݇ ͳሻ ൌ Ͳ ݔିଵ ሺ݇ ଶ െ ݇ െ ݇ ͳሻ ൌ Ͳ
74 M. Kaabar
ݕሺ ݔሻ ൌ ܿଵ ݔ ܿଶ ݈݊ݔሺݔሻ, for some ܿଵ ܿଶ אԸ.
Now, we substitute ݕൌ ݔ ǡ ݕᇱ ൌ ݇ ݔିଵ , and ݕᇱᇱ ൌ ݇ሺ݇ െ ͳሻ ݔିଶ in ݔଷ ݕሺଶሻ ݔଶ ݕᇱ ݕݔൌ Ͳ as follows:
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ݔଷ ሺ݇ሺ݇ െ ͳሻ ݔିଶ ሻ ݔଶ ሺ݇ ݔିଵ ሻ ݔሺ ݔ ሻ ൌ Ͳ ሺ݇ሺ݇ െ ͳሻ ݔାଵ ሻ ሺ݇ ݔାଵ ሻ ሺ ݔାଵ ሻ ൌ Ͳ ݔାଵ ሺ݇ ଶ െ ݇ ݇ ͳሻ ൌ Ͳ ݔାଵ 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 ൌ ݔሺሻ
ሺͳ ή ሺ ݔሻሻ ൌ
ሺሺ ݔሻሻ ݕଶ ൌ ݔሺ ௧ሻ ሺ ݐݎܽ ݕݎܽ݊݅݃ܽ݉ܫή ሺ ݔሻ ൌ ݔሺሻ 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 ݁ݕሺ௫ሻௗ௫ ൌ ܨሺݔሻ݁ ሺ௫ሻௗ௫
Bernulli Method, Separable Method, Exact Method,
Step 2: ݕᇱ ݁ ሺ௫ሻௗ௫ ݃ሺ ݔሻ ݁ݕሺ௫ሻௗ௫ ൌ ൣ 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:
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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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Copyright © 2015 Mohammed K A Kaabar
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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
మ ሺ௫ሻ భ ሺ௫ሻ
ൌെ
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 ܿ)
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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
మ ሺ௫ሻ భ ሺ௫ሻ
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 ݕଵିିଵ 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 ݕି ݀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
82 M. Kaabar
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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
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Then, we substitute ݕൌ
ଵ ξ௪
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in ሺʹሻ as follows:
݀ݓ ሺെʹሻ͵ݔሺݕሻି ݕଷ ൌ ሺെʹሻݔ ݀ݔ ݀ݓ ሺെʹሻ͵ ݕݔଵିଷ ൌ ሺെʹሻݔ ݀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 ሺെʹሻݔ ݀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 ሺെݔሻ ݀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:
85
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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:
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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
87
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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 ൌ െ ிೣ ൌ െ
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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 ܿݏሺݑሻ െ ʹሺݑሻ ݀ݑ ሺݑሻ ൌ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ሺ͵ሻ ሺ ݀ݑ ݏܿ 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 ݏܿ ݔെ ʹሺݑሻ ܿ ݑ ݏെ ʹሺݑሻ ሺݑሻ Now, we need to do a cross multiplication for ሺͶሻ as follows: ܿݏሺݑሻ െ ʹሺݑሻ ݀ ݑൌ ͳ݀ݔ ሺݑሻ ܿݏሺݑሻ െ 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: ܿݏሺݑሻ െ ʹሺݑሻ නቆ ݀ ݑെ ͳ݀ݔቇ ൌ න Ͳ ሺݑሻ
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නቆ
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ሺ ݔ ͳሻ ሺଶሻ ݕᇱ ሺ ݔሻ Ͳ ݕሺ ݔሻ െ ൌ ሺ ݔ ͳሻ ሺ ݔ ͳሻ ሺ ݔ ͳሻ
ܿݏሺݑሻ െ ʹሺݑሻ ቇ ݀ ݑെ නሺͳሻ ݀ ݔൌ ܿ ሺݑሻ
ܿݏሺݑሻ ʹ ሺݑሻ නቆ െ ቇ ݀ ݑെ ݔൌ ܿ ሺݑሻ ሺݑሻ ܿݏሺݑሻ 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 ሺሻ
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 ሺሻ 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
ݕ௨௦ 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 ݕ௨௦ 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 ݕ௨௦ 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 ቀభశೣቁೣ
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
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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 ݀ݏܿ 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 ݕ௧௨ ሺ ݔሻ?
݀ݐ.
݀ݐ.
Thus, the homogenous solution is written as follows: ௫ షయ
ݕ௨௦ 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 ͳͳԬ? 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 ͳͳԬ, 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
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By substituting ܶሺͳͲሻ ൌ ͳ͵Ԭ in ሺͷሻ, we obtain: ܶሺͳͲሻ ൌ ͳͲͶ ͶͲ݁ ఉሺଵሻ
݃ሺݐሻ ൌ െߚ and ܨሺݐሻ ൌ െͳͲͶߚ. Hence, ܫൌ ݁ ሺ௧ሻௗ௧ ൌ ݁ ି ఉ ௗ௧ ൌ ݁ ିఉ௧ .
ͳ͵ ൌ ͳͲͶ ͶͲ݁ ఉሺଵሻ
The general solution is written as follows:
ͳ͵ െ ͳͲͶ ൌ ͶͲ݁ ఉሺଵሻ
ܶሺݐሻ ൌ ܶሺݐሻ ൌ
݁
ܶሺݐሻ ൌ
͵ʹ ൌ ͶͲ݁ ఉሺଵሻ
ܫ ή ܨሺݐሻ ݀ݐ ܫ
ିఉ௧
ή ሺെͳͲͶߚሻ ݀ݐ ݁ ିఉ௧
ሺെͳͲͶߚሻ݁ ݁ ିఉ௧
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ିఉ௧
݀ݐ
݁ ఉሺଵሻ ൌ
By taking the natural logarithm for both sides of ሺሻ, we obtain: ݈݊൫ ݁ ఉሺଵሻ 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 ሺሻ ͶͲ
ߚ ሺͳͲሻ ൌ ݈݊ሺͲǤͺሻ ߚൌ
݈ ݊ሺͲǤͺሻ ൌ െͲǤͲʹʹ͵ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ ሺሻ ͳͲ
Now, we substitute ሺሻ in ሺͷሻ as follows: ܶሺݐሻ ൌ ͳͲͶ ͶͲ݁ ିǤଶଶଷ௧ 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 ൌ ͳͳԬ by
The general solution is: ܶሺݐሻ ൌ ͳͲͶ ܿ݁ ఉ௧ for some
substituting it in ሺͺሻ as follows:
ܿ אԸ.
ͳͳ ൌ ͳͲͶ ͶͲ݁ ିǤଶଶଷሺ୲ሻ
Now, we need to find ܿ by substituting ܶሺͲሻ ൌ ͳͶͶԬ in
ͳͳ െ ͳͲͶ ൌ ͶͲ݁ ିǤଶଶଷሺ୲ሻ
ሺͶሻ as follows:
͵ ൌ ͶͲ݁ ିǤଶଶଷሺ୲ሻ ܶሺͲሻ ൌ ͳͲͶ ܿ݁
ఉሺሻ
ͳͶͶ ൌ ͳͲͶ ܿ݁
ͳͶͶ ൌ ͳͲͶ ܿሺͳሻ ͳͶͶ ൌ ͳͲͶ ܿ
݁ ିǤଶଶଷሺ୲ሻ ൌ
͵ ൌ ͲǤͲͷ 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 ିǤଶଶଷሺ௧ሻ ൯ ൌ ݈݊ሺͲǤͲͷሻ
Thus, ܶሺݐሻ ൌ ͳͲͶ ͶͲ݁ ఉ௧ ǥ ǥ ǥ ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ሺͷሻ
െͲǤͲʹʹ͵ሺݐሻ ൌ ݈݊ሺͲǤͲͷሻ
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ݐൌ
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݈ ݊ሺͲǤͲͷሻ ൎ ͳͳǤͳ െͲǤͲʹʹ͵
Thus, the temperature of the engine will take approximately ͳͳǤͳ minutes to cool to ͳͳԬ 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 ିǤଶଶଷሺଷሻ ܶሺ͵Ͳሻ ൎ ͳʹͶǤͶͻԬ Thus, the temperature of the engine 30 minutes from now will be approximately ͳʹͶǤͶͻԬ.
Copyright © 2015 Mohammed K A Kaabar
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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 ൌ ʹͺǡͺ, 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 ൌ ʹͺǡͺ in ሺሻ 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 ʹͺǡͺ ൌ ቀ ቁ ʹ
ଵ
൬ܹ ିଶ ൰ ܹ݀ ൌ ߚ݀ݐ
ʹͺǡͺ ൌ 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:
ʹͺǡͺ ൌ
ܿଶ Ͷ
ܿ ଶ ൌ Ͷሺʹͺǡͺሻ
ଵ
න ቆ൬ܹ ିଶ ൰ ܹ݀ െ ߚ݀ݐቇ ൌ න Ͳ
ܿ ൌ ඥͶሺʹͺǡͺሻ ଵ න ൬ܹ ିଶ ൰ ܹ݀ ଵ ʹܹ ଶ
ܿ ൎ ͵͵ͺǤͶ
െ 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
ଷଷ଼Ǥସାఉ௧ 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
͵͵ͺǤͶ ߚ ଶ ൰ ʹ
ሺ͵͵ͺǤͶሻଶ ʹሺ͵͵ͺǤͶሻߚ ߚ ଶ ቇ Ͷ
for some ܿ אԸ.
͵ʹǡͲͲͲ ൌ ቆ
Then, we rewrite ሺͷሻ as follows:
ሺ͵͵ͺǤͶሻଶ ʹሺ͵͵ͺǤͶሻߚ 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 ሺሻ ʹ
We square both sides of ሺሻ 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 ሺሻ ʹ
102 M. Kaabar
ʹሺ͵͵ͺǤͶሻߚ ߚ ଶ ൌ Ͷሺ͵ʹǡͲͲͲሻ െ ሺ͵͵ͺǤͶሻଶ ߚ ଶ ʹሺ͵͵ͺǤͶሻߚ െ ͳ͵ǡʹͷͷǤʹͳʹͶ ൌ Ͳ Thus, ߚ ൎ ͵ǡͳǤʹ 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
͵͵ͺǤͶ ሺ͵ǡͳǤʹሻ ݐଶ ൰ ǥ ǥ Ǥ ǥ ǥ ǥ ǥ Ǥ ǥ ǥ Ǥ ሺͳͲሻ ʹ
Then, we need to find the time ݐwhen
103
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Copyright © 2015 Mohammed K A Kaabar
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ܹሺݐሻ ൌ ʹሺʹͺǡͺሻ ൌ ͷǡ͵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 ͷǡ͵ʹ ൌ ൬
͵͵ͺǤͶ ሺ͵ǡͳǤʹሻ ݐଶ ൰ ʹ
minerals is ͷ bounds. Suppose that there is a mixture of minerals containing 0.2 bound of minerals per gallon
ሺ͵͵ͺǤͶሻଶ ʹሺ͵͵ͺǤͶሻሺ͵ǡͳǤʹሻ ݐ ݐଶ ൌ Ͷሺͷǡ͵ʹሻ ݐଶ ʹሺ͵͵ͺǤͶሻሺ͵ǡͳǤʹሻ ݐെ ͳͳͶǡͶ͵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
͵͵ͺǤͶ ሺ͵ǡͳǤʹሻሺʹͲͳͺሻ 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 ൌ
ܹ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
ܹሺݐሻ ൌ
ሺ͵ͷͲ ʹݐሻ ݀ݐ ሺ͵ͷͲ ʹݐሻ
ܹሺݐሻ ൌ
͵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
ଷହ௧
From ሺ͵ሻ, the dependent variable is ܹ, while the
ܹሺݐሻ ൌ ቀሺଷହାଶ௧ሻ ሺଷହାଶ௧ሻ ሺଷହାଶ௧ሻቁ 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 ൌ ͳͷͲ
Since ሺͶሻ is a first order linear differential equation,
ଷହ௧
௧మ
ଵହ
then by using definition 4.1.1, we need to use the
Thus, ܹ ሺݐሻ ൌ ቀሺଷହାଶ௧ሻ ሺଷହାଶ௧ሻ ሺଷହାଶ௧ሻቁ Ǥ ǥ ǥ ǥ ǥ Ǥ Ǥ ሺሻ
integral factor method by letting ܫൌ ݁ ሺ௧ሻௗ௧ , where
The amount of minerals in the tank of WSU Water
ଶ
݃ሺݐሻ ൌ ቀଷହାଶ௧ ቁ and ܨሺݐሻ ൌ ͳ. మ
ܫൌ ݁ ሺ௧ሻௗ௧ ൌ ݁ ቀయఱబశమ
ቁ ௗ௧
ൌ ݁ ୪୬ሺଷହାଶ௧ሻ ൌ ሺ͵ͷͲ ʹݐሻ.
Tower at any time ݐis: ܹ ሺ ݐሻ ൌ ቆ
ݐଶ ͳͷͲ ͵ͷͲݐ ቇ ሺ͵ͷͲ ʹݐሻ ሺ͵ͷͲ ʹݐሻ ሺ͵ͷͲ ʹݐሻ
The general solution is written as follows:
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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 ሺሻ as follows: ܹ ሺ͵Ͷሻ ൌ ቆ
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Appendices Review of Linear Algebra
ଶ
͵ͷͲሺ͵Ͷሻ ሺ͵Ͷሻ ͳͷͲ 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.
Copyright © 2015 Mohammed K A Kaabar
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 ൌ Թൈ ൌ Թൈ , 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͵ሻሺሻ െ 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 ൌ Թൈ , 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 ൌ ͳǤ 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 ൌ ሺʹሻሺሻ െ 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 ൌ ୢୣ୲ሺሻǤ 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 ൌ ୢୣ୲ሺሻ. Since ିଵ ൌ , then ሺିଵ ሻ ൌ ሺܫ ሻ ൌ ͳǤ ሺିଵ ሻ ൌ ሺሻ כሺିଵ ሻ ൌ ͳǤ Thus, ሺିଵ ሻ ൌ
ଵ ୢୣ୲ሺሻ
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 ʹ݊ܽܵ 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 ܵ݊ܽሼሺǡͷሻሽ ് Թଶ where ሺǡͷሻ אԹଶ Ǥ
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We cannot find a number ߙ such that ሺǡͷሻ ൌ 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:
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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 െʹǡͶǡሻ
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 ሺʹǡሻሽǫ 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 ሺʹͲǡͲሻ ൌ ߙሺͷǡͳͷሻ because in this case ߙ ൌ Ͷ where ሺʹͲǡͲሻ ൌ 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 ሺʹǡሻ as a linear combination of each other. ߙଵ ሺͳǡ͵ሻ ߙଶ ሺʹǡሻ ൌ ሼሺߙଵ ʹߙଶ ሻǡ ሺ͵ߙଵ ߙଶ ሻሽ ߙଵ ሺͳǡ͵ሻ ߙଶ ሺʹǡሻ ൌ ሼሺߙଵ ʹߙଶ ሻǡ ͵ሺߙଵ ʹߙଶ ሻሽ Assume ܿ ൌ ሺߙଵ ʹߙଶ ሻ, then we obtain: ߙଵ ሺͳǡ͵ሻ ߙଶ ሺʹǡሻ ൌ 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 ሺʹǡሻሽ. 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 ൌ ݊.
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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.
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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 ܿ.
ܽ ൌͳήܽͲήܾ
ܽ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
134 M. Kaabar
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 ܴݓሺܣሻ. Definition C.1.5 Column Space of a matrix, say ܣis the ܵ ݊ܽof independent columns of ܣ, and it is denoted by ݊݉ݑ݈ܥሺܣሻ. 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 ܴݓሺܣሻ. 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
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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 ݊݉ݑ݈ܥሺܣሻ ൌ ܵ݊ܽሼݏ݊݉ݑ݈ܥ ݐ݊݁݀݊݁݁݀݊ܫሽ
݀݅݉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 ൌ ݀݅݉൫ܴݓሺܣሻ൯ ൌ ݀݅݉ሺ݊݉ݑ݈ܥሺܣሻሻ. 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 ܴݓሺܤሻ.
The basis for ݈݈ܰݑሺܣሻ ൌ ሼሺെͶǡ െʹǡͳǡͲǡͲሻǡ ሺെͳǡ െʹǡͲǡͲǡͳሻሽ.
b. Find ݊݉ݑ݈ܥሺܤሻ.
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, ܴݓሺܣሻ ൌ ܵ݊ܽሼሺͳǡ െͳǡʹǡͲǡ െͳሻǡ ሺͲǡͳǡʹǡͲǡʹሻǡ ሺͲǡͲǡͲǡͳǡͲሻሽǤ It is also a subspace of Թହ . Result C.1.4 Let ܣbe ݉ ൈ ݊ matrix. Then, ܴܽ݊݇ሺܣሻ ݀݅݉൫ܰ ሺܣሻ൯ ൌ ݊ ൌ ܰܣ ݂ ݏ݊݉ݑ݈ܥ ݂ ݎܾ݁݉ݑ. Result C.1.5 Let ܣbe ݉ ൈ ݊ matrix. The geometric meaning of ܴݓሺܣሻ ൌ ܵ݊ܽሼݏݓܴ ݐ݊݁݀݊݁݁݀݊ܫሽ “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, ܴݓሺܤሻ ൌ ܵ݊ܽ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
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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 ݂ ݏ݈ܽ݅݉݊ݕ݈ ݈݈ܽ ݂ ݐ൏ ݊Ǥ
Thus, ݊݉ݑ݈ܥሺܤሻ ൌ ܵ݊ܽሼሺͳǡ െͳǡͲሻǡ ሺͳǡͲǡͲሻሽ.
The algebraic expression of polynomials is in the following from: ܽ ݔ ܽିଵ ݔିଵ 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 ൌ ݀݅݉൫ܴݓሺܤሻ൯ ൌ ݀݅݉ሺ݊݉ݑ݈ܥሺܤሻሻ ൌ 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 ݂ ݏ݈ܽ݅݉݊ݕ݈ ݈݈ܽ ݂ ݐ൏ ʹ (i.e.
Before discussing polynomials, we need to know the following mathematical facts:
x
Fact C.2.1 Թൈ ൌ Թൈ ൌ ܯൈ ሺԹሻ is a vector space.
x
͵ ݔ ʹ ܲ אଶ , Ͳ ܲ אଶ , ͳͲ ܲ אଶ , ξ͵ ܲ אଶ but ξ͵ξܲ ב ݔଶ ). ܲସ ൌ ܵ݁ ݁݁ݎ݃݁݀ ݂ ݏ݈ܽ݅݉݊ݕ݈ ݈݈ܽ ݂ 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 ݔିଵ ՞ 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͵ ݏ݈ܽ݅݉݊ݕ݈ܲ ݐ݊݁݀݊݁݁݀݊ܫǡ ܽ݊݀ ݁݁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 െͳͲǡሻ 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 Ͳthat 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.
C.3 Kernel and Range In this section, we discuss how to find the standard
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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 ൈ ݀݅݉ሺ݊݅ܽ݉ܦሻ 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
153
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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 ݂ݏ݊݉ݑ݈ܥ ݐ݊݁݀݊݁݁݀݊ܫሽ. 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 ൌ ݀݅݉ሺ݊݅ܽ݉ܦሻ.
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 ቄ௦మି௦ାଵଷቅ ൌ ݁ 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 Ͷሻ
ሺʹ 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 ʹݏͳሻቅ ଶ௦
ସ௦
19. ࣦ ିଵ ቄሺ௦మାସሻమቅ ൌ ሺ௦మ ାସሻమ 21. ݓሺݐሻ ൌ ͳ ʹݐand ݄ሺݐሻ ൌ ʹݐ
2.3 Exercises 1. ݕ௨௦ 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 ͲǤͳ
ሺʹݔሻ for some ܿଵ ǡ ܿଶ ǡ ܿଷ אԸ
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Index
7. It is impossible to describe ݕ௧௨ ሺ ݔሻ
A
3.3 Exercises
Applications of Differential Equations, 96
భ
1. ݕ௨௦ ሺ ݔሻ ൌ ݁ ିమ௫ ቂܿଵ
ቀ
ξଵହ ݔቁ ܿଶ ଶ
ቀ
ξଵହ ݔቁቃfor ଶ
some ܿଵ ǡ ܿଶ אԸ య మ
ξଷ
3. ݕ௨௦ 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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