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ISBN 9788130906003
ISBN-10
8130906007
Binding
Paperback
Number of Pages
200 Pages
Language
(English)
Subject
Life Sciences
Maths & Stats : If you are about to study for a degree in the life or medical sciences, you will need some core maths skills. You do not need to be a budding mathematician but you do need to be comfortable handling numbers.
It's hard to believe, but the maths and stats in this book is all presented i a really user-friendly manner to speed your understanding. In addition, the authors provide you with numerous examples to show you the relevance and application of maths and stats to your course. The book also contains lots of questions (and answers), so that you can test your understanding at any time - it really does get easier with practice. About the Author Dr Michael Harris MB BS FRCGP MMEd is a GP and senior lecturer in postgraduate medicine in Bath. Until recently he was an examiner for the Royal College of General Practitioners. He has a special interest in the design of educational materials. Dr Gordon Taylor PhD MSc BSc (Hons) is a senior research fellow in medical statistics at the University of Bath. His main role is in the teaching, support and supervision of health-care professionals involved in non-commercial research. Mrs Jacquelyn Taylor MSc BSc (Hons) PGCE has worked in both secondary and higher education in mathematics and science. Table Of Contents
How to use this book
Handling numbers
Working with fractions
Percentages
Powers
Approximation and errors
Introduction to graphs
The gradient of a graph
Algebra
Polynomials
Algebraic equations
Quadratic equations
Simultaneous equations
Sequences and series of numbers
Working with powers
Logarithms
Exponential growth and decay
Circles and spheres
Differential calculus
Integral calculus
Using and recognising graphs
Sl units
Moles
PH
Buffers
Kinetics
The language of statistics
Describing data: measuring averages
Standard deviation
Checking for a normal distribution
Degrees of freedom
How to use statistics to make comparisons
The standard error of the mean
Confidence intervals
Probability
Significance and P values
Tests of significance
Tests
Analysis of variance
The chi-squared test
Correlation
Regression
Bayesian statistics | 677.169 | 1 |
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Math Olympiad Workshop is a course for secondary school students to cultivate their mathematical analysis and higher order thinking skills abilities. This course is exclusively for advanced mathematics which is not learnt in a classroom and helps them in preparing for international math competitions such as Singapore And Asian School Math Olympiad (SASMO), Olimpiad Matematik Kebangsaan (OMK), Australian Math Olympiad (AMO) and International Math Olympiad (IMO).Our trainers are among the best trainers in Malaysia and they are the coaches for Malaysia team in IMO. | 677.169 | 1 |
Product details
ISBN-13: 9780201251883
ISBN: 0201251884
Publisher: Globe Fearon Educational Publishing
AUTHOR
Barnett, Carne S., Charles, Randall I., Keller, Charles E.
SUMMARY
Now for grades K-8 with a new kindergarten level and newly revised editions of grades 1 and 2. These new editions of this popular program features: -- lessons that use manipulatives to emphasize hands-on learning at each grade level-- assessment guidelinesThe program enhances any math text, teaching students to: -- approach problems systematically-- identify various solution strategies-- realize that more than one strategy may be used on a single problemTeacher Sourcebooks feature easy-to-follow strategies, guidelines for multiple uses of the program, extension problems, and evaluation methods. Each Teacher Sourcebook includes blackline masters with problem clusters to reproduce intact or cut into individual problems. Use them as resources for problem-of-the-day, problem-of-the-week, homework activities, or in a problem-solving resource center. Available in Spanish for grades 1-6.Barnett, Carne S. is the author of 'Problem Solving Experiences in General Mathematics' with ISBN 9780201251883 and ISBN 02012518 | 677.169 | 1 |
* Unique interactive style enables students to diagnose their strengths and weaknesses and focus their efforts where needed * Ideal for self-study and tutorial work, building from an initially supportive approach to the development of independent learning skills * Free website includes solutions to all exercises, additional topics and applications, guide to learning mathematics, and practice material Students today enter engineering courses with a wide range of mathematical skills, due to the many different pre-university qualifications studied. Bill Cox's aim is for students to gain a thorough understanding of the maths they are studying, by first strengthening their background in the essentials of each topic. His approach allows a unique self-paced study style, in which students Review their strengths and weaknesses through self-administered diagnostic tests, then focus on Revision where they need it, to finally Reinforce the skills required.
The book is structured around a highly successful 'transition' maths course at Aston University which has demonstrated a clear improvement in students' achievement in mathematics, and has been commended by QAA Subject Review and engineering accreditation reports. A core undergraduate text with a unique interactive style that enables students to diagnose their strengths and weaknesses and focus their efforts where needed Ideal for self-paced self-study and tutorial work, building from an initially supportive approach to the development of independent learning skills Lots of targeted examples and exercises | 677.169 | 1 |
Evaluating Functions Scavenger Hunt
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I use Scavenger Hunts to review a lot of different subjects in my classes. I use this one to practice evaluating a function at a given value. A variety of types of problems are included: absolute values, very simple radicals, even a couple where the variable is in the exponent.
I use this to review using function notation and evaluating a function in my Algebra 1 classes. I also use it in Algebra 2 classes. Rather than given a lesson to review this topic, I give them this Scavenger Hunt and let the student remember how to evaluate all on their own. I really like this method because it seems like they remember it better, and they have a lot more fun than simply listening to another lesson of a previously learned conceptCommon Core Standards:
F-IF
Understand the concept of a function and use function notation.
2. Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context | 677.169 | 1 |
Product Information
Format: DRM Protected ePub Vendor: Galore Park Publication Date: 2016
ISBN: 9781471849015 ISBN-13: 9781471849015
Publisher's Description
Secure the top marks in 11+ independent school entrance exams and Pre-Tests leading to a better chance at getting into the school of choice with this essential revision guide. Complete coverage of the ISEB 11+ Maths syllabus and stretching extra content ensures that every topic is thoroughly revised ahead of the exams.
- This book covers everything required for the 11+ Maths exam - Prepares pupils for a wide range of independent school exams and pre-tests with challenging extension material - Consolidates revision with all the key information in one place - Features helpful insight in to the exams, with examples, practical tips and advice - Tests understanding and technique with timed, levelled exam-style questions
Revision Guides, Workbooks and Practice Papers are also available for English, Science, Verbal Reasoning and Non-Verbal Reasoning | 677.169 | 1 |
Over the past 10 years, national conferences and committees investigating the state of American mathematics education have advocated an increased emphasis on problem solving and mathematical modeling situations in the secondary school curriculum. However, little effort has been made to prepare secondary school teachers to use mathematical modeling techniques in their classrooms. The document presents a variety of classroom modeling activities that were developed and classroom tested by the Mathematical Sciences Program of the Pennsylvania State University at Harrisburg at the secondary school level. After an introductory chapter explaining the concept of mathematical modeling and how it differs from problem solving, the bulk of the document is a series of 22 classroom activities, followed by an appendix giving teachers' guides for each of the activities, and an extensive bibliography for related mathematical modeling activities. The 22 activities can be broken up into the following subject levels, together with the concepts and skills involved: (1) General Mathematics--involving Ratio and Proportion, Pythagorean's Formula, the Distance Formula, and Probability; (2) Algebra 1--involving Simple Graphing, Area Computation, Inequalities, Functions and Pattern Recognition, Algebraic Operations, and Simple Programming; (3) Algebra 2--involving Linear Inequalities and Graphing, Circle Equation, Permutations and Counting Techniques, Graphing and Programming Linear Parabolic Equations, Basic Trigonometric Functions, Velocity and Acceleration Formulas, Exponential Functions, and Matrix Arithmetic; and (4) Precalculus--involving Transcendental Functions. (MDH) | 677.169 | 1 |
.1 Intro to Logs & Exponential Functions
This video is an introduction to Logarithmic & Exponential Functions…what they look like, common & natural logs, converting from log to exponential (and back), evaluating, and change the base of logs when not 10 or e. | 677.169 | 1 |
Showing 1 to 3 of 3
I would recommend this class for students who do well in any sort of math class, it is a challenging course...
Course highlights:
The highlights of this course were achieving my goals and practicing until I fully understood the problems. I learned that math is super complex and at the age of 25 I will be a perfect square root. :) haha.
Hours per week:
3-5 hours
Advice for students:
In order to succeed in this class one needs to repetitively study and do multiple problems every day. The key is to understand what you're doing and be able to show it through your work. Also, stay organized! | 677.169 | 1 |
MYTUTOR SUBJECT ANSWERS
What's the point of Maths?
Every subject has a paticular career path. Maths is an exception which is at the forfront of most things. Anytime you saw data, results, surveys, polls, predictions, forecasts, returns & any others that was maths.
When your exam results they are sometimes created by combining marks to create a spread of data. Then the grades are given from the percentage levels in the data. | 677.169 | 1 |
Graphing Linear Functions with the Graphing Calculator
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3.84 MB | 18 pages
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This is an Algebra 1 Common Core Lesson on Graphing Linear Functions with the Graphing Calculator. Students will be introduced to using the graphing calculator to create a table of values. Students will be asked follow up questions regarding transformations, domain and range. After a few teacher led examples, students will practice on their own or in groups.
**This lesson is part of the Linear Functions Unit Bundle. Please check it out if you are interested in more resources from this unit! Thank you for visiting Math Masters! We hope that you enjoy the product and would love your feedback. Please be sure to rate our product. We hope you visit Math Masters again | 677.169 | 1 |
Prentice Hall Algebra 2 with Trigonomentry, Teacher's
This article from the NCTM journal Teaching Children Mathematics explains the foundations underlying the process of instructional differentiation for students with learning disabilities in mathematics. This digest was prepared with funding from the Office of Educational Research and Improvement (OERI), U. In the USA are a lot of different programs for in-service and pre-service teachers programs. Try, "That is an interesting way that you sorted your blocks.
Pages: 885
Publisher: Prentice Hall (June 30, 2001)
ISBN: 0130519693
Saxon Math Course 3 Florida: Teacher Manual 2-Volume Set Grade 8
Effectiveness of Cooperative Learning Compared to Lecture-Based Learning in College STEM Classes McDougal Littell High School Math Tennessee: Lesson Plans Algebra 2. When encouraging a student to work through an intelligence that he or she does not prefer, select content that is somewhat familiar Math Homework Assignments, Grade 6. If it does not work for your family, return it for a full refund. The programs have no glossaries or indexes and no search function to locate whatever subjects you may want to find epub. If you wish the student to translate a figure to a given point, rotate it to a new position, and reflect it over a given line, you could use four congruent figures. I would probably want to use magnetic manipulatives or ones with velcro in a confined space, to keep things in place Using Formative Assessment to Drive Mathematics Instruction in Grades PreK-2. There are three major DI mathematics programs. These include: Connecting Math Concepts is a comprehensive developmental mathematics program designed to teach students to compute, solve problems, and think mathematically. It includes levels A-F as well as a Bridge to F program. A pre-publication Level K program also exists for students in kindergarten. Levels A-F typically correspond to grade levels; each level of the program is meant to be taught over an entire school year Harcourt School Publishers Math New York: Teacher's Edition Coll Grade K 2009. The purpose of this web page is to offer some perspectives on mathematics education from an instructional design viewpoint. The authors do this in a somewhat eclectic fashion, beginning with an overview of the ideological "paradigm wars" within the instructional design community. Alternative philosophies of mind, including Vygotsky's emphasis on the social origins of cognition, have implications for the teaching of mathematics, as well as for instructional design generally McDougal Littell Middle School Math California: Transparency Book Chapter 8 Algebra 1. Hiebert, J., Gallimore, R., & Stigler, J. A knowledge base for the teaching profession epub. Here are four very important points that emerge from consideration of the diagram in Figure 3 and earlier material presented in this section: Mathematics is an aid to representing and attempting to resolve problem situations in all disciplines online.
The authority decides or gives the formula and encourages memorization. is 4. He does forget it easily. accepted without reasoning. formulated by the child; therefore he not appreciate its nature and is likely to remembers it with ease. made clear through reasoning. 5 Jost Bürgi's Aritmetische und Geometrische Progreß Tabulen (1620): Edition and Commentary (Science Networks. Historical Studies). We have found 2 articles that may match the URL you entered or followed: by Margaret L download Prentice Hall Algebra 2 with Trigonomentry, Teacher's Edition pdf. Using Multisensory Teaching to teach your child Math concepts through auditory, visual, and hands-on learning is proven by research to improve learning. This is particularly true for children with math-based learning disabilities like dyscalculia. Almost all programs teaching math for kids are text-based, auditory style programs Holt McDougal Larson Pre-Algebra: Teacher's Notetaking Guide. The district ended up choosing Math Expressions (Please share your experience if you use this!) and this decision allowed me to stop teaching Everyday Mathematics "with fidelity" and monkey around with starting Guided Math/Math Rotations/Small Group Math/Math Centers/etc. for the remainder of this year. I know there are many ways of doing this, so I am just sharing what I have been trying and what works so far with my schedule I've Got to Use Words Vl4 (Bk. 4).
Math Intervention 3-5: Building Number Power with Formative Assessments, Differentiation, and Games, Grades 3-5
They could write symphonies and build robots and airplanes. Most people wouldn't think that the students at José Urbina López could do those kinds of things. Kids just across the border in Brownsville, Texas, had laptops, high-speed Internet, and tutoring, while in Matamoros the students had intermittent electricity, few computers, limited Internet, and sometimes not enough to eat. "But you do have one thing that makes you the equal of any kid in the world," Juárez Correa said. "Potential." He looked around the room. "And from now on," he told them, "we're going to use that potential to make you the best students in the world." Paloma was silent, waiting to be told what to do Big Ideas for Small Mathematicians: Kids Discovering the Beauty of Math with 22 Ready-to-Go Activities. Amongst the most advanced in British Higher Education, Aston University recently invested £4m in a major upgrade of its campus IT network. The facilities specifically provided for our computing students include a 100-seat PC teaching lab with audio-visual projection facilities and three other labs containing PCs and Macs. Additional labs provide high performance workstations reserved for project work and specialist teaching Essential Skills: Math Grd 3 (Essential Skills (Teacher Created Resources)). TAI has been incorporated into one portion of the PowerMath program. Existing evaluations of MathWings do not meet PPN criteria, so we are unable to determine whether the entire MathWings program is also promising online. Wilson tiered the remaining class activities based on different tiers and learning projects (see Figure 1.6). Juan and 19 other classmates were assigned to tier 1 and were required to design posters that illustrated each of the steps to take during a tornado *Use 31352x*Smp Book 1 (School Mathematics Project Numbered Books). To present some Information and Communication Technology (ICT) ideas that are important to teaching and learning math at the elementary school level. As you teach math methods, remember that some of your students will eventually become teachers of teachers. All will do this informally in their conversations with fellow teachers. Some will eventually become workshop leaders, make conference presentations, and/or teach a Math Methods course English for the English: a Chapter on National Education / by George Sampson. The Model building approach to solving word problems was developed locally years ago by Hector Chee, a very experienced Mathematics teacher, and has since been widely used in the teaching of kids math in primary schools (or elementary schools if you prefer) in Singapore Harcourt School Publishers Math New York: Teacher's Edition Coll Grade K 2009. During a whole group beanbag game with blue and yellow beanbags, pass out beanbags so that no child in the circle has the same color as the child next to him/her Knowing Mathematics, Intervention Program, Teacher's Guide. Plan lessons and a unit that address appropriate learning goals. Recognize instructional strategies to create data sets or collect, organize, and display relevant data. Analyze instructional strategies to develop and evaluate inferences and predictions that are based on data. Analyze instructional strategies to understand and apply basic concepts of probability Ten-Minute Math Mind-Stretchers (Grades 3-5). | 677.169 | 1 |
Relations and Functions Lesson
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This lesson introduces students to the concepts of a relation and a function. Students are required to explores these ideas as well as identify what a function is from ordered pairs as well as mapping diagrams | 677.169 | 1 |
Uniform convergence and pointwise convergence The aim of this material is to introduce the student to two notions of convergence for sequences of real-valued functions. The notion of pointwise convergence is relatively straightforward, but the notion of uniform convergence is more subtle. Uniform convergence is explained in terms of closed function balls and the new notion of sets absorbing sequences. The differences between the two types of convergence are illustrated with several examples. Some standard facts are also discussed: a uniform Author(s): The University of Nottingham
Dr Jenny Chesters: Gender convergence in paid and unpaid work, at ANU Dr Jenny Chesters gives this lecture: 'Gender convergence in paid and unpaid work time: Assessing the relevance of earlier approaches for explaining current trends' at The Australian National University on 12 October, 2010.
The increase in dual-earner families over the last few decades raises questions about whether men are sharing the homemaker role now that women are sharing the breadwinner role. Theories of the allocation of unpaid work, such as dependency and exchange approaches, time avail Author(s): No creator set
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Lugosi teaches math - convergence of series Béla Lugosi teaches advanced mathematical concepts in this video. This is part of the last lecture of his series. He begins this lecture by talking about the Weierstrauss M-Test and talking about how a series or sequence of sums converges. In his lecture he provides the steps for this test, and proofs as well. A great video for advanced math learners. (Calculus and Calculus II students). Author(s): No creator set
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Lugosi teaches math - convergence of series 2 In part 2 of Béla Lugosi's lecture on convergence of series, he continues to discuss integration of the limit function in these series or sequences. He then goes further into applying these into power series. This is a video for advanced math students who are well into Calculus and Calculus II. Author(s): No creator set
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Lugosi teaches math - convergence of series 3 Béla Lugosi teaches advanced mathematical concepts in this video. How one uses power series to solve varying kinds of problems is discussed, and Lugosi explains how this one application shows the power of power series. It also show that power series have a very important and powerful and analytical technique. A great series for any advanced mathematics student who is working in Calculus and Calculus II. Author(s): No creator set
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Lugosi teaches math - uniform convergance In this video course, Béla Lugosi teaches advanced mathematical concepts. In this part he teaches the concept of uniform convergence in general, which is a difficult and subtle topic. He gives the highlights of what makes up uniform convergence and convergence. A great intro to the concept of convergence for advanced mathematics. Author(s): No creator set
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Lugosi teaches math - uniform convergance2 In this video, Béla Lugosi continues with his teaching of the concept of uniform convergence. Though Lugosi is lecturing, it is as though he is having a discussion with you at times during this lecture, and discusses numerous properties of uniform convergence. Lugosi's passion for advanced mathematics is evident in this video. Author(s): No creator set
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Lugosi teaches math - uniform convergance3 In part 3 of Uniform Convergence, Béla Lugosi teaches advanced mathematical concepts related to uniform convergence. In this video, he shares how to chart, diagram, and graph uniform convergences and functions. Author(s): No creator set
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MIT OpenCourseWare : linguistics and philosophy The Department of Linguistics and Philosophy page of Massachusetts Institute of Technology's OpenCourseWare website offers a variety of freely available course materials within these two disciplines. The courses are divided into those aimed at undergraduates and those for graduates, and range from introductory surveys to more specialist topics. The quantity and type of resources available also varies considerably: some give just a course outline and perhaps a few reading suggestions, whereas oth Author(s): No creator set
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Information technology in forming cognitive skills In the authors opinions, the most important role of education process is forming developed cognitive skills, such as interdisciplinary critical thinking, problem solving, decision making, but enabling role of information technology in forming cognitive skills is not estimated enough. Accomplishing the investigation of the forming cognitive skills authors discovered similarity between study modules method and work with information. Author(s): Zeidmane Anda,Vintere Anna
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Building and Maintaining the Common Ground in Web-Based Interaction In this paper, the main purpose is to explore how participants establish and maintain the common ground in the computer-based conferences. Previous studies assume that before the participants can reach the deeper level interaction and learning, they have to gain an adequate level of common ground (Dillenbourg, 1999; Baker et al., 1999; Veerman, 2000). Subjects were 68 pre-service teachers and 7 mentors from three universities who participated in the web-based conferencing course for eight weeks. Author(s): Mäkitalo Kati,Häkkinen Päivi,Salo Piritta,Järv
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Facilitating collaborative knowledge construction in computer-mediated learning with structuring too Collaborative knowledge construction in computer-mediated learning environments puts forward difficulties regarding what tasks learners work on and how learners interact with each other. For instance, learners who collaboratively construct knowledge in computer-mediated learning environments sometimes do not participate actively or engage in off-task talk. Computer-mediated learning environments can be endorsed with sociocognitive structuring tools that structure the contents to be learned and s Author(s): Weinberger Armin,Reiserer Markus,Ertl Bernhard,Fis
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Cooperation scripts for learning via web-based discussion boards and videoconferencing Computer-supported collaborative learning often means that locally distant learners discuss a task via text-based discussion boards or videoconferencing. Collaborative learning, however, is often suboptimal with respect to how learners work on the concepts that are supposed to be learned and how learners interact with each other. Collaborative learning environments may be improved by scripts that structure epistemic activities and social interactions of learners. Two studies are being reported t Author(s): Weinberger Armin,Ertl Bernhard,Fischer Frank,Mandl
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Fostering collaborative knowledge construction with visualization tools This study investigates to what extent collaborative knowledge construction can be fostered by providing students with visualization tools as structural support.
Thirty-two students of Educational Psychology took part in the study. The students were subdivided into dyads and asked to solve a case problem of their learning domain under one of two conditions: 1) with content-specific visualization 2) with content-unspecific visualization. Results show that by being provided with a content-specific Author(s): Fischer Frank,Bruhn Johannes,Gräsel Cornelia,Mand
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Epistemic and Social Scripts in Computer-Supported Collaborative Learning Collaborative learning in computer-supported learning environments typically means that learners work on tasks together, discussing their individual perspectives via text-based media or videoconferencing, and consequently acquire knowledge. Collaborative learning, however, is often sub-optimal with respect to how learners work on the concepts that are supposed to be learned and how learners interact with each other. One possibility to improve collaborative learning environments is to conceptuali Author(s): Weinberger Armin,Ertl Bernhard,Fischer Frank,Mandl
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A Method for Creating Collaborative Mobile Learning Trails In this paper I report results from recent trials in which students used mobile devices to collaboratively create, edit and share trails. These included 9- to 10-year-olds as well as adult diploma students, in the subject of horticulture in botanic gardens. Findings indicate that a narrow subject focus and a manageable amount of data capture are appropriate in most cases; trails are most effective when framed with structured tasks and a narrow focus. Structure can be introduced through the use o Author(s): Walker Kevin
A QUALITATIVE STUDY OF TEACHERS CERTIFIED BY THE NATIONAL BOARD FOR PROFESSIONAL TEACHING STANDARDS The purpose of this research study was to describe the characteristics and instructional practices of teachers certified by the National Board for Professional Teaching Standards (NBPTS) in the Early Childhood/Generalist category. The problem of the study is to examine if there are common characteristics that exemplary teachers use. The study rests upon the following considerations: 1) literacy learning is important and the early childhood period is prime time for language learning; 2) young chi Author(s): Laverick, DeAnna
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Four Young Women on a Ramp Behind the Lane Home Four young women are standing on the wooden ramp behind the Lane home in Spiceland. The ramp was used to haul carriages to the second floor of the building for repairs. The smiling young ladies are wearing hats and coats. Author(s): Cephas M. Huddleston,Cephas M. Huddleston Collecti | 677.169 | 1 |
Book with Other Items
Click on the Google Preview image above to read some pages of this book!
New Century Maths for the Australian Curriculum Years 7 – 10 is specifically written to meet the requirements of the NSW Mathematics 7-10 syllabus for the Australian Curriculum, to be implemented in Years 7 and 9 in NSW from 2014. These new titles retain all of the successful features of the New Century Maths series, which has been in schools since 1994.
Also available as an interactive NelsonNetBook, either as a supplement to the printed text or as a standalone option for schools seeking a digital-only resource solution.
About the Author
Judy Binns is head teacher of mathematics at Mulwaree High School in Goulburn and has taught at Homebush Boys High School. She has an interest in motivating students with learning difficulties and teaches the Applied Mathematics course at her school, based on the Mathematics General 1 syllabus. Judy has been co-writing the New Century Maths 7-8 series for 20 years, including new editions for the Australian Curriculum.
Gaspare Carrozza is head teacher of mathematics at Homebush Boys High School and was head teacher at South Sydney High School. He has conducted inservices in both mathematics and computing, with a particular interest in the professional development of teachers. He has been involved in HSC and School Certificate marking and trial HSC exam writing.
ISBN: 9780170188777 ISBN-10: 0170188779 Series: New Century Maths 7 for the Australian Curriculum NSW Stage 4 Audience:
Tertiary; University or College
For Grades: 7 Format:
Book with Other Items
Language:
English
Number Of Pages: 576 Published: 7th February 2012 Publisher: Cengage Learning Australia
Country of Publication: AU Edition Number: 1 Edition Type: New edition | 677.169 | 1 |
Introduction to Algebraic Equations - Interactive Notebook Activities
Introduction to Algebraic Equations - 20+ topics/activities included to familiarize your students with the basics of algebraic equations. Use as math journal activities or as scaffolded notes. Some topics include: expressions vs. equations, the addition principle for equations, combining like terms, and more! #algebraicequations #mathjournals | 677.169 | 1 |
This amazing scientific calculator program helps you solve all kinds of mathematical problems, from a simple addition to a complicated geometrical function. Not only does it solve the operation, but also draws a graph based on it if possible, and gives extra information about the generated graph. The program also includes an excellent database with constants and functions.
A software package designed to solve computationally hard problems in algebra, number theory, geometry and combinatorics.
CommercialMacWindowsLinux
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Alternatives to Math Solver II | 677.169 | 1 |
Item Description: Alpha Omega Publications (AZ). Paperback. Book Condition: New. NEW! We pack all items in a protected and padded bubble mailer or a box designed to protect your item! Your item deserves more than just some plastic bag! We pack all items in a protected and padded bubble mailer or a box designed to protect your item! Your item deserves more than just some plastic bag!. Bookseller Inventory # mon0000161285
Item Description: American Education Publishing. Paperback / softback. Book Condition: new. BRAND NEW, Notebook Reference Math Fact Book: Grades 4-8 (2nd), American Education Publishing, The Notebook Reference Math Fact Book offers students everything they need for success in math right at their fingertips! This convenient 144 page fact book is filled with illustrations, formulas, definitions, and examples that children can use to review virtually every type of math problem. Plus, essential information can be found at-a-glance with a section of ready reference charts that cover everything from multiplication to precalculus. The 3-hole punched format allows students to carry this book in a 3-ring binder for quick reference at school, home or on the go! Topics covered include: -Basic number concepts -Operations and computations -Decimals, fractions, and percents -Ratios, proportions, and probability -Pre-algebra -Simple and compound interest and other formulas -Customary and metric measurements -Problem solving -Geometry -Organizing and graphing data -Glossary and mathematical data -And more! The "Notebook Reference series" offers students everything they need for school success - in a convenient, 3-hole punched format. Containing essential information, each comprehensive guide offers subject-specific text that is easy to read and easy to use. The 3-hole punched format allows the guides to be inserted into notebooks for quick and easy reference. Our other titles include Science, Student Planner, Dictionary, Spanish D. Bookseller Inventory # B9780769643403 Createspace, United States, 2013. Paperback. Book Condition: New. 229 x 152 mm. Language: English . Brand New Book ***** Print on Demand *****. The entire Math Quiz series is now available in one e-book. Seven books - 685 questions (The answers can be found in a separate section) - WOW! These quizzes will be extremely helpful to learn the vocabulary that drives the area of math that is linked to each title. Remember, math is not just about solving problems. Find out exactly where your math knowledge lies. Below are the seven books included in the series with information and sample questions for each one. Enjoy! Math Quiz #1 - Whole Numbers and Number Theory Most people know what five plus four equals. But do you know what these numbers are called? What s a composite number? How about the number inside the division box? Do you know or remember what it s called? What is multiplication really? These are just a few of the things you ll need to know to be successful on this quiz. There are 63 questions in this Math Quiz that are all related to Whole Numbers and Number Theory. Math Quiz #2 - Fractions, Decimals, Ratio and Proportion, Percents/Interest When comparing fractions, how can you tell which fraction is larger? What is unique about a fraction whose denominator is twice as large as the numerator? What determines the worth of a decimal number? What is the actual percent in a problem called? These are just a few of the things you ll need to know to be successful on this quiz. There are 89 questions in this Math Quiz that are all related to Fractions, Decimals, Ratio and Proportion, and Percents/Interest. Math Quiz #3 - Customary and Metric Measurement One pint is how many fluid ounces? What is the quantity of matter in an object? How is electricity measured? These are just a few of the things you ll need to know to be successful on this quiz. There are 64 questions in this Math Quiz that are all related to Customary and Metric Measurement. Math Quiz #4 - Geometry Line segments that have the same length are called what (9 letters)? What is a line segment with endpoints on a circle called (5 letters)? This is the point where the x-axis and y-axis meet (6 letters): This reasoning is a way to reach a conclusion based on a pattern (9 letters): These are just a few of the things you ll need to know to be successful on this quiz. There are 134 questions in this Math Quiz that are all related to Geometry. Math Quiz #5 - Pre-Algebra A term that is a number is called a what? When something has the same value or is equal to (10 letters): What is the steepness of a straight line called? This is a number that multiplies a variable (11 letters): These are just a few of the things you ll need to know to be successful on this quiz. There are 102 questions in this Math Quiz that are all related to Pre-Algebra. Math Quiz #6 - Algebra This is the point where coordinate axes cross (6 letters): What does a vertical line test on a graph tell about a set of ordered pairs? These types of lines have the same slope (8 letters): A number or a group of numbers written with operation signs is called what (10 letters)? These are just a few of the things you ll need to know to be successful on this quiz. There are 156 questions in this Math Quiz that are all related to Algebra 1. Math Quiz #7 - Maps, Schedules, Graphs, Charts, Data, Probability, and Statistics This explains what each symbol represents within a graph (3 letters): Sometimes when it s difficult to read data it s helpful to use this plot or diagram (7 letters): This sampling is generally the most accurate method to sample a population (6 letters): These are events for which the outcome of one affects the outcome of the other (9 letters): These are just a few of the things you ll need to know to be successful on this quiz. There are 77 questions in this Math Quiz that are all related to Maps, Schedules, Graphs, Charts, Data, Probability, and Statistics. Bookseller Inventory # APC9781494259846 shortcuts John Wiley and Sons Ltd. Paperback. Book Condition: new. BRAND NEW, Pre-Algebra Essentials For Dummies, Mark Zegarelli, Bookseller Inventory # B9780470618387
Item Description: Paperback. Book Condition: New. 141mm x 10mm x 210mm. Paperback. A series of 9 books based on U.S. Government teaching materials. They make math interesting and fun. The series of books begins with basic arithmetic and extends through pre-algebra, algeb.Shipping may be from multiple locations in the US or from the UK, depending on stock availability. 160 pages. 0.181. Bookseller Inventory # 9780878912001 shortcuts Paperback. Book Condition: Acceptable. The cover is creased on this item. The text of this book has NO notes or markings. This item has minimal damage that does NOT affect text. We ship daily Monday - Friday!. Bookseller Inventory # 1EY7ZG003CJT
Item Description: Cliffs Notes, 2010. Book Condition: New. Brand New, Unread Copy in Perfect Condition. A+ Customer Service! Summary: Basic Math and Pre-Algebra About the Contents: Pretest Helps you pinpoint where you need the most help and directs you to the corresponding sections of the book Topic Area Reviews The basics Whole numbers Fractions Decimals Percents Integers and rational numbers Powers, exponents, and roots Powers of ten and scientific notation Measurements Charts and graphs Probability and statistics Variables, algebraic expressions, and simple equations Appendix Arithmetic and geometric sequences Predicting the next term of a sequence Glossary Defines terminology used in the book Customized Full-Length Exam Covers all subject areas. Bookseller Inventory # ABE_book_new_0470533498 | 677.169 | 1 |
Abstract The nation is facing a crisis in its community colleges: more and more students are attending community colleges, but most of them are not prepared for college-level work. The problem may be most dire in mathematics. By most accounts, the majority of students entering community colleges are placed (based on placement test performance) into "developmental" (or remedial) mathematics courses (e.g., Adelman, 1985; Bailey et al., 2005). The organization of developmental mathematics differs from school to school, but most colleges have a sequence of developmental mathematics courses that starts with basic arithmetic, then goes on to pre-algebra, elementary algebra, and finally intermediate algebra, all of which must be passed before a student can enroll in a transfer-level college mathematics course.
Connie Yarema teaches mathematics courses for non-science majors and pre-service secondary mathematics teachers at Abilene Christian University, Abilene, Texas. She is interested in outreach mathematics - focusing on establishing communities of practice using lesson study, a Japanese professional development model for in-service and pre-service teachers.
Using a Square to Complete the Algebra Student: Exploring Algebraic and Geometric Connections in the Quadratic Formula
David Hendricks teaches mathematics courses and is the chair of the mathematics department at Abilene Christian University, Abilene, Texas. He is interested in cryptography, preparing pre-service teachers, and working with in-service teachers.
Abstract Recommendations and standards from various stakeholders in the mathematical preparation of teachers, such as The Mathematical Education of Teachers ( and Beyond Crossroads ( call for courses that emphasize connections within topics in mathematics, especially those that are most familiar to pre-service teachers. However, these recommendations are applicable for any student taking an algebra course at a post-secondary institution, including developmental algebra courses. One familiar topic to students who take an algebra course in college or who will teach an algebra course in high school is the quadratic formula. Algebra students often see the derivation of the quadratic formula based on the method of completing the square using algebraic procedures. However, the history of mathematics indicates that these procedures originated from geometric concepts. Presenting both algebraic and geometric representations as the quadratic formula is derived helps pre-service mathematics teachers and algebra students, in general, visualize concepts and make sense of algebraic procedures. It also sets up a natural extension for students to make meaning of the quadratic formula itself by connecting its algebraic symbolism to a geometric representation.
Abstract Population models are often discussed in algebra, calculus, and differential equations courses. In this article we will use the human population of the world as our application. After quick looks at two common models we'll investigate more deeply a model which incorporates the negative effect that accumulated pollution may have on population.
Kenneth Kaminsky, Augsburg College Naomi Scheman, University of Minnesota
Abstract At consume eggs containing blood spots. When observant Jews make an omelet or bake challah, they break each egg separately and discard any that are bloody before they add them to the rest of the ingredients. But when you boil eggs, there is no easy way to tell beforehand if the eggs you are cooking have blood spots or not. Once the eggs are boiled and opened, the spots can be seen, but by then it is too late. All the eggs must be discarded, and a pot in which bloodied eggs are cooked is no longer kosher.
Abstract The Motivating Beauty of Mathematics. Most college instructors agree that there is a need to motivate students to learn mathematics. But there is a profound disagreement on how exactly to do this. Can we actually motivate our students to learn basic algebra? Is the instructor to emphasize the usefulness of mathematics in the "real world"? Are we to motivate students by exhibiting our own motivation and passion for the subject? Are we to convey the beauty of mathematics at such an early stage? Needless to say, these questions pose considerable complexity, and to some extent, the answers are yes, yes, yes, and yes. However, as faculty, we need to find a theme, indeed a style in our teaching of developmental mathematics. Although engaging students in the developmental classroom requires all of these considerations, we contend that the coherence, logical structure, beauty, and breadth of mathematics should be kept at the center of our motivational efforts.
Abstract AMATYC's document, Crossroads in Mathematics, encourages mathematics faculty to "foster interactive learning through student writing," among other activities in its section on standards for introductory college mathematics. However, as Meier and Rishel (1998) point out, these student writing assignments must be carefully designed in order to successfully foster student learning and engagement. Without a connection to the class material, a writing assignment will be less en-gaging to students and will be less successful in achieving an increase in student understanding.
Marcus Jorgensen is an assistant professor in the Developmental Mathematics Department at Utah Valley University (UVU). Prior to UVU, he served as Dean of Computing, Mathematics, and Science at Spokane Falls CC. Marc is a retired United States Coast Guard captain where much of his career was spent in the field of education and training.
Suuroji Puzzles: Sudoku-like Logic with a Twist
Marcus Jorgensen, Utah Valley University
Abstract During a break at a conference of the Southwest Association for Developmental Mathematics, I overheard a colleague say that he wished that Sudoku puzzles used more mathematics so that that he could use them in his classes. I was intrigued with the idea and set out to see if it was possible to combine the logic of Sudoku with some mathematics. Kakuro, or cross-sums, puzzles do use some math but I wanted a little more. Eventually, I came up with what I call Suuroji puzzles (Suuroji uses the beginning letters of the Romaji forms of the Japanese words for mathematics and logic). I have used them in my developmental classes as a way for students to practice some basic operations as well as factoring and problem solving.
Abstract As we teach our courses, we often look for new examples, exercises, and projects that illustrate important mathematical concepts. Questions of an applied nature are particularly useful because they help motivate the study of mathematics. A few years ago a student of mine, who happens to be a cabinet maker, posed a question regarding the construction of a chute. The essential ingredients in his question are contained in the following problem:
A construction crew is remodeling the second floor of an office building and will build a chute to guide debris into a truck below. The chute will have four walls, a large square opening at the top (say with side b1 ), a smaller square opening at the bottom (say with side b2 ), and length . Determine the dimensions of the walls of the chute. In particular, determine the miter angles to be cut along the edges of the walls so that they fit together properly.
Herbert Libow has worked in the fields of physics, engineering, and teaching throughout most of his professional career. One of his prime interests is improving the language of algebra. On leave from Santa Monica College in Santa Monica, California, he can be reached at h.rlibow@ca.rr.com. Comments, questions, and conversations are welcome.
Reading and Understanding Expressions: Replacing the Order Of Operations
Herbert Libow, Santa Monica College
Abstract This concept paper presents a system for reading the meaning of expressions that mimics what those who use math think when reading expressions. Concurrently, this system overcomes the difficulties inherent in the Order of Operations.
Michael W. Ecker (DrMWEcker@aol.com or MWE1@psu.edu) is an associate professor of mathematics at Pennsylvania State University's Wilkes-Barre campus. Having taught math since 1972, he received his Ph.D in mathematics from the City College of New York in 1978. He was the founder of The Mathematical Review problem section in 1981, a position he held until 1997. He is the author of over 500 newsletters, columns, reviews, and articles, many computer-related, as well as five books and/or solutions manuals. His other passions include racqutball, sweets, and Renee (Wife 2.0).
Pedagogy on Integral Notation: Defending the Differential
Michael W. Ecker, Pennsylvania State University, Wilkes-Barre Campus
Abstract It seems that at least once every decade, somebody publishes a diatribe against the differential dx in the integral notation, . When I wrote a solution manual for a survey math text back in 1993, I noted that the author refused to use the differential in the text where he had an introduction to integral calculus. Although I would not have included calculus in such a course at all, I would have at least used correct notation. I wound up insisting on including the differential in my solutions, despite the poor editor's understandable plea for consistency with the author.
Abstract The applied calculus course is generally a less theoretical version of the traditional calculus course aimed at non-science majors. In many colleges, students enrolling in this course are business majors, often majoring in economics or finance. A frequent complaint of students is the paucity of realistic applications relevant to these disciplines. The literature study revealed few realistic multivariable applications to these disciplines. The applications found included the usual, hackneyed, maximize profit/revenue, minimize cost, or determine if demand functions are complementary/substitutes exercises. A review of texts in economics and finance found interesting applications and/or interpretations to level curves, homogeneous functions and Euler's theorem. The authors claim no originality in the following discussion; however, these applications need a wider audience in the mathematical community with the goal of ultimately benefitting students.
Abstract Math teachers everywhere are familiar with the common student question, "When am I ever going to use this?" Many students are disinterested and unmotivated in mathematics because they don't see it as relevant to their own lives. Brain research has shown that when new information makes sense and is perceived as relevant by the learner, retention of the new information greatly increases. One way to increase relevance (and student learning) is to teach students how to apply mathematics in meaningful real world situations.
Roxane Barrows is Associate Dean of Arts & Sciences at Hocking College in Nelsonville, Ohio. She has been teaching mathematics for over 20 years. She obtained her BS in Business Administration from Ohio State University, MS in Mathematics at Ohio University, and is currently working on her dissertation in Higher Education Administration at Ohio University.
Mathematics and Democracy: The Case for Quantitative Literacy
Roxane Barrows, Hocking College
Abstract Mathematics and Democracy: The Case for Quantitative Literacy, edited by Lynn Arthur Steen, is a very timely book. Mathematics, mathematics education, and to a lesser degree, quantitative literacy have never been under as much scrutiny as they are today. Many mathematicians and non-mathematicians believe that too much attention has been placed on abstract mathematical concepts and not enough on numeracy. Steen, rightly so, believes that the majority of citizens need an understanding of quantitative literacy to succeed in life and the workplace, not abstract mathematics. To support his idea, he emphasizes that mathematics and numerical literacy are two different subjects and that being able to understand mathematics in the context in which it is being used is essential. The non-mathematical world is slowly coming to the realization that mathematics literacy is important and Steen does an excellent job of illustrating how fundamental quantitative literacy is for virtually everyone in our society regardless of occupation and/or economic status. He reminds the reader that quantitative data are everywhere in our society: (a) increases in gas prices, (b) changes in SAT scores, (c) low interest car loans, and (d) sports reporting (p. 1). Nearly everyone can relate to one or more of these examples, but many in our society do not know how to use the data for meaningful analysis, which is problematic. The first chapter sets the stage for the rest of the book; a collection of articles, by differing authors, about quantitative literacy. | 677.169 | 1 |
Algebra 103 Furthermore, Graphing Calculator by Mathlab displays calculations as it performs them on the high-quality display of the Android device, making it easier for the user to understand the calculations and see them clearly. This app has two great strengths. First, it acts as a fine scientific calculator, but, more than that, it displays the intermediate steps of the calculations as you type. It allows the students to both watch and learn how the calculations are made and how to find the final answer. Second, the graphing ability is absolutely stunning! Not only does the calculator beautifully display the graphs, but it automatically generates the x- and y- values and displays them as well.
The Algebra 2 Tutor is a 6 hour course spread over 2 DVD disks that will aid the student in the core topics of Algebra 2. This DVD bridges the gap between Algebra 1 and Trigonometry, and contains essential material to do well in advanced mathematics. Many of the topics in contained in this DVD series are used in other Math courses, such as writing equations of lines, graphing equations, and solving systems of equations | 677.169 | 1 |
Getting Kids Excited About Mathematics, One Festival at a Time
Writing about Mathematics - Calculus - Limits and Continuity
This packet is part of my series: Writing about Mathematics Calculus. This particular activity is about limits and continuity.I have found that in order for calculus students to really internalize all of the information they are being given, they must write about it themselves.
Writing about Mathematics - Calculus - Secant and Tangent Lines
This packet is part of my series: Writing about Mathematics Calculus. This particular activity is about secant and tangent lines and the concept of the derivative.I have found that in order for calculus students to really internalize all of the information they are being given, they must write about it themselves. | 677.169 | 1 |
Student's Solution Manual Complete, worked-out solutions are given for odd-numbered exercises and chapter review exercises and all chapter test exercises in a volume available for purchase by students. In addition, a practice chapter test and cumulative review exercises are provided for each chapter.
This text is an unbound, binder-ready edition. Barnett, Analytic Trigonometry is a text that students can actually read, understand, and apply. Concept development moves from the concrete to abstract to engage the student. Almost every concept is illustrated by an example followed by a matching problem allowing students to practice knowledge precisely when they acquire it. To gain student interest quickly, the text moves directly into trigonometric concepts and applications and reviews essential material from prerequisite courses only as needed. Extensive chapter review summaries, chapter and cumulative review exercises with answers keyed to the corresponding text sections, effective use of color comments and annotations, and prominent displays of important material all help the student master the subject. Analytic Trigonometry 11th edition includes updated applications from a range of different fields to convince all students that trigonometry is really useful. The seamless integration of Barnett, Analytical Trigonometry 11th edition with WileyPLUS, a research-based, online environment for effective teaching and learning, builds student confidence in mathematics because it takes the guesswork out of studying by providing them with a clear roadmap: what to do, how to do it, and whether they did it right.
Clear explanations, an uncluttered and appealing layout, and examples and exercises featuring a variety of real-life applications have made this book popular among students year after year. This latest edition of Swokowski and Cole's ALGEBRA AND TRIGONOMETRY WITH ANALYTIC GEOMETRY retains these features. The problems have been consistently praised for being at just the right level for precalculus students. The book also provides calculator examples, including specific keystrokes that show how to use various graphing calculators to solve problems more quickly. Perhaps most important--this book effectively prepares readers for further courses in mathematics. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.
ELEMENTARY TECHNICAL MATHEMATICS Eleventh Edition is written to help students with minimal math background successfully prepare for technical, trade, allied health, or Tech Prep programs. The authors focus on fundamental concepts in basic arithmetic including the metric system and measurement, algebra, geometry, trigonometry, and statistics, which are supported by thousands of examples, exercises, and applications surrounding such fields as industrial and construction trades, electronics, agriculture/horticulture, allied health, CAD/drafting, HVAC, welding, auto/diesel service, aviation, natural resources, culinary arts, business/personal finance, and others. For this revision, the authors have added over 150 new exercises, 30 new examples, new applications categories, and a new appendix on simple inequalities. The goal of ELEMENTARY TECHNICAL MATHEMATICS is to engage students and provide them with the math background they need to succeed in future courses and careers. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook versionNOTE: You are purchasing a standalone product; MyMathLab does not come packaged with this content. If you would like to purchase both the physical text and MyMathLab, search for: 013379556X / 9780133795561 Calculus And Its Applications Plus MyMathLab with Pearson397 / 9780321979391 Calculus And Its Applications MyMathLab should only be purchased when required by an instructor. Calculus and Its Applications, Eleventh Edition, remains a best-selling text because of its accessible presentation that anticipates student needs. The writing style is ideal for today's students, providing intuitive explanations that work with the carefully crafted artwork to help them visualize new calculus concepts. Additionally, the text's numerous and up-to-date applications from business, economics, life sciences, and social sciences help motivate students. Algebra diagnostic and review material is available for those who need to strengthen basic skills. Every aspect of this revision is designed to motivate and help students to more readily understand and apply the mathematics.
Peterson's Teens' Guide to College & Career Planning: Your High School Roadmap to College & Career Success is the must-have eBook for middle school and high school students as they prepare for life after graduation. Whether you're heading to a four-year college, a community or two-year college, the military, or the workforce, Teens' Guide to College & Career Planning offers expert advice and tools to help you succeed. Chapters include The Big Jump to High School, The First Steps to a Career, Planning Your Education While in High School, Tackling the Tests (ACT, PSAT/NMSQT, SAT, and TOEFL), The College Search, Applying to College, Financial Aid Dollars and Sense, Other Options After High School, The Military Option, Jump into Work, Survival Skills, and more. Throughout the book, you'll find real-life advice from students, guidance counselors, parents, and college admissions counselors; helpful checklists and worksheets to help keep you organized; essential information to help you decide if the military is right for you; expert financial aid advice and information on scholarships, grants, athletic awards, loans, work-study, and more. Fun graphics along with the informative, easy-to-read chapters make this the perfect guide for the teen on the go. | 677.169 | 1 |
5.7 Properties of Matrices
CLASSROOM EXAMPLE 1 Finding Values to Make Two Matrices Equal
Find the values of the variables for which each statement is true, if possible.
(a)
Adding Matrices
CLASSROOM EXAMPLE 2 Adding Matrices
Find each sum, if possible.
5.5 Nonlinear Systems of Equations
Solving Nonlinear Systems with Real Solutions
The substitution method works well for solving many such systems, particularly when one of the equations is
linear, as in the next example.
CLASROOM EXAMPLE 1 Solving a Nonli
Showing 1 to 3 of 9
Mrs Privette and Mrs Jager are absolutely the best math teachers I have ever taken a course from.
Course highlights:
It fun and easy to learn in this class.
Hours per week:
6-8 hours
Advice for students:
Do your math lab homework everyday.
Course Term:Fall 2014
Professor:Privette
Course Required?Yes
Course Tags:Math-heavyGreat Intro to the SubjectMany Small Assignments
Nov 04, 2016
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
The teacher is very hands on and helps you understand what is going on rather than leaving you hanging.
Course highlights:
I've gained the relative knowledge of mathematics all because of a great teacher. Without the proper guidance I would have never been able to learn this subject the way I have.
Hours per week:
9-11 hours
Advice for students:
Defiantly be prepared to work don't go into the course with arrogance go because you want to succeed. Never forget that you are going to college so that you may better your future.
Course Term:Fall 2016
Professor:Wood
Course Required?Yes
Course Tags:Math-heavyMany Small AssignmentsCompetitive Classmates
May 26, 2016
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
College Algebra was a Dual Credit course that I took during my Senior year at Weatherford High. I would highly recommend any student who may be on the fence about dual credit courses to take them, as they usually in a smaller class environment, are at your own pace, and are beneficial to students in the long run. In addition, Roemer was a very excellent teacher, as he was entertaining, but also helpful.
Course highlights:
Some perks to being in the dual credit courses including the class size being fairly small (about 13 people in the room, including the professor); the homework being at your own pace within reasonable deadlines; and -of course- being able to achieve college credits and being enrolled in a college at the same time that you were enrolled and receiving credit for high school. As for learning, the course itself did not teach me much more on the particular subject that it was on, but more so how to manage my time and how to establish a homework schedule. In addition, It taught me that, no matter how sure you think you understand the material, it is always beneficial in college to study: even just a little bit. Otherwise, you might end up bombing a test that you thought was going to be easy.
Hours per week:
3-5 hours
Advice for students:
No matter how sure you think you understand the material, it is always beneficial in college to study: even just a little bit. Otherwise, you might end up bombing a test that you thought was going to be easy. | 677.169 | 1 |
Types of Sequences
Compressed Zip File
Be sure that you have an application to open this file type before downloading and/or purchasing. How to unzip files.
1.5 MB | 24 pages
PRODUCT DESCRIPTION
This is an Algebra 1 Common Core Lesson on Different Types of Sequences. Using knowledge of sequences and linear equations, students will develop equivalent equations to model the sequence | 677.169 | 1 |
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[color=violet]Please have Adobe Reader installed or another PDF reader. Here is the link to Adobe for the reader Please know if any of my E-Book torrents say there is a CD ROM I don't have it please buy the E-Book to support the creator. Click here to see all my torrents [url] [/color]
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Now, it is easier than ever before to understand complex mathematical concepts and formulas and how they relate to real-world business situations. All you have to do it apply the handy information you will find in Business Math For Dummies. Featuring practical practice problems to help you expand your skills, this book covers topics like using percents to calculate increases and decreases, applying basic algebra to solve proportions, and working with basic statistics to analyze raw data. Find solutions for finance and payroll applications, including reading financial statements, calculating wages and commissions, and strategic salary planning.
Navigate fractions, decimals, and percents in business and real estate transactions, and take fancy math skills to work. You'll be able to read graphs and tables and apply statistics and data analysis. You'll discover ways you can use math in finance and payroll investments, banking and payroll, goods and services, and business facilities and operations. You'll learn how to calculate discounts and markup, use loans and credit, and understand the ins and outs of math for business facilities and operations. You'll be the company math whiz in no time at all! Find out how to:
Complete with lists of ten math shortcuts to do in meetings and drive your coworkers nuts and ten tips for reading annual reports, Business MathFor Dummies is your one-stop guide to solving math problems in business situations.
From the Back Cover
Features tons of formulas and practice problems to hone your skills
The fun and easy way to do financial computations to be successful in business
Are you baffled by business math? Never fear — this easy-to-understand guide explains complex formulas and concepts in plain English and shows you how to use them day to day on the job. From using basic algebra to solve proportions and statistics to analyze data to reading financial statements, calculating wages and commissions, and managing investments, you'll gain the skills you need to conduct real-world business. | 677.169 | 1 |
User ratings
Gilbert Strang is Brilliant. The book is a treasure trove of his insight into the subtleties of Linear Algebra. As another reviewer pointed out, do the review problems. If you do them in sequence they tend to help clarify the content of the chapters, and prompt you into thinking about the material in different ways. The Problems teach as much as the sections explaining them. | 677.169 | 1 |
Journey Into Mathematics: An Introduction to Proofs
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This 3-part treatment begins with the mechanics of writing proofs, proceeds to considerations of the area and circumference of circles, and concludes with examinations of complex numbers and their application, via De Moivre's theorem, to real numbers. "I recommend this as a textbook or supplemental textbook." - Brian Rogers, The Mathematical Association of America. 1998 | 677.169 | 1 |
Hey, buy any book from the internet or book store, but the main thing is that you must learn each concept from book. And if you don't understand it then go with some professional tutor or search related video tutor site for it. | 677.169 | 1 |
Product Description
Help your students transition to pre-algebraic topics such as fractions, decimals, percents, ratios, unit conversions, and graphing; and provides introductions to geometry and discrete mathematics with Saxon Algebra 1/2. Comprehensive lesson instructions feature complete solutions to every practice problem, problem set, and test problem with step-by-step explanations and helpful hints. These user-friendly CD-ROMs contain hundreds of hours of instruction, allowing students to see and hear actual textbook problems being worked on a digital whiteboard. A slider button allows students to skip problems they don't need help on, or rewind, pause, or fast-forward to get to the sections they'd like to access. Problem set questions can be watched individually after the being worked by the student; the practice set is one continuous video that allows for easy solution review. For use with the 3rd Edition. Four Lesson CDs and 1 Test Solutions CD is included.
System Requirements:
Windows:
98, 2000, ME, XP, 7, or Vista with latest updates, 8, 10
450 MHz or faster
128 MB RAM
8x CD-ROM Drive
800 x 600 display
Macintosh:
10.2 or up
G3, 500 MHz or faster
128 MB RAM
8x CD-ROM Drive
800 x 600 display
Please Note: This product is only available for purchase by homeschools, consumers, and public institutions.
My daughter was excited to have this CD-ROM, it has been very helpful for both of us. Each day my daughter completes her assignment, then after I check it she can watch the CD-ROM to see where she went wrong on any problems she did incorrectly. She likes the straight forward format, and it is easy for her to find/view only the problems she missed and not need to review each problem of the assignment. I would recommend this to anyone using Saxon Algebra 1/2.
This is my second year homeschooling my 13 year old. The DVD set for Saxon Math is an absolute must have. The lessons are short and to the point. (10-15 minutes) The program also goes thru the homework sets problem by problem and shows the student how to get the right answer. It's perfect for after my son has made his first attempt at all of the problems and is simply going back to correct the ones that he missed.
I have used Saxon Math products for many years and the addition of the Saxon teacher CDs is a great help. It insures that my sons have help at their fingertips even when I am busy or unavailable. The lessons are read just as you would from the book and the students can watch any or all of the problems worked out for them if they get stuck. Don't worry about them watching them all and not doing the work. The teacher goes slowly and most kids will work as many as they can on their own to save time.
Ask Christianbook
|
Q:
Algebra half CD Rom teachers CD Can I use the CD Algebra half third edition together with the second edition Algebra half text books? Are they the same?
A:
No, there were enough changes in the lesson order to render the editions incompatible. If you are using the second edition of "Saxon Algebra 1/2" you would have to use the DIVE product, which is designed for that edition (CBD stock number WW345242 | 677.169 | 1 |
Isn't it enough just to say that a "periodic function" is one that "repeats itself again and again"? Most normal people can't make heads or tails of a mathematician's definition, so why can't we just stick with something simple?
I like to think of the phrase "a function that repeats itself again and again" as a naive definition of a periodic function. It gets at the main idea in terms that are easy to understand, and when it's given along with an example or two, it is an excellent place to start. The problem is that it's too broad. In our inexperience, we have been too quick to generalize.
Students should begin with informal definitions like these, but they should gradually refine them to make them more and more rigorous. We need to challenge the naive definition by asking questions like, "What do you mean by repeating? How many times does it have to repeat?" Better yet, we should come up with counter-examples.
Would you call x*sin(x) periodic? What about sin(100/x)? By debating these and other functions, mathematicians gradually refined the definition so that it would include only the functions they intended.
Our students have to experience this process for themselves. While we may not arrive at the "formal" definition that a Ph.D. would use, we should leave the naive & intuitive definitions far behind.
I'll kick off class with the question, "What is a radian?" The class should be able to provide a good definition and explain how to convert from degrees to radians, and I'll do my best to call on some of the quieter students for this.
With this very brief refresher, it's time to discuss the solutions to problems 4 - 11 from Radian Practice. I'll have one student come to the board to demonstrate her solution to problem 5, and then have another demonstrate the solution to problem 7. Clearly, arc lengths are easier to calculate if the angle is given in radians!
Next, we'll turn to the area of a sector. Although it isn't given in general form in the problem set, I'll ask a student to come to the board to explain in general how sector area is calculated when the angle's given in degrees. We'll compare this to the general formula in problem 9 to see that, once again, radians have the advantage! In essence, by incorporating the factor, pi, into the angle measure, we can eliminate it from the other formulas.
Finally, I'll ask students what they noticed about the table they were asked to complete. Since they are not used to thinking in terms of limits, they may not have noticed much. In this case, I'll ask, "What would you expect to see if the radian measure continues to get closer to zero?" The answer is that the ratio should get closer and closer to 1. "Is this true when the angle is measured in degrees?" No. Once again, radians are nicer than degrees! This particular fact won't have much use in Algebra 2, but it is fundamental to calculus and simply interesting in its own right. (By the way, when the angle is measured in degrees, the ratio approaches pi/180. Pretty neat!)
Resources (1)
Resources
I'll begin the next section by asking, "Do any of you recall the definition of function in mathematics?" Once we've recalled the essential characteristics of a function, I'll ask, "Can we say that sine and cosine are functions?" To help students see that the answer is yes, we'll briefly consider the following:
What quantities are the "inputs" for the sine function? angles or arcs
What quantities are the "outputs" for the sine function? ratios or lengths in the unit circle
Is there always just one output for each input? Yes
What's the domain? All real numbers.
What's the range? Real numbers from -1 to 1.
Once we've reached this point, I'll take some time to discuss the graphs of the sine and cosine functions. There are many great applets and animations on the web that make the graphs easier to understand (I personally like this one). Using the animation, we'll discuss the key features of the graph and how they relate to the unit circle. Of course, I'll pay special attention to the periodicity of the Sine_Function. I'll need to define periodic function for them, too.
Resources (1)
Resources
"What are some good examples of periodic phenomena in the real world?" (MP 4)
I'll ask my class this question to see what they come up with. I'm expecting things like these:
Planetary orbits, seasons, tides, phases of the moon
Sound waves, water waves, light waves, radio waves, and other waves
A rolling wheel, a mass bouncing on a spring, a swinging pendulum
These are all great examples, but for our first modeling problem we'll begin with something a little different: electricity. Alternating current is an example of a real-life periodic phenomenon. The current (and voltage) fluctuate periodically and may be modeled nicely with a fairly simple sine function.
[N.B. We have not yet discussed amplitude, period, frequency, etc. This problem will serve as a motivation for all of that.]
I'll hand out the AC Generator Problem and ask my students to take a few minutes to read the beginning. Then after an opportunity to ask questions about the physics behind the problem, I'll let them begin working in small groups to complete the graph. As they work, I'll move around the classroom helping individuals and checking the students work so that I can catch & correct their mistakes quickly.
The key is for students to take a systematic approach. Please see this video for details.
Resources (3)
Resources
We've covered a lot of ground today, and I've had many opportunities to check for understanding along the way. My expectation is that by the end of class, my students will be finished with parts (a) through (d) of the AC Generator Problem. For homework, I will likely assign part (e) only.
Developing the equation for current from the voltage equation is a good exercise, and this minor homework assignment will give students who need it the chance to catch up. Also, in case students make mistakes in part (e), I would hate for them to graph the wrong equation it part (f)! | 677.169 | 1 |
This series, well-known for accessibility and for a student-friendly approach, has a wealth of features: Worked Examples, Activities, Investigations, Graded Exercises, Key Points summaries and Discussion Points. To ensure exam success there are plenty of up-to-date exam questions, plus warning signs to indicate common pitfalls. MEI offer full support to schools through their network with newsletters, training days and an annual conference. AS Further Pure Mathematics is the compulsory core component in the AS Further Mathematics qualification (FP1). It is designed so that students can start it, alongside C1 and C2, immediately after higher tier GCSE. The topics covered are complex numbers, matrices, graphs, inequalities, roots of equations, series and proof by induction | 677.169 | 1 |
About this product
Description
Description
The Common Core Math Curriculum is a must-have for all classroom educators. This practical guide is filled with effective learning strategies to help all students succeed in mastering the Common Core standards in the content area of Math. Educators are given the instructional tools and processes to be able to analyze the Common Core standards, develop standards-based lessons, comprehension strategies, building thinking skills, and assessing students through the creation of student projects. | 677.169 | 1 |
Exponential and Logarithmic Functions Unit
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6.36 MB | 50 pages
PRODUCT DESCRIPTION
This unit includes PowerPoint presentations, coordinated guided notes with answers, a mid-unit quiz, and unit test covering Exponential and Logarithmic Functions. The PowerPoint includes warm ups (do-now or bell ringer), key concepts, and examples for students to follow. It also includes a pacing guide with common core state standards and suggested homework and classwork assignments.
I created this to accompany the Prentice Hall Pearson Algebra 2 textbook. This unit covers Sections 7.1-7.5. The lessons are 30 - 45 minutes each.
Unit Essential Questions:
How do you model a quantity that changes regularly over time by the same percentage?
How are exponents and logarithms related?
How are exponential functions and logarithmic functions related | 677.169 | 1 |
Synopsis
This title features: comprehensive and authoritative 'must have' handbook for older primary pupils, secondary and college students, teachers and parents; first part of book that is a dictionary, giving brief and simple explanations of mathematical terms, often with examples and diagrams; second part of book that contains a detailed reference section, providing an overview of key ideas about a wide range of mathematical topics, including tables, number systems, charts, 2-D shapes, 3-D shapes, measures, conversion tables, equivalences, formulas, rules, explanations and symbols. | 677.169 | 1 |
07216648Math for Nurses: A Problem Solving Approach
This book presents dosage calculation and other mathematics problems faced by nursing students. It follows a problem-solving method which parallels the nursing process and provides students with a practical guide for solving all kinds of basic maths problems. Features include a comprehensive review of basic maths principles; ratio and proportion method of dosage calculation; nearly 2000 exercises, with answers to odd-numbered questions given in text; pre-test and post-test for each chapter; the essentials of medication administration and documentation; and an easy-to-learn four step problem-solving method | 677.169 | 1 |
This study guide, designed for use at Oklahoma State University, contains lists of activities for students to perform based on the "mastery of learning" concept. The activities include readings, problems, self evaluations, and assessment tasks. The units included are: Functions, Exponential and Logarithmic Functions, Trigonometric Functions, Polynomial Functions, Trigonometric Equations and Identities, Numerical Trigonometry, Combinatorics, Probability, Linear Algebra, Complex Numbers, and Sequences. (MK) | 677.169 | 1 |
Conic Sections - PowerPoints, Notes, & Examples Bundle
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This unit contains notes and practice to teach students how to graph conic sections, write equations in standard form, and find details about the conic (foci, etc.). There is a PowerPoint Presentation for each of the conic sections (Parabolas, Circles, Ellipses, and Hyperbolas). They include real-life examples, definitions, formulas, visuals, and algebraic examples. There is a Students Notes file that goes with each PowerPoint9.00. | 677.169 | 1 |
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Collections: solving problems from the California Standards Test in Geometry, written by and featuring Sal Khan. It is divided into 4 units. Each unit consists of a number of 10-15 minute videos that can be consumed in one sitting. Sal's manner is approachable, interesting, and fun and is always pedagogically on-target.
This is a first year, high-school level course on Geometry (which is based on Euclid's elements). It revisits many of the basic geometrical concepts studied in earlier courses, but addresses them with more mathematical rigor. There is strong focus on proving theorems and results from basic postulates. | 677.169 | 1 |
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CONCEPT: This algebra workbook has a creative twist: A picture of a Cuddly Bear is used in place of a boring old X for the unknown. The idea behind adding this artistic touch to the equations is to he...
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From signed numbers to story problems — calculate equations with ease Practice is the key to improving your algebra skills, and that's what this workbook is all about. This hands-on guide focuses on...
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Algebra I For Dummies, 2nd Edition (9781119293576) was previously published as Algebra I For Dummies, 2nd Edition (9780470559642). While this version features a new Dummies cover and design, the conte...
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Expanded and revised, the Second Edition of the Algebra Survival Guide unleashes its power for a new generation of students. Now that Âthe Common Core StandardsÂhave changed how math is taught, this... | 677.169 | 1 |
GRE Subject Mathematics
The test consists of approximately 66 multiple-choice questions drawn from courses commonly offered at the undergraduate level.
Approximately 50 percent of the questions involve calculus and its applications — subject matter that can be assumed to be common to the backgrounds of almost all mathematics majors.
About 25 percent of the questions in the test are in elementary algebra, linear algebra, abstract algebra and number theory. The remaining questions deal with other areas of mathematics currently studied by undergraduates in many institutions.
The following content descriptions may assist students in preparing for the test. The percents given are estimates; actual percents will vary somewhat from one edition of the test to another.
CALCULUS — 50%
Material learned in the usual sequence of elementary calculus courses — differential and integral calculus of one and of several variables — includes calculus-based applications and connections with coordinate geometry, trigonometry, differential equations and other branches of mathematics.
ALGEBRA — 25%
Elementary algebra: basic algebraic techniques and manipulations acquired in high school and used throughout mathematics
Other topics: general topology, geometry, complex variables, probability and statistics, and numerical analysis
The above descriptions of topics covered in the test should not be considered exhaustive; it is necessary to understand many other related concepts. Prospective test takers should be aware that questions requiring no more than a good precalculus background may be quite challenging; such questions can be among the most difficult questions on the test. In general, the questions are intended not only to test recall of information but also to assess test takers' understanding of fundamental concepts and the ability to apply those concepts in various situations | 677.169 | 1 |
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0.4 MB | 9 pages
PRODUCT DESCRIPTION
Students use Algebra Tiles to model and solve one variable two step equations and graph the solutions on a number line. This includes notes for an interactive notebook, practice page and ticket out.
TEKS 7.11A, 7.11B, 7.10B | 677.169 | 1 |
Mathematica is an impressive piece of software, and the range of things which can be done with it is one of its strengths, especially for researchers. In an educational setting, however, Mathematica's generality and comprehensiveness can be intimidating to students. This is a problem faced by anyone who uses Mathematica in an undergraduate course/ The following article by John Costango and Barry Tesman is the result of a student project at Dickinson College, and describes how parts of Mathematica can be modified to be more suitable for classroom instruction. | 677.169 | 1 |
Math to Build on62
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$11Since everything assembled consists of either straight lines, curved lines, or a combination of both, the ability to calculate circles and right triangles is essential for anyone who works in a building trade. This simple and straightforward book explains the basic math used in construction, manufacturing, and design. Starting with fractions and decimals and moving to mitered turns and arcs, these principles are presented with detailed illustrations, practical applications, and in larger print for easy reading. The result is increased efficiency, productivity, and confidence in one's work from initial design to final product. | 677.169 | 1 |
This paperback edition is not available in the U.S. and Canada.This text offers a comprehensive presentation of the mathematics required to tackle problems in economic analysis. To give a better understanding of the mathematical concepts, the text follows the logic of the development of mathematics rather than that of an economics course. The only prerequisite is high school algebra, but the book goes on to cover all the mathematics needed for undergraduate economics. It is also a useful reference for graduate students. After a review of the fundamentals of sets, numbers, and functions, the book covers limits and continuity, the calculus of functions of one variable, linear algebra, multivariate calculus, and dynamics. To develop the student's problem-solving skills, the book works through a large number of examples and economic applications. This streamlined third edition offers an array of new and updated examples. [Some of the lengthier proofs and examples have been moved to the book | 677.169 | 1 |
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Abstract
Understanding the concept of mathematical variables gives an opportunity to expand and work on high-level mathematics. This study examined college students' comprehension of variables as well as variable use in well-known mathematics formulas. These formulas consist of the Pythagorean Theorem, slope, as well as the y-intercept. Students were asked to complete a ten-problem quiz in a twenty minute time frame. Immediately following the quiz, students were asked to complete a five question survey in which they described their reactions to the quiz and their knowledge of variables. Similarly 15 high school mathematics teachers were given a survey on their reflection of their students' knowledge of variables. The results of the quiz and surveys were collected and analyzed to determine if any correlations existed. The data collected showed that there was a strong indication of variable misunderstanding by college level mathematics students. | 677.169 | 1 |
Karl Smith book reviews
Mathematics: Its Power and Utility
Mathematics- It's Power and Utility
Excellent material for pre-high school studants, or for people who want to advance their understanding of math. Includes Alegebraic problem solving, Geometry,Statistics, Logic and Sets.
Well | 677.169 | 1 |
ELEMENTARY STATISTICAL CONCEPTS: The primary objective of this book is to introduce some basic statistical concepts that can be easily understood by college freshmen by minimizing the use of mathematical formulas and statistical symbols.
About the Book
We're sorry; this specific copy is no longer available. Here are our closest matches for ELEMENTARY STATISTICAL CONCEPTS: The primary objective of this book is to introduce some basic statistical concepts that can be easily understood by college freshmen by minimizing the use of mathematical formulas and statistical symbols. by WALPOLE, RONALD E..
Description:
B000MNJHY MACMILLAN PUBLISHING CO., INC. Hardcover. Book Condition: VERY GOOD. Very Good: Cover and pages show some wear from reading and storage. May have light creases on the cover and binding. Bookseller Inventory # 2619967125 | 677.169 | 1 |
Contemporary Abstract Algebra, by Joseph A. Gallian, Provides a solid introduction to the traditional topics in abstract algebra while conveying that it is a contemporary subject used daily by working mathematicians, computerMore...
The seventh edition of Contemporary Abstract Algebra, by Joseph A. Gallian, Provides a solid introduction to the traditional topics in abstract algebra while conveying that it is a contemporary subject used daily by working mathematicians, computer scientist, and chemists. The text includes numerous theoretical and computational exercises, figures, and tables to teach you how to work out problems, as well as to write proofs. Additionally, the author provides biographies, poems, song Lyrics, historical notes, and much more to make reading the text an interesting, accessible and enjoyable experience. Contemporary Abstract Algebra will keep you engaged and gives you a great introduction to an important | 677.169 | 1 |
This is part one of a two-part SMSG text for grade eight students whose mathematical talents are underdeveloped. The reading level of this text has been adjusted downward, chapters shortened, and additional concrete examples included. Nevertheless, the authors warn that the text may not be appropriate for the very slow non-college-bound student. Chapter topics include: linear measurement, area and volume, angles and parallels, polygons and prisms, circles, and statistics and graphs. (MN) | 677.169 | 1 |
Linear Algebra and Its Applications29.58
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About the Book
Linear algebra is relatively easy for students during the early stages of the course, when the material is presented in a familiar, concrete setting. But when abstract concepts are introduced, students often hit a brick wall. Instructors seem to agree that certain concepts (such as linear independence, spanning, subspace, vector space, and linear transformations), are not easily understood, and require time to assimilate. Since they are fundamental to the study of linear algebra, students' understanding of these concepts is vital to their mastery of the subject. David Lay introduces these concepts early in a familiar, concrete "Rn" setting, develops them gradually, and returns to them again and again throughout the text so that when discussed in the abstract, these concepts are more accessible. | 677.169 | 1 |
Sequences linear and quadratic - Common Core Math 2
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2.91 MB | 22 pages
PRODUCT DESCRIPTION
At least 2 lessons worth of content. Basics through to quadratic sequences. Includes examples and practise questions. Includes calculating the nth term. Fibonacci sequence starter and a chess board puzzle at the end (related to sequences | 677.169 | 1 |
CHAPTER 12
Symbolic Programming with Structured Data
Symbols are everywhere. Numbers are symbols that stand for quantities, and you can add, multiply, or take square roots of numbers that are so small or large that it's hard to imagine the quantity they represent. You can solve equations, multiply polynomials, approximate functions using series, and differentiate or integrate numerically or symbolically—these are just a few everyday examples of using symbols in mathematics.
It would be a mistake to think symbols are useful only in mathematics and related fields. General problem-solving can't do without symbols; they provide the ability to abstract ...
The best content for your career. Discover unlimited learning
on demand for around $1/day. | 677.169 | 1 |
ISBN 9788181282156
ISBN-10
8181282159
Binding
Paper Back
Number of Pages
384 Pages
Language
(English)
Subject
Discrete mathematics
DESCRIPTION This introduction to discrete mathematics is aimed primarily at undergraduates in mathematics and computer science at the freshmen and sophomore levels. The text has a distinctly applied orientation and begins with a survey of number systems and elementary set theory. Included are discussions of scientific notation and the representation of numbers in computers. Lists are presented as an example of data structures. An introduction to counting includes the Binomial Theorem and mathematical induction, which serves as a starting point for a brief study of recursion. The basics of probability theory are then covered. Graph study is discussed, including Euler and Hamilton cycles and trees. This is a vehicle for some easy proofs, as well as serving as another example of a data structure. Matrices and vectors are then defined. The book concludes with an introduction to cryptography, including the RSA cryptosystem, together with the necessary elementary number theory, e.g., Euclidean algorithm, Fermat's Little Theorem. Good examples occur throughout. At the end of every section there are two problem sets of equal difficulty. However, solutions are only given to the first set. References and index conclude the work. A math course at the college level is required to handle this text. College algebra would be the most helpful. CONTENTS Preface Properties of Numbers Sets and Data Structures Boolean Algebras and Circuits Relations and Functions The Theory of Counting Probability Graph Theory Matrices Number Theory and Cryptography Bibliography Answers to Odd Number Exercises Index. | 677.169 | 1 |
Applications of Greens Theorem
Let us suppose that we are starting with a path C
and a vector valued function F in the plane. Then as
we traverse along C there are two important (unit)
dy
vectors, namely T, the unit tangent vector h dx
ds , ds i,
dy
dx
an
Coordinate systems
There are three main ways to describe a point in
space. We have already discussed the Cartesian coordinate system (also known as rectangular coordinates). The other two are generalizations of polar coordinates in the plane, so before ju
Cross product
Cross product, denoted with , gives a way to
multiply vectors together and get a new vector, i.e.,
(vector)(vector)=(vector). But there is a big catch,
namely it only works in three dimensions (conveniently though we live in three dimensions
Algebra
Algebra is the foundation of calculus. The basic
idea behind algebra is rewriting equations and simplifying expressions; this includes such things as factoring, FOILing (i.e., (a+b)(c+d) = ac+ad+bc+bd),
adding fractions (remember to get a common d
Second Derivative Test
1. The Second Derivative Test
We begin by recalling the situation for twice dierentiable functions f (x) of one variable.
To nd their local (or relative) maxima and minima, we
1. nd the critical points, i.e., the solutions of f (x)
Calculus 1
Lia Vas
Concavity and Inflection Points. Extreme Values and The
Second Derivative Test.
Consider the following two increasing functions. While they are both increasing, their concavity
distinguishes them.
The first function is said to be concav | 677.169 | 1 |
Solve Word Problems in Algebra, 2nd Edition
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Solving word problems has never been easier than with Schaum's "How to Solve Word Problems in Algebra"This popular study guide shows students easy ways to solve what they struggle with most in algebra: word problems. "How to Solve Word Problems in Algebra," Second Edition, is ideal for anyone who wants to master these skills. Completely updated, with contemporary language and examples, features solution methods that are easy to learn and remember, plus a self-test | 677.169 | 1 |
Lecture 2: Introduction to Linear programmingContents1Defnition and geometryA mathematical program is alinear programif it has(i) continuous variables(ii) one linear objective function(iii) all constraints are linear equalities or inequalitiesA generic linear programming (LP) problem is deFned as follows.maximizecTx,subject toaTix≥bi,i∈G,aTix=bi∈E,aTix≤bi∈L,xj≥0,j∈P,xj≤0∈N.The last two sets of constraints although inequalities are special and are usually treated separately. Thevectorcis called theobjective vectorand the numbersbi,i∈G∪E∪L, are called the RHS coefficients. Wecan, of course, stack up the constraints of each kind and rewrite the above problem in the following form:maximizecTx,subject toAgx≥bg,Aex=be,Alx≤bl,xj≥0∈P,xj≤0∈1.1Why bother with LPsOne part of the answer lies in the geometry of the feasible region of an LP. Since equalities simply statethat the problem lies in a lower dimensional space, we will consider LPs with only inequality constraints,i.e.E=∅. Recall that a linear inequalityaTx≤bdivides the entire space into two parts, i.e. the set{x:aTx≤b}is ahalfspace. Thus, the feasible region of an (inequality constrained) LP is an intersection ofhalfspaces. The set obtained by taking the intersection of halfspaces is called apolyhedron.aT1x≥b1aT2x≥b2aT3x≥b3aT4x≥b4aT5x≥b5P
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Introduction to Linear Programming2Polyhedra are often good Frst approximations to more complicated feasible sets.FLIn the Fgure above,Lis alinear approximationfor thenonlinearfeasible setF. Therefore, linear programstend to provide a good approximation to more complicated optimization problems.Second part of the answer to why one should bother with LPs is that LPs can be solved rather efficiently.1.2Geometrical solution of LPsContinuing further with this geometrical approach, lets try to examine optimal solutions of LPs. Lets startwith a one-dimensional LPmaximizecxsubject toa1≤x≤a2The optimal solutionx∗of this LP is given by(a)c>0:x∗=a2(b)c=0: anyx∈[a1,a2] optimal ... in particularx=a12(c)c<0:x∗=a1Moral: The optimal solution is alway at the boundary of the feasible set.Push this geometric approach a little further and consider general (inequality constrained) LPs.maximizecTx,subject tox∈P(polytope)±or a given scalarz,thesetofpointsxwith costcTx=zis a plane perpendicular toc(ahyperplane).Thus, an algorithm for solving the LP is to increasez, or equivalently slide the hyperplane in thecdirection,until the plane is at the boundary of the feasible region.Lets try our algorithm on the LPmaximizec1x1+c2x2,such that−x1+x2≤1,x∈R2+The feasible region of this LP is given byP=nx:−x1+x2≥0,x1≥02≥0o
Introduction to Linear Programming3x1x2c1c2c3c4The results of our algorithm on variouscvectors is given by(a)c=(
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Software for teaching and learning of linear equations with two unknowns. Solutions by the elimination and substitution methods. Worked examples of solutions with each method shown side by side for easy comparison. Explanations provided for important steps in the solution to help understanding. It can generate hundreds of sums (up to 200 per session) for practice, but users may also choose to key in their own equations.
The interactive guide can help the user do his homework if he keys in the sums. Marks automatically and can print a detailed report after each session. Version 3 includes New feature that allows two player to compete with each other on one computer Synchronised start for both players Each player can do the sums at his own pace Prints two separate detailed reports, each showing the sums done by the individual player Each report includes the scores of both players and the winners name Other minor improvements and bug fixes.
Limitations:
30-use trial
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Other Windows Software of Developer «EleMaths Software»:AlgeBasics Software for the teaching and learning of basic algebra including the addition, subtraction, multiplication, and division of algebraic terms, algebraic fractions and indices. It includes order of calculations, simple indices and removal of brackets. The tuFractionsPractice Software for the teaching and learning of fractions and decimals. Capable of generating thousands of different sums for practice. Four different levels of difficulty on fractions cater for students at different stages of the topic. The Basic level includesPanIntegers Software for the teaching and learning of integers including addition, subtraction, multiplication, division, order of calculations, removal of brackets, and simple indices. Tutorial sessions include graphics and animation for easy understanding of the conMoNooN Grapher From the Developer: ""MoNooN2DR v1.1 is the simple software that draws and shows the graph of the 2-dimensional mathematical functions in rectangular coordinates. If user input equations, choose colors, and choose line types, user can see the graph. MoNooNClinTools ClinTools is a statistical suite of evidence-based decision making tools. It augments traditional statistical software by conducting specialized analyses that are required for evidence-based practice, whether that be in the health system, education system,
FisherFaces for Face Matching FisherFaces for Face Matching 1 Matlab Source Code. We develop a face recognition algorithm which is insensitive to large variation in lighting direction and facial expression. Taking a pattern classification approach, we consider each pixel in an image as
FC Win FC-Win (tm) is a front-end program for FortranCalculus (tm). The FortranCalculus (FC) language is for math modeling, simulation, and optimization. FC is based on Automatic Differentiation that simplifies computer code to an absolute minimum; i.e., a math
The Ultimate Calculator This is The Ultimate Calculator designed to suit all your needs. It has abilities to calculate the area, volume of all solids like cube, cuboid. Compute area and volume in the fill version.Requirements:Windows 95/98/2000/XPLimitations:See description | 677.169 | 1 |
MAT 150/ Quiz #3/ 15 pts./ 2.53.1,Ch3Skills/Sp13 NAME
Show all work!
1) Let it) be given by the table below. Use successive averag_rates of change to determine if the graph of
this function will be concave up or concave down. (2 pts)
2) in fall 2008 t
MAT 150/ Exam #3/ 100 pts./ Ch5&6,10.1/ (Sp13)
4 bonus point included
NAME _
Show all work! Try to relax! Go with first instincts!
1)
Let f(x) be given by the graph below. Sketch the graph of each transformation listed on the grid provided.
Write in w
MAT 150/ Exam #1/ 100 pts./ Ch1-2.3/ Sp13
3 bonus points included
NAME _
Show all work! Smile and relax! Do not change any answer unless you are 100% sure the change is correct.
First instincts are nearly always correct!
1)
The following table shows t
MAT 150/ Exam #2/ 100 pts./ 2.5,Ch3,Ch4/ Sp13
1 bonus point included + bonus problem
NAME
Show all work! Remember to go with first instincts as they are almost always correct!
1)
Without your calculator, match each equation to one of the graphs gi
Showing 1 to 3 of 3
I would recommend this course because the instructor is very helpful and when there is a misunderstanding among the students, he finds another way for us to interpret the material. He also connects real life situations to the mathematical problems and shows us that the material s useful.
Course highlights:
I learned how to effectively but also efficiently use math to solve everyday math issues I will inevitably come into contact with. From basic financial lessons to even understanding proportions I have gained a better appreciation for math.
Hours per week:
6-8 hours
Advice for students:
I would advise prospective students to practice the problems and ask questions regardless of the feeling of embarrassment.
This professor was excellent at explaining ways to remember the various kinds of equations. On lab days, students are encouraged to call out which math problems they were unable to complete due to difficulty and the professor showed the students how to work it out on the board.
Course highlights:
I learned how to calculate compound interest and to solve logarithms. The highlights were being able to apply real life events to math problems.
Hours per week:
3-5 hours
Advice for students:
Make sure to stay on top of the online assignments, the more problems that you get right, the better it will be for your overall grade at the end of the semester.
Course Term:Spring 2016
Professor:Dontoe Tomei
Course Required?Yes
Course Tags:Math-heavyMany Small AssignmentsA Few Big Assignments
May 12, 2016
| Would recommend.
This class was tough.
Course Overview:
I took College Algebra online due to my work schedule, and because of this it was extremely challenging. Professor Rodski did a wonderful job of communicating regularly with the class and offering support. I realized early on that the course would be difficult, and implemented a strict study schedule, which I adhered to throughout the semester. This was the main factor that contributed to my overall success in the class. My main recommendation would be to learn and make use of the graphing calculator. I was not familiar enough with the functions to utilize it properly.
Course highlights:
The obvious educational objective was to learn and apply complex math, including linear, quadratic, polynomial, rational, exponential, logarithmic and piecewise functions; systems of equations; and an introduction to analytic geometry. . Outside of the course objectives I learned that I can overcome challenges and succeed.
Hours per week:
9-11 hours
Advice for students:
My advice would be to set a strict study schedule and adhere to it. Dedicate time every day, free of interruptions and distractions, to practicing problems. I also suggest regular communication with your instructor, and most certainly ask for help! Learn to use the graphing calculator that is allowed in the course as it will be infinitely helpful. | 677.169 | 1 |
The Facts On File Dictionary of Mathematics, Fourth Edition
Encompassing every mathematical term and concept of interest, The Facts On File Dictionary of Mathematics, Fourth Edition conveys information to students and general readers in a proven, accessible format. This fourth edition contains approximately 320 new entries, dozens of new line illustrations, new pronunciation symbols, a list of websites, and a bibliography. Current entries and back matter have been revised as needed. Relating significant information in a non-specialist manner, this dictionary is an invaluable resource.
John Daintith is editor and contributor to many reference books, including eight recent dictionaries in the Facts On File Science Dictionary series.
Richard Rennie was formerly a research worker in theoretical physics. He is a science writer and editor and has contributed to a number of reference books, including the Oxford Dictionary of Physics and The Facts On File Dictionary of Atomic and Nuclear Physics, from Facts On File. He lives in Cambridge, England. | 677.169 | 1 |
Publisher Description
3DMath Explorer is a computer program that pilots 2D and 3D graphs of mathematical functions and curves in unlimited graphing space. It has many useful feature such as;
1-3D curve ploting in real time,
2-perspective drawing,
3-graph scaling (zooming),
4-active graph rotation,
5-fogging effect,
6-cubic draw,
7-unlimited space ploting,
8-four view plot screen,
9-auto rotate animation,
10-single coordinate system defination,
11-additional parameter function and loop variable definations,
12-3D surfaces with 3D volumes
13-curve line length and surface area calculation,
14-full control on all graphical elements,
15-drawing many curves in the same screen,
16-working with many graph screen in the same time,
3DMathExplorer is a very useful program for students to make experiment and observation, for teachers to teach the subjects more interesting and comfortable, for writers to select graphs for their books within more suitable, beautiful and comprehensible graphs, and for all people that interests in this subject to use with ease. | 677.169 | 1 |
Math Lab
The RVC Math Lab is staffed by math faculty and serves all RVC math students. The study side, JCSM-0210, has some computers and tables for studying and doing math homework. The classroom side, JCSM-0212, has an additional 24 computers for computer-assisted math classes and for students to use for math-related work such as online homework. | 677.169 | 1 |
Summary and Info
Barnett, Ziegler, Byleen, and Sobecki's "College Algebra with Trigonometry" text is designed to be user friendly and to maximize student comprehension by emphasizing computational skills, ideas, and problem solving as opposed to mathematical theory. The large number of pedagogical devices employed in this text will guide a student through the course. Integrated throughout the text, students and instructors will find Explore-Discuss boxes which encourage students to think critically about mathematical concepts. In each section, the worked examples are followed by matched problems that reinforce the concept being taught. In addition, the text contains an abundance of exercises and applications that will convince students that math is useful. There is a MathZone site featuring algorithmic exercises, videos, and other resources accompanies the text
More About the Author
Thomas William Hungerford (March 21, 1936 – November 28, 2014) was an American mathematician who worked in algebra and mathematics education. | 677.169 | 1 |
The 2 Sigma Problem: In 1984, Benjamin Bloom discovered a method for drastically improving educational efficiency, that improved results by a factor of two standard deviations (two sigma). This is known as Bloom's 2 Sigma Problem.
Bloom's method means that the "average" student within a given class could now perform better than 49 out of every 50 students within a traditional classroom setting.
What was his secret sauce?
It is the combination of two education approaches: mastery learning and personalized instruction.
productive disposition—habitual inclination to see mathematics as sensible, useful, and worthwhile, coupled with a belief in diligence and one's own efficacy.
Prescription
Discover Innovative strategies that compliment your learning style and needs. Our progress-based curriculum adapts to your performance, ensuring that you are implementing the most optimal methods at all times! Foundation:
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It can identify your academic strengths and weaknesses and understand your learning style. It is the ultimate guide for discovering the most effective methods and strategies that make you learn faster and succeed in less time.
The Smart Scoring System is also a learning tool. Its User Interface emphasizes tactile inputs, helping you to increase your concentration and retention. It measures your time management and cognitive efficiency.
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The Gateway Series: In today's digital world, on-site one-to-one tutoring is often not feasible and in many cases not desirable. We therefore offer a cloud-based online service that connects students to tutors, especially for those who cannot make it on-site. Regardless, the student gets one-to-one mentoring. | 677.169 | 1 |
About this product
Description
Description
Maximize student use of the TI-Nspire while processing and learning algebraic concepts with this resource. Lessons provided delve into the five environments of the TI-Nspire including calculator, graphs and geometry, lists and spreadsheets, tes, and data analysis. | 677.169 | 1 |
discrete mathematics
This text aims to unify and inter-relate mathematical topics and explain how to design, run and analyse better algorithms. Many of the less common algorithms are included and are actually run so that students can see the importance of working through each step of an algorithm by hand.Read more... | 677.169 | 1 |
A First Course in Mathematical Modeling
4.11 - 1251 ratings - Source
Offering a solid introduction to the entire modeling process, A FIRST COURSE IN MATHEMATICAL MODELING, 4th Edition delivers an excellent balance of theory and practice, giving students hands-on experience developing and sharpening their skills in the modeling process. Throughout the book, students practice key facets of modeling, including creative and empirical model construction, model analysis, and model research. The authors apply a proven six-step problem-solving process to enhance studentsa€™ problem-solving capabilities -- whatever their level. Rather than simply emphasizing the calculation step, the authors first ensure that students learn how to identify problems, construct or select models, and figure out what data needs to be collected. By involving students in the mathematical process as early as possible -- beginning with short projects -- the book facilitates their progressive development and confidence in mathematics and modeling. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.To facilitate an early initiation of the modeling experience, the first edition of this
text was designed to be taught ... In the third edition we included solution
methods for some simple dynamical systems to reveal their long-term ... The
student investigates meaningful and practical problems chosen from common
experiences encompassing many academic disciplines, including the
mathematical sciences, operations research, engineering, and the management
and life sciences. xiii Thisanbsp;...
Title
:
A First Course in Mathematical Modeling
Author
:
Frank Giordano, William P. Fox, Steven Horton, Maurice Weir
Publisher
:
Cengage Learning - 2008-07-03
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
6.EE.9 Review Practice of Function Tables, Rules, and Figure Patterns
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
0.57 MB | 4 pages
PRODUCT DESCRIPTION
Aligned to Common Core State Standards 6.EE.9
This 2-page document with answer key can be used as practice or a review / study guide of function tables, rules, and figure patterns.
The following are required of students:
Complete function tables using the given rule.
Find the rule and then complete the function table.
Use a word problem to create a function table and find the rule (containing a constant, i.e. y = 20 + 15x), and then answer a question about a later term. For example, how many months would it take for Sue to save up $214?
Use a pattern of figures to create a function table, find the rule, and give the nth term.
There are 8 total problems in the document. Please download the preview to see a sample | 677.169 | 1 |
Whether you're brushing up on pre-Algebra concepts or on your way toward mastering algebraic fractions, factoring, and functions, CliffsQuickReview Algebra I can help. This guide introduces each topic, defines key terms, and carefully walks you through each sample problem step-by-step. In no time, you'll be ready to tackle other concepts in this book such as
Equations, ratios, and proportions
Inequalities, graphing, and absolute value
Coordinate Geometry
Roots and radicals
Quadratic equations
CliffsQuickReview Algebra I acts as a supplement to your textbook and to classroom lectures. Use this reference in any way that fits your personal style for study and review — you decide what works best with your needs. Here are just a few ways you can search for topics:
Use the free Pocket Guide full of essential information
Get a glimpse of what you'll gain from a chapter by reading through the Chapter Check-In at the beginning of each chapter
Use the Chapter Checkout at the end of each chapter to gauge your grasp of the important information you need to know
Test your knowledge more completely in the CQR Review and look for additional sources of information in the CQR Resource Center
Use the glossary to find key terms fast.
With titles available for all the most popular high school and college courses, CliffsQuickReview guides are a comprehensive resource that can help you get the best possible grades. | 677.169 | 1 |
Carol Schumacher, Books, Science and Nature, Chapter Zero: Fundamental Notions Of Abstract Mathematics Books>Science and Nature, Pearson Education
Schumacher, Carol:
[ED: Taschenbuch], [PU: ADDISON WESLEY PUB CO INC], This
Versandfertig in über 4 Wochen, [SC: 0.00], Neuware, gewerbliches Angebot
This book?is designed for the sophomore/junior level Introduction to Advanced Mathematics course. Written in a modified R.L. Moore fashion, it offers a unique approach in which?readers construct their own understanding. However, while?readers ? Logic, Sets, Induction, Relations, Functions, Elementary Number Theory, Cardinality, The Real Numbers ? For all readers interested in abstract mathematics. education and reference,mathematics,science and math,science and math,textbooks Mathematics, Pearson
Chapter Zero: Fundamental Notions of Abstract Mathematics Schumacher, Carol, Addison Wesley Longman
Schumacher, Carol
Titel:
Chapter Zero: Fundamental Notions of Abstract Mathematics
ISBN-Nummer:
0201437244
Chapter Zero is designed for the undergraduate level Introduction to AdvancedMathematics course. Written in a modified R.L. Moore fashion, it offers a unique approach in which students construct their own understandings. However, while students are called upon to write their own proofs, they are also encouraged to work in groups. There are few finished proofs contained in the text, but the author offers 'proof sketches' and helpful technique tips to help students as they develop their proof writing skills. This book is most successful in a small, seminar style class. *NEW - Coverage of Isomorphisms and Graph Theory. *Exercise sections have been improved by smoothing out the grade of difficulty. *Proof Sketches. Woven throughout the early chapters of the text, these sketches assist students with proof techniques. *Logic is used as a tool for analyzing the content of mathematical assertions and for constructing valid mathematical proofs. *Rigorous axiomatic treatment of set theory is introduced in Appendices A and B (which are written in the same style as the texts chapters.) | 677.169 | 1 |
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Unformatted text preview: SHENS CLASS NOTES-191 1 Chapter Two The Language of Mathematics In this chapter, we discuss sets, functions, and sequences. These are very fundamental notions or language to model mathematical problems as well as tools for solving mathematical problems. 2.1 Sets Definition 2.1 A set is an unordered collection of objects. The objects in a set are also called elements or members. Example 1 A = {1, 2, 3, 4} is a set that contains four numbers. Example 2 V= { a , e , i , o , u } is the set of all vowels in English alphabet. One way to define a set is to list all members of the set as shown in Example 1 or 2. SHENS CLASS NOTES-191 2 Another way is to use the set builder notation. For example, R = { x | x is a real number}, B = { x | x is a positive, even integer}. The following are some commonly used sets: N = {0, 1, 2, 3, }, the set of natural numbers. Z = {, -1, 0, 1, 2, }, the set of integers. Z + = {1, 2, 3, }, the set of positive integers. Q = { p / q | p Z , q Z , q 0}, the set of rational numbers. R , the set of real numbers. { } or , the empty set. The size of a set S is called its cardinality , denoted by |S|. If S is a finite set, consisting of n elements, than | S | = n . For example, |{1, 2, 3, 4}| = 4. Definition 2.2 Two sets are equal if and only if they have the same elements. SHENS CLASS NOTES-191 3 Example 3 (2.1.1) Let A = { x | x 2 + x-6 = 0} and B = {2, -3}, prove A = B. Proof. If x A, then x satisfies x 2 + x- 6 = 0, which means that x is a root of the equation. Notice that x 2 + x- 6 = ( x - 2)( x + 3). This equation has two roots, 2 and -3. Therefore, x must be either 2 or -3. Obviously, any x that is in set A must be in set B also. Conversely, any element of set B must be either 2 or -3 which are the roots of x 2 + x- 6 = 0. It must belong to set A also. Therefore, A = B. Definition 2.3 Suppose that X and Y are sets. If every element of X is an element of Y, we say that X is a subset of Y , denoted by X Y . Y is also called a super set of X . Example 4 (2.1.2) C = {1, 2}, A = {1, 2, 3, 4}, C A. We see that A B if and only if x ( x A x B ). Definition 2.4 Suppose that X is a subset of Y but X Y , then SHENS CLASS NOTES-191 4 we say that X is a proper subset of Y . denoted by X Y . Definition 2.5 The power set of a set X is the set of all subsets of the set X , denoted by P ( X ). Example 5 (2.1.5) If A = { a , b , c }, then P ( A ) = { , { a }, { b }, { c }, { a , b }, { b , c }, { a , c }, { a , b , c }}. Theorem 1 (2.1.6) If | X | = n , then | P ( X )| = 2 n . Proof. Let P ( n ): If | X | = n , then | P ( X )| = 2 n . We prove the claim P ( n ) by induction on n ....
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Product Overview
Numerical Methods for Roots of Polynomials - Part II along with Part I (9780444527295) covers most of the traditional methods for polynomial root-finding such as interpolation and methods due to Graeffe, Laguerre, and Jenkins and Traub. It includes many other methods and topics as well and has a chapter devoted to certain modern virtually optimal methods. Additionally, there are pointers to robust and efficient programs. This book is invaluable to anyone doing research in polynomial roots, or teaching a graduate course on that topic.
First comprehensive treatment of Root-Finding in several decades with a description of high-grade software and where it can be downloaded
Offers a long chapter on matrix methods and includes Parallel methods and errors where appropriate | 677.169 | 1 |
Resource Added!
Type:
Description:
Relationships Between Quantities and Reasoning with Equations
Project: Farmer's Market
Subjects:
Mathematics > General
Education > General
Education Levels:
Grade 6
Grade 7
Grade 8
Grade 9
Grade 10
Keywords:
algebra linear equations variables inequalities exponents
Language:
English
Access Privileges:
Members
License Deed:
Creative Commons Attribution Non-Commercial
Collections:
None
Update Standards?
CCSS.Math.Content.HSN-Q.A.3: Common Core State Standards for Mathematics
Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.
CCSS.Math.Content.HSA-CED.A.1: Common Core State Standards for Mathematics
Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
CCSS.Math.Content.HSA-CED.A.2: Common Core State Standards for Mathematics
Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.CCSS.Math.Content.HSA-REI.B.3: Common Core State Standards for Mathematics
Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
MA.9-12.CCSS.Math.Content.HSN-Q.A.3: Mathematics
Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.
MA.9-12.CCSS.Math.Content.HSA-CED.A.1: Mathematics
Create equations and inequalities in one variable and use them to solve problems.
MA.9-12.CCSS.Math.Content.HSA-CED.A.2: Mathematics
Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
MA.9-12.CCSS.Math.Content.HSA-REI.A.1: Mathematics
Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
MA.9-12.CCSS.Math.Content.HSA-REI.B.3: Mathematics
Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters20.
Component Ratings:
Technical Completeness: 3 Content Accuracy: 3 Appropriate Pedagogy: 2
Reviewer Comments:
This resource is a group project that is a culminating activity based on four lessons on writing and solving equations and inequalities. The activity includes a list of materials needed, instructions and an example project. Students may need some background instruction on economics (profit, loss, mark-up) The project leaves a bit of room for teacher modifications. Overall, a useful resource that a teacher can use to reinforce multiple lessons.
Not Rated Yet.
Table of Contents
Unit 1 Project: Farmer's Market
In this project students apply the skills they learned and practiced in the first four lessons. Students apply their ability to write, interpret, and translate various forms of linear equations and inequalities, and use them to solve problems. Students apply their mastery of solving linear equations and apply related solution techniques and the laws of exponents to the creation and solution of simple exponential equations. These applications are grounded on understanding quantities and on relationships between them. | 677.169 | 1 |
East Coast Computer Algebra Days
The East Coast Computer Algebra Day (ECCAD) is an annual one-day meeting for those interested in computer algebra and symbolic mathematical computation. It provides opportunities to learn and to share new results and current work in progress. The schedule includes prominent invited speakers along with contributed posters and software demonstrations. Plenty of time is allowed for unstructured interaction among the participants. Researchers, teachers, students, and users of computer algebra are all welcome!
Previous ECCADs
Fuller history can be found here (conference sites, organizers, dates, invited speakers and their titles). | 677.169 | 1 |
Offering a solid introduction to the entire modeling process, A FIRST COURSE IN MATHEMATICAL MODELING, 5th Edition delivers an excellent balance of theory and practice, and gives you relevant, hands-on experience developing and sharpening your modeling skills.
This text provides an introduction to the entire modeling process. Throughout the book, students practice key facets of modeling, including creative and empirical model construction, model analysis, and model research. The authors apply a proven six-step problem solving process to enhance a student's problem solving capabilities.
Mathematical Analysis (often called Advanced Calculus) is generally found by students to be one of their hardest courses in Mathematics. This text uses the so-called sequential approach to continuity, differentiability and integration to make it easier to understand the subject.Topics that are generally glossed over in the standard Calculus courses are given careful study here.
How many dimensions does our universe require for a comprehensive physical description? In 1905, Poincare argued philosophically about the necessity of the three familiar dimensions, while recent research is based on 11 dimensions or even 23 dimensions. | 677.169 | 1 |
ISBN 9789351419457
ISBN-10
9351419452
Binding
Paperback
Edition
5th
Number of Pages
986 Pages
Language
(English)
Subject
Entrance Exam Preparation
Cracking an examination like GATE (Graduate Aptitude Test in Engineering) demands extensive knowledge and revision of the concepts covered under the GATE 2015 curriculum. GATE tests the in-depth knowledge and practical knowledge of varied concepts. Arihant understand the needs of the students which require comprehensive chapter-wise study for in-depth understanding followed by thorough practise of the syllabi to master the concepts.
The present GATE tutor for GATE computer science and Information Technology has been divided into three sections namely general aptitude, engineering, mathematics and computer science and information technology, each sub-divided into number of chapters. The general aptitude section covers verbal ability, critical reasoning, numerical ability and previous years general aptitude questions. The Engineering Mathematics section has been divided into linear algebra, calculus, differential equations, complex variables, probability and statistics, numerical methods and transform theory whereas the Computer Science and Information Technology section has been divided into 12 chapters namely digital logic design, computer organisation and architecture, data structure, programming in C, design analysis and algorithm, theory of computation, compiler design, operating system, database management system, software engineering, computer network and web technology.
The chapters cover unit wise and topic-wise collection of theory and question answers. The whole coverage of the syllabi in the book has been kept identical to the format of the exam including relevant study material and variety of questions with their complete solutions. Each topic in the book has been categorised into numerous sub divisions. Also each unit ends with intro exercise for practising the concepts discussed in the unit. The exercises contain questions as per the GATE pattern i.e. 1 mark questions, 2 marks questions and common data and linked answer questions. It has been kept into account that the concepts illustrated in the book are self-sustaining and the students do not feel the need to go through any other book for any missing topic. The book also contains last four years (2011 - 2014) solved papers of original online GATE Computer Science and Information Technology to give the aspirants an insight into the trends of questions asked in GATE Examination. This book contains focused study material for General Aptitude and Engineering Mathematics. For final and thorough revision, three practise sets for GATE Computer Science and Information Technology with full solutions have also been given in the book.
As the book contains ample study and practise material, it for sure will help the aspirants achieve success in the upcoming GATE 2015 computer science and information technology exam. readers TABLE OF CONTENTS
Solved Paper 2014 (Online Set 1)
Solved Paper 2014 (Online Set 2)
Solved Paper 2013
Solved Paper 2012
Solved Paper 2011
General Aptitude
1. Verbal Ability
2. Critical Reasoning
3. Numerical Ability
4. Previous Years General Aptitude Questions Asked in Previous Years Examination
Engineering Mathematics
1. Linear Algebra
2. Calculus
3. Differential Equations
4. Complex Variables
5. Probability and Statistics
6. Numerical Methods
7. Transform Theory
Computer Science and Information Technology
1. Digital Logic Design
2. Computer Organisation and Architecture
3. Data Structure
4. Programming in C
5. Design Analysis and Algorithm
6. Theory of Computation
7. Compiler Design
8. Operating System
9. Database Management System
10. Software Engineering
11. Computer Network
12. Web Technology
Practise Sets (1 - 3) | 677.169 | 1 |
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Synopses & Reviews
Publisher Comments
Master topology with Schaum's--the high-performance study guide. It will help you cut study time, hone problem-solving skills, and achieve your personal best on exams!
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Confusing Textbooks? Missed Lectures? Not Enough Time?
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Fully compatible with your classroom text, Schaum's highlights all the important facts you need to know. Use Schaum's to shorten your study time-and get your best test scores!
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Algebra Connections: Mathematics for Middle School Teachers
Description
For one-semester undergraduate courses in algebra for the middle grades. Strong mathematics performance in the middle grades is more important than ever--and teachers entering the field need to prepare for this endeavor in new and innovative ways. This new approach introduces some basic concepts of number theory and modern algebra that underlie middle grade arithmetic and algebra, with a focus on collaborative learning combined with extensive in-class and out-of-class assignments. The primary goal is to help future teachers (both in-service and pre-service) gain a fundamental understanding of the key mathematical ideas that they will be teaching so that, in turn, they can help their students learn important mathematics. This text presents is designed to equip future middle grade mathematics teachers with the skills needed for teaching NCTM (National Council of Teachers of Mathematics) Standards-based curricula. Throughout the text, the reader will find a number of Classroom Connections, Classroom Discussions, and Classroom Problems.
These instructional components are designed to deepen the connections between the college-level abstract algebra and number theory the students are studying now and the algebra they will teach. Other titles in the Prentice Hall Connections in Mathematics Courses for Teachers include: * Geometry Connections: Mathematics for Middle School Teachers * Algebra Connections: Mathematics for Middle School Teachers * Data and Probability Connections: Mathematics for Middle School Teachers * Calculus Connections: Mathematics for Middle School Teachers | 677.169 | 1 |
Second Edition of this engaging text for the one-semester finite mathematics course uses intriguing, real-world applications to capture the interest of business, economics, life science, and social science majors. This practical approach to mathematics, along with the integration of graphing calculators and Excel spreadsheet explorations, exposes students to the tools they will encounter in future careers. A wealth of pedagogy includes the following distinctive features: detailedWorked-out Examples with Annotationshelp students through more challenging concepts;Practice Problemsare offered to help students check their understanding of concepts presented in the examples;Section Summariesbriefly restate essential formulas and key concepts;Chapter Summary with Hints and Suggestionsunify chapter themes, give specific reminders, and reference problems in the review exercises suitable for a practice test; andCumulative Review Exercisesappear at the end of groups of chapters to reinforce previously learned concepts and skills. Graphing Calculator ExamplesandExerciseslocated throughout the text explore new topics, guide students through "messy" calculations, or show technology pitfalls. These may be omitted without disrupting the flow or cohesion of the text. Application Previewsplace mathematics in a real-world context and motivate students' interest in the material. Annotationsbeside many formulas and solution steps emphasize the importance of being able to "read mathematics" by restating much of the mathematics in words.
Table of Contents
Note: Each chapter concludes with a Summary with Hints and Suggestions | 677.169 | 1 |
This is one in a series of manuals for teachers using SMSG high school supplementary materials. The pamphlet includes commentaries on the sections of the student's booklet, answers to the exercises, and sample test questions. Topics covered include sets, definition and graph of a function, constant, linear and absolute-value functions, composition, inversion, one-to-one functions, ordered pairs, circular motion, graphs of sine and cosine, angles, vectors, addition formulas, tables of circular functions, and waves. (MP) | 677.169 | 1 |
1 MATLAB FUNDAMENTALS 1 Dr. Vyas Overview: • About Matlab • Basic Matlab operations (open/exit) • Basic arithmetic with Matlab • Basic algebra with Matlab About Matlab 1. The software name Matlab is a truncated version of, Mat rix Lab oratory. 2. Matlab is a mathematical software package that serves as a programming language as well. 3. Matlab was (and still is) predominantly used by engineers who design/analyze the controlling process of physical systems. However, since the late 90s, Matlab is increasingly used by engineers/scientists from other disciplines as well. A1. Starting Matlab from Mac OS X in classroom Ewing 207 • Click on the Matlab icon on the Launch bar at the bottom of the screen A2. Starting Matlab from Mac OS X at public computing site • Click on the Matlab icon on the Launch bar at the bottom of the screen if present on the Launch bar, else • Use Finder to locate the Matlab icon and double click on that icon. A3. Starting Matlab from Windows at public computing site • Double click on Matlab icon shortcut on the desktop screen if you see one • Else, left click on Start (lower left hand corner of screen), select All Programs and click on the Matlab icon/shortcut in the list of programs. • Please be patient , some Windows machines at public computing sites take anywhere between 5 to 10 minutes before Matlab is launched. A4. Starting Matlab from Sun OS on public computing sites • On the top taskbar on the screen, left click on Applications . Select Programming and then select and left click on xterm on Strauss . • Thereafter, type matlab at the "strauss.udel.edu%" prompt • Alternatively, you can also select "Other" from taskbar and proceed to Matlab. Quitting Matlab To exit Matlab there are following options: 1. type exit at the Matlab command prompt, >> 2. From File menu on the toolbar select Exit Matlab 3. Close the Matlab program window via the "x" button on the top right 4. Use the keyboard shortcut, Ctrl+Q (Not available in Mac OS X)
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2 Following are the screenshots that are intended to aid students in accessing Matlab on Unix terminals. This is the upper right half of the Matlab program window launched from Sun OS. Note the input commands are written in Command Window. Also, note the Current Directory indicated in the tool bar and the contents are displayed in the Current Directory window . Matlab program files must be placed or saved in the same directory as indicated in the toolbar for programs to work .
3 To know the current directory you are in, type >> pwd ans = C:\Program Files\MATLAB704\work The above answer is the way it is because I am using Matlab on Windows XP to prepare notes. You will have a different answer depending on whether you are using Linux or Mac OS X. This shows the lower left hand side of the Matlab program window. Note the
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Faces 484pp. Index. Wear boards.
Priceless Books
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$13 primarily for the Liberal Arts or Survey of Mathematics course, this straightforward, non-technical introduction to mathematics shows students how to think like mathematicians. The text focuses on the "four faces of mathematics"-solving problems, finding order, building models, and abstracting from the familiar-that form the four parts of the text. | 677.169 | 1 |
step-by-step guide to computing and graphics in regression analysis In this unique book, leading statisticians Dennis Cook and Sanford Weisberg expertly blend regression fundamentals and cutting-edge graphical techniques. They combine and up- date most of the material from their widely used earlier work, An Introduction to Regression Graphics, and Weisberg's Applied Linear Regression; incorporate the latest in statistical graphics, computing, and regression models; and wind up with a modern, fully integrated approach to one of the most important tools of data analysis. In 23 concise, easy-to-digest chapters, the authors present:?
This text book contains a collection of six high-quality articles. In particular, this book is devoted to Linear Mathematics by presenting problems in Applied Linear Algebra of general or special interest. | 677.169 | 1 |
Wadsworth Publishing Company Belmont, California
A division of Wadsworth, Inc.
ABOUT THE COVER: "Dirichlet Problem" (Miles Color Art A25) is the work of Professor
E. P. Miles, Jr., and associates of Florida State University using an InteColor 80501 computer. Programming by Eric Chamberlain and photograph by John Owen. The function graphed is the discrete limiting position for the solution by relaxation of the heat distribution in an insulated rectangular plate with fixed temperature at boundary positions. This is an
end-position photograph following intermediate positions displayed as the solution converges from an assumed initial average temperature to the ultimate (harmonic function) steady-state temperature induced by the constantly maintained boundary conditions.
All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transcribed, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher, Wadsworth Publishing Company, Belmont, California 94002, a division of Wadsworth, Inc.
Preface
This book is designed for an introductory, one-semester or one-year course in differential equations, both ordinary and partial. Its prerequisite is elementary calculus. Perusal of the table of contents and the list of applications shows that the book contains the theory, techniques, and applications covered in the traditional introductory courses in differential equations. A major feature of this text is the quantity and variety of applications of current interest in physical, biological, and social sciences. We have furnished a wealth of applications from such diverse fields as astronomy, bioengineering, biology, botany, chemistry, ecology, economics, electric circuits, finance, geometry, mechanics, medicine, meteorology, pharmacology, physics, psychology, seismology, sociology, and statistics. Our experience gained in teaching differential equations at the elementary, intermediate, and graduate levels at the University of Rhode Island convinced us of the need for a book at the elementary level which emphasizes to the students the relevance of the various equations to which they are exposed in the course. That is to say, that the various types of differential equations encountered are not merely the product of some mathematician's imagination but rather that the equations occur in the course of scientific investigations of real-world phenomena. The goal of this book, then, is to make elementary differential equations
more useful, more meaningful, and more exciting to the student. To accomplish this, we strive to demonstrate that differential equations are very much "alive" in present-day applications. This approach has indeed had a satisfying effect
in the courses we have taught recently. During the preparation and class testing of this text we continuously kept in mind both the student and the teacher. We have tried to make the presentation direct, yet informal. Definitions and theorems are stated precisely and rigorously, but theory and rigor have been minimized in favor of comprehension of technique. The general approach is to use a larger number of routine examples to illustrate the new concepts, definitions, methods of solution, and theorems. Thus, it is intended that the material will be easily accessible to the student. Hopefully the presence of modern applications in addition to the traditional applications of geometry, physics, and chemistry will be refreshing to the teacher. Numerous routine exercises in each section will help to test and strengthen the student's understanding of the new methods under discussion. There are over 1600 exercises in the text with answers to odd-numbered exercises provided. Some thought-provoking exercises from The American Mathematical
X
Monthly, Mathematics Magazine, and The William Lowell Putnam Mathematics Competition are inserted in many sections, with references to the source. These should challenge the students and help to train them in searching the literature. Review exercises appear at the end of every chapter. These exercises should serve to help the student review the material presented in the chapter. Some of the review exercises are problems that have been taken directly from physics and engineering textbooks. The inclusion of such problems should further emphasize that differential equations are very much present in applications and that the student is quite apt to encounter them in areas other than mathematics. Every type of differential equation studied and every method presented is illustrated by real-life applications which are incorporated in the same section (or chapter) with the specific equation or method. Thus, the student will see immediately the importance of each type of differential equation that they learn how to solve. We feel that these "modern" applications, even if the student only glances at some of. them, will help to stimulate interest and enthusiasm toward the subject of differential equations specifically and mathematics in general. Many of the applications are integrated into the main development of ideas, thus blending theory, technique, and application. Frequently, the mathematical model underlying the application is developed in great detail. It would be impossible in a text of this nature to have such development for every application cited. Therefore, some of the models are only sketched, and in some applications the model alone is presented. In practically all cases, references are given for the source of the model. Additionally, a large number of applications appear in the exercises; these applications are also suitably referenced. Consequently, applications are widespread throughout the book, and although they vary in depth and difficulty, they should be diverse and interesting enough to whet the appetite of every reader. As a general statement, every application that appears, even those with little or no detail, is intended to illustrate the relevance of differential equations outside of their intrinsic value as mathematical topics. It is intended that the instructor will probably present only a
few of the applications, while the rest can demonstrate to the reader the
relevance of differential equations in real-life situations. The first eight chapters of this book are reproduced from our text Ordinary Differential Equations with Modern Applications, Second Edition, Wadsworth Publishing Co., 1981.
provides a thorough introduction to its respective topic. We feel that the chapters on difference equations and partial differential equations are more
extensive than one usually finds at this level. Our purpose for including these topics is to allow more course options for users of this book. We are grateful to Katherine MacDougall, who so skillfully typed the manuscript of this text. A special word of gratitude goes to Professors Gerald Bradley, John Haddock, Thomas Hallam, Ken Kalmanson, Gordon McLeod, and David Wend,
Preface
xl
who painstakingly reviewed portions of this book and offered numerous valuable suggestions for its improvement.
Thanks are also due to Dr. Clement McCalla, Dr. Lynnell Stern, and to our students Carl Bender, Thomas Buonanno, Michael Fascitelli, and especially Neal Jamnik, Brian McCartin, and Nagaraj Rao who proofread parts of the material and doublechecked the solutions to some of the exercises. Special thanks are due to Richard Jones, the Mathematics Editor of Wadsworth Publishing Company for his continuous support, advice, and active interest in the development of this project.
N. Finizio G. Ladas
that is.. however. a2u/ate and a2u/axe are the second partial derivatives of the function u(t.
y' + xy = 3
(1)
y" + 5y' + 6y = cos x
Y" _ (1 + Y'2)(x2 + Y2)
(2) (3)
(4)
d2u82u _ 0
ate
axe
are differential equations. or can be put. of the function y(x) with respect to x.." the term "differential equations" (aequatio differentialis) initiated by Leibniz in 1676 is universally used. In Eqs.. The terms y' and y" in Eqs. Although such equations should probably be called "derivative equations. x). (4) the unknown function u is assumed to be a function of the two independent variables t and x.. For example. in the form
y"I = F(x. y = y(x). In Eq.Y.
(5)
. x) with respect to t and x. respectively. (1)-(3) are the first and second derivatives.CHAPTER 1
Elementary MethodsFirst-Order Differential Equations
1
1
INTRODUCTION AND DEFINITIONS
Differential equations are equations that involve derivatives of some unknown function(s)..y. that is.y. respectively... y("
where y. Equations (1)-(3) involve ordinary derivatives and are ordinary differential equations. y("> are all evaluated at x. In this book we are primarily interested in studying ordinary differential equations. respectively. Equation (4) involves partial derivatives and is a partial differential equation. an introduction to difference equations and an introduction to partial differential equations are presented in Chapters 9 and 11.. u = u(t..
DEFINITION
1
An ordinary differential equation of order n is an equation that is.. The argument x in y(x) (and its derivatives) is usually suppressed for notational simplicity. (1)-(3) the unknown function is represented by y and is assumed to be a function of the single independent variable x.
that is. y = V is a solution of the first-order ordinary differential equation y' = 112y valid only in the interval (0. and c2.
2..y(x). Eq. As we have seen. Also. any solution y(x) of Eq.is also a solution and moreover
y = c.y = 0.y'(x). .. equations of the form
c.
3. +x). e is a solution of y" .2x)/2y valid only in the interval (0. and c2. . (5) should have the following properties: 1.2
1
Elementary Methods-First-Order Differential Equations
The independent variable x belongs to some interval I (I may be finite or
infinite).e + cZe-' is a solution of this equation for arbitrary values of the constants
0 can be obtained from the function c.y(x). y"I(x) = F(x.
Clearly. Indeed. y = e is a solution of the ordinary differential equation
y" . y"-'"(x)) for every x in J. . . (2) and (3) are ordinary differential equations of order 2.
DEFINITION 2
A solution of the ordinary differential equation (5) is a function y(x) defined over a subinterval J C I which satisfies Eq. the function F is given.
Clearly. +x) and y = x( is a solution of the ordinary differential equation y' = (1 . and the function y = y(x) is unknown. By the general solution we mean a solution with the property that any solution of y" .. that is. In this chapter we present elementary methods for finding the solutions of some first-order ordinary differential equations. y). 1). In fact.
Y"(x)-Y(x)=(e')'. The differential of a function y = y(x) is by definition given by dy = y'dx.y = 0.e + c2e-' is the "general solution" of the ordinary differential equation y" . As an illustration we note that the function y(x) = e' is a solution of the second-order ordinary differential equation y" . As another example._e'=e -e'=0. Thus. (5) identically over the interval J. y"-"(x)) should lie in the
domain of definition of the function F.
y"(x) + y(x) = (cos x)" + cos x = . + x). cos x + c2 sin x for arbitrary values of the constants c.y =
y' = F(x. In each of the illustrations the solution is valid on the whole real line
On the other hand.. the function y(x) = cos x is a solution of y" + y = 0 over the interval ( --. For every x in J the point (x. (1) is an ordinary differential equation of order 1 and Eqs. in Chapter 2 we will show that the general solution of the ordinary differential equation y" + y = 0 is given by y(x) = c. For the most part the functions Fand y will be real valued.y = 0.e + cZe-' for some special values of the constants c. and c2.y'(x).
.cos x + cos x = 0. We further observe that y= e.y = 0 valid for all x in the interval
(--.
(6)
together with some interesting applications. y should have derivatives at least up to order n in the interval J. It will be shown in Chapter 2 that y = c. F should be defined at this point.
Both appear frequently in applications. we usually assume that the actual situation is governed by very simple laws-which is to say that we often make idealistic assumptions. the next step is to solve the differential equation and utilize the solution to make predictions concerning the behavior of the real problem. For
example. and many other types of differential equations are reducible to one or the other of these types by means of a simple transformation. Let the function y = y(x) represent an unknown quantity that we want to study. in chemistry as reaction rates. It is the case with many mathematical models that in order to obtain a differential equation that describes a real-life problem. For example. Many natural laws and hypotheses can be translated via mathematical language into equations involving derivatives.1.y)dx or in an algebraically equivalent form. the differential equation (6) sometimes will be written in the
differential form dy = F(x.
and linear equations. First-order ordinary differential equations are very useful in applications. Of all tractable types of first-order ordinary differential equations. by
. that is.
In case these predictions are not in reasonable agreement with reality. in economics as rates of change of the cost of living. We know from calculus that the first derivative y' = dy/dx represents the rate of change of y per unit change in x. and in finance as rates of growth of investments. that is. derivatives appear in physics as velocities and accelerations.1
Introduction and Definitions
3
With this in mind. in geometry as slopes. in psychology as rates of learning. the differential equation
3x2
x'+1(Y
Y
+1)
can be written in the form
dy=rx31(y+1)Jdx
11
or
Y
x'+ly x'+
3x2
3x2
There are several types of first-order ordinary differential equations whose solutions can be found explicitly or implicitly by integrations. If this rate of change is known (say. equations that can be put into the form
y' + a(x)y = b(x).1. in biology as rates of growth of populations. Once the model is constructed in the form of a differential equation.1
Differential equations appear frequently in mathematical models that attempt to describe real-life situations.
APPLICATIONS 1. two deserve special attention: differential equations with variables separable. equations that can be put into the form
y' = QTY)
or
P(x)dx = Q(y)dy. the
scientist must reconsider the assumptions that led to the model and attempt to construct a model closer to reality.
let
N = N(t) be the number of bacteria present at any time t. then the quantity y satisfies the first-order ordinary differential equation y' = F(x.1
It should be emphasized that Eq. if the number of bacteria is very large. it is not continuous and so not differentiable.
law: It grows at a rate proportional to the number of bacteria present at any time t. perhaps oversimplified. (7) N stands for dN/dt. (As is
customary. Then.
Biology
It has long been observed that some large colonies of bacteria tend to grow at a rate proportional to the number of bacteria present. R. derivatives with respect to x will he denoted by primes and deriv-
atives with respect to t by dots.I.
. if k is the constant of proportionality. Y). Malthus observed in 1798 that the population of Europe seemed to be doubling at regular
time intervals. and so he concluded that the rate of population increase is proportional to the population present. On the other hand. For such a colony.3. Equation (7) is a separable differential equation and its solution N(t) = N(0)e"' is computed in Example 3 of Section 1.) In this instance it is the time that is the
independent variable.
N(r)
N(r) . Here N(O) is the number of bacteria present initially. that is. (7) is a mathematical model describing a
colony of bacteria that grows according to a very simple. T. since the changes in the size of the population occur over short time intervals. the function N = N(t) satisfies the first-order ordinary differential equation'
IV=kN. y).
(7)
This equation is called the Malthusian law of population growth.
'Since the function N(r) takes on only integral values. at time t = 0.
We next give some specific illustrations. However. we can assume that it can be approximated by a differentiable function N(r).N(0)e"
N(0)
0
Figure 1. The solution. of Eq.4
1
Elementary Methods-First-Order Differential Equations
experience or by a physical law) to be equal to a function F(x. (7)
provides us with an approximation to the actual size of this colony of bacteria. The solution N(t) can be represented graphically as in Figure I. assuming this very simple law of growth leads us to
a very simple differential equation. N(t) = N(0)e". In Eq.
and consequently some simplifications and modifications of real-life laws are often necessary in order to derive a mathematically tractable model. Sci. S. It goes without saying that a mathematical model that is impossible to handle mathematically is useless. To achieve this it is necessary to give the patient an initial booster dose yo of the drug and then at
equal intervals of time. However. 1. in many cases it is necessary to maintain (approx-
imately) a constant concentration (and therefore approximately a constant amount) of the drug in the patient's body for a long time.2
to zero as t -> oc. (9) (see also Figure 1.
'This model. Math. The negative sign in (8) is due to the fact that y(t) decreases as t increases. and the like.3)
y(t) = Yoe-r`. Of course.
(8)
Pharmacology
Drug Dosages
where k > 0 is the constant of proportionality. Technol. is discussed by J.
(9)
where yo = y(O) is the initial amount (initial dose) of the drug.
limitations of food. as well as other mathematical models in medicine.
.
0
Figure 1. As we see from Eq. say every T hours. a more realistic mathematical model for the growth of this colony of
bacteria is obtained if we take into account such realistic factors as overcrowding.ky.1
Introduction and Definitions
5
Clearly. Educ. give the patient a dose D of the drug. then the rate of change y(t) of the drug is proportional to the amount present. 2 (1971): 193-203. For each drug the constant k is known experimentally.2). Rustagi in Int.
to It is well known in pharmacology2 that penicillin and many other drugs administered to patients disappear from their bodies according to the following simple rule: If y(t) is the amount of the drug in a human body at time t. and hence the derivative of y(t) with respect to t is negative.1.
The solution of the differential equation (8) is (see Example 3 of Section 1. the amount of the drug in the patient's body tends
Y
0
Ype -A. y(t) satisfies the separable differential equation
y = .. That is. the differential equation will then become more complex.
If we want to maintain the initial amount yo of the drug in the body at the times T. .. the dose D should satisfy the equation
yoe-' + D = yo. Recall that the momentum of a body is the product my of its mass and its velocity v. the desired dose is given by the equation
D = yo(1 .6
1
Elementary Methods-First-Order Differential Equations
Equation (9) indicates the amount of the drug in the patient's body at any time t.
(11)
where x denotes the education of an individual at time t and the constant k is the rate at which education is being made obsolete or forgotten. the greater the cost of the
instruction]. Psycho!." implies immediately that the motion of any body is described by an ordinary differential equation. If F is the resultant force acting on the body. Operations Res. then
dt (mv) = kF. 2r.
. In fact.)
Operations Research
(10)
Southwick and Zionts' developed an optimal control-theory approach to the education-investment decision which led them to the first-order linear (also separable) differential equation
x=1-kx.' Here G is known as the characteristic learning function and depends on the characteristics of the learner and of the material to be learned.
Psychology
In learning theory the separable first-order differential equation
P(t) = a(t)G(p(t))
(12)
is a basic model of the instructor/learner interaction. the amount of the drug present in the body is
y(r) =
yoe-'". it is simple to determine the amount of the dose D. but also.
Hence. p(t) is the state of the learner at time t. 3T.e-k. Chant. at time T.. G. hence. and a(t) is a measure of the intensity of instruction [the larger the value of a(t) the greater the learning rate of the learner. Zionts.
Mechanics
Newton's second law of motion. which states that "the time rate of change
of momentum of a body is proportional to the resultant force acting on the body
and is in the direction of this resultant force. V. . 22 (1974): 1156-1174. Math. J.
(13)
'L. Southwick and S. and before we administer the dose D. 11 (1974): 132-158.
F can be constant.
(14)
Figure 1. We want to compute another family of curves such that each member of the new family cuts each member of the family (15) at right angles.1. (16).
where I = 1(t) is the current in the circuit at time t.dy=0. Thus. are the partial derivatives of F with respect to x and y.
(17)
The general solution of Eq.
F. we want to compute the orthogonal trajectories of the family (15).y) = c. In view of Eq. or even a function of t and v.3 gives rise to the first-order linear differential equation (see also Section 1. respectively. The mass m can be constant or a function of t. the slope of the orthogonal trajectories of the family (15) is given by [the negative reciprocal of (16)]
dyFr dx F
(15). Also.
(16)
gives the slope of each curve of the family (15).
Kirchhoff's voltage law states that. we obtain
(15)
Orthogonal Trajectories
Fdx+F.
Computing the differential of Eq. (17) gives the orthogonal trajectories of the family
. that is." This law applied to the RL-series circuit in Figure 1. a function of t.
dy dx
_ _ E.
where F and F.1
Introduction and Definitions
7
where k is a constant of proportionality.3
Consider the one-parameter family of curves given by the equation
F(x.4)
LI + RI = V(t). Equation (13) is an ordinary differential equation in v whose particular form depends on m and F. "the algebraic sum of all voltage drops Electric Circuits around an electric circuit is zero. (15).
y = (1/2x) e'2 is a solution of the differential equation xy' + y =
xe'2. y = cos 2x is a solution of the differential equation y" + 4y = 0. The emphasis there is twofold: first.to low-pressure areas. y = sin 2x is a solution of the differential equation y" . We have given only a few of the many applications of first-order ordinary differential equations.
3. Many others are developed in detail in subsequent sections.
2. In meterology the orthogonal trajectories of the isobars (curves connecting all points that report the same barometric pressure) give the direction of the wind from high.
4.y = 0. and second. Y = e' + 3e-` is a solution of the differential equation y" . Additionally.8
1
Elementary Methods-First-Order Differential Equations
Figure 1. to expose the reader to the diversity of models incorporating ordinary differential equations. to sharpen the
reader's skill at solving differential equations while simultaneously emphasizing
the fact that many differential equations are not products of the instructor's imagination but rather are extracted from real-life models. y = 5 sin x + 2 cos x is a solution of the differential equation
y"+y=0. 5.
. In electrostatic fields the lines of force are orthogonal to the lines of constant potential.4y = 0.
2. In two-dimensional flows of fluids the lines of motion of the flow-called streamlines-are orthogonal to the equipotential lines of the flow (see Figure
1. we present a number of applications in the exercises.
EXERCISES
In Exercises 1 through 8.
1.4).
3.4
There are many physical interpretations and uses of orthogonal trajectories:
1. answer true or false.
If more than one function is listed.
7. a differential equation and a function are listed.
8. show that both functions are solutions of the differential
equation. 2x .0.xe-' + I is a solution of the ordinary differential equation y" + 2y' + y = 1 for any values of the constants
c.e2'. y" + y' = 2.is a solution of the differential equation y' .1
13. Y = e.e-' + c. y"-(tan x)y'.1. Show that the functions y. are solutions of the ordinary differential equation
y.X y=XY. Y"-2Y'+Y= z(Y-y').. Show that the functions y. 16. Show
that the function is a solution of the differential equation.)
.(x) = 0 and y2(x) = x2/4.
9. ms
tan x
1
g.
1
17. xsecx
11. and c2. y"'-5y"+ 6y' =0. = yin
18. 2x.c)2
forx<-c
forx > c . is a solution of the differential equation
Y' = ya
(Warning: Don't forget to show that the solution is differentiable everywhere and in particular at x = c. y"'-5y"+ 6y' =0.
for any real number c.< _ :8t 24mt
1
2
d2s)
s
dt'JJ
14.(x) = e-' and y2(x) = xe-' are solutions of the ordinary differential equation y" + 2y' + y = 0. Y = e' is a solution of the differential equation y' + y = 0.e'Inx
15. Show that the function
Y(x) _
10 (x .e'' 12.
is
a
solution of the differential
equation
In Exercises 9 through 14. Show that the function y(x) = c.y = 0.1
Introduction and Definitions
9
6. x >. y = 2 In x + 4
x2y"-xy'+y=2Inx.3
10.
y = -1 + e.10
1
Elementary Methods-First-Order Differential Equations
19. y = 28. Prove that if the family of solutions of the differential equation
y' + p(x)y = q(x). 24. Math. Verify that each member of the one-parameter family of curves
x2 + y2 = 2cx
cuts every member of the one-parameter family of curves
x2 + y2 = 2ky
at right angles and vice versa.cp(k)](x . that is. y = 4
3x
is a solution of the differential equation y' _ X2 + y2
Y
X2 +
3xx is a solution of the differential equation y'
y2
Y
29.
p(x)q(x) 0 0
is cut by the line x = k.]
'From the William Lowell Putnam Mathematical Competition. c) is y . y = x + 1 is a solution of the differential equation yy' . Y = sine x is a solution of the differential equation y" + y = cost X.k). Show that the function
x(t) =
1
k
+ k x o .
20.2y = e'(1 . q(k)lp(k). 25.is a solution of the differential equation y" = y'(y' + y).
27. is a solution of Eq.' [Hint: The equation of the tangent
line to the solution at the point (k.
26. y = xe is a solution of the differential equation y' . What is the meaning of the constant c? 21. (7).1 e-"' k
'°
where xo and to are constants.
22. 1954.
In Exercises 22 through 28. x(to) = xo.x). Y = x= is a solution of the differential equation y"' = 0. Show that N(t) = cek' for any constant c is a solution of Eq. Show also that this solution passes through the point (to)xo). (Hint: Show that slopes are negative reciprocals. Using your geometric intuition.) 30.
and this equation passes through the point [k + 1/p(k). the tangents to each member of the family at the points of intersection are concurrent. Find the differential equation of the orthogonal trajectories of the family of straight lines y = cx.
23. See Amer. Monthly 61 (1954): 545.y2 = x2.
. answer true or false. (11).c = [q(k) . guess the solution of the resulting differential equation.
y). do there exist solutions to the differential equation? For example.
. Such proofs can be found in almost any advanced text on differential equations. that is. and in fact it should have only one solution (uniqueness). Starting at time xo the particle will move and will move in a unique way.y). Often the proofs of these theorems require mathematical sophistication which is customarily beyond the scope of an elementary treatment. where y is the position (the
state) of the particle at time x. Before we attempt to discover any ingenious
technique to solve a differential equation. the differential equation (y')2 + y2 + I = 0 has no real solution since the left-hand side is always positive. the motion that we are observing. We shall now state (see Appendix D for a proof)' a basic existence and uniqueness theorem for the IVP (1)-(2) which is independent of any physical considerations and which covers a wide class of first-order differential equations. the differential equation (1) has a solution that satisfies the condition (2). The general form of a first-order ordinary differential equation is
y' = F(x. That is. That is. and moreover it has only one solution.1. if
Y(xo) = Yo
(2)
we should be able to find its position at a later time x. it would be very useful to know
whether the differential equation has any solutions at all.
(1)
For the sake of motivation.
THEOREM 1
Consider the IVP
y' = F(x. The term "initial" has been adopted from physics. (2) yo is the initial position of the particle at the initial time
xo. The condition (2) is called an initial condition and Eq.y). assume that Eq.
y(xo) = yo. Since we imagine that we are observing the
motion of the particle. it is plausible that if we know its position y6 at time xo. the models are useful if the resulting differential equation can be solved explicitly or at least if we can predict some of the properties of its solutions.2
Existence and Uniqueness
11
1 2 EXISTENCE AND UNIQUENESS
We saw in the last section that first-order differential equations occur in many diverse mathematical models.
On the basis of the motivation above we may expect that an IVP which is a reasonable mathematical model of a real-life situation should have a solution (existence).
'This theorem and many others that appear in this text are truly important for an over-all appreciation of differential equations. Naturally. In Eq. (1) together with the initial condition (2) is called an initial value problem (IVP). that is. The reader can rest assured that these theorems are accurately presented and that their proofs have been satisfactorily scrutinized. (1) represents the motion of a
particle whose velocity at time x is given by F(x.
Here F(x. in the notation of Definition I of Section 1. Then F(x. 0).
REMARK 1
For linear equations such as Eq.xo s A is contained
in the interval I. Show that the IVP
Y' = x' + Y2 Y(0) = 0
(3) (4)
EXAMPLE 1
has a unique solution in some interval of the form .1
1=jx:ix-xol<AI and J= Ix: x-xo.
EXAMPLE 2 Assume that the coefficients a(x) and b(x) of the first-order linear differential equation
y' + a(x)y = b(x)
(5)
are continuous in some open interval 1.
Solution
Choose a number A such that the interval x . one can prove that its solutions
exist throughout any intervai about xo in which the coefficients a(x) and b(x) are continuous.yo !5. the IVP
.a(x)y and FF.12
1
Elementary Methods-First-Order Differential Equations
Assume that the functions F and aFloy are continuous in some rectangle
I(X.shj. (5) has a unique solution through any point (xo. B> 0
xo
A
about the point (xo.h. yo) where xo E 1. that is. Then there is a positive number h :5 A such that the P/
has one and only one solution in the interval x .xo ' < It.yo). (5) has a unique solution satisfying the initial condition y(xo) = yo [in other words. By Theorem 1 there exists a positive number h such that the
Proof
IVP (3)-(4) has a unique solution in the interval
-h<_x:5 h.y) = x' + y' and aFlay = 2y are continuous in any rectangle 9t about (0.
I x .h s x :5 h. for any real number
Yo.(x.Y):
x
y . Eq. y) = -a(x) are
continuous in
x
Ix-x0I:r Al
By Theorem 1.
Let us illustrate this theorem by a few examples. yo)]. through the point (xo. Show that Eq.
Thus.0 <. (5). y) = b(x) . For example. B' A> O.
these solutions are given by
10. Uniqueness of solutions of initial value problems is very important. In fact. Since FY is not continuous
at the point (0.1
has a unique solution in the interval (2. y) = y12and FY(x. x).1.
(7)
Is this in violation of Theorem 1?
Solution
Here F(x.c)2 4
x'5 c
c<x
'
and their graphs are given in Figure 1. and the IVP
Y'+x-2 1 y=InIxI
y(-1)=5
has a unique solution in the inverval ( --.5.
Y(x) = (x .
REMARK 2 As we have seen in Example 3 for the IVP (6)-(7). the IVP (6)-(7) has infinitely many solutions through the point (0. 0). in addition to y. 0). When we know that an IVP has a unique solution we can apply any method (including guessing) to find its solution. For example. and consequently there is no guarantee that the IVP (6)-(7) has a unique solution. 0).2
Existence and Uniqueness
13
y' +x I2y=1nx
y(l) = 3
has a unique solution in the interval (0.2'v=Inx
y(3) = .(x) ° 0 and y2(x) = x2/4 are two different
solutions of the IVP
(6)
Y' = Y12
Y(0) = 0.
EXAMPLE 3 The functions y. we know that the IVP
Y. + xy = 0
Y(0) = 0
(8) (9)
. the IVP
y'+ x. y) = }y-12. 2). one of the hypotheses of Theorem I is violated. For each c > 0. we have existence of solutions but not uniqueness. and Y2.
Such line segments are called linear elements.
(10)
For a better understanding of the differential equation (10) and its solutions. yo). yo) in the domain D of F. Just by inspection we see that y = 0 satisfies the IVP. yo).
Theorem 1 gives conditions under which a first-order differential equation
possesses a unique solution. Knowing that a solution exists. Hence. the
differential equation (10) assigns a slope y' equal to F(xo. we desire to develop
methods for determining this solution either exactly or approximately. its slope is F(xo. Thus.14
1
Elementary Methods--First-Order Differential Equations Y
Y
0
C. The construction of the direction field can
.5
has a unique solution. y). then (without knowing the solution explicitly) we know immediately the equation of the tangent line to this solution at the point (xo. it is useful to interpret y' as the slope of the tangent line drawn to the solution
at the point (x. The totality of all linear elements is called the direction
field of the differential equation (10).
X
0
C=>CJ
X
Y
0
Figure 1. called the method of isoclines. if we know that (10) has a solution through a given point (xo. yo). graphically. Consider again the first-order differential equation
y' = F(x. an
approximation to the true solution of (10) near the point (xo. is useful in finding an approximate solution. Then at each point (xo. The following method. b) E D with slope F(a. b). y(x) = 0 is the only solution of the IVP (8)-(9) and no further work is required in solving this IVP. In particular. yo) will give us. yo). A short segment along this tangent about (xo. Thus for
graphical approximations of the solutions of (10) it is useful to draw a short line
segment at each point (a. Y). yo).
.6).. and each one of them has slope c.1. y) = c and realizing that for any constant c all points on the (level) curve F(x. the linear elements of the direction field for all these points are parallel. IC. 16) and linear elements which give a good idea of the direction field of the differential equation (11). centered at the origin (0. At each point on the circle x2 + y2 = c the differential equation
defines the slope c.7
. 9.YO)
" C2
Figure 1. Therefore. corresponding to the values c = c c2. In Figure 1.2
Existence and Uniqueness
15
Y4
I
IIIIVII'
Figure 1. For example. y) = c. Through the point (xo. First we
draw a few isoclines C C2. y) = c have the same slope c.. c
.6 we have drawn a few isoclines (for c = 1. see Figure 1. is the curve F(x.. 4.6
be carried out more efficiently by setting F(x. yo).. This is the method of isoclines (equal slopes).7]. yo) we
(xo. C . the isoclines of the differential equation
y2
y' = x2 +
(11)
are the circles x2 + y2 = c. 0) with radius \ (see Figure 1. Now suppose that we want to find a graphical approximation of the solution of the differential equation (10) which goes through the point (xo.
Proceeding in this fashion we construct a polygonal curve through (x0. Astronomy In an article' concerning the accumulation processes in the primitive solar nebula. yo) and extend it until it meets the isocline C.Bt)3a
where a. answer true or false.
3. Note that.y'+xy=3
Y(O) = 0
4. (-1. and (3. Find all points (t(.1. W.
. In Figure 1. G. b. 5).16
1
Elementary Methods-First-Order Differential Equations
draw the linear element with slope ! x0.6 we have drawn an approximate solution of the differential equation (11) through the point (0. x0) through which this differential equation has unique solutions. at a point A.
EXERCISES
In Exercises 1 through 6. from Example 1.
until it meets the isocline C2 at a point A2. Cameron. (1.
9. Review Remark 1 of this section and apply it to the differential equation
xy
+2x+3y=In:x-2
for its solutions through the points (-3. the following first-order differential equation was
obtained:
dx dt
ax`
(b . 0).y'= x+y
y(o) _ -1
x y
6. and B are constants..
In Exercises 10 through 13. Through A.
8. the differential equation (11) has a unique solution through (0.
1. 0). Icarus 18 (1973): 407-450. 0). we draw the linear element with slope c. 0). -7). check whether the hypotheses of Theorem 1 are
satisfied. Y
y' = Y'
Y(1) = I
Y' = Yv3
Y(0) = 0
x + y
Y(l) _ -I
7.
'A. Show that the only solution of the IVP
xy'-y=1
y(2) = 3
is y(x) = 2z .xy'+y=3
y(o) = 1
5.
2. yo) which is an approximation of the true solution
of the differential equation through that point.
3
Variables Separable
17
10.
.
12. The only solution of the IVP
y'-xy=I-x2
Y(o) = 0
is Y(x) = X.
13. (2) gives implicitly the general solution of Eq. Q(y)dy = P(x)dx. (2). more explicitly.
11. Algebraic manipulations enable us to write separable differential equations in the form y' = P(x)/Q(y) or.
(2)
where c is an arbitrary constant of integration.1.x Y(O) = 0
15.3 VARIABLES SEPARABLE A separable differential equation is characterized by the fact that the two variables of the equation together with their respective differentials can be placed on opposite sides of the equation. Use the method of isoclines to find graphical approximations to the solutions of the following IVPs. The differential equation y' = yJ4 has a unique solution through the point (0.
14. 0). it is desirable to solve the resulting expression for the dependent variable y. In such equations the equality sign "separates" one variable from the other.
y' = y .
(1)
To obtain the general solution of a separable differential equation we first separate the two variables as in Eq. (1) and then integrate both sides. y' = xy y(1) = 2
16. thereby obtaining y explicitly in terms of x. (1). to obtain
f P(x)dx = f Q(y)dy + c. then Eq. 0) but not through the point (2. After performing the integrations
in Eq. If this is not possible or convenient. The unique solution of the IVP
y'-xy=1-x2
Y(0) = 0
is y(x) = -x. The unique solution of the IVP
xy' + y2 = 1
y(-2) = I
is y(x) = 1.
y' = x2 + y2
y(0) = 1
1.
where the variables x and y are separated.1)(y + 1) = ±e`.
lnIx2. y # . (4) can be separated as in Eq. In what follows we shall use this convention frequently. (3). Integrating
both sides of Eq. (4').
Z x2
= y` + 3y 4 c
is an implicit representation of the general solution of Eq.
Thus. "it follows.l
(4')
e"'dy = e-'sinxdx. (3).
EXAMPLE 1
(5')
Find the general solution of Eq. (4') yields
lnj x2. y # O. Integrating both sides of Eq.1
Thus. different than zero.
EXAMPLE 2
Solution The variables of Eq. we obtain
Solution
fxdx=J(5y4+3)dy+c. Since c is an arbitrary constant. (3) is separable because it can be written in the form (3')." and will be used when
. (3').(5y' + 3) dy = 0
2x(y2 + y) dx + (x' .I)(y + 1) 1 = e` (x2 . into the forms (3').
stands for "imply" (or "implies"). and for economy in notation we can still denote it by c with c * 0.
forx
1.'
-In I y + I1+c. we have
i (x2 .
The symbol appropriate. (4').
As we have seen.11+InI y+11=c'In I (x2-I)(y+1)1=c. respectively. Eq. and (5'):
x dx = (5y' + 3)dy
(3')
x22 l dx = -
y I
dy.
(3)
(4)
(5)
In fact. clearly ±e` is again an arbitrary constant.1) y dy = 0
e`-'y' = sin x. they can be brought." "then.
Taking exponentials of both sides and using the fact that e'" ' = p.
Solve the differential equation (4).18
1
Elementary Methods-First-Order Differential Equations
The following are examples of differential equations that are separable:
x dx .
= c.
. and the proof is complete.
EXAMPLE 3
Prove that the solution of the IVP
y = ky.ky.8.1 will be contained in the formula
y = -1 + [cl(x2 . we find that y. do not forget that Eq.. Using the initial condition. Sometimes such curves are called singular solutions and the one-parameter family of solutions
Y
C = -1 + x2-1'
where c is an arbitrary constant (parameter). where c is an arbitrary constant. for y # 0.
Y(0) = Yo
is
k is a constant
y(t) =
Yoe-"'
Proof Clearly. Thus. that is. Similarly. we require that x finally examine what happens when x = ± 1 and when y = 0 or y = -1.1 also satisfy the differential equation (4). (4) by dividing through by the expression (x2 . (4) we see that the four lines x = = 1.1)(y2 + y).
k is a constant
is y(t) = yoek'. we have
InIyI=kt+cz> I y I
=ek-=e`e"''y=
± e'ek'
'y=cek'. and y 0.. Since y = dyldt. it follows that the general solution of y = ky is y = ce". (4') is obtained from Eq.
The following initial value problems are frequently encountered in applications of exponential growth and decay. Going
back to the original Eq. and y = . The solutions of the above IVPs for k > 0 and k < 0 are represented in Figure 1.
Since y = 0 is also a solution of y = ky. show that the solution of the IVP
y = . separating the variables y and t we obtain.e". Thus. no matter what the value of c is.
Now. y * -1. However.
Y
Integrating both sides. y(t) = y.
c#0. If we relax the restriction c * 0.1. Hence. the curves x = ± 1 and y = 0 are
not contained in the same formula. We must is not zero.1)] for c = 0. y = 0. the curve y = . and readers are advised to familiarize themselves with the solutions. it suffices to solve the first IVP. we should ensure that this expression t 1.3
Variables Separable
19
(x2-1)(Y+1)=c=> Y=-I+x2c#0.
dy=kdt. the solution to the second IVP follows from the solution to the first IVP by replacing the constant k by .k.
Y(0) = Y. is called the general solution.
Psychology
In 1930.3. respectively. Review also the applications in Section 1. If y(t) denotes the state of a learner at
time t. then Thurstone's equation is the following separable differential equation: dy _ 2k dt.1 that involved separable differential equations.1
One frequently encounters separable differential equations in applications. Psychol.
(7)
. For the solution of the differential equation (6). how long will it take for the bacteria to triple?
Let N(t) be the number of bacteria present at time r. while learning
a specific task or body of knowledge. Thurstone' obtained a differential equation as a mathematical model describing the state of a learner.1. Gen. Then the assumption that this colony of bacteria increases at a rate proportional to the
Solution
number present can be mathematically written as dN = kV. or the learning curve of a learner. see Exercise 16. 3 (1930): 469-493.
Biology
Assume that a colony of bacteria increases at a rate proportional to the
number present. We shall mention a few instances here.J.8
APPLICATIONS 1. If the number of bacteria doubles in 5 hours. (6) y)311 y °(1
-
Here k and m are positive constants that depend on the individual learner and the complexity of the task. dt
'L.20
Y
1
Elementary Methods--First-Order Differential Equations
Y
Y = Yoe"
Y(0)
Yo
Y(f) -
Yoe-"
r
I
Figure 1. Thurstone. L.
945. and if m is a constant.89hours.
The time t that is required for this colony to triple must satisfy the equation N(i) = 3N(0)
#>
N(0)e"' = 3N(O)
1
e" = 3
1.1.1 we observed that according to Newton's second law of motion a moving body of mass m and velocity v is governed by the differential equation
Mechanics
dt (mv) = kF. the constant k is equal to 1. In the cgs (centimeter-gram-second) system and in the English system. then Eq. Its solution (by Example 3) is
y(t) = y(0)ec"loon
Since y(0) = 5000 (the initial amount invested). Then the rate of change of the money at time t is given by
dt
100
Y.28.3 Variables Separable
21
where k is the constant of proportionality.
(10)
where F is the resultant force acting on the body and k is a constant of proportionality.1.
. Since the number of bacteria doubles in 5 hours.
The sum of $5000 is invested at the rate of 8% per year compounded continuously. we have
N(5) = 2N(0) N(o)es" = 2N(0)
es' = 2
z' k=5In 2.69=7.
In 3
Finance
(9)
Equation (9) is clearly a separable differential equation. What will the amount be after 25 years?
Solution Let y(t) be the amount of money (capital plus interest) at time t. From Example 3 we have N(t) =
N(0) is the initial number of bacteria in this colony.
In Section 1. (10) is a separable
differential equation. If it happens that F is a function of the velocity v and does not depend explicitly on time.09
t=kln3=51n2-50. we find that
y(25) = 5000e(&100) u = 5000e2 = $36.1. whose units are given in Table 1.
The resultant force acting upon this falling body is
F = (weight) . Eq (10). Find the velocity of the body after i = 10' sec. In addition to its weight. and so
v(103) = 9 Medicine
8
e
105 cm/sec =
I9 -
/
8
e
km/sec = 6.2vo)e-:`]. (Take g = 900 cm/sec'.
dv = dt.
dt
(mv) = mg . by Newton's second law of motion.05 kmisec.1
Table of Units
EXAMPLE 4
From a great height above the earth.22
1
Elementary
Differential Equations
Cgs system
Distance
centimeter
English system
foot
Mass
Time
Force
gram
second
slug
second
dyne
pound
Velocity
Acceleration
cm/sec
ft/sec
cm/sec'
ft/sec'
Table 1.(mg . air resistance is acting upon this body.
.(air resistance) = (mg .
Thus.
v(0) = 105. which is numerically (in dynes) equal to twice its speed at any time.
From the practical point of view this is a reasonable model. a body of mass m = 2
kilograms (kg) is thrown downward with initial velocity vo = 105 centimeters per second (cm/sec).2v) dynes. in spite of the many simplifications that we introduce.)
Solution Let v(t) be the velocity of the body at time t.
Here we shall derive a separable differential equation that serves as a mathematical model of the dye-dilution procedure for measuring cardiac output.2v
'
mdt=mg-2v
2v
mgm
Also.
(12)
The solution of the IVP (11)-(12) is v(t) = 1 [mg .
(15)
. We also assume that the heart is a container of constant volume V liters.
The quantity (concentration) D(t)/V is useful in measuring the cardiac output. The dye is mixed with the blood passing through the heart.
With these data and assumptions.
We also have the initial condition D(0) = Do. The solution of the IVP (13)-(14) is
D(t) = De-("v)`. Thus. then
Radioactive Dating
Archaeology. the amount
of dye per liter of diluted mixture). Thus. if N(t) is the number of atoms of the substance present at any time t.1. To derive our model we shall assume that the mixture of blood and dye inside the heart is uniform and flows out at
a constant rate of r liters per minute.
It is known. Geology. This is a special case of mixture problems that we will study in the next section. Arts
N(t) _ -kN(t).3
Variables Separable
23
In the dye-dilution method an amount of dye of mass D. and at each stroke of the heart the diluted mixture flows out and is replaced by blood from the veins. that radioactive substances decay at a rate proportional to the amount of the substance that is present. we have the
equation
dDdlt)
= rate in . from experiments.
where D(t)/V is the concentration of dye in the heart (in other words. Here
rate in = 0
as no dye runs into the heart. milligrams (mg) is injected into a vein so near the heart that we shall assume that the heart contains
D. milligrams of dye at time t = 0.
(13)
(14)
where Do is the (initial) amount of dye injected at time t = 0. and
rate out =
r.rate out.
Paleontology.V D(t). the differential equation describing this mixture problem is
ddt) _ . we can now formulate an initial value problem for the amount D(t) of dye in the heart at any time t. Since dD(t)/dt
represents the rate of change of the dye in the heart at time t.
where "rate in" is the rate at which dye runs into the heart at time t. and "rate out" is the rate at which dye runs out of the heart at time t.
No . then
2=e-kT
k=1T
that is. we utilize the half-life of the radioactive substance. To find the decay constant k. In fact. N(t). See also the articles in Science 155 (1967): 1238-1241 and Science 160 (1968): 413-415. and so on. If we know the values of k. Thus. Using Eq. from Eq. The value of N(t) is easily computed from the present amount of the
substance.24
1
Elementary Methods-First-Order Differential Equations
where the positive constant k is called the decay constant of the substance. For example. (17) we obtain e-` = N(t)IN. rocks. [The negative sign is due to the fact that N(t) > 0 and decreasing. it is known that each decaying uranium 238 atom gives rise to a single lead 206 atom.N(t) = N(t)(e"
and solving for t we find the age of the mineral to be
t = k In
I l + N(t)J
'
Those with some interest in radioactive dating of old paintings should read the work of Coreman.] If N(0) = No
(16)
is the (initial) number of atoms of the substance at time t = 0. the solution of the IVP (15)-(16) is given by (17) N(t) = N°e-". fossils.)
"P.. 1949). old paintings. (17). (17) we can find the time t that it took the radioactive substance to decay from its initial-value No to its present value N(t). (Vermeer and De Hooghs are famous seventeenth-century Dutch painters. Half-life is easily measured in the laboratory. we can utilize some additional information known about some radioactive substances to compute their age. and (taking logarithms of both sides)
t=-
k In
N(i)
No
=
I
In N(f)
(18)
Equation (17) is the basic idea behind radioactive dating when we try to determine the age of materials of archaeology. the time required for half of a given sample of a radioactive
substance to decay. and N°. Although the value No is in general unknown. the decay constant of a radioactive substance is obtained by dividing the natural logarithm of 2 by the half-life of the substance. then from Eq. and tables are available that give the half-life of many radioactive substances.10 where it is proved beyond any scientific doubt that the paintings sold by Van Meegeren are faked Vermeers and De Hooghs. Van Meegeren's Faked Vermeers and De Hooghs (Amsterdam: Meulenhoff. that is.
. so it has negative "derivative" NV(t). the number P of lead 206 atoms found in a uranium mineral containing N(t) atoms of uranium 238 is given by
P . Coreman. we see that if T is the half-life of a radioactive substance.
This is the case (rather it is very close to being the case.1/m = ep.p/m.
(19)
This equation was first derived by Daniel Bernoulli (in 1760) in his work on the effects of smallpox. that is. all the individuals born in one specific year. so we can write this relation for its solution:
dyldr
oPY .(N/S2)dS/dt. That is.
dy = a. Y(0) . j. If we multiply both sides by N/S2 and regroup terms.
Since N/S occurs as a variable. If p is the probability of a susceptible's getting the disease (0 < p < 1).P/m.1/m
= a. The equation is actually less formidable than it looks. which is now believed to be completely eradicated.3
Variables Separable
25
Certain diseases" that affect humans can be considered to impart immunity for life.1.
. and substitute in the lefthand side of this last equation to get
d(N/S)ldt = pN/S . and
Medicine
1/m the proportion of those who die due to the disease.p/m
So.P/m
Or
°
dt = a.1/m
P log y(0) . then the following
relation can be derived:
dS(t)ldt = -pS(t) + (S(t)/N(t))dN(t)ldt + pS2(t)ImN(t). we get
(1/S)dN/dt . In trying to assess the effect of such a disease.1/m
The authors wish to thank Stavros Busenberg for bringing this model to their attention. since there are exceptions to all rules in these matters) for such common diseases as measles.
From this we get
y(a) . an individual who has had the disease and survived is protected from ever having that disease again. define y = N/S and rewrite the equation as
dyldt = py .tot PY . and define N(t) to be the number who have survived to age t. and chicken pox. mumps.p/m.
y(a) . as well as for smallpox.
(21)
Equation (21) is a differential equation with separable variables. we can consider one cohort group.(N/S2)dS/dt = pNIS .
(20)
Recall that (d/dt)(N/S) = (1/S)dN/dt . and S(t) the number who have not had the disease and are still susceptible to it at age t.
let us examine what its terms mean. [M(t)/dl + pS(t)lm](S(t)/N(t)) is indeed the rate of decrease of susceptibles due to death for all reasons other than the disease. yielding y(O) = 1. There are special cases where we can have reasonable expectations for the ratio S(a)/N(a). S(a)/N(a) would be about 0.26
1
Elementary Methods-First-Order Differential Equations
Solving for my(a). where dN(t)/dt is the rate of change of the whole cohort group due to death for all reasons. recalling that y(a) = N(a)/S(a).
£22)
Since. (23). pS(t)/m is the rate of death due to the particular disease. only people who have not had the disease survive). the number of susceptibles at age a among a cohort group with N(a) survivors of that age would be
S(a) = 8N(a)/(1 + 7e°'r). we can use differential equations to get
relation (23). (19).1)e°^]. Equation (22) now gives my (a) = 1 + (m . Now we need an expression for the rate of removal of susceptibles due to death from all other causes. we get
S(a) = mN(a)l[1 + (m .1). The first.056. This is given by the second group of terms [dN(t)ldt + pS(t)lm](S(t)/N(t)). every member of the cohort group is susceptible.1)e°r. and S(t)/N(t) is the proportion of susceptibles among the cohort group. So. we have
my(a) = 1 + (my(0) . at birth. dS(t)/dt. if the mortality rate of the disease is very high (m . Bernoulli estimated p = 1/8 and m = 8 for the case of smallpox in Paris of the 1760s. and we should make sure that (23) fulfills these expectations. is the rate at which we expect susceptibles at age i to be infected with the disease. The values of these parameters in the right-hand side of the equation can be determined from census records and statistics on the disease. (19) are reasonable on the intuitive level
and lead to the mathematical formulation. In other words. but certainly not fpolproof. Thus.
So.
Formula (23) was derived on the basis of what one believes is a reasonable. To see that (19) is reasonable. Once we have Eq. of the situation we are examining. The terms on the right can be collected into two groups.
REMARK 1
All these assumptions leading to Eq. is the rate at which the number of susceptibles is changing (decreasing). This term includes those who will die from the disease.1)e'°. For example. Finally. Here we would expect that S(a)IN(a)
. we could expect just about all survivors N(a) to also be susceptible (that is. (19) is an equality between two different ways of expressing the rate of change of susceptibles. Using these constants in Eq. only one in about eighteen of those twenty-four years old would not have had smallpox. -pS(t). Eq. The left-hand side. or model. by age 24 (a = 24).
EXAMPLE 5 Using the population data that were available to him. S(0) = N(O).
(23)
This expression gives the number of susceptibles in terms of the number of survivors to age a of the cohort group. and the two basic constants m and p. relation (19).
k. A colony of bacteria increases at a rate proportional to the number present.
23. Chemistry Solve the differential equation
dx
dt
= k(a .»ntial Equations
18. What will the size of this population be at any time t? What will the population be after a very long time. that
is. If the number of bacteria triples in 4 hours. and the constants A and B are the vital coefficients of the population. for radium 226.x).) (a) Write the differential equation that describes the decay of the substance.
20. (b) Compute the decay constant of the substance. (This is true.088(2.-?
'Australian J. What will the value of x be after a very
long time. Compute the orthogonal trajectories of the one-parameter family of curves
y
cx. as t .
.Y8
1
Elementary Methods-First-Order Dif.Bt)32
given in the astronomy application in Exercise 8 of Section 1. that is. Assume that the rate at which a radioactive substance decays is proportional to the amount present.1). (c) What percentage of the original sample will disappear in 800 years? (d) In how many years will only one fifth of the original amount remain?
19. as t .4 . 4-5 (1956-57): 159.co?
21. Botany Solve the differential equation
d!
dw
= 0. In a certain sample 50% of the substance disappears in a period of 1600 years.2." and find
the value of I as w .
where N = N(t) is the population at time t.
22.-.b>0
which occurs in chemical reactions.x)(b .
which seems to fit the data collected in a botanical experiment.BN2.a. Biology Assume that a population grows according to Verhulst's logistic law of population growth dN
T = AN . Bet. for example. how long will it take for the bacteria to be 27 times the initial number?
24. Solve the differential equation
dx _
dt
ax 6'6
(b .
A person borrows $5000 at 18% annual interest compounded continuously. C > 0. runs out of the tank at the same rate as the inflow. and 10 seconds.
35. B = D = 0. kept uniform by stirring. C. In how many years will the money double?
Statistics The solutions of the separable differential equation
A-x
Y
B+Cx+Dx'Y
give most of the important distributions of statistics for appropriate choices
of the constants A. In the medicine application dealing with cardiac output assume that V = 500 milliliters (2 liter).
30. air resistance is acting upon this body which is numerically (in pounds) equal to three times its speed at any time. B = 0.
Pure water runs into the tank at the rate of 3 gallons per minute.1.
(exponential distribution).1)/C < 1. The sum of $15. C A/C > -I
(gamma distribution). B > 0. A tank contains 50 gallons of brine in which 5 pounds of salt is dissolved. For how long will the body be moving upward?
26.000?
. 221 seconds. Solve the differential equation in the
following cases. and A arbitrary 29. r = 120 liters per minute. 2 seconds. and D. A sum of $1000 is invested at 1% annual interest compounded continuously.
32. B. C = D = 0. From the surface of the earth a body of mass 2 slugs is thrown upward with initial velocity of vo = .
36. How long does it take to become a millionaire if $1000 is invested at 8% annual interest compounded continuously? What if the initial investment
is $10.
28. and D. and 31. Repeat Exercise 34 if it is desired that the money be tripled at the end of
10 years.30 ft/sec (the convention adopted here is that the positive direction is downward). = 2 milligrams. In addition to its weight.
33.000 is invested at the rate of 6% per year compounded continuously.
(beta distribution). How much will this person owe the lender at the end of 1 year?
37.5% annual interest compounded continuously?
34. A = B = D = 0 and C > 0
(normal distribution). How much salt is in the tank after 15 minutes?
27. How long will it take for $2000 to grow to $8000 if this sum is invested at 5.
Find the amount of dye in the heart after: 1 second. The
mixture.3
Variables Separable
29
25. and A > -C D. Find i if it is desired that the money be doubled at the end of 10 years. (A .
The general
solution of Eq.[ c + fb(x)efcxidx dx].(x) * 0 for all x in 1. If we divide both sides of the differential equation by a. (1) can be found explicitly by observing that the change of
variables
w = yefa(x)dx
(2)
transforms Eq.(x) and set a(x) = ao(x)/a.
This is a separable differential equation with general solution
w(x) = c + f
yefacxwx = c + f b(x)efacx d dx
y = efacxwx [c + f b(x)efacxwx dx].
REMARK 1
.)d. (1) into a separable differential equation. and the function f(x) are
continuous functions in some interval I and that the leading coefficient
a. then the general solution of the differential equation y' + a(x)y = b(x)
is given by the formula
y(x) = e-M. ao(x). we obtain the equivalent differential equation y' + a(x)y = b(x).
w' = y' efaixwx + ya(x)efa(x)dx
= [y' +
a(x)y]efaawx
(3)
= b(x)efa<xW.(x).
THEOREM 1
if a(x) and b(x) are continuous functions in some interval 1.
(4)
If you read the method above carefully once more. you will realize that the general solution of a first-order linear differential equation is found in three steps as follows. In fact (recall that (d/dx)[f a(x)d-] = a(x)).(x) and b(x) = f(x)/a.30
1
Elementary Methods-First-Order Differential Equations
1 4 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS
A first-order linear differential equation is an equation of the form
a.(x)y' + ao(x)y = f(x)
We always assume that the coefficients a. We summarize our results in the form of a theorem.(x). (1)
where a(x) and b(x) are continuous functions of x in the interval 1.
which is the general solution of Eq. and
Example 2. The reader can verify by direct substitution (see Exercise 18) that the unique solution of the IVP is given by the formula
oa ( :)d: + J
b(s)e -
y(x) = yoe
'
a(r)dr
ds. 2 and 5. (1) by e1''. together with Example 2 of Section 1. It will be helpful in computations if you recall that for any positive function p(x) the following identities hold:
REMARK 3
em p(X) = P(x)
e-I°P(-) =
I
P(x)
From Remark 1 of Section 1.
Before you apply the method described above. That is the case in Exercises 1. (1). has a unique solution which exists throughout the interval I.. we obtain
[yell"']' = b(x)er°(')dx
[This is a restatement of Eq. E I.2. make sure that the
REMARK 2
coefficient of y' is 1.
(5)
.
Step 3 Dividing both sides by e5°(')d' yields
y(x) = e-m(x)'[c + fb(x)ef°(')dxdx].2 the solutions of the differential equation (1) exist throughout the interval I.
where x. This. we find
yef'(')d' = c + f b(x)ef°(')d dx. (3). you have to divide by the coefficient of y' before you apply the method. implies that the
REMARK 4
IVP
y' + a(x)y = b(x)
y(xo) = yo.
Step 2 Integrating both sides.4
First-Order Linear Differential Equations
31
Step I
Multiplying both sides of Eq. In general.] The term el("d' is called an integrating factor.1. where the functions a(x) and b(x) are continuous.
Here a(x) = -2/x and b(x)
lies in the interval (0.32
1
ElementaryMethods-First-Order Differential Equations
EXAMPLE 1
Find the general solution of the differential equation d dx + (tan x) y = sin x
in the interval (0.=
we obtain
1
1
cosx '
sinx
ycosx
Integrating both sides.
Dividing both sides by 1/cos x (in other words. Although we could use formula (5) to immediately obtain the solution of the IVP. rr/2). the interval I. Therefore. where a(x) and b(x) are continuous. it is perhaps more instructive to proceed directly by using the steps described in the text. Multiplying both sides by
ea..-.(2/x)y = -x. 00). it follows from Remark 4 that the IVP has a unique solution that exists in the entire interval (0.
e-f(2J.
EXAMPLE 2
Solve the IVP
xy' .. Since xo = 1
Solution
-x with x + 0. cc).In cosx. ir/2). x). = e-m
we have
f1/
=
1
X2. = e-2 1.
Solution
This is a first-order linear differential equation with
a(x) = tin x and b(x) = sinx. 0) or (0.2y = -x2
YO) = 0. we find
ycos x
1
cosx'
= c .
1
y X2
x I
.)d. Multiplying both sides by the integrating factor.=a-1.
both continuous in the interval (0. is either (-so.d. multiplying both sides by cos x) yields the solution y(x) = (cos x)(c .In cosx).
The differential equation can be written in the form y' .
0<x<2.
and a generator that supplies a voltage V(:) when the switch
Circuit Theory
V(! )
Figure 1.
Consider the RL-series circuit of Figure 1.In x).9. Using the initial condition y(l) = 0.10
.1.4
First-Order Linear Differential Equations
33
Integrating both sides.
yl
x
Figure 1.Inx.
and therefore the solution of the IVP is
y(x) = -x2lnx. some of which we illustrate below. we obtain
y
1
=c .9
APPLICATIONS 1.
The graph of the solution is given in Figure 1.10.4. which contains a resistance R. we find
0=12(c-In1) c=0.1
In the literature there are many occurrences of the applications of linear differential equations.
So. an inductance L. the general solution is
y(x) = x2(c .
Across the inductor the voltage drop is L(dl/di).2
Table of Units
.[c +
F
JV(t)etRcudt
.
Quantity
Resistance
Inductance
Unit
ohm
henry
Symbol
R
L
C
Capacitance
farad
Voltage
volt
V
Current Charge
Time
ampere coulomb
seconds
I Q
sec
Symbols of circuit elements
V
Generator or battery
JW/VN1r
L
R
Resistor
_
Inductor
Capacitor
Switch
Table 1.34
1
Elementary Methods-First-Order Differential Equations
is closed. Across the resistor the voltage drop is RI.1. The current I = 1(t) in the circuit satisfies the linear first-order differential equation
L dl + RI = V(t).2 lists the units and conventional symbols of circuit elements that will be used in this book in connection with circuit-theory applications. V(t) is the only voltage increase in this circuit. 3.1) and the following facts: 1.L). (6) is given by (see
Exercise 19)
I(t) = e-(R. 2.
electric circuits application in Section 1.
(7)
Table 1.
(6)
This differential equation is obtained from Kirchhoff's voltage law (see the
The reader can verify that the general solution of Eq.
y) is given by 2 .11
. = e"' = x. it follows that the solutions of the differential equation
y' = F(x. The differential equation can be written in the
form
I )' + X y=2
and so is a linear first-order differential equation.1. whose slope at the point (x.
(8)
(9)
Let us solve this IVP (8)-(9).
EXAMPLE 3
Solution The differential equation that describes the curve is
y'=2-y. y) are curves y = y(x) whose slope at any point (x.
Figure 1.4
First-Order Linear Differential Equations
With the geometric interpretation of the concept of derivative in mind. passing through the point (1. and soy = x + (l/x) is the curve with the desired properties (Figure 1.)e. 2). y).yix. we should have
y(l) = 2. Multiplying both sides by eIt . we obtain
(xy)' = 2x xy = c + xz
'
From y(l) = 2 it follows that c = 1.
Geometry
Find a curve in the xy plane. y) is equal to F(x. x
Since the curve passes through the point (1. 2).11).
the price of the commodity P(t)
Mixtures
First-order linear differential equations arise as mathematical models in rate problems involving mixtures.
It is left to the reader (see Exercise 26) to verify that the solutions of Eq. Since pounds of salt in the tank at time t concentration = gallons of brine in the tank at time t y(t)
-81+(5-2)t'
. S= -c + dP
(a.
D = a . Let us denote by P the "equilibrium" price.
where "rate in" is the rate at which salt runs into the tank at time t. Clearly." we should first find the concentration of salt at time t. The mixture. To compute the "rate out. Let us illustrate this by means of a typical example. In the simple case where D and S are given as linear functions of the price. Thus.
that is. How much salt is in the tank at the end of 37 minutes?
Let y(t) be the amount of salt in the tank at any time t. and "rate out" is the rate at which salt runs out of the tank at time t.F. c. Then y'(t) is the rate of change of the salt in the tank at time t. Brine containing 3 pounds of dissolved salt per gallon runs into
the tank at the rate of 5 gallons per minute.bP. that is. (10)
is given by
P(t) = [P(0) . kept uniform by
stirring.
(10)
where the constant a is called the "adjustment" constant. runs out of the tank at the rate of 2 gallons per minute. EXAMPLE 4 A large tank contains 81 gallons of brine in which 20 pounds of salt is dissolved. d positive constants). (10) is a linear first-order differential equation. that is. b. that is.S of the demand D and supply S in the market at any time t.
Eq.
Solution
y'(t) = rate in .S). (11) that as t -* -. 15 pounds of salt per minute flows into the tank.P]
+. the price at which
a-bP=D=S= -c+dP. Here
rate in = (3 lb/gal)(5 gal/min) = 15 lb/min. dP =
dt
a(D .rate out.
(11)
Observe from Eq. the amount of salt per gallon of brine at time t.36
I Ebmentary Methods-First-Order Differential Equations
Economics
Assume that the rate of change of the price P of a commodity is proportional
to the difference D .
y 9. Verify formula (5) and apply it to the IVPs of Examples 2 and 3. sin wt. Find the solution of the differential equation (6) subject to the initial condition 1(0) = 4. y _ 1+x2
Y
x
__
X
1 +x2Y
13.xy'+6y=3x+1
10.R" le teen
where 1o is the current at time t = 0. R = 5 ohms. V = 8 volts.
20. where Vo and
w are given constants. y2dy + y tan x dx = sin' x dx
7. Verify Eq. (7) of this section. Assume that the voltage V(t) in Eq.
21. and
1(0) = 0 amperes.y = 2x2
6. y) is given by
2-y+e X x
16. xy' . e dx + x' dy + 4x2y dx = 0
1x y-y2=0 sin
11.
18. In an RL-series circuit assume that the voltage V(t) is a constant Va. Find the current at the end of 0. Find a curve in the xy plane that passes through the point (1.1 seconds.
where
IX0. L = 4 henries. 1 + e) and whose slope at any point (x. Find the solution of the IVP
y'-y=b(x)
Y(O) = 1. ydx+x=Y
15. What will the current be after a very long time?
22. (6) is given by V. Show that the current at any time t is given by 1(t) = e° +
(1o
.
17.38
1
Elementary Methods-First-Order Differential Equations
5.
.xy'+y=x5
12.
b(x) =
x<0
x >_ 0. y'+
3
X-y=(x-1)'
8.yy'-7y=6x
14. In an RL-series circuit.
19. Compute the orthogonal trajectories of the family of concentric circles
x2+y2=R2. .
In an RL-series circuit [see Eq.
28.n)b(x). A large tank contains 40 gallons of brine in which 10 pounds of salt is
dissolved. 27.2x
y(e) = -1
Y(0) = -1
36.Ordar Unaar Difbnntlal Equations
39
23. (a) How much salt is in the tank at any time t? (b) Find the amount of salt in the tank at the end of 1 hour. The mixture is kept uniform by stirring and the well-stirred
mixture flows out of the tank at the same rate as the inflow. runs out of the tank at the rate of 3 gallons per minute.xy = (1 . V(t) = 3 sin t.2xy = 4x y'°
Y
In x
=y
2
In Exercises 32 through 37 solve the initial value problem. The mixture. xy' + y = 2x
y(2) = 2
34.
n * 0. R = 6 ohms.
32. and 1(0) = 10 amperes. kept uniform by stirring.
Show that the transformation w = y' -" reduces the Bernoulli differential equation to the linear differential equation
w' + (1 . y' + Xy= -2xy2
31 .n)a(x)w = (1 . (x-1)y'-3y=(x-1)'
y(-1) = 16
.
In Exercises 28 through 31. Verify Eq. Compute the value of the current at any time r.4
Flrst. Starting at time t = 0 brine containing 3 pounds of salt per gallon flows into the tank at the rate of 2 gal/min.1. A tank initially contains 10 gallons of pure water. (11) of this section. (1-x)y'+xy=x(x-1)2
y(5) = 24
37. The Bernoulli equation is the differential equation
y' + a(x)y = b(x)y". (6)] it is given that L = 3 henries.1.
24. How much salt
is in the tank after 5 minutes? How much salt is in the tank after a very
long time?
26. y' . xy ' -
1
y ' . xy'
In x = 0
35. y' 30 .x2)e°11i'2 y(0) = 0
33. Brine containing 2 pounds of dissolved salt per gallon runs into the tank at the rate of 4 gallons per minute. (1 + x2)y' + 2xy = .
XI
y= -
2y
1
29.
25. solve the Bernoulli differential equation (see Exercise 27).
and B are constants.T. Show that the general solution of the differential equation
y'-2xy=x'
is of the form
y = Y0(x) + Ae'.
where Y. S.
41. Monthly 63 (1956): 414. That
is. A body of temperature 70°F is placed (at time t = 0) outdoors.
. Show that the general solution of this differential equation is
T(t) = T + ce-"'. 25 (1963): 251 72.
where k > 0 is a constant of proportionality." Bradner and Mackay. Math. This model gives a good
approximation of the true physical situation provided that the temperature difference is small. A.
(a) How long will it take the body to cool to 50°F? (b) What is the temperature of the body after 5 minutes?
"H. Bradner and R. Physics Newton's law of cooling states that the time rate of change of the temperature T = T(t) of a body at time t is proportional to the difference T . Bull.
dT
dt
_ _ k(T. "Amer. YO (x)
[Hint: Find the general solution of the differential equation and observe that the improper integral jo t'e-"dt converges.
b
39. Math.40
1
Elementary Methods-First-Order Differential Equations
38. Show that the general solution of this differential equation is given by
B_
where-c is an arbitrary constant. Biophys.T j.(x) is a solution of the differential equation that obeys the inequalities
-x
2
1
4x
-Y(x) _ -x xx ?2. Mackay.
where a. After 3 minutes the temperature of the body has fallen
to 60°F. Biophysics In a study related to the biophysical limitations associated with deep diving. in temperature between the body and its surrounding medium. where the temperature is 40°F.'S
40. obtained the following first-order linear ordinary differential equation
y'-Ay=B+be-°`. b. generally less than about 36°F.
Math.
Mass Behavior In a research article on the theory of the propagation of a single act in a large population (in other words. A Riccati equation is a differential equation of the form
y' = f(x) + g(x)y + h(x)y'. Z. Show
that if y. "For a detailed study of the differential equation (14). where the functions f. y)dx + N(x.03.
. Solve this Riccati differential equation in each of the following cases.
(2)
A. such as suicide). x (t) is the external influence or stimulus and the product by is the imitation component.y)[x(t) + by]. Rapoport.(x) is a particular solution of the Riccati differential equation. 17 (1965): 7-13."] 43. Bull.4
First-Order Linear Differential Equations
41
42.
1. see the paper by J. [Hint: y = 1 is a particular solution. Hearon. Biophys. x (t) =
t
10'b
10
44.1. that is (suppressing the arguments x and y)
df=Mdx+Ndy. 14 (1952): 159-69. y) whose total differential is equal to M(x.b=0.
(14)
Here y = y (t) is the fraction of the population who performed the act at time t. Bull.(x) +
w(x)
reduces it to the linear first-order differential equation
w' + [g(x) + 2y. Rapoport" derived the following Riccati differential equation:
y = (1 . x(t)= -tlI t
-0. an act that is performed at most once in the lifetime of an individual.03.(x)h(x)]w = -h(x). and h are continuous in some interval I. then the transformation
1
Y(x) = Y. y)dx + N(x. Biophys. Math. y)dy.5
EXACT DIFFERENTIAL EQUATIONS
A differential equation of the form M(x. y)dy = 0
(1)
is called exact if there is a function f(x. g.
is there a systematic way to solve it?
Answer 1
The differential equation (1) is exact if and only if
M.
Thus. For example.xy + 2y2)dy
= (2x .y)dx + (-x + 4y)dy = 0
is exact because
d(x2 . N. (4) is given (implicitly) by
x2-xy+2y'=c.xy + 2y') = d (x2 .. conversely. (1) is exact.
QUESTION 1
(5)
Is there a systematic way to check if the differential equation
(1) is exact?
QUESTION 2
If we know that the differential equation (1) is exact. it suffices to prove that
. (1) is
given by
fix. then Eq.) If Eq. then [because of (2) and (1)] it is equivalent to
df=0. if the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x. yo) in the region where the functions M. y)dy = c
o
J
o
(7)
gives (implicitly) the general solution of the differential equation (1). Y) = c. (1) is exact. the general solution of Eq. if Eq. the differential equation
(3)
(2x . then (6) holds. Y) = jF M(x. To justify the answers above.xy + 2y2)dx +
ay
(4)
(x2 . provided that the partial derivatives of the function f with respect to x and y
exist. the function f(x. y0)dx + r N(x.
Thus. are continuous.
Answer 2 Choose any point (x(. (1) is exact.y)dx + (-x + 4y)dy.
The following two questions are now in our minds. = N. y) is a constant and the general solution of Eq.42
1
Elementary Methods-FirstOrder Differential Equations
(Recall that the total differential of a function f is given by df = fdx + frdy. N and their partial derivatives Mr.-
(6)
That is. Then
AX.
y = M y and f . N and
My. (7) should be
f(x. . (7) is useful to establish the existence of the function
f and also to verify the test for exactness. yo) to (x. Students sometimes find the presence of the "arbitrary" point (xo. that is Eq. then (6) holds. exist and are continuous. y)dy
yo yo
= M(x. y() in Eq.
EXAMPLE 1
Solve the differential equation
(3x2 + 4xy2)dx + (2y .
Eq. which proves the "if" part of Answer 1. yo)dt + j y N(x. yo) and the judicious selection of
values for xo and yo confusing. there exists a function f such that df = M dx + N dy f .
Solution
Here
M(x. yo) to (x. In fact. y) = 3x2 + 4xy2
''
and N(x.
Ni. y() in the first integral of (7) and the evaluations of both integrals at the lower points of integrations. yo)
and from (x. = 8xy and N.3y2 + 4x2y M. y)dy = M(x.Y)
0
and similarly fy = N(x.3y2 + 4x2y)dy = 0. Consequently alternative systematic methods for the determination of f are given in Examples 3 and 4.1. y) = 2y . = N. .. df = f dx + fdy = M dx + N dy.
REMARK 3 As shown. Therefore. N.. y) lie in the region where the functions M. (7) we have
f = M(x.
. To prove the converse. (7) may be chosen judiciously for the purpose of simplifying M(x. = N
M. yo) + Jy My(x. =f .
REMARK 1
REMARK 2 The precise form of Eq. Y) =
JM(t.Y)Iiy = M(x.y). since
f = M and f . (6).Y0) + M(x. = N and since f. we have
The point (xo.
where f is the sum of the two integrals in (7) and conversely if Eq. from Eq. (1) is exact. Yo) + Jy N (x. observe that. s)ds = c. Eq.
implies that
df = Mdx + N dy. (1) is exact.5
Exact Differential Equations
43
M. = 8xy My = N.
o
(T)
and the assumption should be made that the segments from (xo.
= N.
and the general solution is given (implicitly) by
x3 + 2x2y2 + y2 .
EXAMPLE 4
Solve the differential equation (3x2 + 4xy2)dx + (2y . = N with respect to y. = 4x2y + h'(y). = N.
From Example 1 we know that this equation is exact.3y2 + 4x2y)dy = 0. y) = x3 + 2x2y2 + y2 . Regrouping terms in accordance with Remark 5.y3
f(x. = M and f.3y' + 4x2y)dy = 0. This omission is permissible since otherwise it would be combined with the constant c when we set f = c.
As an alternative construction we could have integrated the expression f. Consider
Solution
f..3y2 + 4x2y).
The "constant" of integration is a function of y due to the partial integration with respect to x. Taking the partial derivative with respect to y. we have
[3x2 dx + (2y .
Note that the solution for h(y) omitted a constant of integration which normally would appear.y3.
Solution From Example 1 we know that this equation is exact.) = c.1.
. this last expression yields
f.5
Exact Differential Equations
45
of the form p(x)dx and q(y)dy and the other the remaining terms-and recognizing that each group is (in fact) a total differential of a function.
Integrating with respect to x we obtain
f = x3 + 2x2y2 + h(y).
EXAMPLE 3
Solve the differential equation
(3x' + 4xy2)dx + (2y .3y2
Thus. and g(x) would be determined by setting f = M. To determine h(y) we utilize the fact that f.
Setting this equal to N (N = 2y .
h(y) = y2 .y3) + d(2x2y2) = 0
d(x3+y2-y3+2r2y2)=0 x3+y2-y3+2x2y2=c. and therefore there exists a function f such that f.3y2)dy] + (4xy2 dx + 4x2y dy) = 0
d(x3 + y2 . In this case the "constant" of integration would be a function of x. say g(x). = M = 3x2 + 4xy2. we have
h'(y) = 2y .
4AC =
(2)2
. (13) is equivalent to (the total differential)
(13)
Zd(-x2+y2+4x +4y+Zxy)=0
. the rotation angle 0 is
"G.x + y + 2)dx + (x + y + 2)dy = 0. (3y2 .y2 . or ellipse depending on whether
B2 . the slope of the orthogonal trajectories of (10) is
x-y-2
Y
x+y+2'
11
)
Equation (11) is the differential equation of the orthogonal trajectories of the curves (10).
we recognize that Eq. B2 . 1972). Differentiating with respect to x.46
1
Elementary Methods-First-Order Differential Equations
APPLICATION 1. Thus.4x .2d(x2 .aye = a p. p.: Addison-Wesley.4y = const. B2 .4x . parabola. and xy = y with at > 0.4(1)( -1) = 8. R > 0.4AC is positive.2)2 -. Mass. The
standard forms for hyperbolas are axe . and y arbitrary.2xy . In general. or negative.4y .1
Geometry
Compute the orthogonal trajectories of the one-parameter family of curves
x2+2xy-y2+4x-4y=c. Equations (10) and (14) can be
written in these forms by the transformations of variables known as rotation of axes and translation of axes (in that order).
and so
(10)
Solution First. = k. This equation is not separable or linear. and for the family (14). B..
(14)
The curves given by Eq. both of these families consist of hyperbolas.y2 . Alternate Edition (Reading. 508. Calculus and Analytic Geometry.4AC = (. we obtain
y'=
x+y+2
x-y-2
Thus.5.
Thus.4y' = 0. but after writing it in the
form
( .
2x + 2y + 2xy' ." For the family (10). (12) in the form
(-xdx + y dy + 2dx + 2dy) + (y dx + xdy) = 0. we should compute the slope of the curves (10).
(12)
we recognize it as an exact differential equation. Thomas. We recall from analytic geometry that an equation of the form Axe + Bxy + Cy2 + Dx + Ey + F = 0 represents a hyperbola. (14) are the orthogonal trajectories of (10).axe = a(3.
.2xy) = 0. the general solution of the differential equation (13) is
x2 . zero. Regrouping the terms in Eq. Jr.2yy' + 4 .4(1)(-1) = 8.
(x-y)dx+(-x+y+2)dy=0
y(1) = 1
19. The function µ is then called an integrating factor of the differential equation (25).
Y'=
y(1) = -1
y-x+1 -x+y+3
(25)
y(1) = 2
21.5
Exact Difarential Equations
49
15. = (µN). . If it happens that the expression
N (M. Integrating Factors If the differential equation
M(x.Mµ. . M. and E are arbitrary (but fixed) coefficients. that is. (25).
.1. y) is an integrating factor of the differential equation (25) if and only if it satisfies the partial differential equation
Nµ.. we
can solve it. that is. it is very difficult to solve the partial differential equation (27) without some restrictions on the functions M and N of Eq. (27) into a first-order linear differential equation whose solutions can be found explicitly.M dx + µN dy = 0 (26) is exact. Show that a function µ = µ(x.By . (x2 + y2)dx + 2xy dy = 0
18. C. y) dy = 0
is not exact.
(24)
16. (µM).. D. In this and the following exercise the restrictions imposed on M and N reduce Eq.. (a) Under what conditions will the differential equation (24) be exact?
(b) Using the condition obtained in part (a).)µ. (x+y)dx+(x-y)dy=0
y(0) = 2
20. find the equation of orthogonal trajectories and show that they and the family (23) are hyperbolas. as a function of x only. Show that the differential equation for the orthogonal trajectories of this family is
(Bx + 2Cy + E)dx + (-2Ax . y) dx + N(x.
. = (M.D)dy = 0...
(27)
22. B. Since (26) is exact. * N we can sometimes find a (nonzero) function µ that depends on x or y or both x and y such that the differential equation
p. and its solutions will also satisfy the differential equation (25). Solve the following initial value problems:
17. In general.
(23)
where c is the parameter and A.N)
is a function of x alone. Consider the one-parameter family of curves
Axe + Bxy + Cy2 + Dx '+ Ey = c. it is always possible to choose p.N. Show that with these assumptions the function
W(x) = -r(11N)(My-Ns)d:
is an integrating factor of the differential equation M dx + N dy = 0..
is an arbitrary constant. dy
(9)
which is linear.x' + c.d = F (y. (3) is of order 3. Eq. Sometimes this latter equation can be solved by one of our earlier methods. y')
transformation
(6)
(they should not contain x) can be reduced to first order by means of the
w=y'.x + c. Eq. (3) is
y(x) = c.e-Y + c and so
y'(x) = w = -y .. from (7) we obtain (using the chain rule)
(7)
y" = dx =
Thus.
In fact. Eq. For example.I + c. Multiplying both sides of (9) by a-Y we obtain (d/dy)(we-Y) _ ye-Y. w). (8) becomes
Solution
dw _ w = Y.
Y
which is first order with independent variable y and unknown function w. its general solution (5) contains the three arbitrary constants c c2. and as we have seen.56
1
Elementary Methods-First-Order Differential Equations
Another integration yields y = b c.
.
(5)
The general solution of an ordinary differential equation of order n contains n arbitrary constants. As c.
B.
W.
EXAMPLE 2
Solve the IVP
y"=Y'(y+Y)
Y(O) = 0
(8)
Y'(0) = -1.eY. In fact. (6) becomes
dw
dy
dw dy _ dw dx . the general solution of Eq..
Here the differential equation (8) does not contain x and therefore can be reduced to a first-order differential equation by means of the transformation w = y'.x' + czx + c.
REMARK 1
and c3. Integrating with respect to y we find we-Y = -ye-Y .dy w. Second-order differential equations of the form
y' = F (Y.
1. Find the concentration in each compartment at any time t.y(4) = 0
. =
k
V. Dividing by the volumes of the compartments. and
between
Two
Compartments
c.(t) = k(c. (8) is
y(x) _ . we find that
k
(1
1V2)
k
or
//
11
i
V+i
V.c.
(c2 -. we find that c.
V2II/
(11)
Equation (11) is a differential equation of form A and can be reduced to a firstorder differential equation. from the higher concentration to the lower. respectively. = 0.c2). + k l
\V. .
l
+
l l c.c2)
y'
V2
Differentiating both sides of the first equation and using the second equation. = 0. Through the barrier a solute can diffuse from one compartment to the other at a rate proportional to the difference c. at time t. we find that c
1.
y'= -y -l
and y(x) _ -1 + ce '. Thus. Using the initial condition y(O) = 0. and A2. . and A2.
yzt)
are the concentrations in compartments A.(t) = V. y' + y' = 3
2.)
and
c2 =
k
2
(Cl . and V2. and A respectively. of volumes V. and the solution of Eq.l + e-'.(t) be the amount of the solute in the compartments A. The differential equations describing the diffusion are
y.c2 in concentration of the two compartments. we get
c. and 7.7.
where k is a constant of proportionality (k > 0).(t) and y.
APPLICATION 1.) and Y2 = k(c.c. 6. y(5) .1
Assume that two compartments A.
EXERCISES
Compute the general solution of the following differential equations. Then
ca(t) =
ye(t) V. We leave the computational details for Exercises 5.
v'
(10)
e. respectively.
Solution Let y.1. .
Chemistry
Diffusion
are separated by a barrier.7
Equations Reducible to First Order
57
Using the initial conditions.
14. Show that
1(t) __ R2 V
+ww2L2 e' .14
28. A colony of bacteria increases at a rate proportional to the number present. is an angle defined by
t sin .60
1
Elementary Methods--First-Order Differential Equations
25. How long does it take this colony of bacteria to double?
30.
where
26. Find the solution of the IVP
Y'+1y=b(x)
y(1) = 0. Compute the orthogonal trajectories of the one-parameter family of curves x2 + b2y2 = 1.y =
wL
(R2 + w2L2)" 2 '
cos W = (R2
.1
R
+ w2L2)1R
Note that the current 1(t) is the sum of two terms. Circuits In the RL-series circuit shown in Figure 1. A colony of bacteria increases at a rate proportional to the number present. For this reason the first term is called the transient and the second term the steady state.} (R2
VO
+
sin (wt -
where 4. as t --o m.
27. What will the amount be after 6 years?
. The sum of $4000 is invested at the rate of 12% compounded continuously. sin wt and 1(0) = 0. assume that
V(t) = V.
Figure 1. If the number of bacteria doubles in 3 hours. In 1 hour their number increases from 2000 to 5000. how long will it take for the bacteria to triple? 29. the first term dies out and the second term dominates.
Harvey. 0 1971 Addison-Wesley Publishing Company. A large tank contains 30 gallons of brine in which 10 pounds of salt is dissolved. 2nd ed. What was the age of the tomb? 36. Reprinted with permission of Addison-Wesley Publishing Company.
"This is Exercise 5-33 in J. In Figure 1. In the circuit20 shown in Figure 1. Find a curve in the xy plane. 33.62 disintegrations per min.15. Brine containing 2 pounds of dissolved salt per gallon runs into the tank at the rate of 3 gallons per minute. Kuh. G. It is known that the half-life of radiocarbon is 5568 years. H. Inc. A. A. this '°C decays with a half-life of 5568 years. passing through the origin and whose slope at any point (x. Reprinted by permission of McGraw-Hill Book Company. Radioactive "C is produced" in the earth's atmosphere as a result of bombardment with cosmic rays from outer space. and C2 is zero. [Note:
1µf = 10-6 f. Shearer.
. (a) How much salt is in the tank at any time t? (b) How much salt will be in the tank after 10 minutes?
34. Basic Circuit Theory (New York: McGrawHill Book Co. Desoer and E. The radioactive carbon is taken up by living things.
"This is Exercise 9 in B. kept uniform by stirring. 169.. knowing that i(0) = -10 mA and given R = 500 fl. L. calculate and sketch the current i for
t ? 0. per g. On their death.: Addison-Wesley.16. [Note: ImA = 10-'A and 1mH = 10-' H]
Figure 1. 54. Reprinted by permission of Prentice-Hall. Mass. S. The mixture. y) is given by y + e`. and H. with the result that they contain 10 disintegrations per minute of "C per g of carbon. and no fresh '°C is absorbed.15
35. Richardson.: Prentice-Hall. Initial voltage across C. N.
(Englewood Cliffs. 1971). L = 10 mH. p.Review Exercises
61
31. 1969). Introduction to
System Dynamics (Reading. Murphy. How old is a wooden archaeological specimen which has lost 15% of its original radiocarbon?
32.]
-This is Exercise 12a in C. Introduction to Nuclear Physics and Chemistry.J. T. find and sketch" the voltage v2 across C2 as a function of
time after the switch is closed. and e = 8 volts. Inc. p. 1 kilohm = 10' fl. runs out of the tank at the rate of 2 gallons per minute. 1969). 143. Carbon derived from wood taken from an Egyptian tomb proved to have a 10C content of 7. p.
Publishers. Burghes and A.1)
(Hint: The first has (x + y)
as an integrating factor. integrate this equation
and show that
dv
v2=u2+ 2f . eagle. Modern Introduction to Classical Mechanics and Control (West Sussex. 1975).
41. 109. If v = u at x = 0. and sparrow are in the air. Math. A ball" is thrown vertically downwards from the top of a tall building.4 in D.28 The eagle is 50 feet above the sparrow and the hawk is 100 feet below the sparrow. p. 31. Downs. England: Ellis Horwood.]
'"This is Problem 2. A hawk. "This is Example 5. A particle26 is moving on a straight line with constant acceleration f.x. The sparrow flies straight forward in a horizontal line. p. Reprinted by permission of Ellis Horwood Limited. Monthly 50 (1943): 572. M. "From Amer. show that the equation dtxldt2 = f can be written as
vd-=f
where v is the particle's speed. The hawk flies twice as fast as the sparrow. How far does each fly and at what rate does the eagle fly?
43. Monthly 40 (1933): 436-37. Solve the differential equations39
x2-y2+1
Y
x2 . N. "From Amer. show that if the building is sufficiently tall. the ball's velocity on hitting the ground is approximately independent of its initial speed.17
40. 42. Reprinted by permission of Ellis Horwood Limited.
Assuming a model with constant gravity and air resistance proportional to its speed.
. Publishers. the second is exact. The hawk and eagle reach the sparrow at the same time. If x is the distance from some fixed point on the straight line.Review Exercises
63
Switch
Figure 1.y2
1
'
y
y(2x+y-1)
x(x + 2y . Classical Mechanics.1 in Burghes and Downs. Both hawk and eagle fly directly towards the sparrow. Math.
A cat30 is running along a straight edge of a garden.17 in Burghes and Downs. Find the time that elapses before the cat is caught and show that the cat runs a distance (2/3)b before being caught. A dog.
. L = 2 henries.volts. Reprinted by permission of Ellis Horwood Limited. Classical Mechanics. p. 34. Find the current at any time t. In an RL-series circuit. The dog immediately chases the cat with twice the cat's speed in such a way that it is always running towards the cat.64
1
Elementary Methods-First-Order Differential Equations
44. and 1(0) = 5 amperes. sees the cat when it is at its nearest point.
'This is Exercise 2. R = 4 ohms.
45. V(t) = e. sitting in the garden at a distance b from the edge. Publishers.
. ( x ) . . y'.(x)Y' + au(x)Y = f(x)
(1)
. y")... tion f(x) are continuous functions in some interval I and that the leading coef0 for all x E I. Eq.y = 0. If all the coefficients we speak of Eq. (1) is called
a. (1) as a linear differential equation with constant coefficients..
. the second
because of the term yy'. .
The first differential equation is nonlinear because of the term y2. The interval I is called the interval of definition ficient of the differential equation.(y'/3!) + (yb5!) . otherwise.(x). aside from applications.. The following are nonlinear differential equations:
y"+y2=sinx
y"+siny=0..
x
0
(2)
y" + 2y' + 3y = cos x
Y") . When the function f is identically zero. (1) is homogeneous.. Equation (4) is a homogeneous linear differential equation of order 4 with constant coefficients.1. and the third because of the term sin y [recall that
sin y = y . . Equation (3) is a nonhomogeneous linear differential equation of order 2 with constant coefficients.. -. ...CHAPTER 2
Linear Differential Equations
2.
(3)
(4)
Equation (2) is a nonhomogeneous linear differential equation of order 1 with variable coefficients. we say that
Eq..(x)y") +
a.2y = x'.ao(x) and the funcWe always assume that the coefficients a"(x).. The following are examples of linear differential equations:
xy' . is to develop the elements of the theory of solutions of linear differential equations and to discuss methods for obtaining their general solution. When f(x) is not identically zero.
Our main concern in this chapter. it is a linear differential equation with variable coefficients.1
INTRODUCTION AND DEFINITIONS
A linear differential equation of order n is a differential equation of the form a.. nonhomogeneous.(x) are constants. a. . The term linear refers to the fact that each expression in the differential equation is of degree one or degree zero in the variables y.
(2) is easily found to be
y(x) = x' + cx'. = -a sin at ' y. the variable coefficients of a differential equation can be approximated by constant coefficients. let us consider the second-order linear differential equation
y+a'y=0.
Y(0) = v0. is a solution. no matter what the order of the differential equation is. (6) reduces the differential equation to an identity.
(6)
(7)
As is customary. In fact. For the sake of motivation. In fact. (6) is Eq. This was done in Section 1.4. in "small" intervals. (6) yields -a' cos at
+ a' cos at = 0. which will give a better approximation. Another way. we shall learn in Sections 2. called "qualitative theory of differential equations. For example.4 of Chapter 1.8). Hence. _ -a' cos at. A direct substitution of the power-series solution into the differential equation will enable us to compute as many coefficients of the power series as we please in our approximation. However.66
2
Linear Differential Equations
Of course. Furthermore (subject to some algebraic difficulties only). there is no way to find explicitly the general solution of a second-order (or higher) differential equation with variable coefficients unless the differential equation is of a very special form (for example. It will be shown in the applications that the initial value problem (JVP) (6)-(7) describes the motion of a vibrating spring [Eq. 2.9 and Chapter 5).
(5)
where c is an arbitrary constant. For example. y. = cos at y. dots stand for derivatives with respect to time t. A practical way out of this difficulty is to approximate the solutions of differential equations with variable coefficients.11 how to find the general solution of any linear differential equation with constant coefficients. (10) with a' = K/m]. and 2. the general solution of Eq. the reader should be told that in more advanced topics on differential equations." one can discover many properties of the solutions of differential equations with variable coefficients directly from the properties of these coefficients and without the luxury of knowing explicitly the solutions of the differential equation.(t) = cos at when substituted into Eq. It is easy to verify that the function y. of the Euler type. In a similar fashion we
can verify that the function y2(t) = sin at is also a solution of the differential
. and substitution of the results into Eq. is to look for a power-series solution of the differential equation with variable coefficients (see Section 2.
together with the initial conditions
Y(0) = Y..7) or unless we know (or can guess) one of its solutions (see Section 2.5. matters are not so simple when we try to solve linear differential equations with variable coefficients of order 2 or higher. y. Section 2. or 0 = 0. and hopefully the solution of the differential equation with constant coefficients is an "approximation" to the solution of the original differential equation with variable coefficients. we already know how to find the general solution of any linear differential equation of order I with variable or constant coefficients. Finally.
a solution to the initial value problem is
y(t) = yo cos at + ° sin at.
APPLICATIONS 2. It would be natural to ask what motivated us to try cos at as a solution.1
Linear differential equations occur in many mathematical models of real-life situations.1
Introduction and Definitions
67
equation (6).2. for example. Newton's second law of motion.y. The preceding comments serve to distinguish linear homogeneous differential equations from other types of differential equations. then y(t) as given in (8) will be a solution of the initial value problem (6)-(7). the linear combination
Y(t) = c1Yi(t) + c2Y2(t)
(8)
of two solutions is also a solution.y2 also solves the initial value problem (6)-(7). in general. (9) is the only solution of the IVP (6)-(7). although they satisfy the differential equation (6). and y2(0) = a. Thus. + c. + c2y2 is also a solution.. It seems reasonable. Now y(t) satisfies the differential equation (6) and will satisfy the initial conditions provided that c. do not. the sum of the two solutions is also a solution. =
e'°'. c2 = 2'y + (i/2a)v.
Note that the functions y. and c2? [For example. to suspect that if the constants c..
vo = ac2. For linear homogeneous differential equations we have the principle that if y. this system takes the form
Yo = c.
a
(9)
In Section 2. c. involves a second
. and c2 satisfy the system of algebraic equations
Yo = Y(0) = c1Y1(0) + c2Y2(0)
ve = y(0) = c'yi(0) + cryz(0)
Since y. then c. and any constant multiple of any one of these solutions is again a solution.
Hence. and c2 are chosen properly. where i2 = . if y.3 we establish that Eq. satisfy the initial conditions. then the linear combination c. Or we might ask whether there are simple solutions other than cos at and sin at.2. y2 = e-'°'.yo .(i/2a)v.] The answers to these questions will be obtained as we proceed through the remainder of this chapter.1. [The reader can verify that e`°' and e-'". and c2. and c2 are arbitrary constants. are solutions of Eq. since y. and y2 produce different values for c. or would some other choice
of y. y. = . This result and its proof are presented as Theorem 1 of Section
2. y. y2(0) = 0. however. = cos at and y2 = sin at. Furthermore. for any constants c.(0) = 0( # vo) and y2(0) = 0 (# yo). and y2 are solutions and if c. (6)] How many "different" solutions are there to the differential equation (6)? Is (9) the only solution to the initial value problem.(0) = 0.(0) = 1.1.
Mechanics
The Vibrating Spring
A light coil spring of natural length L is suspended from a point on the ceiling [Figure 2.
(10)
Recall that the mass was set to motion by first displacing it to an initial
equilibrium
(c)
Figure 2.1
. It is this motion that we wish to investigate. the acceleration of the mass). K.
The differential equation that describes the motion of the mass m is
my = . Assume that at the equilibrium position the length of the spring is L + 1. with the constant of proportionality. According to Hooke's law. the force needed to
produce a stretch of length I is proportional to 1. and the
positive direction for the vertical (y) axis is downward. There are two forces acting upon the mass: one is the force of gravity. we have
my=F.Ky. and at that point we give it a vertical (initial) velocity (directed up or down). Let y = y(t) be the position of the mass at time t. equal to mg. and consequently linear differential equations of second order have been of primary interest in motion problems. we choose the origin 0 at the equilibrium point.1(b)]. Then by Newton's second law of motion. which by Hooke's law is K(1 + y). That is.68
2 Unear Differential Equations
derivative (the acceleration).
where the double dots denote second derivative with respect to time (in other words. Naturally the stretch 1 is caused by the weight mg of the mass m.
or
y+mKy=0. setting the mass in a vertical motion [Figure 2. known as the spring constant. For convenience.1(a)].
F=mg-K(l+y)= -Ky. Next we displace the mass vertically (up or down) to a fixed (initial) position. mg = Kl. Thus (since mg = Kl).1(c)]. and the other is the restoring force of the spring.
We attach a mass m to the lower end of the spring and allow the spring to come to an equilibrium position [Figure 2.
The IVP describing the motion of the mass is [from Eqs. Also.. and then giving it an initial velocity v0. homogeneous. dyn/cm = 30 dyn/cm.. (11) yo is positive if the initial position is below the equilibrium and negative if it is above. The differential equation (10).3
y(0) = From Eq. (10) and (11)J
y+. say y. together with the initial conditions (11).2.
Solution
Using Hooke's law we compute the spring constant:
K = . (9) the solution of this IVP is
y(t)
5.1
Introduction and Definitions
69
position.
3 cos tt + 5
sin vl0
t
. constitute an initial value problem.. vo is positive if the initial velocity was directed downward and negative if it was directed upward. Equation (10) we recognize to be a second-order.1 we list the units used in the two most common systems of
measurement.'n''a.1
EXAMPLE 1
Units of Measurement
A spring is stretched 12 cm by a force equal to 360 dyn.
Cgs system
English system
Distance
Mass
Time
centimeter (cm) gram (g)
seconds (sec)
foot (ft)
slug
seconds (sec)
Force
dyne (dyn)
cm/sec
pound (lb)
ft/sec
Velocity Acceleration
cm/sect
ft/sec'
Spring constant Gravity (g)
dyn/cm
980 cm/sec'
lb/ft
32 ft/sect
Table 2. linear differential equation with constant coefficients./=0
y(0) _ . A mass of 300 g is attached to the end of the spring and released 3 em above the point of equilibrium with initial velocity 5 cm/sec directed downward. In Eq. together with the differential equation (10). In Table 2. we have the two initial conditions
Y(0) = vo (11) Y(0) = Yo. Thus. Find the position of the mass at any time t.
To compute G(x) and G(x + Ax). Then the amount G(x) of gas diffusing into S at the point x minus the amount G(x + Ax) diffusing out of S at the point x + Ax must be equal to the amount of gas disappearing in S by the chemical
reaction. given that we know the concentration at the points 0 and 1. Assume that the gas reacts chemically with the liquid and that the amount H(x) of gas that disappears by this reaction is proportional to y(x). at any point x.
G(x) = -Dy'(x).
which states that the amount of gas that passes through a unit area per unit time is proportional to the rate of change of the concentration. we use Fick's law of diffusion. Similarly. where the vibrating spring is discussed in greater detail.
(16)
[ . Therefore. Suppose that this process takes place for such a long period of time that the concentration y(x) of gas in the pipe depends only on the distance x from some initial point 0 (and is independent of time).
(15)
Equation (12) gives the amount of gas that disappears at the point x because of the chemical reaction of the gas with the liquid.4.
(14)
where D > 0 is the diffusion coefficient and the negative sign justifies the fact that the gas diffuses from regions of high concentration to regions of low concentration.
(12)
We want to compute the concentration y(x) of gas in the liquid.
H(x) = ky(x). Thus.Dy'(x)] . That is.Dy'(x + Ax)] =
Jky(s)ds.[ .
(13)
To solve this problem we consider a small section S of the pipe between the point x and x + Ax (Figure 2.
Diffusion
Suppose that a gas diffuses into a liquid in a long and narrow pipe. say
y(O) = A
and
y(I) = 0. That is.2
.
G(x + Ax) = -Dy'(x + Ax).
(17)
Figure 2. the total amount of gas that disappears by the chemical reaction in the section S is given by
Jf * ky(s)ds.2).70
2
Linear Differential Equations
The reader is urged to read Exercise 40 of Section 2.
we obtain the second-order linear differential equation
y
Dy=O. with constant coefficients or with variable coefficients.xy = 1. lowercase roman numerals. Use Eq. (1 + x2)y" .2. x > 0 19. (sin x)y' + e''y = 1
9.yy' =3
18.
EXERCISES
In Exercises I through 19. Taking limits on
both sides of Eq. 7y' . classify the linear ones as homogeneous or nonhomogeneous. y"+xy=x
12.y"'-2=0
15.
.6y = 0 17. and state its order.
y'(x+Axx)-y'(x)=
fly (x+ 2x). e'(y')' + 3y = 0
13. xy" + 4xy" .xy = 0
20. The system is then set to motion by pulling the mass 4 cm above the point of equilibrium and releasing it. y"+xzy=e' 4.)
1. classify the differential equation as being either linear
or nonlinear. with an initial velocity of 2 cm/sec directed downward.2 = 0
11. Furthermore. (18) as Ax-b 0. y"+cosy=0
6.4y' + xy = 0
10. ys) . x2y" . (17) becomes
Ax.2y(4) + y = 2x' + 3
2. A spring is stretched 2 cm by a force equal to 16 dyn.
(18)
where y[x + (Ox/2)] is the value of y at the midpoint between the limits of integration and Ax the length of the interval of integration. y' + y" = 3x
7.1
Introduction and Definitions
71
A good approximation to the integral in (16) is given by ky(x + Ax/2) Thus after some rearrangement Eq. y" + 5y' .xy' + y 0 8. 2y"' + 3y" . (9) to find the solution of this IVP. y'+yy"=5
5. y.e'y' . at time r = 0. 7y"+3xy' . A mass of 32 g is attached to the end of the spring.4y 0
16.2y-3xy'+4y' =0
14.
k
(19)
The problem consisting of the differential equation (19) and the "boundary" conditions (13) is referred to as a boundary value problem (BVP).+xy=0 3. or by arabic numbers in parentheses. Boundary value problems for ordinary differential equations are discussed in Chapter 6 and boundary value problems for partial differential equations are discussed in
Chapter 11. (Note that derivatives are denoted by primes. Derive an IVP that describes the motion of the mass.
f(x) <_ M. 23.
y2(T) + [y'(p)]2 = y2(o) + [y'(o)P. If we recall a technique from trigonometry [that an.).[y'(0)]2. vo = 0 in. by a 4-lb weight./sec
23.
y(t) = yo cos
1K
t+
vo1/m
sin
fK
m r.. K = m. at time t = 0. vo = 4 cm/sec 25. below the point of equilibrium and releasing it.
The solution of the vibrating spring is given by Eq. The system is then set to
motion by pushing the mass (note that mg = weight and that g = 32 ft/sec')
6 in.5 in. A function f is said to be bounded' as x -> x if there exists a positive
number M such that lim. 7 (1967). consequently. ya = 5 in. Use Eq. = 10V cm/sec 24.expres-
sion of the form a cos t + p sin t can be written as A cos (t .
Jo
Hence.
. with
an initial velocity of 4 ft/sec directed upward.
22. and phase 4. (9) to find the position of the mass at any time t.. Hint: Multiply the differential equation by e-y' and integrate from 0 to T.. where a2 + Rz and 4. yo = 0 cm. Monthly 74. K = m._. we must have
21re-y'y'dx = 2
o
y.4. = cos-'(a/ a2 -+P') = sin ' ((3/ a2 + (32)]. the motion of the mass repeats itself after every time interval of length
2ir vA/K. (9).72
2
Linear Differential Equations
21. period 2. obtaining r
y2(T) + 2
0
e-'y'y"dx = y2(0). K = 5m.y"dx = [y'(p)]2 . Show that all solutions of the differential equation y" + e'y = 0 are bounded as x -> x. Amer. A spring is stretched 1. the A=
solution can be written in the form
y(r)=Acos( mrwith A = y + mvo/K and 4 = cos-' (y/A ). yo = 3 cm. Math. Hence.rr\/N/K.
Note that for some p such that 0 s p s T. and 24.
'Problem 25 and its method of solution are taken from the 1966 William Lowell Putnam Mathematical Competition.
The graph of y(t) is periodic of period tar \/1/K.
In each of Exercises 22. we say that the motion is oscillatory of amplitude A. no. For this reason we say that the motion is oscillatory or that
vibrations occur. v(. evaluate the amplitude and the period of the oscillation and sketch the graph of y(t).
Consequently. Since it is possible to express every solution in terms of the solutions y.y.
2 = y'(1) = c. (1). we obtain
y' = c
have
y = c. Regardless of the values given to the constants c. From Eq. (1) with respect to x. is a constant of integration. if y is a solution of Eq. = 2 and c2 = -1. Integrating Eq. then there exist constants c. and as a consequence the general solution of such a differential equa-
. (3) shows that they play a special role in that every solution can be expressed
as a linear combination of these two solutions.x + C21
(2)
where c. and.(1) + c2. In this section and the next we concentrate on answering the question: How many solutions can a linear differential equation have? To gain some insight into this question.
Equation (3) is referred to as the general solution of Eq. c. = 1 and c2 = 0. we consider an example.
EXAMPLE 1
Solve the second-order homogeneous linear differential equation
y' = 0. Thus. (1). we suspect that to solve an nth-order equation requires n integrations. (3) we have
1 = y(1) = c. Hence. suppose that we want a solution of Eq. (1). we speak of y. Notice that y. + c2y2. and the solution we seek is
y=2x-1.2. (1) that satisfies the initial conditions y(l) = 1. (1) and Eq. if we try to formulate conditions that reduce the infinity of solutions to a single solution. and if y = y2. = x and y2 = 1 are solutions of Eq. and c2. we
(3)
where c2 is another constant of integration. = 0 and c2 = 1. and y2. differentiating Eq.. the expression for y given in (3) is a solution of the
differential equation (1). (2) with respect to x. c.2 LINEAR INDEPENDENCE AND WRONSKIANS
In Section 2. y'(1) = 2. For example. then c. and c2. (3). and y2 as constituting a fundamental set of solutions for Eq. and c2. Integrating Eq.
Intuitively. there are an infinity of solutions to the differential
equation (1).2
Linear Independence and Wronskians
73
2. we need two conditions which serve to determine c. That is.
(1)
Solution Because of the simplicity of this equation.1 we posed certain questions concerning the number of solutions to a linear differential equation and to the manner of finding solutions to linear differential equations. and c2 such that y = c. For this differential equation the general solution contains two arbitrary constants. Of course.
if y = y then c. we can integrate it directly to obtain solutions.
= a2 = 0. Hence. Furthermore. ..x <. .1.. .1 derivatives are continuous on the interval a s x <.
...
(b) Show that the functions f. Before we state this test. and f2 are linearly independent
on the interval -1 <.74
2
Linear Differential Equations
tion involves n constants of integration.. and a2. .x s b is said to be linearly dependent on a s x s b if there exist
constants a a2. am. This is possible if we choose for example.
(b) Otherwise f. there exist constants a. The Wronskian o f f f2. . f. In what follows these notions will be made more precise.
EXAMPLE 2
on the interval -1 s x s 1. f. and f2 are linearly dependent on the interval -1 <.. f . f are solutions to a linear homogeneous differential equation. not all of which are zero. and f2 are linearly dependent. . That is. not both zero..x 1.b. each defined and continuous on the interval a <..
DEFINITION 2
Let f f2.(x) = 3x + (12/5) and f2(x) _ 5x + 4 are linearly dependent on the interval ( -_).
(4)
for every x in the interval a <. such that
a. and f2 are linearly dependent we must show that there exist constants` at.. + a2 = 0
and
-()(I
The only solution of this system is a. f be n functions which together with their first n .3.b. and a2. we need the following definition..
DEFINITION 1
A collection of m functions f f2.< x < x.
a.(x) = x and f2(x) = x2 are linearly independent
Solution (a) To show that the functions f. not both zero. such that
=0
linearly independent on this interval. But for x = I and x = -1 we find
a.x :5 1. there is a simple test that serves to determine whether they are linearly independent or not. = 5 and a2 = . such that
a. . we suspect that in this case there should be a set of n special solutions having the property that every solution can be expressed as a linear combination of these special solutions.x + a2x2 = 0
for all x in the interval -1 s x <. If the functions f f2. the functions are said to be
(a) Show that the functions f.. .. Otherwise. This contradicts our assumption that f. (3x + 5 I + a2(5x + 4) = 0
for all x in the interval .
x) =
All
fz
f
. x).2.. f2(x) = cos x.... their properties.2xcosx. and their application to linear systems of algebraic equations.sinx. which distinguishes linear from nonlinear differential equations. is embodied in the following theorem.
THEOREM 1
If each of the functions y.
EXAMPLE 3
Given f.
In the introduction to this chapter we mentioned that linear homogeneous differential equations have the property that linear combinations of solutions are also solutions.
a^(x)Y`"i + a"-. + a.(x) = x2.
cosx _ -x'sinx .
Solution
From Definition 2 and the functions given we compute
W(x2. is denoted by W(f f.
L
W(f. Nevertheless.
then for every choice of the constants c c2. ..
(f + f 2 +
+f)'=f+f2
-
+ fk
'For a brief discussion of determinants. y.x) _
xz
. That is.f2. The proof of this theorem is neither difficult nor sophisticated...... find W( f f2. This format will be followed in many places in the text.. c".(x)Y("-') + .cosx. y2. the linear combination
ciY1 + c2Y2 + .2x
. see Appendix A. we choose not to give a proof under the most general circumstances but rather demonstrate a proof for a specific version of the theorem.. f..
. ..
fo. 'In Chapter 1 we expressed the philosophy that difficult or sophisticated proofs would be omitted in this text. the theorem will be stated in its general form and the proof given for a specific case. f . .. x) and is defined to be the determinanl2
fI
f2
. . .(x)Y' + ao(x)Y = 0. ... with the implication that the general proof follows along similar lines of reasoning... This principle.2
Linear Independence and Wronskians
75
evaluated at x. + cmYm
is also a solution." are solutions to the same linear homogeneous differential equation.-Ii
ft"-tt
Each of the functions appearing in this determinant is to be evaluated at x.
Proof'
The proof depends on two basic properties of differentiation.
it follows from Section 1. 9 (1961): 847-56.y = 0 are linearly independent because their Wronskian
e`
e
_
-e°-e°= -2
is never zero.
In Example 2 we demonstrated that the functions x and x2 are linearly independent on the interval -1 <_ x s 1. + a. which is called Abel's
'For a discussion of the connection between the Wronskian having the value zero and linear de-
pendence of functions. Math.. their Wronskian
REMARK 1
X
1
x21 2x
I
x2
vanishes at the point x = 0 of the interval -1 s x s 1. .
For example.(x)y + au(x)y = 0
defined on an interval a <._. = x and y2 = x2 are solutions of the second-order linear differential equation
x2y" . we conclude that
Theorem 2 is not contra icted by
n interesting property of the Wronskian is that it satisfies a first-order linear differential equation. Meisters.1 < x s 1 that has x and x2 as solutions..2.(x) = cos x and y2(x) = sin x of the differential equation y" + y = 0 are linearly independent because their Wronskian
cosx
-sinx
sin x = cost x + sine x = I cos x
is never zero. However.2
Linear Independence and Wronskians
77
true: if the functions are linearly independent. Equation (8). H. y2. y" of the nth-order differential equation
a"(x)y"i +
a.
THEOREM 2
n solutions y.(x)yi"-" + . Rather. Thus. "Local Linear
Dependence and the Vanishing of the Wronskian..
. y x) = 0 for every x in the interval a < x s b. . the reader is referred to the article by G. any interval of definition I for this differential a uation must e d x = .2xy' + 2y = 0. Therefore we have the following criterion for testing linear dependence (and linear independence)..2 that the Wronskian is either identically zero or never vanishes. there must not exist a second-order linear differential equation with interval of definition .. . the solutions y.
However. Also.. since we have assumed that a2(x) # 0 in L Thus. no. then their Wronskian is nonzero for some x. the solutions e` and a-' of the differential equation
y" ..x < b are linearly dependent if and only if W(y y2. The reader can verify that y." Amer. Does this contradict
Theorem 2? No. Monthly 68.
qy2) . +
a
... the functions e' and
... ! en et er may.(x) # 0.
p(x) = a. .
Suppose also that these functions are linearly independent on the interval of definition of this differential equation..
Y2
= Y1Y. the functions cos x and sin x constitute a fundamental set of solutions for the differential equation y" + y = 0. .
For example.
Using Eq..(x)Y1"-u + .8). x vanishes identically on a <_ x<_ b or W(y ..78
2
Linear Differential Equations
formula is also the basis for determining a second linearly independent solution of a second-order linear differential equation when one solution is already known (see Section 2.
THEOREM 3
I f y y2. y. For simplicity we show the details for the second-order linear differential equation. ..(-pyi ..
DEFINITION 3
Suppose that y y .. + a.Yiy2
= y. we find that the differential equation W' = -pW has the solution
W(x) = W(xo) a%p [
-
Jp(s)ds]
(8)
Since the exponential function is never zero. x)..(x) is defined and continuous on a s x s b and a (x) # 0 on s x <.(x)Y' + ao(x)Y
= 0. + a. then it is identically zero.(x). y are solutions of the differential equation
a"(x)Y") +
a"_.
Y. and q(x) = ao(x)/a2(x). .(-py2 .y.)Y2 = -p(Y.
W' = Y1A' + Y1Y2 .(x)Y' + ao(x)Y = 0.gy. it follows from Eq. y x is never zero on a s x
6
Proof The proof is for the case n = 2. (8) that if W is zero at a point x0 on the interval a s x s b..
where each a.Y1Y2)
= -pW. with a(x) = p(x) and b(x) = 0. (5) of Section 1. but we state the theorem for the more general case.Y"Y2
Y. . and if W is different from zero at a point . y" are n solutions of the differential equation
a"(x)Y(".
Y.(x)/a. Also. Note that both p and q are continuous functions since a. We then say that these functions form a
fundamental set (or fundamental system) of solutions for the differential equation.(x)Y"-' + ..(Y1Y2 + YiY2) = Y1Y2 .r..4. Now
w=
Therefore. then it is never zero. For brevity we set W = W(y.
x"..
EXERCISES
integer n such that f has 2n continuous derivatives on R and
In Exercises 1 through 15. ai # X2
16. Eggan and A. y"' = x'
19. 2x
10. . y" form a fundamental set of solutions for a linear homogeneous differential equation of order n. 3 (1973). In x. then f is a solution of an nth-order homogeneous linear differential equation with constant coefficients. Math. any solution of a homogeneous linear differential equation is a linear combination of the solutions in the fundamental set. as we shall prove in Section 2. Monthly 80.
11. c are arbitrary constants. e". e'. 14x2. if y y2. Amer. Inselt. cos x2
4.
REMARK 2
Our use of Wronskians is primarily as a test to determine whether
or not a given collection of solutions to a differential equation are linearly
independent. then the expression
Y=c1Yi+c2Y2+.2
Linear Independence and Wronskians
79
e-' form a fundamental set of solutions for the differential equation y" y = 0.
1.x
5.
THEOREM 4
If f is a real-valued function defined on the real line R and if there exists a positive
W ( f .. J. xe' x2e'
14. 6x2. e
7. f (x). More precisely. x"2. . Y' = ez
17. e'. 2 .. e"
9..
16. sin 3x.3x. C. is the general solution of the
differential equation. x2.
1.. . ft"I) = 0 on R. compute the Wronskian of the set of functions. sin x'. not all of which are zero. Eggan and Insell' present a different use of the Wronskian concept.-+c"Y"
where c c2. cos 3x
1 + x. f. c a constant
In Exercises 16 through 23. a". . a"2"..2. Their results are embodied in the following theorem. x2 In x
S. . e"
12. .
Discovering a fundamental set of solutions for a homogeneous linear differential equation is of great importance because. x312
8. 4 .. x12.3.
3. no. xe'
13.
. x. cf (x). Y" = sinx
"L. integrate the differential equation directly to find the general solution. e'. x2
2. y"' = 0
18.
'See E. y( -')(x0) = 0. The main result of this section is Theorem I below. 30 (1968): 663-80. ai(x).. .:
Prentice-Hall.(x)yt"-.. .. Willett.
(1)
(2)
Using Theorem 1 it is easy to construct a fundamental set of solutions for the homogeneous differential equation
a"(x)yl"1 +
a"-. Barton and S. Although we choose not to present the proof. This existence and uniqueness theorem is.y'(xo) = a.(x)y' + ao(x)y
= 0. and obtain the general solution.3 EXISTENCE AND UNIQUENESS OF SOLUTIONS
As we pointed out in Chapter 1. + a. Define y2 to be the unique
'C.. .
the initial value problem
a"(x)y".. which states conditions on the coefficient functions that guarantee the existence and uniqueness of solutions to the IVP containing an nth-order linear differential
equation... + a"-..(x)y' + ao(x)y = f (x)
y(xo) = (3o.. y'(xo) = 0. 2 (1968). An Introduction to Ordinary Differential Equations (Englewood Cliffs. the solution y is defined in the entire interval a s x <_ b. Coddington. (3) satisfying the initial conditions y(xo) = 1 . a Ax) and f (x) be defined and continuous on a s x 5 b or a <_ x t xo e suc t t a xo with a . there is only one solution (uniqueness) to the differential equation. in their study of peristaltic flow in tubes . 1961). in addition." Amer. it is reassuring to know that there is (existence) a solution to the differential equation that one hopes to solve..
. Biophys. The situation is
even better if. Monthly 5..
Show that p(v) = 1 and p(v) = In v are linearly independent solutions of this differential equation. Raynor. . Flows Barton and Raynor... y"(xo) = 0.3
Existence and Uniqueness of Solutions
81
43. "The Existence-Uniqueness Theorem for an nth-Order Linear Ordinary Differential Equation.2. or D.. Math. x b and let 30.."
THEOREM 1
EXISTENCE AND UNIQUENESS
Let a x).
dp+vdp=0.
(3)
We define y. be any constants. to be the unique solution of Eq.J. Math. no.(x)yl"-" + . Bull. it is possible to prove this theorem using arguments based on a knowledge of calculus only. "-. + a.
2. Then there exists a unique solution y satisfying
. A.$ obtained the following linear homogeneous differential equation with variable
coefficients. N. from one point of view. yl"-O(xo) = R"-r
Furthermore. the most important theorem associated with nth-order linear differential equations.1 + .
v>0.
82
2
Linear Differential Equations
solution of Eq. there always exists a fundamental set of solutions. . . then given (3a.
0
0
0
0
0
0
0
.. c" such that yl"-')(x0) = P .. It remains to show that these functions are linearly independent. The observations of the preceding paragraph emphasize the fact that for any linear differential equation whose coefficients satisfy the hypotheses of Theorem 1. Thus. the existence and uniqueness of the IVP (10)-(11) of that section simply means that the mass m moves (existence) and. + c.y.Y2 + . 0. (3) . + c. are not always easy to construct in practice. y"(xo) = 1. (3) satisfying the initial conditions y(x. moves in only one way (uniqueness)..
.. by Theorem I of Section 2.1.) = 0.y. although guaranteed by the theory. (3) satisfying 0.. in fact. . 1 each of these functions exists and is uniquely determined. By Theorem the initial conditions y(x. We
Proof
.
. Define y. . These solutions. . y( '"(xo) = 1. it follows that the function Y is a solution of the differential equation (3). In this case
1
0
1
0
0
0 0
1
..
y"(xo) = 0. y"^-')(xo) = 0.) = 0. .12.1.
The existence and uniqueness theorem for the IVP (1)-(2) is something that should be expected to be true when the IVP is a "good" mathematical model of a real-life situation.) = 0.
y("-')(x(. Define y" to be the unique solution of Eq. These forms of equations are discussed in Sections 2.. satisfying the initial conditions
.. Nevertheless.y2 + + c. there are some forms of linear differential equations for which systematic construction of a fundamental set of solutions can be described...
.. y. .) = 0. satisfying the initial conditions there exist unique constants c c2. y'(xo) = any other solution.
REMARK 1
THEOREM 2
If y y2.
P
y = c. and so on. For example.. y'(x. y" constitute a fundamental set of solutions of Eq.Y.
1
Thus. . (3). y'(x. can be expressed uniquely as a linear combination of the fundamental set of solutions.
. . . . to be the unique solution of Eq. y. . this other solution.
That is.2. in the case of the vibrating spring discussed in Section 2.) = 1.4 through 2. + c"Y..
Set Y = c.. these functions constitute a fundamental set of solutions.
are arbitrary constants is the general solution of Eq. On the basis of Theorem 2.. + c"Yl"-" (xo) = k .
pression
y = c. At this point we realize that the problem of finding all solutions of Eq.3
Existence and Uniqueness of Solutions
83
calculate Y'.
.
+ c"Y.
EXAMPLE 1
Consider the IVP
y" . see Appendix A) there is a unique solution for the unknowns c.
(4)
c. This process leads to the system
c. (3) boils down to finding a fundamental set of solutions. and y = c.e' + c2e. cos x + c2 sin x is the general solution of the differential
equation y" + y = 0. Y".yi"-"(xo) + c2y2 -') (xo) + .. (5).Y. . then the ex. Now the functions Y and y are both solutions of
the differential equation (3).is the general solution of the differential equation y" . .(xo) + c2Y2(xo) + .. By Cramer's rule (for details..
COROLLARY 1
If y y2.. For example. . (c) Find the unique solution of the IVP (5)-(6)..
Since these functions form a fundamental set of solutions. with the same initial conditions.Y. the following result is well justified. The proof is complete. (3).y = 0.and y2(x) = ek form a fundamental set of solutions for Eq.Y'(0) = 8.-
Equations (4) constitute a nonhomogeneous system of linear algebraic equa-
tions in the unknowns c c2. + c2y2 + . by the unique-
ness part of Theorem 1.(xo) + c2Yi(xo) + . .
(5)
(6)
(a) Show that the functions y.8 are devoted to discussing methods associated with the problem of finding a
fundamental set of solutions.(x) = e . Sections 2. their Wronskian is not zero.. xo).4 through 2.Y.
. .(xo) = P. (3). .
. y = c. C. y" form a fundamental set of solutions for Eq. + c. y". .y. 1't"-" and then evaluate Y and each derivative at
x = xo. Y = y.
+ c"Y"(xo) = ao
c. and.2y = 0
Y(0) = 1. (5). . (b) Find the general solution of Eq. where the c..2..y' . whose determinant of coefficients is
W(yr
.
2(-2e. (9) is a solution of Eq (5).c.1 we encountered the differential equation
y+aay=0. (5) is
y(x) = c.+ 12e) . and y2 form a fundamental set of solutions for the equation.2(e-r) = e.
(9)
If we want to check that (9) is indeed the solution of the IVP (5)-(6).
The unique solution of the IVP (5)-(6) is therefore
y(x) = -2e.
Therefore y.
y(O)
2e° + 3e° = 1.e -r + 2c2e'
Setting x = 0 in (7) and (8) we find.+ e-' .
EXAMPLE 2
In Section 2. = cos at and y2 = sin at form a fundamental set of solutions for Eq. Third.e-' + c2ea'.(-2e-' + 3e')' . In fact. we find
(e-')" . Their Wronskian
e_.84
2
Linear Differential Equations
Solution
(a) By direct substitution of y. it follows that y.y'(x) = 2e-' + 6ea' => y'(0) = 2e° + 6e° = 8. and find its general solution.
(7)
(c) To find the unique solution of the IVP (5)-(6).+ 6e1') .
(-2e" + 3ea')" .c. Since Eq. Second.
a#0. c2 = 3. we must check three things: First.(ear)' .
(10)
Show that y.
y'(0) = 8. we must use the initial conditions (6) in (7) and
y' (x) = . (5) is second order. (5).+ 3e'.(e-')' . + 2C2 =
i
-2. (5).(2e. (10). and y2 are solutions of (5).2e-' = 0
and
(es')" . and y2 are linearly independent solutions of Eq. and y2 into Eq. y(O) = 1.2(-2e-' + 3e') =
(
2e. (b) By Corollary I the general solution of Eq.
(8)
.
. Hence y.
I
ear
I
I
-e-'
=2e'+e'= 3e`
2ea'
I
is never zero.2ea' = 0.2(ea') = 4ear .+ 3ea=) =
-2e'+12ea'-2e-'-6ea'+4e-r-6ea'=0.2e' .
at time t = 0. y'(xo) = y where xp is to be a point of the interval. y(t)) of the mass m at time r satisfies the two initial value problems
mx=0
x(0) = 0
x(o) = Va cos 0
my= -mg
and y(O) = 0
y(0) = v.7(cos x)y' + y = e15. the IVP
my+Ky=0
(11)
(12) Y(0) = Yo. and that the x axis is the horizontal (Figure 2. (x .1. that the projectile was fired at the origin 0.1) y' + 3y = 0
18. (cos x)y" + y = sin x
+X41y=0
17.1.xy = 0
16.
12.
(a) Show that
Y1(t) = cos m t. (x2 .
20. (c) Find the unique solution of the IVP (11)-(12). the mass is located at the position yo and has initial velocity
VO. (x2 + 1)y" + (2x .86
2
Linear Differential Equations
For each of Exercises 12 through 17. Assume that the motion takes place on the xy plane. y" + 4(tan x)y' .
Find the solution to the IVP first by physical intuition and then by applying Theorem 1. Mechanics A projectile of mass m leaves the earth with initial velocity vo in a direction that makes an angle 0 with the horizontal.1)y" + 3y' = 0 14. Use Newton's second law of motion to show that the position (x(t).. Give a physical interpretation of the IVP
y+4y=0
Y(O) = y(0) = 0.3).
Show (by direct integrations) that the solutions of these initial value problems are given by x(t) = (v0 cos 0)t and y(t) = -. y(0) = vo describes the motion of a spring of mass m and spring constant K. determine an interval in which there exists a unique solution determined by the conditions y(xo) = yo.gt2 + (v0 sin 0)t.
Y2(t) = sin
mt
form a fundamental set of solutions of the differential equation (11). As we explained in Section 2. (b) Find the general solution of the differential equation.
.4)y' + 3x3y'
13. Initially.
19. 2xy" . sin 0.
297. we need to find n linearly independent
solutions. Substitution of y = e' in
Eq. with a a constant. = e' and y2 = e-'°' form a fundamental set of
solutions for the differential equation
y+a2y=O.X"-'e' +
or
.a . we see that in a sense any solution of this equation must have the property that derivatives of this solution merely produce constant multiples of themselves.a)(x .
a*0.
. Find the general solution of the differential equation
(x .4 Homogeneous Differential Equations-The Characteristic Equation
87
Figure 2.he' + aoe' = 0
+ alK + ao) = 0.a"-' +
"See also Math.
As we examine Eq. as a solution of Eq.
22. y" = Me..b)y' + 2y = 0
[Hint: Compute" the second derivative of the expression (x . y") = X"e.b)y" + 2(2x . (1) yields
a"We' + a"_. then y' = Xe. (1) carefully (especially in the case n = 1).
(2)
e''(a"a" + a"-.2. .. This obser-
vation motivates us to try e"'. Show that the functions y. 56. To find the general solution of Eq. (1)
where each coefficient a. Magazine 44 (1971): 18. (1).3
21 Show that the functions y. If
y = e. is a constant and the leading coefficient a * 0.a)(x .
23.
a#0. = cosh at and y2 = sinh at form a fundamental set of solutions for the differential equation
y-a2y=0. (1).b)y.
4
HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS-THE CHARACTERISTIC EQUATION
The linear homogeneous differential equation with constant coefficients has the form a"y"' + a1y' + aoy = 0.J
2
.
+ a.
Grove and G. If n = 2. the characteristic equation is a cubic equation. the only way that y = e' can be a solution of the differential equation is that k be a root of the polynomial equation
a"k"+a"-. and the formulas for its solutions (called Cardan's formulas) were discovered by Scipione del Ferro in 1515.k"-'+ +a.
DEFINITION 1
(3)
We conclude. the characteristic equation is the quadratic equation a2k' + a. If n = 4.
"For the solutions of the cubic and quartic equations. the problem of solving homogeneous linear differential equations with constant coefficients can be reduced to the problem of finding the roots of the characteristic equation for the differential equation. then.
In Section 2. we can develop some information about linearly independent solutions to Eq. (3). (1). when n 5.k + ao = 0 and its two roots are given by the quadratic formula
REMARK 1
z
2az
If n = 3. the interested reader is referred to the text
by E. A.k + as is called the charThe polynomial f(X) acteristic polynomial for Eq.k+a0=0.. the problem of finding the roots of the characteristic equation may be far from trivial.k"-' + + a.
.88
2 Linear Differential Equations
The exponential function never takes the value zero. y = e' is a solution of the
a"k" + a. therefore. (1). Ladas. counting multiplicities. the characteristic equation is a. if n is large._. that if k is a root of Eq. the characteristic equation is a quartic equation. and its solution was discovered by Ludovico Ferrari in the sixteenth century. However. In spite of the possibility of algebraic difficulties surrounding the solution of
the characteristic equation.
If n = 1. (1). Some elementary methods for finding roots of polynomial equations are given in Appendix C." For n ? 5.5 we will demonstrate that knowledge of the roots of the char-
acteristic equation is all that is needed to construct the general solution of
homogeneous linear differential equations with constant coefficients. The roots of the characteristic equation are called
characteristic roots. 1974). Consequently. and the equation f(k) = 0 is called the characteristic equation for Eq. This fact was proved by Abel and Galois early in the nineteenth century. The characteristic polynomial is of degree n and the fundamental theorem of algebra states that every polynomial of degree n (n a 1) has at least one root and. Introduction to Complex Variables (Boston: Houghton Mifflin. the characteristic equation cannot be solved (by radicals) in general
unless it is of some special form. indeed. exactly n roots.k + ao = 0 and has the root k = -ao/a.
differential equation (1). Thus.
and Remark I in Section 2.1.i sin 0). These two solutions are linearly independent.ok = e--. (4) and recall that 1/ --K = i %/K-]
«=
262 .. see Example 2 below. Suppose now that the coefficients a.
.A + bo for which b. The real number a is called the real part of k. For further discussion of this case. since their Wronskian has the value .
with a2 = Kim. Thus.4 Homogeneous Differential Equations-The Characteristic Equation
89
We emphasize once more that if. where a and R are real numbers and i2 = . to say that \ is a complex number means that A is of the form a + i(3.
EXAMPLE 2
Solve the differential equation of he vibrating spring example of
Section 2. a. then e" is a solution of the linear homogeneous differential equation with constant coefficients. = e-`. Then the only way that a characteristic root can be a complex number is that the characteristic polynomial contain a quadratic factor of the
form b2X2 + b.2(# 0).i(3. In the special case that X is a complex number.1.e' = e°'(cos Ox + i sin px). it is conventional to write the corresponding solution in an alternative form.
EXAMPLE 1
Solve the differential equation y" . = e. a.2. and the real number p is called the imaginary part of X. any solution y of this differential equation is of the form
y = c. We then write e' = eca.y = 0. Exercises 21 through 29.
b
a=
4bu 262
In addition to the solution (5) there would also be the solution e-(cos Ox i sin (3x).*e. and c2 are arbitrary constants.1 =0=>
0
' A 'A= t1. where [refer to Eq. = -1) k must satisfy the characteristic equation
)` 2 . of the differential equation are real numbers. Thus.\ is a characteristic root. y. two characteristic roots would be a + i(3 and a .5.
We have two solutions.iR is called the conjugate of the complex number a + i1). e' = cos A + i sin 0 (' e-i° =
cos 0 . First. In this case the complex characteristic
roots occur in conjugate pairs (a .
where c. = 0.a.e' + c2e-'. < 4b2bo. = e' and Y. (5)
The last form follows from the Euler identity. then (since a2 = 1.1:
+a2y=0.
Solution Set y = e'..
Write the general solution of this differential equation in the form y(t) =
c. Domar.]
39. Econ.y"-4y=0
33.y
= 0 can be written in the form y = c. however. provided that the velocity of the mass is small. it is realistic to assume that the spring is subjected to a damping force. The definitions of the hyperbolic cosine and hyperbolic sine are e` + e-`
cosh x =
2
and
sink x =
e'
-e'
2
(a) Show that these definitions can be manipulated algebraically to yield
e` = cosh x + sinh x and a-' = cosh x .y'-81y=0 37." Amer. air resistance) which act to retard (dampen) the motion and eventually cause the system to come to rest. y"-25y=0
35.y"-9y=0. [Hint: y2 + 13y + 36 = (y + 4) (y + 9).
. Thus.
In Exercises 31 through 37 write two equivalent versions of the general solution.
"See E. cosh x + c2 sinh x. y"-49y=0 36. "The Burden of the Debt and the National Income.4 Homogeneous Differential Equations-The Characteristic Equation
91
29.
32. D.
1944): 798-827. Apply the method of Example 2 to write the general solution of
y" + 13y" + 36y = 0
in terms of trigonometric functions. (Hint: See Example 1 and Exercise 30.)
31. experiments have shown that the magnitude of the damping force is approximately proportional to the velocity of the mass.e"' + c2e`x and also in terms of the hyperbolic sine and hyperbolic cosine. Naturally. a vibrating spring is most often subjected to frictional forces and other forces (for example.2. Rev. the damping force is difficult to formulate precisely.12 Find the characteristic roots associated with this differential equation. y" + 25y = 0 30.16y=0 34. The Vibrating Spring with Damping In practice.y"-121y=0
38. (Dec. where the income grows at a constant relative rate 0 (0 < a < 1) and D is the total public debt. Generally speaking. Economics Second-order linear differential equations with constant coefficients of the form
b-(3D=0
occur in the Domar burden-of-debt model.y".sinh x.
40. (b) Show that the general solution of the differential equation y" .
THEOREM 1
If all the roots of the characteristic equation are distinct. 1968). 2nd ed. the damping force is negative when dyldt is positive and positive
when dy/dt is negative. 140.
0'z . as described.y2+x)
eA"
X1eA. show that the equation of motion in this case
is
my+ay+Ky=0. i = 1.4 we demonstrated that linear homogeneous differential equations with constant coef-
ficients will have solutions of the form e' provided that X is a root of the
characteristic equation associated with the differential equation.. and m which guarantee that the characteristic roots will not be complex numbers. This we accomplish in the present section via theorems and a number of illustrative examples. Using Newton's second law. W. In fact. = e ". The proof for the general case would follow along similar lines of reasoning. then the nfunctions y. . In Section 2."
"For the statement and evaluation of the van der Monde determinant. i = 1. .. we can express the damping force as -a(dy/dt). K.i
ek2'
X2eA2'
X2e(X1.
. n.
w(yl. p.
1
Thus. the evaluation of the corresponding Wronskian would not be trivial. Linear Algebra. say X = X. Describe conditions on a.
(6)
Write the characteristic equation for this differential equation. the basic problem is to find a fundamental set of solutions. the Wronskian reduces to a determinant commonly called the Vandermonde determinant. Consequently.. 2. . What conditions will guarantee that the characteristic roots are complex?
2-
5
HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS-THE GENERAL SOLUTION
In Sections 2. However. see C.92
2
Linear Differential Equations
the damping force. 2.XIe(AI.X)e(Ala"2>' # 0.. (Boston: Allyn and Bacon. Curtis.3 we demonstrated that for linear differential equations. where a > 0 is called the damping constant. Thus.A2). n constitute a fundamental
set of solutions.2 and 2. The loose end that still persists is to relate the solutions e' with a fundamental set of solutions of the differential equation. y. and y2 form a fundamental set of solutions.. .. acts in a direction opposite to that of the
mass.A2).
Proof The proof is for n = 2.
If the roots of the characteristic equation are not all distinct.e'.y'. 3 and the fundamental set of solutions is y.e' + c2e' + c.
This last differential equation is a first-order linear differential equation and can be solved by the methods of Section 1.1)(x2 . 2.
. We introduce
a new unknown function v by the relation v = y' . Thus we obtain (after long division or synthetic division)
V . x = 1.
where y.
That is.
Hence. Whenever this
happens it is always true that x = 1 is a root.x + c2) = c.
Y = c1Y1 + c2Y2. then we cannot generate a fundamental system of solutions by considering just the exponentials e'"`. To try to obtain a feeling for the form of the general solution in this case.2.(xe') + c2e'. and y2 are
solutions of Eq.
EXAMPLE 2
Solve the differential equation
y"-2y'+y=0.6 = (x .1)2.
Therefore.e'. (1).
Solution
(1)
For the differential equation (1) the characteristic polynomial is x2 . = xe' and Y2 = e'.5x + 6)
1)(x .5 Homogeneous Differential Equations-The General Solution
93
EXAMPLE 1
Find the general solution of the differential equation
y'"-6y"+Ily'-6y=0.
Solution
For this differential equation the characteristic equation is
x'-6x2+11x-6=0.3).
Note that the coefficients in this polynomial add up to zero. Furthermore. y3 = e''.
which.2x + 1 = (x .e'. if the characteristic equation has at least one repeated root. = e'.C x+1
1
=e2`(-1)#0.
(x+1)e'
e'
-?. takes the form
v'=(2y-y)-y'=y'-y=v v'=v 'v=c.6x2 + I lx . using the differential equation (1). Thus x = 1 is a root of multiplicity 2. Then v' = y" .y.e'dx + c2) = e'(c.4 to yield y = e'(fe-' c. that is. we first consider some special examples. The reader can verify that both y.2)(x .
v=y'-y=c. y2 = e2s. The general solution is y = c.
x"-' e°' sin (3x.
"For a proof. of multiplicity m. 1961)
..2 + a..ip we find 2m complex valued solutions.at. are linearly independent solutions. that the real and imaginary parts of the solutions x. ai. such as A = a + ip. . xe°' cos px.a>r
.y' + ay = 0. xe".\ + a."
THEOREM 2
If X = A.J. Then the form of the general solution y(x) of Eq. then there
are m linearly independent solutions associated with X = X. . x`e( -e)-namely
the 2m functions e°' cos px. and A2 are roots of the characteristic equation a21.ip as a root again of multiplicity m.
REMARK 1
When the coefficients a.
where c. a # 0. we know from the theory of polynomial equations that if the characteristic equation has a complex root. see Coddington.. (3) is described by the
following cases:
CASE 1
Real and Distinct Roots X.y' + aoy = 0.2. = 0. X2ex"x"-'eh'.
(3)
Assume that t.
COROLLARY 1
Consider the second-order linear differential equation with real coefficients
a2y" + a.eca.5 Homogeneous Differential Equations-The General Solution
95
The results embodied in Examples 2 and 3 are capable of generalization. PrenticeHall. Applying Theorem 2 first to the root a + ip and then to a . then the characteristic equation also has the complex conjugate a .
xe'"-. is a root of multiplicity m of the characteristic equation associated with
the differential equation a"y'"' + a"_. of the differential equation are all real numbers.. It can be shown however. . e°' sin (3x. From this and the previous results it follows that the general solution of a linear differential equation with real and constant coefficients can always be expressed as a linear combination of n real valued functions.. N.y1`) +
+ a. and this generalization is stated without proof in the following theorem. and c2 are arbitrary constants.. xe" sin px. These solutions are
as follows: e". x" -' e'' cos px. Ordinary Differential Equation (Englewood Cliffs.e"" + c2e"2+.40% xec" t'
and
e(a-. * X2
y(x) = c.
The motion of the vibrating spring subjected to damping (retarding) forces is governed by the differential equation [see Eq. Section 2. and
m > 0 is the mass. Exercise 40. however. the characteristic roots are
X = -a + V a' . the solution looks as shown in Figure 2.1. m m
where a > 0 is the damping constant./(c2. Thus. (6) is
(6)
k2+2ak+b2=0. K > 0 is the spring constant. it is convenient to set 2a = a/m and b2 = K/m. The three possible cases are discussed separately.
Solution it follows from Theorems 1 and 2 and Remark 1 that the general solution of the differential equation in question is
y = c. + 2
and
.e + c. To avoid fractions.o is the coefficient of y10' in the original differential equation.5 Homogeneous Differential Equations-The General Solution
97
where a.
Graphically. the equation of motion has the form
y+2ay+b2y=0.
A = 1 cr.xe'` cos 2x + c. = cos-'[c.4 before studying this application.
CASE 1
b2 >
b > a) The general solution of Eq. This situation is referred to as being underdamped
"As in Exercise 21. (6) is
y(t) = e-0f (c.a2 t + c.
Consequently.)"].4.1
It is suggested that the reader review Exercise 40 of Section 2.
APPLICATION 2.
X.2.4>] d. b > 0. We note. + c. and for this reason we term the motion oscillatory.e
+ c3xe ' +
+ c7e
c4x2e-2.b2. (6). We observe that the motion [that is y(t)] oscillates about the t axis. that the "amplitude" of the motion15 decreases as time goes on. (6) depends very heavily upon the nature of the characteristic roots. / V
: . _
-\/a2
We realize that the form of the general solution of Eq.a2 t).axe ` sin 2x. + cse2+ cos 3x + c6e2z sin 3x
cos 2x + c8e . Section 2.5.4]
Mechanics
The Vibrating Spring with Damping
y+aY+-Y=0. the solution in this case can be written in the form
y(t) = Ae-°' cos [ .
The characteristic equation associated with Eq. Write down the general solution of this differential equation. cos
b2 .sin 2x + c. sin V b2 .
and therefore the motion
is classified as critically damped. However.
. (6) is
y(t) = c. (6). This is Case 2.98
2 Linear Differential Equations
Figure 2. (6) is
y(t) = e-°' (c.
CASE 2 b2 = a2 (z' b = a)
The general solution of Eq.e"" + c2e"2`.
Classify this case as being underdamped. For this reason. and X2 negative real numbers. critically damped. we note that there are no vibrations (oscillations). this case is called critically damped. in reference to Case 1. + c2t). and hence there is not enough damping to overcome the
oscillations. hence a = b = 2. the damping force is so strong that it causes the system to "slow down" very quickly. In this case there are no oscillations and the graph of the solution approaches the t axis very quickly as time goes on. This case is referred to as being overdamped.
Since y(t) contains no sine or cosine terms.4
in the sense that the amount of damping (or the damping constant) is less than the spring stiffness.
CASE 3
b2 < a2 (=> b < a) Here the general solution of Eq.
with both \. we note that
2a = 4 and b2 = 4. That is.
EXAMPLE 7 It is known that the motion of a certain damped vibrating spring is governed by the differential equation
y+4y+4y=0.
Solution Comparing this differential equation with Eq. we realize that even the slightest decrease in the damping will produce vibrations. or overdamped.
assume that.
Find the solution of each boundary value problem in Exercises 30 and 31.
yI q) = 1.) The problem that is proposed here is the following: Verify that the substitution z = x + iy enables one to collapse the given system into the single differential equation
z-ii+6z=0
with initial condition i(0) = 0.y+9y+4y=0 38.6y+4y+y=0
41. Classify the resulting motion as being underdamped. or overdamped. Section 2.
y(O) = 3.1
Assume that each of the differential equations in Exercises 34 through 43 governs the motion of a certain damped vibrating spring. Monthly 80._. Find the position of the mass at any time t. y" + 3y' + 2y = 0.100
2
Linear Differential Equations
27.5 cm by a force of 4 dyn.y+20y+64y=0 37. in addition. Find the position of the mass as a function of time.
'Problem 44 appeared on the 1971 William Lowell Putnam Mathematical Competition. 2 (1973). Discuss the behavior of the motion as t . air resistance acts upon the mass.i +y+6x=0. y" .
28.
yO)=0. 33. A mass of 4 g is attached to the end of the spring. Solve this differential equation. y(0) = 0. Amer.
. at time r = 0.y+8y+ 16y=0
43. critically damped. This air resistance force equals m times the velocity of the mass at time t.
y'(0)=5
Y'
y" (0) = 2
(4) = 0
.
30.
y-x+6y=0
which satisfy i(0) = y(0) = 0 are hypocycloids. y+5y+y=0
42. (A hypocycloid is the path described by a fixed point on the circumference of a circle which rolls on the inside of a given fixed circle.
29. y"_ y"_y'+y=0. Show16 that the graphs in the xy plane of all solutions of the system of
. The system is then set into motion by pulling the mass 6 cm above the point of equilibrium and releasing it. y'(0) = 0. y" + y = 0. y+2y+y=0 40. In Exercise 32.5y+10y+20y=0
39.2y" + 2y' = 0. 4y+ 8y + 4y = 0
differential equations
35. [Hint: See Exercise 40. Math. with an initial velocity of 5V/6 cm/sec directed downward. y"+4y=0.4. y+y+y=0
44.
y
(2) _ 2
32.y+69+12y=0
36.v'"(0) = 4
y(0) = -2. y(-1) = e
31. A spring is stretched 1.
34. no.
Magazine 47." Amer." Math Magazine 48. Math.
is
Y'(0) = 1
Yo(x) = 1 [et' Of
-e
w2)_)
(b) Show that lim.a)I=J
(b) Show that lim. For further reference.
It is often possible to extract information about solutions to a differential equation without explicitly solving the differential equation. Leighton. One can motivate the form of the general solution in the case of equal roots of the characteristic equation in the following manner.(x) = xe". (a) Show that the solution of the IVP
y"-2(r+(3)y'+r2y=0
Y(0) = 0. 53. Birkhoff and G-C. "The outline presented in Exercises 48-53 is extracted from D. p. Find" all twice differentiable functions f such that for all x
f'(x) = f(-x). no. (Hint: Differentiate the expression. Monthly 65.2. C'(0) = 0. y. 5 (1974). 1952). Hastings.a y=o
Y(o) = 0. Baslaw and H.-. no.ea B)I=
-
. Monthly 65. Rota. M. 2 (1975).
Define S(x) to be the solution of the IVP
y"+y=0
S(0) = 0. and the result will be a differential equation that you can solve. See also G. and W. 1962). 8 (1958).)
46. "See R." Amer.. 41. see R.5 Homogeneous Differential Equations-The General Solution
101
45. Larsson. Math.
.
"Problem 45 appears in Math. y8(x) = xe". An Introduction to the Theory of Differential Equations (New York: McGraw-Hill. S. use the expression itself."
(a) Show that the solution of the IVP
y"-2ry' + r2. no. S'(0) = 1
Define C(x) to be the solution of the IVP
y"+y=0
C(0) = 1. A. "General Solutions of Linear Ordinary Differential Equations. "On the Critically Damped Oscillator. 47. Kearns. D.
Solve this problem. p. Ordinary Differential Equations (Boston: Ginn & Company. "An Analytic Approach to Trigonometric Function. 7 (1958). Perhaps one of the best examples of this idea19 is the trigonometric differential equation
y"+y=0. no.
is
y'(0) = I
YO(x) =
2\/(2r
1
eb-e*.
2. 51. and 2. Once we solve the homogeneous differential equation the nonhomogeneous equation presents no difficulties (at least in theory). and homogeneous). the initial conditions. for which we know already one solution. In the last two sections we learned that (subject to some algebraic difficulties only) we can solve any linear homogeneous differential equation with constant coefficients.12 we will see that the nonhomogeneous linear differential equation with constant coefficients can be handled in a straightforward manner.)
50. is to try to approximate their solutions by assuming that the solutions are power series. then. The supposition that the solution is a power series is valid only if the coefficient functions a. G HOMOGENEOUS EQUATIONS WITH VARIABLE COEFFICIENTS-
. explicitly.
S(x + a) = S(x)C(a) + S(a)C(x). Show that [S(x)]2 + [C(x)]2 = 1. and the uniqueness theorem.6
OVERVIEW
Throughout this book we strive to find solutions of differential equations.(x) satisfy certain differentiability requirements. Show that for arbitrary a. (Hint: Multiply S' + S = 0 by S'. In Chapter 1 we learned how to solve a few types of differential equations of order 1 (such as separable.10. the Euler-type differential equation and differential equations of order 2. In general there is no way to solve. it is difficult if not impossible to solve linear differential equations with variable coefficients.
49.
(Hint: First show that S(x + a) is a solution. a linear homogeneous differential equation with variable coefficients unless the differential equation is of a very special form. Show that for arbitrary a. at this level. as we will see in Sections 2. for example. Unfortunately.
2 . then use the fact that S(x) and C(x) form a fundamental set.S(x)S(x). The following two sections deal with these two special cases. linear. and integrate. But what do we do if we have such a differential equation and we desire information about its solution(s)? Perhaps the best we can do with differential equations having variable coefficients. Show that C(x) and S(x) are linearly independent.102
2 Unear Differential Equations
48. Apparently. Show that S'(x) = C(x). In Sections 2.)
53. exact.11 and 2. matters are not so simple when we try to solve linear differential equations with variable coefficients of order 2 or higher.11. When these requirements are met. the supposition that the solution is a power series is nothing more than looking at the Taylor-series representation of the solution.)
52. Show that C'(x) = -S(x).12.
C(x + a) = C(x)C(a) . A direct
. (Hint: Use the differential equation.
a2)Y + a.
(3)
where dots denote derivatives with respect to t.. The Euler differential equation is probably the simplest type of linear differential equation with variable coefficients.x"y(") + a"-.r Differential Equation
103
substitution of the power-series solution into the differential equation enables us. Since the leading coefficient should never be zero. . a and a2 given constants.7 EULER DIFFERENTIAL EQUATION
An Euler differential equation is a differential equation of the form
a. is reduced to the differential equation
a2Y + (a.. We include a brief discussion of power series here to round out the treatment within this chapter.
Proof
First. in many cases.. . 0). That is.xy' + a0Y = 0.7
Eul. the differential equation should be solved for either x > 0 or x < 0. We illustrate this fact for the second-order case. We have devoted an entire chapter (Chapter 5) to this approach. By the chain rule for differentiation.
2.. The reason for this is that the change of independent
variable
__
1e`
x
-e'
if
if
x>0 x<0
produces a differential equation with constant coefficients. to compute as many coefficients of the power series as we please for our approximation.
(2)
with a0.
azx2Y" + a.
EXAMPLE 1
Show by means of the change of independent variables above that the Euler differential equation of second order. oo) or the open interval ( . assume that x > 0.2..a"_ . Then the transformation x = e' implies that dx/dt = e.xy' + ay = 0. and so dt/dx = e-'.
(1)
where a". we have
y
and
dx
dt dx
-yeYXz'xy`=Y
dy' dt_d
dt
dx
dt (ye
')e'
y
_
_(Ye-`-Ye `)e `=(Y-Y)e
. + a.y = 0. a ao are constants and a" # 0.M.x"-'yc"-n + . the interval of definition of the differential equation (1) is either the open interval (0.
.x+2>0
39. is a known solution of the linear differential equation
a"(x)y(") +
a"-i(x)yc"-n + .I for u'. then the change of dependent variable
y = you
(2)
produces a linear differential equation of order n . x'y"' + 3x2y" .. Perhaps the most interesting application of the reduction of order is
that we can find a second linearly independent solution of a differential equation of order 2 with variable coefficients provided that we know one nontrivial (not identically zero) solution of it.
5
1Y' + (x 4 1)2Y = 0.(x)Y' + au(x)y
= 0.x2y'"-xy'"=0.. (x-4)y"+4y' -X 4y=0.x<0
36.
5
5
0. The procedure associated with reduction of order is somewhat systematic and hinges upon our having specific knowledge of at least one solution to the original differential equation. x > 0
Solve each of the differential equations in Exercises 33 through 40.x>4 38. + a.x<1
40. reduction of order is a device whereby the problem of solving a (linear) differential equation of certain order is replaced by a prob-
lem of solving a (linear) differential equation of lower order.108
2
linear Differential Equations
32.x<3
2. (x+2)y"-y' +x+2y=0.x>0
X
37.
THEOREM 1
If y.2y = 0.xy"'-x6y=0.x<0
35. The method is embodied in the following theorem.x>0
34.
(1)
which has the property that it is never zero in the interval of definition of the differential equation. 3xy"-4y'+5y= 0.
33)iy=0.2xy' .8 REDUCTION OF ORDER
As the terminology implies.
use Abel's formula to obtain a second linearly independent solution.
Many theoretical studies in physics utilize Laplace's equation. and m is the mass of the particle whose wave function is 4) (4) is related to the probability that the particle will be found in a differential element of volume at any particular time).2)
J
(4t2 .
Since the integral cannot be evaluated in terms of elementary functions.2)
exp[.2. known as Hermite's equation.0. V is the potential energy.
Solution We leave it to the reader to verify that y.2)2
e2
= (4x2 . How one obtains the first solution is another question altogether.
In theoretical physics a basic equation of quantum mechanics is the
Schrodinger wave equation
:
Y1_r
Quantum Mechanics
z
z
z
8m rt2 C -ax. Given that y. Most often at least one of the linearly independent solutions (very often both) of these equations is an infinite series. A more elaborate and detailed treatment of the question is presented in Chapter 5.2xy' + 4y = 0.2s) ds]dr
fj
(4t2 . It is not uncommon in these circumstances to find one solution and then express the second solution in integral form by means of Abel's formula.2 is a solution of
Eq. = 4x2 . It is not our intention to discuss these considerations but rather to consider a special instance (for more elaboration and a few other considerations see Section 11.f'( .2xy' + 2py = 0. (15).
(15)
where p is an arbitrary constant. (14)] we have
Y2 = (4x2 .2 is a solution of y" .
8z
z
Mathematical
Physics
:
az2
.2)2
di. y. and there are a number of considerations that govern which technique is most efficient for a given situation. z)&
where h is a constant known as Planck's constant. = 4x2 .
+
_ay
y
a
+ V(x. A partial answer to this question is presented in Section 2. (15) (with p = 2). we
content ourselves with having y2 in integral form. The search for acceptable solutions of Schrodinger's equation for the special case of a harmonic oscillator leads to the differential equation
y" .
There are various techniques associated with the problem of obtaining solutions to this equation. Equation (15).1.e
Reduction of Order
113
in such mathematical studies as approximation theory and probability theory. If Laplace's equation is expressed in terms of spherical coordinates and then
.9.X2 + 8y2
. Applying Abel's formula [Eq. will be studied in detail in Section 5.
EXAMPLE 5
Let p = 2 in Eq.8).4.
2x)y' + (x .
Show that the reduction-of-order method leads to the differential equation
xv"+(3-x)v' . 31.
32.
2. 33. Solve the IVP
xy"+(1 .
given that y.1)y = 0
y(1) = 2e
y'(1) = -3e. (8) to obtain
[(jx' + 1)x]w' + [3(gx' + 1) . (8) of this section. Perform the integrations required to simplify Eq.116
2 Lines Differential Equations
28. (13) of this section.y' + ay = 0 be a differential equation with constant coefficients. (17). where the method is discussed in detail.
. Show that y. Discussion of these
ramifications is postponed to Chapter 5. Verify Eq.
34.
29. is equal to a constant.(x) = e' is a solution of the differential equation.4x']w = 0. (9). = xe' is a solution of the differential equation
y" -4y"+5y'-2y=0. Show that the reduction-of-order method leads to a (first-order) differential equation with constant coefficients if and only if y/y. Verify Eq.
9
SOLUTIONS OF LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS BY THE METHOD OF TAYLOR SERIES
In this section we provide a brief introduction to the method of finding Taylorseries representations for solutions of differential equations. is a nontrivial solution of the differential equation. (7) of this section. Set w = v' in Eq. Verify that v2 = 1 is a solution of the differential equation (8). There are many ramifications associated with the Taylor-series technique.
Solve this differential equation by separation of variables and then integrate
the result to obtain Eq. Assume that y. This method is perhaps the most widely applicable technique for solving (or approximating the solutions of) differential equations with variable coefficients.2v = 0. Let a2y" + a. We include the section here simply to round out our treatment of linear differential equations.
35.
36.
30. Verify Eq.
xo)".2.
Solution The Taylor (Maclaurin) expansion has the form
y(X)
= y(O) + Y'(O)x +
Y2o)
xZ +
Y2°) x3 + .3 guarantees us that this IVP possesses a unique solution. we have
y.) +
or. solve the IVP (1).f'(xo)(x .
(3)
Differentiating (3). y(O) = 1.
.X(."(o) = 0. recall from calculus that a Taylor-series representation for a function f is of the form
f(x) = f(xo) +. and we illustrate each approach via an example.(0)(0) _ -1. Consider the initial value problem
y" + xy = 0
Y(O) = 1.
Similarly.
f(x) = i f
(x .
EXAMPLE 1
Using a Taylor-series approach...
y(5) _ -3y" _xYy("(o) = 0
Y(6) _
-4y".
(2)
Now. First. and from (1) we have
Y' = -xY
Y"(0) =
0. .
y" = -2y' . None of the methods that we have discussed can be applied to this differential
equation.xo)3 + .9
Solutions of Linear Homogeneous Differential Equations
117
We reiterate the assumption made earlier in this chapter that all the coefficient functions are continuous and that the leading coefficient function does not vanish in the interval of definition of the differential equation.)' + fx(x
n(
(xo)
.
y'(0) = 0.x(. compactly.xy'
--t>
y(b)(O) = 4.
(1)
Theorem 1 of Section 2.xy"
Differentiating (4).
We will seek a Taylor expansion for the function y defined by the IVP (1). y'(0) = 0..
f 2(x . There are essentially two approaches that one can follow. . we obtain
'
(4)
y"'(0)
_ -1 .
(In general. The first of Eqs..12a4 + a. therefore. . Thus. we obtain
y' = a.118
2
Linear Differential Equations
We could proceed in this fashion to generate as many terms of the Taylor
expansion as we want.)x + (24a. .x' +
we have
) = 0. are constants to be determined. + 2a2x + 3a3x2 + 4a.x'+. . . since there are an infinity of terms. (8) yields a2 = -7ao and the second yields
a3 = tae . = 0.x' +
and
y"=2a2+6a3x+12a. solve the differential equation
(2-x)y"+y=0.x' + 5a.
(5)
Solution In this example we are not given the initial values.
EXAMPLE 2
Using power series.
Substituting these results in the differential equation (5) yields
(2-x)(2a2+6ax+ 12a4)? +20a. we must have that the coefficient of each power of x must vanish.) Substituting our results into Eq. .. From (6).
. (5) has an expansion of the form
y = ao + a.x' +
.
(6)
where the coefficients a.oao .x3+ )
+ (ao + a.x2+20a. (2)
yields
y=I-3f X+61x6+.fl2a. = 0
(8)
24a. Here we assume that the solution y of Eq.
(4a2 + ao) + (12a3 .
(7)
In order that Eq. which in light of the result for a2 reduces to a3 = .x + ax2 + a9x3 + a4x' + a.. . we are led to the infinite system of equations
4a2+ao=0
12a.2a2 + a. + a3)X' +
= 0. .r3 + ax' + a. we cannot compute the coefficients in the Taylor expansion in the manner of Example 1.6a3 + a2)x2
+ (40a.
Doing the multiplication and gathering together terms with like powers of x. the number of terms to be computed would have to be prescribed at the onset.2a2 + a.
and so on. we cannot find them all in this manner.12a. therefore. (7) hold for every x in the interval of definition of the differential
equation.6a3 + a2 = 0
40a.x + a2x2 + a.'r1a1.
2<x3 + 960' +
+ al(X
ix3
.
y = a0 + a.5. y'(0) = 1
4. Example 2 illustrates the Taylor-series method most often used.In (1 + x)y = 0.axz . Eqs.)x5 +1 . y'(0) = 1. we obtain the coefficients a" from this infinite system. (10) converge for every x in the interval of definition of the differential equation? These conjectures and observations will be discussed in Chapter 5.
ao(1 .960a0
-
14l0a1+
and so on.9
Solutions of Linear Homogeneous Differential Equations
119
Analogous manipulations apply to each subsequent equation. y"(O) = 0 3. y(O) = 0. these
expressions are infinite series and therefore their convergence must be considered. y" + xzy = 0. The next step is to set the coefficient of each power of (x . since the differential equation (5) is of second order.nao)x2 + (-. (10) are the two linearly independent solutions to the differential equation (5).x0) can be combined.. y' . do the expressions in parentheses in Eq.Izal
a4 _ -°0a. y" + (3 + x)y = 0.zaa0 . use the Taylor-series method of Example 1 to solve the initial value problem. y"(0) = 0. y"'(0) = 0 5. We assume a"(x . thus obtaining an infinite system of equations [for
example. Finally. y(0) = 1. compute the coefficients out to the fifth power of the x term. (8)].x0) equal to zero.
2. y'(0) = 1.x0)".
(10)
We note that the solution (10) depends on two arbitrary constants. y'(0) = 2
. y(0) = 1. _ -z°ao . In each exercise.ix'
z°ozS + . y'(0) = 0
6. y(O) = 2.
1.
EXERCISES
In Exercises 1 through 10. y" + y' + ey = 0. ao and a as it should.x + (. We also conjecture that the parenthetical expressions in Eq. We substitute this expression into a solution of the form y =
the differential equation and then rearrange terms so that like powers of
(x . and we are led to the results
a2 = . Thus.. y" .°ao
a. Furthermore.
1
i
(9)
a5 .iza')XI + (+ (960a0
°18a1)x°
240a.2.).(sin x)y = 0. That is. y(0) = 0.
the general solution of this equation is
c. a.
(4)
where the c.(x)Yi" = f(x).
EXAMPLE 2
Find the homogeneous solution of the differential equation
2x2f .3y' + 2y = cos x.e' + c2e'`.3xy' .. + a.
-o
(2)
The function f is called the nonhomogeneous term for the differential equation (1).5. the homogeneous solution of the differential equation (5) is
Ye = c. In some texts the homogeneous solution is referred to as the complementary
solution...
Using the methods of Section 2.e + c2e2r. defined by
Yn = c1Yi + c2Y2 + .(x)yro-n + .10 Nonhomogeneous Differential Equations
121
The general nonhomogeneous linear differential equation has the form
a"(x)Y'"' +
a"_.. (2).3y = e'. there is an associated homogeneous differential equation defined by
r-o
i a. briefly..
(6)
.(x)Y = f(x)
(1)
or. ]
EXAMPLE I
Find the homogeneous solution of the differential equation
y" ..
(5)
Solution The homogeneous equation associated with the differential equation (5) is
y"-3y'+2y=0.. is called the homogeneous solution for Eq.2. It is the general solution of the associated homogeneous differential equation. + c"Y". then the function y. y" constitute a fundamental system of solutions for the homogeneous differential equation (3).
x > 0. are arbitrary constants. (2). Thus.
(3)
DEFINITION 2
If the n functions y y2. [Note that the homogeneous solution is not an actual solution of Eq.(x)Y' + a.. .(x)Y(') = 0. .
DEFINITION 1
With every nonhomogeneous differential equation (2).
We should prove that Y can be written in the form of Eq. (2) can be written in the form
y = Yi. we refer to this function as a particular solution of Eq...
i a. (7) satisfies the differential equation (2). we have
i-o
I ai(x)Yt1 = I a. and its solution (given in Example 2 of Section 2.i a (x)y(' = f(x) .vD]
= c. (2).7) is c.(x)[c. (7). I ai(x)Y.' + c2yi' + .(x)(Y .2 + c2x'. .3y = 0. = c1Yi + c2Y2 +
Y.x-"2 + c2x'.. Substituting into the differential
equation (2).y. (7).
That is. (2) and denote it by yo. In fact.122
2
Linear Differential Equations
Solution (6) is
The homogeneous equation associated with the differential equation
2x2y" . =
c. (3). and if y. This is equivalent to proving that the function Y . + c"A°
.y. . + C. + Y.
If.° + I ai(x)Y( )
s-0
i-O
i-o
= f(x). (2). ai(x)y("
i-o i-o
+ . Thus...f(x) = 0. + y0i)
i-o
_ i ai(x)[c.
is any particular solution of Eq.3xy' . + c2Y2 + .-o
+ Y..Y. + C.
i-0 i-0
i-0
..
THEOREM I
If y y2. then the general solution of Eq.y.
This latter differential equation is an Euler differential equation. Now let Y be any solution of the differential equation (2). I a.
Yr. the function y as defined by Eq.Y.. we find a function that satisfies Eq.x. satisfies the homogeneous differential equation (3).(x)Y°' + c2 7. somehow or other.. y form a fundamental system of solutions for Eq.
(7)
It suffices to demonstrate that y as given in the expression (7) satisfies the differential equation (2) and that any solution Y of the differential equation
Proof
(2) can be written in the form of Eq.)u' = i ai(x)Yt) .
x. 3. 5.
The key observation that makes the method of undetermined coefficients work is the fact that not only f(x) but also any derivative (zeroth derivative included. where a is a positive integer or zero. 2.
For example. 1 span the derivatives of the function 3x2. which are all of type
1.. We use the symbolism
3x2
Ix'. e)
. sin 2x1
xe -. (cos 2x. On the other hand.xe' sin x -. a particular solution of Eq. (xe...
(1)
where the coefficients a. . The coefficients of this linear combination are the undetermined coefficients (hence the name attached to the method) that are to be determined by substituting the assumed particular solution into the differential equation (1) and equating coefficients of similar terms. (1) is a linear combination of those functions of types 1-5 that span all the derivatives of f(x). where y is a nonzero constant. A (finite) product of two or more functions of types 1-4. 1 on the right. a . and 1.111
5 cos 2x -.. The following example contains the typical features of the method of unde-
. e2s cos x). x°. recall that the zeroth derivative off is f itself) of any term of f(x) is a linear combination of functions of types 1-5. es'. 4. (xe2' sin x. where S is a nonzero constant. the functions 1/x and log x are not of these types. 1)
to denote that any derivative of the function 3x2 on the left is a linear combination of the functions x2. x. e2' sin x. the function
f(x)=3x2-2+5e' -x(sinx)e'x+5eos2x+xe
is a linear combination of functions of types 1-5. are constants and f(x) is a linear combination of functions of the following types:
1. xe' cos x. x. For example. where R is a nonzero constant. any derivative of 3x2 is a linear combination of the functions x2. we have 2e` -.11
THE METHOD OF UNDETERMINED COEFFICIENTS
The method of undetermined coefficients is used when we want to compute a particular solution of the nonhomogeneous differential equation
ay') + an y'
+ ..
Roughly speaking. sin &x. As further illustrations. We also say that the functions x2. x.124
2 Unear Differential Equations
2. (e'')
-5-. cos yx. + a. a.y' + apy = f(x).
and 2. then the particular solution of Eq.y = 0. C.y = 0.of the differential equation
y" . (1). see Exercises 65 and 66. we should
multiply e' by x. and so we must multiply the function in this set by the lowest
integral power of x. respectively). then all the elements of that set should
be multiplied by the lowest integral power of x. that is. We first compute the functions that span the derivatives of each one of the three terms of the function f. 1. 1. B.11
The Method of Undetermined Coefficients
125
termined coefficients. = Axe + Bx + C + Dxe'.2x2 .y = 0.=Axe+Bx+C+De'.y = 0.(Ax2 + Bx + C + Dxe') _ -2x2 + 5 + 2e'.2.
(2)
Solution The method of undetermined coefficients is applicable to this example
since the differential equation (2) is of the form of Eq. 1} is a solution of y" . 11 and {e'} is a solution of the associated homogeneous equation for (2).)
EXAMPLE 1
Compute a particular solution. and so we leave this set as it is. Thus.. x. x. With our previous notation. we have
. so that the resulting new set does not contain any function that is a solution of the associated homogeneous equation. The set {1} in (4) is omitted because it is contained in the larger set (3). obtaining {xe').
if any one of the functions in any one of the sets {x2. a solution of y" .2x2 + 5 + 2e'. 1} and {e'} is a solution of the associated homogeneous equation. x. C.{x2. so that the resulting function is not a solution of y" . a particular solution of Eq. (For motivation of the method. On the other hand.
To obtain the undetermined coefficients A. but xe' is not. and f(x) =
-2x2 + 5 + 2e' is a linear combination of the functions x2. 1} and {e'}. x. the derivatives of f are spanned by the functions in the sets
{xi. B. x. we compute
(6)
yo=2Ax+B+De'+Dxe' y'=2A+2De'+Dxe'
and substitute these results into the differential equation (2). Hence. e' in the set {e'} is a solution of y" .y = .y = 0.
where A.
(4) (5)
Therefore.
. y" . obtaining
(2A + 2De' + Dxe`) . (2) is of the form
y. Now if none of the functions in the sets {x2. 1} and {xe'}.
y. and D. and the reader is urged to study this example very carefully. Since e' is. no function in the set {x2. On the other hand. (2) is a linear combination of the functions in the sets {x2. x. and D are the undetermined coefficients. Now. and e` (which are of types 1. 11
(3)
5-{1}
2e'-> {e'}.
Alternatively. C = -1. Ix.
Thus.126
2
Linear Differential Equations
Equating coefficients of similar terms. we obtain the following system of equations:
2A . A = 2. So
yo=Axe+Bx+C+Dxe
is the correct form of a particular solution of Eq. we suggest that the explanations that lead us to the form (6) for a particular solution of Eq.{xe')
(e is a solution of the associated homogeneous equation and xe is not).Ix"e. x2e. we have
Solution
3x2e'-> (x2e. Should we assume an inappropriate form for a particular solution. a contradiction will occur in the resulting system of equations when we attempt to compute the undetermined coefficients. xe'}
e2'. cos x}
2--. (2) is
y.(11
2e'Ie'I. (2) be abbreviated in accordance with the following format:
-2x'->Ix2. D = 1.x. 1l. and a particular solution of Eq.lei'}
x sin x -> {x sin x. it may happen that the coefficients of unnecessary terms will be found to be equal to zero. (2).C = 5
2D=2
-B=0
-A = -2. e} .[1l
3x -.=2xz-1+xe. sin x. 1}
5 -. xe.
Since e is a solution of the associated homogeneous equation but xe is not.
. To save space.
EXAMPLE 2
Find the form of a particular solution of the differential equation
y"+2y' -3y=3x2e'+e'+xsinx+2+3x.
The crucial aspect of the method of undetermined coefficients is that we assumed the proper form for a particular solution (such as. B = 0. x cos x. xe instead of e in Example 1).
1
The electric circuit shown in Figure 2. we use the initial conditions to determine c c2.-1= -1. c. it suffices to compute Q. Here a generator supplying a voltage of V(t) volts is connected in series with an R-ohm resistor. sin x
.
APPLICATIONS 2.c.
Now
y(0)=0=> c1+c2=0
Y'(0)=1. By definition
I = dQ
(11)
Thus.z x sin x.11
The Method of Undetermined Coefficients
129
Finally.1xcosx
2 2
y" = . We would like to find the current I as a function of time.
Thus.1sinx .11.cos x + i x sin x. c.cosx + 2 .c. = 0. = 0. an inductor of L henries. It is known from physics that the current I produces a voltage drop across the resistor equal to RI. When the switch is closed. and a capacitor of C farads. We also want to find the charge Q = Q(t) coulombs in the capacitor at any time t. and a voltage drop across the capacitor equal to
Figure 2.C2 COS x . c3 _ -1. and c3. we have
y' = . a current of I = 1(t) amperes flows in the circuit.2.5
.c3+2=1 y"(0)= -1=> -c.sin x + 2x . a voltage drop across
the inductor equal to L(dI/dt). and the solution of the initial value problem
is
y(x) = .sinx + c. Differentiating Eq.5 is customarily called an RLC-series Electric Circuits circuit. (10).
(12).
Mechanical
Mass Friction (damping)
Electrical
m F.5 (with the switch closed). C
1
Reciprocal of capacitance
Displacement External force Velocity
y-Q
F(r) H V(r)
)).130
2
Linear Differential Equations
(1/C)Q. (16) of Exercise 57. the latter condition being obtained from the differential equation (12). we impose two initial conditions Q(O) = Q0 and Q(0) = 0.
(12)
Using the relation (11). Applying this law to the circuit of Figure 2. (12) can be written in the form
L dQ+R
z
dQ+CQ=V(t).(11C)Q(0)1. These analogies
are very useful in applications. For example. because at time t = 0 there is no current flow in the circuit. this is one of the basic ideas behind the design of analog computers. Initial conditions associated with this differential equation are 1(0) = 0 and 1(0) = (11L)[V(0) . To find the current 1(t). L
aHR
Inductance Resistance
Spring stiffness
K. Eq.2 Mechanical-Electrical Analogies
. which is usually easier to study. we observe the analogies Analogies between mechanical and electrical systems given in Table 2. we can use the relation (11) or the differential equation
Ldt2+R dt + C 1
dt dV(t)
(14)
which is obtained by differentiating Eq. In order to find the charge Q(t) in the capacitor. (13) with Eq. we must find the general solution of Eq. (13).I
Charge on capacitor Electromotive force Current
Table 2. Often the study of a mechanical system is converted to the study of an equivalent electrical system. To determine these two constants. we have
L d1
+ R1 + I Q = V(r).
Electromechanical If we compare Eq.
(13)
This is a second-order linear nonhomogeneous differential equation.2. This general solution involves two arbitrary constants. Q0 is the initial charge on the capacitor and Q(0) = 1(0) = 0. Kirchhoff's voltage law states that the voltage supplied is equal to the sum of the voltage drops in the circuit..
Danziger and G. Find xr for this differential
equation. for Eq.
.4. and C are constants. K. V.
where r. Herfindahl and A. Bull.
54. and (3 = 1. Medicine: Periodic Relapsing Catatonia In the development of a mathematical theory of thyroid-pituitary interactions. 1974). 18 (1956): 1-13. Social Behavior Second-order linear differential equations with constant coefficients of the form
y-Ay=B. find 0. Find the general solution of this differential equation. Bull. Solve the nonhomogeneous differential equation of Exercise 51 if T = . Merrill. b = 6.bx= -b. in this case. Here
0 = 9(t) is the concentration of thyroid hormone at time t and a a a3. Referring to Exercise 51. Herfindahl and Kneese26 presented the following second-order differential equation: r(3 rK
x-rx. a = 1. Prove. "O.4b < 0.
and
e < C
(15)
a30+a20+a10+0=0. Elmergreen. Biophys. Math. and K are nonzero constants.
These equations describe the variation of thyroid hormone with time. p. Math. If K # -1. Forced Motion of the Vibrating Spring In addition to the damping force assumed in Exercise 40. Danziger and Elmergreen29 obtained the following third-order linear differential equations:
a36 + a26 + a18 + (1 + K)6 = KC.
A>0. Math.
0>C. C. a.2. Minerals In finding the optimum path for the exploitation of minerals in the simple situation where cost rises linearly with cumulated production. Here B is an arbitrary constant.
were obtained by Rashevsky27 in his study of riots by oppressed groups.11
The Method of Undetermined Coefficients
133
52. suppose that an external force equal to f(t) acts upon the mass of the spring. b. (This phenomenon was observed in the actual experiment with rats. Rashevsky.
"L. (15). See also N. Economic Theory of Natural Resources (Columbus. show that when (1/T2) . 30 (1968): 735-49. 30 (1968): 501-18. Ohio: Charles E. Biophys. the motion is called a free motion. "N. Section 2.. 182. Bull.
55. Kneese. L. Biophys. Rashevsky. that the differential equation of the vibrating spring is given by
my + ay + Ky = f(t)
(16)
Such a motion is called a forced motion..) 53. the solutions contain oscillatory terms (linear combinations of sines and cosines). 57. When the external force is identically equal to zero.
56. Discuss the behavior of the solution at t -> oo.
= f(x).J.
e"
(x)=cex+c zex+ (a . B. f(x) = x'
The portion of the method of undetermined coefficients associated with the case wherein f(x) has terms that are solutions of the homogeneous equation may seem unmotivated. Timoshenko. consider the following exercises.134
2
Linear Differential Equations
58. For the forced motion of Exercise 57.
is
1
a* 1. 1928).Yo)xex +
e + (1 (a .3
y(o) = 5.1)2
a)xe
"S.
In Exercises 59 through 64. (a) Show that the general solution of the differential equation
y"-2y'+y=e°x. we did not give any indication where the proper form for yo came from. assume that m = 4 g. f(x) = e-x
63. Thus.
Solve this IVP.=Yo-(a-1)2
hence
yoex
and
cz=Yi . To this end. The mass is also subjected to an external periodic force equal to 5 cos t. N. and cz in the solution of part (a) are
c.
58. "On the Solutions of Linear Differential Equations. (17) for the
given f." Amer. The mass is raised to a position 3 cm above the equilibrium position and released with a velocity of 5 cm/sec downward. f(x) = sinh x
64.
Elasticity: Static Deflection of a Beam In the consideration of the static deflec-
tion u of a beam of uniform cross section (for example. E being Young's modulus of elasticity and I the moment of inertia of a cross section of the beam about some fixed axis. K = 3 dyn/cm. f(x) = xz
61. and that air resistance acts upon the mass with a force that is 7 times its velocity at time 1. assume that El = I and'solve Eq. Monthly 54. an I beam) subjected to a transverse load f(x). Vibration Problems in Engineering (Princeton. f(x) = cos x 62.: Van Nostrand."
65.1)z x
(b) Given that y(O) = yo and y'(0) = y show that c.Yo -A e-
Y(x) =
+ (y. Certainly. no. Farnell. . 3 (1974). there arises the differential equation's
EIdX. 11rhese exercises are extracted from A.
(17)
where E and I are constants. Describe the limiting motion at t --> x. Math. the motion of the mass satisfies the IVP
4y+7y+3y=5 cos t
y(0) = .
. f(x) = 1
60.
wo is referred to as resonance.12 VARIATION OF PARAMETERS
The method of variation of parameters. + a. (16) of Ex.x cos x).
67. If f(t) . sin x + yo cos x + {sin x .a sin x
y(x) = y.12
Variation of Parameters
135
(c) Show that
lira y(x) = yoe + (y. = y. (b) If w is "close to" wo. . hence sin ax ..yo) xe + rr2e.2. is large and gets even larger if w is chosen closer to wo. ..
(x)yc"-n + . sin x + yo cos x +
1-a
2
(c) Show that
lim y(x) = y. 57)
y + w2y = Fcos wt.
(a) If w # wo.. is used to compute a particular solution of the nonhomogeneous differential equation
a. requires that the coefficients
. as w . (a) Show that the general solution of the differential equation
y" + y = sin ax
is
sin ax y(x) = c.-. sin x + c2 cos x + 1 . however.(x)y' + ao(x)y = f(x)
(1)
The method of undetermined coefficients. Resonance When an undamped vibrating spring is subjected to a periodic external force of the form f(t) = M cos wt the differential equation takes the form (see Eq.a/(1 . For this reason wo is called the natural frequency of the free vibration.
w2 0
2. That is. show that -' =
F N2 cos wt. Find y.(x)yl'i +
a. the solution in part (a) shows that the amplitude of y.a2) and c2 = yo. for Eq. This phenomenon associated with the unbounded growth of the amplitude of y.
(18)
where wo = K/m and F = M/m. resonance is said to occur when the frequency of the external force matches the natural frequency of the system.. (18) in the case w = w. and c2 in the solution
of part (a) are c.0 (z>M = 0) the spring is said to be free.a2
(b) Given that y(O) = yo and y'(0) = y show that c. like the method of undetermined coefficients.
1
66.
and y2 form a fundamental
system of solutions for the equation a2(x)y" + al(x)y' + ao(x)y = 0.11.136
2 Unbar Differential Equations
a0. u" satisfy the system of equations
y... which in theory "always works. . + a.u.
i" qu + Y(2"-.
and that u..uR
= 0. y1'u. (See Appendix
. form a fundamental system of solution for Eq. y. which we present in Theorem 1.. and y2.)u2 + . + y2uZ +
+ y. .
=0. but for simplicity of presentation we prove it for the case n = 2. and that the function f be one of the types described in Section 2. and u2 satisfy the system of algebraic equations
Y1ui + Y2u2 = 0
(4)
Y'X +Y2u2 = a2(x)
f(x)
The determinant of the coefficients of the nonhomogeneous system (4) is the
Wronskian of the linearly independent solutions y.. and if the
functions u u2. + y("-2)un = 0.. ." evolves from the following consideration... and a of the differential equation be constants. (1).. Suppose that we know the general solution
of the associated homogeneous differential equation
a"(x)Y(") +
a"-. + u2y2 +
Proof
.u. we use the variation-of-parameters method.Y..y. (2). + u2y2 + . Y..(x)Y' + ao(x)y
= 0.u
y. This theorem is stated for the general nth-order equation... When we encounter differential equations for which we cannot apply the method of undetermined coefficients. + U. as functions (variables) u.
is a solution of the nonhomogeneous equation (1)? Such an approach has been
shown to lead to meaningful results.y.
. and to impose appropriate conditions on these functions so that the expression
u.
THEOREM 1
If y y2..(x)y("-') + .+y2uZ+ +y"u.
(Special case n = 2) By hypothesis. u"y" is a particular solution of Eq. the system has a unique solution u.
(2)
Is it possible to treat the constants c. +
(x) a"(x)
then y = u.
(3)
yi"-2)ul + Y( -2)u2 + . This method.. u2. Since this determinant
is different from zero..+y2u2 + +y.
= 0. . a .
' and u2 we determine u.. the proof in the general case follows in exactly the same manner. Hence. u. and in some cases differentiation is easier than integration. Indeed.. we then solve Eqs.
iY. The reader should observe that Theorem 1 and its proof show us how to construct the particular solution y.. Next. .(x)y + ao(x)y. + u2Yz + u2y2 = u. + u2Y2
(5)
is a solution of the differential equation.(x)Y' + ao(x)Y = a2x)[uiY + u2Yz + a (x)]
+ a. (2). (In these integrations. And. we integrate each of the u.Yi + u2y2. Using these functions. The method of
undetermined coefficients involves differentiation. We note that the variation-of-parameters method applies to any nonhomogeneous differential equation no matter what the coefficients and the function f happen to be. while the method of variation
of parameters involves integration. there is no loss in generality if we
set the constant of integration equal to zero.. u2.
Y
11
= u"+u '+u "+u =u '+u "+f(x) 1Y12Y2
.y.. + u.Y. . respectively.(x)y2 + ao(x)Y2] + f(x)
= f(x). . from (5) we have
Y' = u.) From u. u2. for otherwise the term corresponding to a nonzero constant of integration would be absorbed into the homogeneous solution when we wrote out the general solution). .. (4).]
+ u2[a2(x)y' + a.12
Variation of Parameters
137
A.[a2()Yi + a. Now
a2(x)Y" + a. (4). finally. Therefore.
The last expression is obtained by applying the first of Eqs. (3) for u. The reason for introducing the method of undetermined coefficients is that it is sometimes quicker and easier to apply. + is the desired particular solution y. and u2 via integration. . We first find the fundamental system of
solutions for Eq.2.
2Y2
2'Y2
a2(x)
Once again the last expression follows by applying the second of Eqs. y satisfies Eq.y.
utyt + u2y2 + . (1) (for n = 2) and the proof is complete..
.
EXAMPLE 1
Solve the differential equation
y" + y = csc X. u. to find. All that remains to be shown is that the function
Y = uy..u. u u2..y. Other than the tedious notation.' + u2Yi] + ao(x)[uy.(x)[u. + u2y2]
= u..
What velocity is attained at this time and what is its height?
"This is Exercise 8 in David Halliday and Robert Resnik.001v2. (New York: McGraw-Hill. "This is Exercise 11. Reprinted by permission of John Wiley & Sons. 68. p.
. Part I. 1977).001 v2. How high above the target must the gun be aimed so that the bullet will hit the target? Take g = 32.21.0 ft above ground so that its angle of projection is 45° and its horizontal range is 350 ft. p. ibid. 454.. Dynamics (New York:
McGraw-Hill. p.]
31. 3rd ed. p.28. and g is the acceleration due to gravity at the surface of the earth. which is travelling in the opposite direction at a constant speed of 45 ft/sec.3-2 in Eduard C.2 ft/sec'. the minimum velocity with which a particle should be projected vertically upward from the surface of the earth if it is not to return to the earth. "This is Exercise 6. Russell Johnston.Review Exercises
143
27. A rifle" with a muzzle velocity of 1500 ft/sec shoots a bullet at a small target 150 ft away.16 in Ferdinand P. In a vertical takeoff. 67. A short time later it is passed by truck B. Thompson. Reprinted by permission of McGraw-Hill Book Company. It has been determined experimentally" that the magnitude in ft/sect of the deceleration due to air resistance of a projectile is 0. Reprinted by permission of John Wiley & Sons. Jr. Beer and E. Physics.2 ft/sec'.
"This is Exercise 1. Inc. p.]
32. where r is the distance from the center of the earth to the particle. 447. (a) How long does the projectile remain in the air? (b) At what horizontal distance does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground? Take g = 32. determine when and where the vehicles pass each other. A projectile is fired horizontally" from a gun which is located 144 ft (44m) above a horizontal plane and has a muzzle speed of 800 ft/sec (240 m/sec). If the projectile is released from rest and keeps pointing downward. Reprinted by permission of McGraw-Hill Book Company. Reprinted by permission of McGraw-Hill Book Company. Pestel and William T. Inc. 1977). 30. 3rd ed.' rocket engines exert a thrust that gives it a constant acceleration of 2. where g = 32. 17.0 g for 40 sec. Copyright m 1977 by John Wiley & Sons. "This is Exercise 11.
28. A batter" hits a pitched ball at a height 4. "This is Exercise 11.2 ft/sec2. p. "This is Exercise 14. 446. Reprinted by permission of McGraw-Hill Book Company. ibid. Copyright ® 1977 by John Wiley & Sons. Reprinted by permission of John Wiley & Sons. Will the ball clear the fence? Take g = 32. 1968). The acceleration due to gravity36 of a particle falling toward the earth is a = -gR2/r2. R is the radius of the earth. Derive an expression for escape velocity. The ball is fair down the left field line where a 24 ft high fence is located 320 ft from home plate. ibid. Inc..2 ft/sect.. 68. ibid.. determine its velocity after it has fallen 500 ft. [Hint: The total acceleration is g .0. Automobile A starts" from 0 and accelerates at the constant rate of 4 ft/sec'. Copyright ® 1977 by John Wiley & Sons. p. (New York: John Wiley & Sons. where v is expressed in ft/sec. Vector Mechanics for Engineers: Statics and Dynamics.
33. Knowing that truck B passes point 0 25 sec after automobile A started from there. [Hint:
v=0forr=x.
29..
"This is Exercise 1.4-13. 34. as shown in Figure
2.
"This is Exercise 1. ibid. A telephone pole" is raised by backing a truck against it. 39. where w2 = s/m. Show that the steady-state solution (the particular solution) is given by
X =
FO sin wt
m(wa .3-18. 2nd ed. and m is the mass. The displacementa2 of a simple harmonic oscillator is given by x =
a sin wt. Reprinted by permission of McGraw-Hill Book Company. [Hint: The differential equation is my + ay + Ky = 0. Show that this ellipse represents a path of constant energy. i) is an ellipse.6.. and the time of flight. Reprinted by permission of John Wiley & Sons.11 in H. Inc. Pain. Reprinted by permission of McGraw-Hill Book Company. in putting a shot. J. with a velocity of 34 ft /sec.. find the least time required for the train to go between two stations 82 miles apart. Determine the range. An overdamped spring mass system is displaced a distance A from its
equilibrium position and released from rest. the maximum height attained by the shot. If the angular velocity of the pole is to be a constant. at an angle of 40° above the
horizontal.w2) '
1117his is Exercise 1.. then stop
within a distance of 1/2 mile. If the values of this displacement x and the velocity x are plotted on perpendicular axes. Copyright ® 1976 by John Wiley & Sons.]
38. Inc. (New York: John Wiley & Sons. It is estimatedi0 that an athlete. Assuming that acceleration and deceleration are uniform. p.
. p.144
2 Uneer Differential Equations
h
}
Barrier
Figure 2. p. releases the shot at a height of 7 ft..
35.
36. 68. p. ibid.] 39. p. Copyright C 1976 by John Wiley & Sons.4-42. The Physics of Vibrations and Waves. 1976). s is the stiffness.
37.3. Find the position of the mass at any time t. [Hint: Energy of a simple harmonic oscillator equals 2 (mass)(velocity)2 The basic differential equation for the mo+i tion is z + w2x = 0. The equation mz + sx = FO sin wt describes" the motion of an undamped simple harmonic oscillator driven by a force of frequency w. eliminate t to show that the locus of points (x. A train" can attain its maximum speed of 60 mph in 4 min. Draw a velocity-time diagram. 'Ofhis is Exercise 1. Reprinted by permission of McGraw-Hill Book Company. 19.6
34. Reprinted by permission of John Wiley & Sons. ibid. show that the required speed of the truck is i = h9/sin2 0. ibid.
"This is Exercise 2. 43..
p.=
1
18q+27q. p. ibid. Reprinted by permission of McGraw-Hill Book Company. The circuite5 shown in Figure 2.]
(b) Calculate the current through the source for the charge found in part
(a).
0. 333. Reprinted by permission of McGraw-Hill Book Company. 332.7
where wa = s/m.8
"This is Exercise 4 in Charles A. calculate
the complete response.145
L=1H
Figure 2.7. "This is Exercise 8.
2
.
where A and B are constant. [Note: the charge oscillates at one-third of the frequency of the source.
The input is e and the response is v. 41. Kuh. and that at time t = 0 the state is iL = 2 amp and v.w2) 0
+ A cos wot + B sin Wat.5f
T
Figure 2. Basic Circuit Theory (New York: McGraw-Hill. Sketch the behavior of x versus w and note that the change of sign as to passes through wo defines a phase change of IT radians in the displacement..8 is made of linear time-invariant elements. Now show that the general solution for the displacement is given by
X=
Fo sin wt
nt(w2 . a response q(t) = cos
(0) coulombs satisfies the differential equation.
. Consider" the nonlinear time-invariant subharmonic generating circuit shown in Figure 2.
(a) Verify that for an input e.(t) = sin 2t volts. Knowing that e. = 1 volt. = (1/54) cost volt. Desoer and Ernest S. 1969). 40. The inductor is linear and the capacitor has a characteristic
v.
Reprinted by permission of Prentice-Hall. N. Intermediate Classical Mechanics (Englewood Cliffs.2 in J.SO
47. Let' a sinusoidal voltage e.J. 51. A'
V
-CC'v . 2nd cd. 102. Reprinted by permission of Prentice-Hall. Norwood.J.
[Hint: Differentiate both sides of the equation with respect to x and factor. 334. Monthly 60 (1953): 264.3y(y')2 + 4y' = a. Determine01 the general solution for the motion of an underdamped harmonic oscillator having mass m. Jr.
2
46. (Englewood Cliffs.
43. "See Amer. Introduction to Nuclear Physics and Chemistry. "This is Exercise 3-3.
. Math.146
2
Linear Differential Equations
1NH
I P1
T
Figure 2. N. Solve" the differential equation
(y')' . ibid. 26. What° is the time of flight of a projectile fired with a muzzle velocity vo at an angle a above the horizontal if the launching site is located at height h above the target?
44. and spring constant K when the driving force (the external force) is a constant FO. Reprinted by permission of Prentice-Hall.9. p. By differentiation. "See Amer.49 confirm that 4)(4) =
lution of
1 d2(b
__
1
2. Inc.9
42. 1969). Monthly 36 (1929): 112-13. Inc. Given i(0) = 1 mA and v(O) = 0.uv')2
U
and c is an arbitrary constant. p. Inc. "This is Exercise 14 in Bernard G.]
"This is Exercise Ila. where A and B satisfy the equations B.. damping constant b. ibid. a JPd. Harvey. "This is Exercise 2-2. p..: Prentice-Hall.(t) = 3 cos (106t) volts be applied at time
t = 0 to the linear time invariant LC-circuit shown in Figure 2. calculate and sketch i(t) fort ? 0.. Reprinted by permission of McGraw-Hill Book Company. Math.6. If u and v are linearly independent particular solutions of the differential equation
y"'+Py"+Qy'+Ry=0.: Prentice-Hall. 1979).
45. Solve the nonhomogeneous equation by assuming a form for the particular solution and then determining the constant coefficients.Tr
(cos m4i + i sin m4)) is a so-
4)
-m.
prove' that the general solution is given by y = Au + By. p.
CHAPTER 3
Linear Systems
3. . = a11(t)x1 + au(t)x2 + . we shall restrict our attention to systems of linear differential equations. These definitions and theorems
extend easily to systems of n linear differential equations in n unknown functions of the form
X. + a1 (r)x + fi(t) X2 = a21(t)x. Nonlinear systems. systems
of nonlinear differential equations. a21. This notation has obvious advantages when we
are dealing with general systems. that is..2(t)x2 + ... it is custom-
ary to use the symbols x x2. by approximating each nonlinear equation of the system by a linear one. a22. in each case up to now.. one is interested in studying systems of n differential equations in n
unknown functions where n is an integer >_ 2.
REMARK 1
In the treatment of systems of differential equations.. Consequently.1
INTRODUCTION AND BASIC THEORY
The differential equations that we studied in Chapters i and 2 are ordinary
differential equations involving one unknown function.... However. in special cases alternative notation may be used. a72. the dots over x. and the functions fl. including applications and generaliza-
tions. As usual. Since in this book we are mainly
interested in obtaining solutions. f2 are all given continuous functions of t on some interval I and x x2 are unknown functions of t. + a.. we had to solve one differential equation which involved one unknown function. are usually (and successfully) studied in more advanced courses.. linear systems. = a11(t)xl + au(()x2 + f1(t) X2 = a21(()x1 + a22(()x2 + f:(t). x to denote unknown functions and to use t as the independent variable. and x2 in (1) indicate differentiation with respect to the independent variable t.(t)x + f2(t)
(2)
X=
a. For many reasons.(t)x + fi(t). For some results on nonlinear systems. In this chapter we mainly consider systems of two linear differential equations in two unknown functions of the form
X. see Chapter 8. .
. In this section we present a few definitions and state without proof some basic
theorems about the solutions of system (1).
(1)
where the coefficients a. + a32(t)x2 + + ay..
as the reader can easily verify.(t) = y1(t) and x2(t) = y2(t). These methods.
For example. a21. Sometimes it is convenient (and suitable for studying more general systems) to denote a solution of (1) by the column vector
For example. are all known constants. each being
differentiable on an interval I and which.
z. For the reader's sake we mention here that the method of elimination discussed in Section 3.0.148
3
Linear Systems
or. n. more compactly.
J-1
f(t).
xl(t)
3t . and an are all known constants.
DEFINITION 1
A solution of the system (1) is a pair of functions x1(t) and x2(t).
is a solution of system (3)..
In subsequent sections we shall present two elementary methods for finding explicit solutions of system (1) when the coefficients a.. when substituted into the two equations in (1). can be extended to solving system (2) when its coefficients a.
.3 is recommended for system (2) when n z 3. makes them identities in t for all t in I.. This is an elegant method. but its detailed presentation and justification requires a good knowledge of matrix analysis and linear algebra. x2(t) = -t . a12. with varying degrees of technical difficulties. 2.3
x1 = x2 + t
(3)
is a solution of the system
X2 = -2x1 + 3x2 + 1
for all t. the easiest method to solve system (1).
i = 1.2 is.. The
matrix method outlined in Section 3.
REMARK 2 The algebra of column vectors that is used in this chapter is embodied in the following statements:
(a) Equality x (t) =
-
[t]
if and only if
x. perhaps. . .
(t)l rc. system (3) is nonhomogeneous. + 3x2
cal n is homogeneous. = x2
(4)
X2 = .
DEFINITION 3
22e-]
The column vector
[0]
'
that is.. the theory of linear systems resembles the theory of linear differential equations. each of the two column vectors in (5) is a solution of system (4). Let us consider the homogeneous system
z. x1(t) 3 0. the linear combination
cl I eJ + c2I 2e2]
. and f2 in (1) are equal to zero. is a solution of (6) for any choice of the coefficients.(t) fIrx. The reader(caln verify that each of the two vectors
I
e
12e-]
(5)
is a solution of system (4).2x.2(t)x2. we have the following. the system is called
homogeneous. + a.
. On the other hand.3. = an(t)x1 + au(t)x2
z2 = a21(t)x.x. and c ry. To a great extent. Otherwise it is called nonhomogeneous.
(6)
where the coefficients a. x2(t) = 0..
For example.
Ly2(t)J . For any constants c.Lclx2(t) + c2y2(1)
DEFINITION 2
When both functions f. This solution is called the trivial solution. the system
X.1
Introduction and Basic Theory
149
(b) Linear Combination.. a. a21.
THEOREM I
Any linear combination of solutions of (6) is also a solution of (6). Any other solution of (6) is called a nontrivial solution..(t) + c2y.[c. As in the case of homogeneous linear differential equations. therefore. and a22 are all continuous functions on some interval 1.(t) Lx2(1)] + c.
For example. and c2.e +
is also a solution of (4). for any constants c.
. which establishes our claim.
c2x12(t) = 0 and c. we find that c2e' = 0. = a11(t)x.(t)x. can be used to check the linear dependence or independence of solutions of system
(6). The reader can verify that
Ie
J
and
[6e2'.
then c.150
3 Uneer Systems
DERNRION 4 Two solutions
and
x21(1)
X12(t)xzz(t)
JJ
(8)
ndent of system (6) are called linearly independent oon an interval 1 if the statement
c+ c2
= [g]
forall t in 1. we can demonstrate that the two solutions (5) are linearly
independent solutions of the system (4).e' + 2c2e' = 0. In fact.e' = 0 and so c.
For example.
(9)
are linearly dependent solutions of (4). + a22(t)x2
are linearly independent on an interval I if and only if the Wronskian determinant
X. Subtracting the first equation from the
second. Thus c2 = 0. = c2 = 0. the first equation becomes c. Therefore. + a12(t)x2
X2 = a2. Consequently. if
C1 [e"] + C2 L2?2'2'1 = [o
. c1 = c2 = 0.x21(t) + c2x (t) = 0
for all t in I implies that c.
THEOREM 2
The two solutions
and
Lx21(t)J
X"(1)]
x22(1)
of the system
z. Otherwise the solutions (8) are called
linearly dependent. reminiscent of the Wronskian determinant. The following criterion.
that is.e' + c2e2i = 0 and c. (t)
x21(t)
x12(1)
x22(t)
is never zero on I. = 0.
+ a22(t)x2.u and x. the reader can verify (by direct substitution) that the unique solution of the IVP
X. the two solutions (9) of system (4) are linearly dependent
because
I
e2' 3eu 2eu 6e" =6e"-6e"=0. = a11(t)xl + a12(t)x2
(6)
x2 = a21(t)x.
Furthermore.t .2x..
THEOREM 4
There exist two linearly independent solutions of the system
x.2.) = x10.
On the other hand. Then the IVP consisting of system (1) and the initial conditions
x1(t.
THEOREM 3
Assume that the coefficients a. The reader should recognize the strong resemblance between this theorem and the existence-uniqueness theorem of Chapter 2. a2 a22 and the functions f.
.1
Introduction and Basic Theory
151
For example.
The following basic existence and uniqueness theorem for linear systems is stated here without proof.
is
x1(t) =i e' -2 t . Let to be a point in I and let x.(t)
(1)
X2 = a2.3.
x2(to) = xm
has a unique solution
(x.(t)1 lx2(t) . this unique solution is valid throughout the interval I. the two solutions (5) of system (4) are linearly independent
because
I'E`
e'
2e21
e'I=2e' -
*0
forallt.
Using this theorem.e
x2(t) = a e .(t)x1 + a22(t)x2 + f2(t)
are all continuous on an interval 1.. and f2 of the
system
z. + 3x2 +
1
x1(0) = -1
(10)
x2(0) _ -. a12. =
a12(t)x2 + f. = x2 + t
x2 = . be two given constants.
Linear systems of differential equations are utilized in many interesting mathematical models of real-life situations. Here we present a sampling of these
applications. Consider two species which coexist in a given environment and which interact between themselves in a specific way. For example, one species eats the other, or both species help each other to survive. Let us denote by N,(t) and N2(t) the number, at time t, of the first and second species, respectively. If the two species were in isolation from each other, it would be reasonable to assume that they vary at a rate proportional to their number present. That is, N, = aN, and N2 = dN2, where the constants of proportionalities a and d are certain constants (positive, negative, or zero, depending on whether the population increases, decreases, or remains constant, respectively). However, we have assumed that the two species interact between themselves. Let us assume for simplicity that the rate of change of one population due to interaction of the two species is proportional to the size of the other population. Then we obtain the following linear system of differential equations:
Ecology
IN,=aN,+bN2
N2 = cN, + dN2, where b and c are certain constants (positive, negative, or zero). In Turing's theory of morphogenesis,' the state of a cell is represented by Morphogenesis
the numerical values of a pair of substances x and y which are called morphogens. As an illustration, Turing proposed that these morphogens interact chemically according to the kinetic equations
x=5x-6y+1
y=6x -7y+ 1.
The general solution of the system (14) is given by
have been used to describe periodic fluctuations of the concentrations of hormones in the bloodstream associated with the menstrual cycle.' In 1957, Danziger and Elmergreen' proposed mathematical models of endocrine systems that are linear systems of differential equations. For example, reactions of the pituitary and thyroid glands in the regulation of the metabolic rate and reactions of the pituitary and ovary glands in the control of the menstrual cycle can be represented by simple linear systems of the form
X1 = -471x, - 412x2 + a,o
x2 =
a21x, - a22x2 + a20,
(17)
where all the av are positive constants with the exception of am, which is equal to zero. Recently, Ackerman and colleagues' obtained system (17) with am > 0 as a linear approximation of a mathematical model for the detection of diabetes.
Circuit Theory
Consider the electric circuits shown in Figures 3.1 and 3.2. Assume that at time t the current flows as indicated in each closed path. Show in each case that the currents satisfy the given systems of linear differential equations.
These systems are obtained by applying Kirchhoffs voltage law (see Section 1.1.1) to each closed path. For example, the first equation in (19) is obtained by applying Kirchhoffs voltage law to the closed path in Figure 3.2, where the current 1, flows. In fact, we have
Solution
V(t) - L,%, - R,1, + R,1, - R211 + R,12 = 0,
(20)
which after some algebraic manipulations leads to the first differential equation
in the system (19). Clearly, the sign in front of any term in Eq. (20) is + or - according to whether the term represents voltage increase or drop, respectively.
Lewis F. Richardson, in a classic monograph,' devised a mathematical model of arms races between two nations. In this model two very simplistic assumptions are made:
Arms Races
The Richardson Model
1. There are only two nations involved, denoted by A and B. 2. There is only one kind of weapon or missile available. Let MA(t) and Me(t) denote the number of missiles available to countries A and B, respectively. Then M4(t) and Me(t) are the time rate of change of the missile stocks in the two countries. The Richardson model for a two-nation
'Lewis F. Richardson, "Generalized Foreign Politics," British J. Psycho!. Monograph Suppl. 23
(1939).
156
3
Linear Systems
armament race is given by the following system of nonhomogeneous linear
differential equations
MA = -a,M + b,MB + C,
Mt+ = a2MA - b2MB + c2,
where the coefficients a b a2, and b2 are nonnegative constants. The constants b, and a2 are the "defense coefficients" of the respective nations; the constants a, and b2 are the "fatigue and expense" coefficients, respectively; and the constants c, and c2 are the "grievance coefficients" and indicate the effects of all other factors on the acquisition of missiles.'
Economics
The economics application described in Section 1.4.1 can be extended to
interrelated markets. For simplicity we assume that we have two commodities. Assume that the rates of change of the prices P, and P2 of these commodities
are, respectively, proportional to the differences D, - S, and D2 - S2 of the demand and supply at any time t. Thus, Eq. (10) of Section 1.4.1 is replaced
by the system of equations
d = a,(D, - S,)
and
d
= a2(D2 -
S2)
In the simple case where D D2, S and S. are given as linear functions of the prices P, and P2, the system above takes the form of a linear system in the unknowns P, and P2. That is,'
dt
EXERCISES
=
a21P, + a22P2 + b2.
1. (a) Show that the column vectors
[e3,]
and
[3e"]
are linearly independent solutions of the linear system
s,=2x,+x2
X2 = - 3x, + 6x2.
'For more information on arms races, the interested reader is referred to M. D. Intriligator, Proceedings of a Conference on the Application of Undergraduate Mathematics in the Engineering, Life,
12. Consider the electric network shown in Figure 3.3. Find a system of differential equations that describes the currents I, and I2. 13. Use Kirchhoff's voltage law to verify the equations in system (18). 14. Use Kirchhoff's voltage law to verify the equations in system (19).
3.1
Introduction and Basic Theory
159
V(r)
Figure 3.3
15. Professor Umbugio,' a mythical mathematics professor, has invented a remarkable scheme for reviewing books. He divides the time he allows himself for reviewing into three fractions, a, 0, -y. He devotes the fraction a of his time to a deep study of the title page and jacket. He devotes the fraction E3 to a spirited search for his name and quotations from his own works. Finally, he spends the fraction y of his allotted time in a proportionally penetrating perusal of the remaining text. Knowing his characteristic taste for simple and direct methods, we cannot fail to be duly impressed by the differential equations on which he bases his scheme:
dx=y-z,
dt
dy=z-x,
dt
dz=x-y.
dt
(a)
He considers a system of solutions x, y, z which is determined by initial conditions depending on a (small) parameter E independent of t. Therefore, x, y, and z depend on both t and e, and we appropriately use the notation:
x = f(t, E),
f (O, e) = 3 - E,
y = g(t, f),
g(O, e) = 3,
z = h(t, E).
h(0, e) = D + E.
(b)
The functions (b) satisfy the equations (a) and the initial conditions
(C)
Deflate the professor! Find a, p, and -y without much numerical compu-
+ z2
reduces the system (6) to the single differential equation
z2
-
r
`°2, _ yylz a.2) z2.
This problem is taken from Amer. Math. Monthly 54 (1974): 223.
160
3
Linear Systems
Furthermore,' the function z, is given by
a,
Z,
y,
Z2-
17. Given that [11 is a solution of the system
X, =X,+(1 -t)X2
X2 = - x, - x2,
t>0
use reduction of order to compute the general solution of the system.
18. State Definitions 1 through 4 of Section 3.1 for system (2) and the corresponding homogeneous system
X,=ia9(t)xl,
J.1
i=1,2,.. 2,.(21)
19. State Theorem 2 of Section 3.1 for n solutions of system (21).
20. State Theorem 3 of Section 3.1 for system (2). 21. State Theorem 4 of Section 3.1 for system (21). 22. State Theorem 5 of Section 3.1 for system (2).
In Exercises 23 through 30, answer true or false.
Thus. = S2 = . given in (3) to find the other unknown function x2 = X. + 2x.2x.(t) = c.2x.
The first equation in (1) states that x2 = x. we have succeeded in eliminating the function x2 from system (1).162
3
Linear Systems
30. = 0. . obtaining in the process the differential equation (2).5e2'. (2) is
x. + 3x2. we obtain
X. + x2 + 7x. + 3x2. The general solution of Eq. s.3s.
(4)
Equations (3) and (4) give the general solution of system (1). hence
z. Other ways to eliminate one of the unknown functions and arrive at a single equation for the remaining unknown are the following: Solve the first equation
.
EXAMPLE 1
Find the general solution of the homogeneous linear system
X.2
THE METHOD OF ELIMINATION
The most elementary method for solving a system of linear differential equations in two unknown functions and with constant coefficients is the method of elimination. = -10x. = . + X2 + 12X3
X1(0) = 5.
3. which contains one single function and which can be solved. + 3s.
X3(0)
2
x. = X2
(1)
X2 = .
x2 = -9x.
is
x2(0) = 2. In this method the aim is to reduce the given system to a single differ-
ential equation in one unknown function by eliminating the other dependent
variable. = 2e' + 3e2.
(2)
Thus.e' + c2eu.
x2(t) = c. + 4x2 + 5x3
x3 = -17x.2x.
Solution Differentiating both sides of the first equation with respect to t and using the second equation.
x2 = e + e.
(3)
At this point we use the first equation in system (1) and the value of x. The unique solution of the IVP
z.
x3 = 3e .e' + 2c2e2i..
and
B = -.e + c2eu.+2x.
Adding (7) and (8). as the following example illustrates.P = At + B
za = A.(t) = c. or. and substitute the value of x. The method of elimination can also be applied efficiently to nonhomogeneous systems.
x2(0) = 3. we obtain
z. into the first equation of the system.+3x2+2.t..(0) = -1.2 The Method of Elimination
163
of the system for x2 and substitute the value of x2 into the second equation of the system.
Using the first equation in (5) yields
Thus. (6): x.2x. we have succeeded in eliminating the variable x2 from system (5).= -3t+2.=z2+1=-2x. so
x2(t)=c. = -2x.h = c. (6) is x. .
. + 3(X. differentiating both sides of the first equation with respect to t and using the second equation.3.
(5)
Solution As in the previous example.e' + c2e' . The solution of the homogeneous equation associated with Eq.
(9)
Now from the first equation in (5) we have x2 = z..-3z.=x2+t
C2 = .Zt .
EXAMPLE 2
Find the general solution of the nonhomogeneous system
z.
(6)
tion and can be solved. (6) is of the form x. (7)
A particular solution of Eq.1 t-1 .
. + 3x2
(11)
(12)
(13)
x. obtaining the nonhomogeneous differential equation (6).tip =0' -3A + 2At + 2B= -3t+2
and
2A = -3
(8)
-3A + 2B = 2=>A = -. .
EXAMPLE 3
(10)
Solve the IVP
z. we obtain the general solution of Eq.e'+2c2e'-t-22. solve the second equation of the system for x. which contains one single func-
'
X. x. = -3x.
Equations (9) and (10) give the general solution of (5). + 3x2 + 1.t) + 2
z.o = . + 4x2 z2 = -2x.
ce-' .+zc2=3. however.3i. (11) we have
4x2 = i. + 3x2)
_ . For n = 2. we obtain
c.
x. .164
3
Linssr Systems
Solution Differentiating both sides of (11) and using (12) and then (11) again.1.(t) = c. + 3x)
#'
thus.
Thus. = 7.3i. + 4(. two arbitrary constants.
(14)
x. (14) and (15).e' + .e' + 2c2e-'.=0. we can find x2(t) either from the first
equation of the system or from the second.c2e-' + 3c.
x2(t) = c.8x.
(15)
Using the initial conditions (13) in Eqs. c. .e + 3c2e"' = 4c.(t).
REMARK 1
EXAMPLE 4
Find the general solution of the system
x.. = c.x. ' x. + 4i2 = .
Thus.3i.8e-'
x2(t) = 7e . =x2+ 1
x2 = x. + 12X2 = -3i.x. as
expected. + c2 = -1
c.8x. from Eq.
It can be shown that the general solution of a system of n firstorder linear differential equations in n unknown functions contains n arbitrary constants.c2e-'. if we find x2(t) from the second
. We illustrate this by the following example.e' + c2e-'. we find x.e' + c2e-'. The extraneous constants can be eliminated by substituting the solutions into the system and equating coefficients of similar terms. we obtain
. + 3x.4e"'. Now that we have x.2x. However. + 3(i.
Solution
using the second equation. this follows from Theorem 5 of Section 3.
and so. If we use the first equation we find
Differentiating both sides of the first equation with respect to t and
x2(t) = c. . and the general solution of the system contains. the method of elimination introduces extraneous constants.e' . and the solution of the IVP (11)-(13) is x1(t) = 7e . = 0 * X1(t) = c. c2 = -8. Sometimes. = z2 = x.e .1.
represents the number of grams of albumin in the extravascular fluids. 43 (1959): 415. Reeve and J. E.c2e-' + c3. breakdown. and synthesis of albumin in animals has led to systems of ordinary differential equations. which introduces a third arbitrary
constant. We refer to the model above as being associated with unlabeled albumin. y. k and k.4
In this figure. x. Some albumin is transferring to the extravascular fluids (or.1. Gen.(t) = c. Part of this albumin is assumed to be present in the animal's vascular system (plasma) and the remainder in the extravascular fluids (lymph and tissue fluids)
Biokinetics
(Figure 3.
1 k. and we therefore assume that some breakdown occurs in the plasma and some in the extravascular fluids. "Kinetics of I"'-Albumin Distribution and Breakdown.
APPLICATIONS 3.
Vascula
x
Y
k2
Extravascular
k.(t) = c. we find that c3 should be -1. k3. A certain amount of radioactive albumin. to the plasma).c2e-' . k. x represents the number of grams of albumin in the plasma. Thus.4). respectively. and this catabolized protein is replaced by newly synthesized
protein. Thus the general solution is
x. The newly synthesized protein is shown reentering the system via the plasma. are constants associated with the rate at which the various processes take place. Roberts.e + c2e-'. It is not completely known exactly where this breakdown takes place. B. and k. The albumin in the plasma (and in the extravascular fluids) is changing in the following manner. and k.e' + c2e-' and x2(t) = c. and k2 are associated with the rates of interaction. we have to integrate. In order to study the albumin process.2 The Method of Elimination
165
equation of the system. in its natural state the study of albumin can be viewed in terms of the compartment model given as Figure 3.4.1
Some investigations into the distribution. and some albumin is being broken down. referred to as I13'-albumin.
x2(t) = c.e' .
Flgure 3. and k k2. Physiol.e' . is injected
1OThe models and developments presented here have been extracted from E. are associated with the breakdown process.
. In this case." J. some investigators proceed as follows.
Synthesis
k.10 Under normal conditions an animal is assumed to have a constant amount of albumin.2.3. If we substitute these functions into the system and equate coefficients of similar terms.
The first equation is arrived at by noticing in the model that x is changing due to gaining key g/day and losing k. y = z = u = 0. Initially. all the radioactive substance is in the
plasma. The quantities x. we must use derivatives
with respect to time. is continuously being broken down as a result of the liberation of radioactive breakdown products.x-(k2+k. Since there is only one injection.166
3
Linear Systems
into the vascular system. The process described by this compartment model can be analyzed by studying the following system of differential equations: dx
dt=k. being radioactive. no new protein is being synthesized in this model.
k. The solution of this system is left to the exercises.5. After "infinite" time. and certain of the breakdown products are being excreted by the animal.5
. all the
x = y = z = 0. This system also has initial and terminal conditions associated with it. At any time t thereafter. z. u = 1. This solution depends. The other equations are obtained by means of similar reasoning.
z
Breakdown Products
u
I Excretion
Figure 3. determines a unique solution. however. We note. where it mixes with the vascular fluid. The compartment model associated with V`-albumin is
shown in Figure 3. y. there is a certain amount of radioactive substance x in the plasma and a certain amount y in the extravascular fluids.x and kax g/day.ksz di du
dt = kz.y . and u in the figure represent fractions of the total amount originally injected. thus. All these actions are continuous in nature. t = 0 ' x = 1.
Vascular
x
Y
Extravascular
k. and therefore to speak of the rates of change. After a few minutes its behavior is assumed to follow that of the unlabeled albumin.)y
(16)
dz = kax + k. together
substance has been excreted. It"-albumin. t --> 0
with the initial conditions. that (16). thus.
Theoret. Substituting (2) into (1).eu + a12A2e"
A2ke"` = a21A. J. b. we find that A A2. The method of this section. and k must satisfy the system of equations A. Mimicking the method of solving linear homogeneous differential equations with constant coefficients.
Illustrative examples will be given for systems of n equations in n unknown functions for n = 2. we present the main features of this method in the special case of linear homogeneous systems with constant coefficients of the form
x. which was discussed in Sections 2.
.4 and 2. For the sake of simplicity. and f are positive constants. = A.170
3
Linear Systems
31. A. Grossberg" obtained the following IVP:
x=a(b-x)-cfy d(x-y)-cfy-ay
x((1) = b. (1969): 325-64. known as the matrix method. and 4. has the advantage that it can be easily extended to systems with constant coefficients of n linear differential equations in n unknown functions.e'' and x2 = A2e' satisfy system (1). 3. and A are to be determined by demanding that
x.3 THE MATRIX METHOD
As we saw in Section 3. Show that for every fixed f the unknown function y decays monotonically to a positive minimum. Grossberg.
3. we have no difficulty solving systems with constant coefficients of two linear differential equations in two unknown functions. + a22.2.he' = a. c. Learning Theory In an article with applications to learning theory. let us look for a solution of system (1) in the form
Lx2i (2)
LA2euJ
The constants A A2.5.. Biol. d.e` + a22A2eu
"S. The method of elimination is a simple method to use for such systems. =
a12x2
(1)
x2 = a21x. y(O) = a
db
d'
where a.
Substituting these values of A. * k2).
a. this means that there is a solution (A A2) * (0. + a12A2 = 0
(3)
a2. and X2 be the characteristic roots of system (1). 0) gives rise to the trivial solution x1(t) = 0.x
0. in
. and there are three unknown quantities involved: X. we obtain a nontrivial
solution of system (1).X)A.3
The Matrix Method
171
Dividing through by a" and rearranging terms. (4') has two distinct roots (in other words. Naturally. Nevertheless.
(4)
that is.
System (3) consists of only two equations. Thus.2
a22
(5)
and the roots of the characteristic equation are called the characteristic roots or eigenvalues of the matrix (5).3. 0) if and only if
a11 . If Eq. we obtain the equivalent system
(au . if Eq. (4') does not have two distinct roots (in other words.
X2
. 0).(a + a22)X +
a12a2. On the other hand. (4) is called the characteristic equation of the matrix
an
a2. the corresponding system (3) has a nontrivial solution (A.)
Let X. We view system (3) as a linear homogeneous system in the unknowns A. if X. A2). it can be shown that these two
solutions are linearly independent. the above procedure gives. and A2. + (a22 . . (A A2) should be different from (0. since the solution (A A2) = (0. (4) is obtained from the determinant
a au
a2.X)A2 = 0. and the roots of the characteristic equation are called the characteristic roots of the system. Furthermore. one can solve this system by taking the following approach. and the other from K = K2.A.
a22
of the coefficients of system (1) by subtracting X from the main diagonal. [In matrix analysis. In the case of system (3).
It will be useful to the reader to note at this point that the determinant in Eq. if K1 = X2). = 0. Eq. we seek a nontrivial solution of this
system. A and A2. x2(t) = 0 of system (1). If the h in (3) is replaced
by a characteristic root. We now recall (Appendix A) that a homogeneous system of algebraic equations has a nontrivial solution if and only if the determinant of its coefficients is equal to zero.
(4')
DEFINITION I
Equation (4') is called the characteristic equation of system (1).X
aZl
a12
I
a. and A2 in (2). we can obtain two nontrivial solutions of system (1): one from K = K.
r + b. As we are interested in finding the general solution of system (1).=A2.] When A = A2 = 5.t + b2)e"" '
(6)
where the coefficients a b a2. = 3. A. we still need a second linearly independent solution of system (1). one nontrivial solution (and its multiples) of system (1). The trick in this case is to look for a second linearly independent solution of (1) in the form
lx.+A2=0
A2 = 3A1. the constants A. = A2 = 1. and A.)e"" (a. = -3x.=0
-3A.
6-\
1
0. See Remark 4. a2.
that is.+A2=0 -3A.=2x1+x2
X. of the solution (8) should satisfy the homogeneous system (3). that
-A. = 3 and A2 = 5.
A2 . and a22 = 6)
2-\ -3
is. + 6x2.
. system (3) becomes
-3A.8A + 15 = 0.
z2
__
(a. When A = A. and b.+A. are to be determined in such a way that (6) is a solution of (1) which is linearly independent to the first solution
LxzJ IA2eI:I
This is always possible.
EXAMPLE 1
Find the general solution of the homogeneous system
z. = -3.
The characteristic roots are A. a13 = 1.=0
=>A. =
A2 = 2) would just give us another solution with which (9) is linearly dependent.
Choosing A.
Solution Let us look for a solution of system ](7) of the form
(7)
f
f
(8)
LxzJ = LAze"`1
Then A should be a root of the characteristic equation (note that a = 2. any other nontrivial solution of this system (for example.+3A.172
3
linear Systems
general. we have the solution
(9) [e ] [Clearly.
a. + b.t + b. = c.)e = (alt + b2)e'
a2e' + (alt + b2)e' = -(a. = 0.
where c. system (15) does not even have a
nontrivial solution of the form
[x2J
[az1e j
. = (a.
(19)
Now any choice of a b a2.)e and x2 = (a2t + b2)e into (15). = a2 = 1.l = [(a.
a2 + b2 = . it is not enough to multiply (16) by t as in the case of linear homogeneous differential equations with multiple characteristic roots. Substituting x. (Why?) Thus. we look for a second linearly independent solution of (15) of the form (6). and b2 = 1 yields the solution
re'
(t + 1)e
'
(20)
which is linearly independent of (16).+b.b.e + (a. = b2.1te'el
is not a solution of system (15).t + b.=b2.e + c2te'
x2 = c. and b2 that satisfies (19) and gives a solution
which is linearly independent to (16) is acceptable. = a2.3
The Matrix Method
175
Since the roots of the characteristic equation are equal. we obtain
a. + 2a2
or (since the last two equations in (18) are equivalent to the first two]
a.
Notice that for a second linearly independent solution of (15).e + c2(t + 1)e'. b. and c2 are arbitrary constants.=a2
(18)
a2 = _a.
[e e'
te'
+ C2
ll
and
[(t + 1)e'.
Dividing by e' and equating coefficients of like powers of t. the general solution of (15)
is
[x2j = that is.t + b. Indeed.
x. and b2 are to be determined in such a way that (17) is a solution of (15) that is linearly independent of (16).
a. we have
a.)e' + 2(a2t + b2)e. + 2b2. For example.)e`l
x2j
(alt + b2)e'j '
where the constants a b a2.3.
REMARK 3
t
[e'i
. (17) [x . In fact.t + b. the choice a.
we obtain the two
poi
Thus. the matrix
method has the distinct advantage that it can be easily extended to homogeneous
systems of n linear differential equations with constant coefficients in n unknowns.=5x1 +2x2+Zx3
x2 = 2x. That is.4x3
(22)
X3=2x. a21 = 0. + 2x2 . = x1
Xz = x2
(21)
Solution The characteristic equation of system (21) is 1 . a system of the form (1) with a double characteristic root may have two linearly independent solutions of the form (2).
As we explained in the introductory paragraph of this section. This is also true for more general linear homogeneous systems. and A2 are arbitrary constants. = 1. We shall now present two examples with n = 3 and n = 4. the constants A.
Choosing first A. as the example below indicates.
. A2 = 1. am = 1. of the solution (8) satisfy the homogeneous system (3) with a = 1.
0A1+0A2=0
linearly independent solutions
'A. = c. 0
I
10
The characteristic roots are \. and A. unlike the case of linear homogeneous differential equations with multiple characteristic roots. See Example 5 and Remark 4.178
3 LinearSystems
Also. and k = 1. Therefore. A2 = 0 and then A. )\2 .
yl
and
C]
f
. the general solution of system (21) is [ell
x2 J
that is.
EXAMPLE 5
Find the general solution of the system
8.
EXAMPLE 4
Find the general solution of the system
z. but we will not go into the details in this book. a12 = 0. = a2 = 1. that is. -4x2+2r3. respectively.1\ 0 = 0.cl
+ c2 LeJ
x. The method of elimination is highly recommended in the case n = 2. and show how to apply the matrix method to compute their general solution. = 0.2X + 1 = 0.e'
and
x2 = c2e.
(See Section 2.]
Lx2p = u. The main idea in the method of variation of parameters is to seekf a particularf solution of (37) of the form particular[
x. + u2(t)e5' =
f(t)
til(r)e3r + 3i 2(t)e5' = g(t).
Using Cramer's rule.182
3
Linear Systems
geneous system and a particular solution of the nonhomogeneous system.] +
c2
[e]
. That is.
=
3e5'!
[g(t) -f(t)le s`
(41)
From (40) and (41) we find u.
xlp = ul(1)e3r + u2(t)es' x2p = u. Consider the nonhomogeneous linear system
zl = 2rl + x2 + f(t)
(37)
z2 = -3x. The general solution of (37) is given by
.3 we found that the general solution of the homogeneous system corresponding to (37) is
j
Ix2h = Cl [. and u2 are functions to be determined.g(t)]e-3r
(40)
3e5r
u2(t) _
`
lea'
e3'
f(r)I
g(t).(t) and u2(r) by integration. and c2 are arbitrary constants. To find a particular solution of the nonhomogeneous system we can use a variation of parameters technique similar to the corresponding method for linear nonhomogeneous differential equations.(t)e3..(t)e3. + 6x2 + g(t). + 3u2(t)e3'
Substituting into (37) and simplifying we obtain
u.
In Example 1 of Section 3. (t) = g(t)
1e3+
eSil
s'
le'
and
e
= i [3f(t) . and.
(38)
where c. substituting the results into (39). we obtain a particular solution of (37).12.(t)
+ u2(t)
[3e3. we find
f(t)
ti.) We illustrate the technique by means of an example.]
(39)
where u.
. It is called the method of Laplace transform.a 3
l
I
o
)
=1lm(3e16-3)=x.e-"
b-. Then by definition
(t)dt = lim g(t)dt.
0
ob
1
D
o 1
¢e-"dt = Jim b--
e'3'dt = lim . the
improper integral j edt converges but ledt diverges. This method resembles in some ways the use of logarithms for solving exponential equations. By this method an initial value problem is transformed into an algebraic equation or system of equations which we can solve by algebraic methods and a table of Laplace transforms.
but
1
e"Ib
1
me"dt = Jim
0
b--
e"dt = lim
1".
.2 THE LAPLACE TRANSFORM AND ITS PROPERTIES
Assume that the function g is defined for 0 s t < x and is bounded and integrable
in every finite interval 0 s t s b.
4.
b-.1
INTRODUCTION
In this chapter we present another method for solving linear differential equations with constant coefficients and systems of such equations.. -'g f-b We say that the improper integral on the left converges or diverges according
to whether the limit on the right does or does not exist.
3
o
=lbim(-3e '6+3)=3. For example.CHAPTER 4
The Laplace Transform
4. In fact.
G(s).192
4 The Laplace Transform
bolism T[f(t)] = F(s). Thus. the integral (1) converges for s > a. and k denote arbitrary constants.g(t)] = c.
. From Examples I and 2 we see that
-w
ll=1
and
Y'Is12J=eu.
We note that for a given f the corresponding F is uniquely determined (when it exists). we will state Theorem 1. then F(s + k) = `. f is of exponential order a if there exists an a such that
Iimlf(t)Ve' = L. then f is unique.and right-hand limits. and if f is of exponential
order a as t tends to infinity.
DEFINITION 3
A function f is of exponential order a as t tends to infinity if there exist numbers
M.
We now quote without proof a theorem that guarantees the convergence of the integral (1). We also write f(t) = `.F(s) + c.
DEFINITION 2
-
A function f is said to be piecewise continuous on an interval I if I can be
subdivided into a finite number of subintervals.C[e-`'f(t)]. The proof of the first theorem is a simple consequence of the definitions and will therefore be omitted (see Exercise 60).
THEOREM 3
If F(s) = `. Furthermore. if f and g are piecewise continuous functions whose Laplace transforms exist and satisfy e[f(t)] = 2[g(t)].o). Concerning inverse Laplace transforms. then f = g at their points of continuity.
THEOREM 1
If f is piecewise continuous on every finite interval in [0. In what follows the symbols c c2.
THEOREM 2
X[c. and T such that
If(t)I s Me"'
when
t ? T.
The following theorems provide us with useful tools for manipulating Laplace transforms. but
first we give the following definitions. f(t) + c. if F(s) has a continuous inverse f.Li[f(t)]. where L = 0 or L is a finite positive number.
Alternatively. in each of which f is continuous and has finite left. a.P`[F(s)].
k')'
Solution
Using formula (xv) of Table 4.s > k
k.k > 0
(\.s > k k =
sinh kt
k. In much the same way as one uses a table of integrals or formulas for differentiation.ekr] = y[ek"] .
. one can also use a table of Laplace transforms. .k. one must develop a capability for expressing
the functions in the forms found in the tables.k ?
(s .1 is presented in Theorem 2 of Section 4.4. Formula (xxii) of Table 4.. we do not resort to the definition every time we wish to determine a particular Laplace transform.f-'f(0)
P10)
Our approach thus far has been to apply the definition of the Laplace transform to obtain the transform of certain specific functions or classes of functions.k)(s .1.k2
(s .T[ek"
.r)g(-r)d-t
r
G(s)F(s)
F(S)
sl-' F(s)
fo f(t)dt .
ekv .
EXAMPLE 6 Show that
.)(0)
(xxi)
(xxii)
-s"-'f(0) .)(s
2ks
. if one
is to be proficient at using the table. .ekr
(s' + k')'
'S
>0
t sin kt
t cos ki
Ctf(t) + cig(t)
(s2+k '
)
s
>0
(xv) (xvi)
c1F(s) + c'G(s)
F(s + k)
F(ks)
P*)(s)
e "f(t)
(xvii) (xviii)
(xix) (xx)
f(k). are continuous and of exponential order. Laplace transforms. -
s"F(s) .. we have
.t)"f(t)
fog(t .f fu f(u)du dT
P-. That is. Therefore. We take notice of the fact that all the specific functions that appear in Table 4.1.T)f(v)dr = fof(( ..3.2[ek9'].T[ek" .ek"] =
k.2 The Laplace Transform and Its Properties
S
195
s>
k
cosh kt
(xi ) (xii ) (xiii )
(x i v)
s_
k
k'.)
. the reader realizes that not every function will appear in such a table. Naturally. but
rather we prefer to look up the transform in a table.
0
66 Show that C[ek' sin mtl _ (s-k)2+ m2. In order to apply Laplace transforms to differential equations.P[( . show that ds F(s) = `.
43
.
.
Proof
21RIA = I. Use Theorem 5 to compute T[sin kt].e-'f'(t)dt. If F(s) = _T[f(t)].
THE LAPLACE TRANSFORM APPLIED TO DIFFERENTIAL EQUATIONS AND SYSTEMS
At the beginning of this chapter we emphasized that Laplace transforms provide
us with a useful tool for solving certain types of differential equations. or higher) of a function is.
d'
64. Show that T[e" cos mt] =
s-k
m1*
?-s-. Show that e[t cos kt] _
65.Z[sin t]. Use Theorem 5 to compute `. show that dsF(s) = `. Use Theorem 5 to compute T[cos t]. In both theorems we assume that all functions of t that appear satisfy the hypotheses of Theorem I of Section 4.k)2 + 68. 71.
67. If n is a positive integer. The next two theorems provide us with this information.F[-tf(t)].4.3 The Laplace Transform Applied to Differential Equations and Systems
199
62.2. we need to know what the Laplace transform of a derivative (or second derivative. 70.
69.
THEOREM 1
2[f (t)) = sF(s) . The
theorems of Section 4.f(O).t) f (t)]. so that their Laplace transform exists.
63. Use Theorem 5 to compute Z[cos kt]. Show that
m
4[j l f(u) du d r] = s F(s).2 are helpful to us for the purpose of manipulating transforms.
we assume that f and its
derivatives are of. Hence.
In the boundary term. Therefore. 4v = f'(t)dt. it is assumed that lim. we "take the Laplace transform of it" (in other words.
Note In the proof of Theorem 2 just given. we apply the Laplace transformation to each term in the differential equation). Hence. the method of solving differential equations by Laplace transforms
is outlined as follows: Given a differential equation.
THEOREM 2
"'(t)) = s"F(s) . This results in an equation (usually a linear algebraic equation) for the Laplace transform of the unknown function. J (n . Formula (xxii) of Table 4. for k = 1.
f(t)( -se-"dt)
= -f(0) + s 09 e-'f(t)dt J
= sF(s) . we must have an initial value
.s"-'f(0)
-s"-Zf'(0)
f("-a(0)..exponential order as t tends to infinity.e-"f(t)=0. we know that lim_e-"f(t) = 0. That is.sf(0) . we find the inverse of this transform. in the final step._.
ARIA = j
Integrate by parts. 2. .e-"f")(t) = 0.1..200
4 The Laplace Transform
Integrate by parts setting u = e-". e-"f(t) I ..1 indicates that in order to obtain explicit results.. since f is of exponential order. n . f'(0). Thus.
Y(f"(t)] = e-"f'(t)io -
f f (t)( -se-"dt)
= -f'(0) + slme-'f'(t)dt
0
= -f'(0) + s.f(0)) = s2F(s) . one must know f(0). For systems of differential equations the procedure is similar and leads to a system of algebraic equations. setting u = e-" and dv = f"(t)dt.f'(0). Then du = -se-" dt and
v = f(t). Basically. We then solve this equation for this transform. .
Proof We prove the result for n = 2 and leave the generalization to the reader.T[f'(t)] = -f'(0) +s(sF(s) . in the general case it must be assumed that
lim. Similarly. .')(0). du = -se-"dt and
v = f(t). and.f(0)..
y(0) = 0
15.1) + 10h2(t .4y = 20h. y(0) = 1
13.3) y(0) = 1.3
4. y(O) _ -5
12. it is reasonable to assume that during this period the force is a constant and that this constant is large in magnitude. the ordinate would be zero up to the time the hammer struck the nail. y(0) = 4 y(O) 14. the integral of the force with respect to time) is unity.(t . y+6y+9y=6h.
f(t)
I
t
1
2
3
4
5
6
7
8
Figure 4. If the graph of this force were drawn.
. Use Theorem 5 of Section 4.2)
Y(0) = 0. happens not to be a function as the word is used in mathematics.5) -7.y . y + by + 26y = 52h. there are advantages to thinking of the unit impulse function as if it were a function.(t .(t .3. Use Theorem 5 of Section 4. say at t = t0 (to > 0).2 to compute T[f(t)] where f(t) is the "sawtooth" function of Figure 4. it then has some value for a very short period of time and is zero thereafter. one might be interested in describing the force involved when a hammer strikes a nail.(t) + 40h. It is conventional that the impulse of the force (that is.3y . Since the time interval of application of the force is small.2 to compute T[f(t)] where f(t) is the squarewave function of Example 1. y . With this convention we can approximate the force graphically in Figure 4. Y . The idea behind the delta function is to provide an analytical model
for certain types of physical phenomena that act for extremely short periods of time.(t) + 26h. For example.1)+9h. however.210
4 The Laplace Transform
11. or the Dirac delta function.6y = 50 cos t + 30h.4.(t.5 THE UNIT IMPULSE FUNCTION
The unit impulse function.(t-3)
Y(0) = 0.
16. For a first exposition.
11.11.05 farad.
Electric Circuits
In Section 2.
(1)
Equation (1) is referred to as an integrodifferential equation for i. C = 0. the equation
L di + Ri + C
o
i(T)dT = v(t). we can solve Eq.2y . (1).1 we differentiated Eq.214
4 The Laplace Transform
13. In Section 2. (1) is Eq. Here we have used lowercase i and v for current and voltage. y(0) = 4
14. this requires that v(t) be a differentiable function.6
APPUCATIONS
In this section we illustrate a few practical applications of Laplace transforms.1)+38(t-2)
y(O) = 0. Note that Eq. in which we have used the relation (11) of that section. Naturally. to conform to the convention of labelling functions of tin lower case and their Laplace transform in upper case. (1) to obtain a differential equation for i. y(0) = . (1) directly. (12) of Section 2.3
4.
EXAMPLE 1
Solution Applying the Laplace transform to Eq. and v(t) = sin t. they transform a given problem into a simpler problem.11.8y = 85 cost + 68(t .5) y(O) = -9. y+2y+y=S(t)+b(t. We do not restrict the treatment to differential equations exclusively.
. respectively.
Find i(t) for the electric circuit with L = 20 henries.1. Using Laplace transforms. via Kirchhoffs voltage law.s2 + (R/L)s + 1/LC V(s)
= G(s)V(s). R = 40 ohms. y .1 we obtained. we obtain [recall that
i(0) = 0]
LsI(s) + RI(s) + Cs I(s) = V(s)
I(s}=Ls+R
1
+1/CsV(s)
s/L . It is worth emphasizing that Laplace transforms have the nice feature that when they are applicable.
and v(t) into Eq. Then it would be simply a matter of substituting given values for the input and determining the values of the output via this relation. The phenomenon illustrated by the electric circuit can be viewed in the following way. use of tables of Laplace transforms. chemical. Eq.+ asin t. Equation (3) is frequently called the input-output relation. Of course. C. we obtain
!(s) = sz+2s+1 s2+1 = (s+l)2(52+1) _ A B Cs+D
s/20
1
s/20
s+I+(s+ 1)'+ s2+1
Now (see Appendix B) A = 0. a system acts upon an input to produce an output. Biologists observe input and output and try to determine the transfer function from this information (for more discussion in this direction. but unfortunately the relationship is not simple enough to easily obtain output from input. In an abstract way many processes (biological.5
. Engineers and biologists take the attitude that knowledge of the transfer function provides complete information of the system. it has the advantage that given an input v(t) to obtain the output i(t) requires some algebraic manipulation.
1(s)
40 40
(s+1)Z+-i+1
°° `-e
t
i(t)
[(s _+11 )2] +
_ -40te. Following this model.t. physical) can be viewed in this fashion. Thus. and D = . On the other hand. it would be convenient if one could obtain an equation relating output to input. engineers know the transfer function and wish to analyze the output. Equation (1) is such a relation.
Nevertheless. (2) relates
REMARK 1
the Laplace transform of the output to the Laplace transform of the input. R. For the most part. B = -4`-0. A voltage (input) is applied to the circuit (system) and a current (output) is produced. see Exercises 14 through 17). The function G(s) is called the transfer function or system function.4.
System
Input
V(s)
G(s)
Output
t(s) . and in some cases performing integrations.5.G(s)V(s)
Figure 4. (2).6
Applications
215
Substituting the given values for L. Relation (3) is viewed diagrammatically as in Figure 4. C = 0. that is. Eq. (2) is better suited for these purposes.
1967).
with initial values x(O) = y(O) = z(O) = 0 and x(0) = u. Optimality Principles in Biology (New York: Plenum Press. Y. i(0) = w. Planck.216
4 The Laplace Transform
It is in this context [Eq.
. taking into account the rotation of the earth. Hirzel.2w(x sin l3 + i cos
(4)
i = 2wy cos
g.
X(s) = u
I
4w2a2
s2 + 4w2 + s2(s2 + 4W2)) +
4w2ab
4w2ab
v
2wb
s(s2 + 4w2)
(7)
-W
s2(s2 + 4w2) + g s3(s2 + 4w2)
Y(S) = . We choose the origin of a three-dimensional
coordinate system to be at the point from which the particle was projected. The origin is taken to have latitude P and the angular velocity of the earth is denoted by w. the positive y axis east. Set cos P = a. 'See B.'
Mechanics
A particle is projected from a point on the surface of the earth with a given initial velocity. p.M. and apply the Laplace transform to system (4) to obtain the following system (after rearrangement) (see Exercise 1):
s2X .
(6)
Our restrictions on s guarantee that .g. (3)] that the Laplace transform is utilized in presentday investigations.
S
i
System (5) can be solved for X. sin (3 = b. see R. Thus (see Exercises 3. and the positive z axis points opposite to the direction of the acceleration due to gravity. and 6). Neglecting the mass of the particle. Einfithrung in die aligemeine Mechanik. y(O) = v. (Leipzig: S. 4.2wbsY = u
2wbsX + s2Y + 2wasZ = v
(5)
-2wasY + s2Z = w . Rosen.u s(s2
2wb + 4w2) +
1
V s2
+ 4w2
-w
2wa
s(s2 + 4w2)
(8)
2wa + g s2(s2 + 4w2)
'For a more complete discussion.5 # 0. The positive x axis is to point south. and Z since the determinant of coefficients has the value (see Exercise 2)
A = s`(s2 + 4w2). it can be shown2 that the equations of motion are as follows:
X=2wysin(3
. 1928). 4th ed. We wish to analyze the motion of this particle. 81.
and 9).4.
y=vt.w a sine wt
w
+ 4w2 a(2wt .
(13)
Equations (13) would represent the motion of a particle relative to a still earth.) of these individuals are placed into the population at time t.). then at a future time t > t.
Furthermore. if m(t.
z=wt-1gt2. all of a certain age. Then n. we assume that at time t = 0. where
m(t) = nt(t. Guide to the Applications of the Laplace and Z. with the use of a survival function f (t) by the relation
Population Growth
n.) f(t .. there will be m(t) of these individuals left in the population. we consider that individuals of age zero placed in this population at some time t. 1967).sin 2wt) + v a sine wt
+ Zw (2b2wt + a2 sin 2wt) (b2w2t2 + a2 sine wt).
x(t) = u (2a2wt + b2 sin 2wt) + v b sine wt
2w
w
. For convenience. 8.(t) of these specimens still in the population.
'For a more complete discussion of this illustration and other applications of the Laplace transform.Transforms (London: Van Nostrand.ab(2wt .'
In studying the population growth of some species.sin 2(ot) + 2w2 ab(w2t2 y(t)
sin 2 wt)
(10)
w b sine wt + v sin 2wt .
2w2
Note that if one considers the limit as w tends to zero in Eqs. there results
x=ut. we refer to this age classification as zero age. (11).
. At a future time t there are n.t.sin 2wt) z(t)
2tn ab(2wt . say n(0). and (12).6
Applications
4w2ab s2(s2 + 4w2)
Ir
1
217
Z(s)
u
2wa
+ v s(s2 + 4w2)
4w2b2
+ WLs2
+ 4w2 + s2(s2 + 4w2) 1 + 4w2b2 s(s2 + 4w2) s'(s2 + 40)
(9)
Consequently (see Exercises 7.(t) = n(0) f(t).. > 0 also survive according to this rule.(t) and n(0) are connected. (10). there are a certain number of them. That is. see Gustav Doetsch.
(14)
Equation (14) is a model for studying the growth of populations governed by the considerations above. In fact.000e-' +
-(-Jroe'')di
= 34.000.218
4 The Laplaee Transform
Consider now that age-zero individuals are placed in the population at a rate r(t).T.
Suppose that in Example 2 it is desired to determine a rate function r(t) such that the wild rabbit population is a linear function of time. Thus.
n(t) = n(0)f(t) + j
r() f(t -
)dT.e-). r(TI)AT f (t . At time t = 0 there are n(O) zero-age individuals of a certain species given.000e-' + roe-' I e' \
ICJ
o
= 34.
Suppose that in 1976 there were 34. According to the survival law.) of these individuals will still be present in the population at time t. and that wild rabbits are introduced into the population at a constant rate ro. We wish to determine the total
population n(t) at some later time t.)OT.
n(t) = n(0) + ct. that
EXAMPLE 3
is. Naturally.
It is easy to see that if ro = 34.000e-' + roe-` e d r
f
/
= 34. then n(t) = 34. must be such that T s
T + AT.
placed into the population at the rate r(t). How many wild rabbits will be present at time t? What value should ro have if the population is to stay constant?
EXAMPLE 2
Solution
n(t) = 34. individuals are placed in the population.000 for all t.000 wild rabbits in Rhode Island. As time goes on. zero-age individuals are
T. the approximation is made more accurate by making the time interval AT smaller. T. we can say that the approximation tends to n(t) as AT tends to zero. The consideration is as follows.
. that the survival function for these rabbits is a-'. t] into intervals of length AT and adding up the number of survivors for each interval. Certainly one could approximate the answer for n(t) by splitting the time interval [0.000e-' + ro(1 . commonly called the replacement rate. r(T. In the time interval from T to T + AT.
Controlled
Error Signal
Controller
GI (s)
Control Signal
System
Output
G2(s)
r
Ye(s)
Y. (16) in each of the following cases:
(a) w0 = a2
(b) w0 > a2
(c) w0 < a2
Feedback Systems Feedback systems are comprised of two separate systems:
a controlled system and a controlling system. Suppose that v(t) = vp sin wt (vo.(s)G2(s) is called the open-loop transfer function.6
14.(s).(s) . Using the definition of transfer function.(s)G2(s)
Y.(s)G2(s)
Y.(s)
Y2(s)
= G. Determine i(t) from Eq.(s)G2(s)Y. Such functions are performed by varying the input to the controlled
system in amounts that depend on the discrepancy between the controlled system and its desired state at each time t. w # wo).(s)G2(s)'
r
Hint:
a_a bl
c
bc
. A controlled system is one in which
a specific activity is executed so as to maintain a certain fixed activity of the system. Suppose that v(t) = vo (a constant). Determine i(t) from Eq. From the transfer functions
G.6.(s)Y. Show that
YO(s)
=
G. (16) in each
of the following cases:
(c) wo < a2 (b) wo > a2 (a) wo = a2 13.Y0(s) and Y.(s) = Y. The amount of input to the controlled system is administered by another system whose function is to monitor the deviance of the controlled system and to supply the controlled system with inputs in such a way as to minimize the deviance. a number of other transfer functions can be derived which
characterize the behavior of the entire feedback system.(s)
1 + G.220
4 The Laplace Transform
12. we can write
Y0(s) = G. This other system is the controlling system. where Y.(s) = G.(s)
G. w constants.(s)
Feedback Loop
Figure 4. Such a feedback system can be represented diagrammatically as in Figure 4.(s) and G2(s).
The transform of this signal is denoted by Yt (s).'
REMARK 2
REVIEW EXERCISES
In Exercises 1 through 4. Show that
Y"(s) = G (s)
Yf (s)
s
is the open-loop transfer function.(s)G=(s)*
Regulators For a "pure regulator" the input is a constant (thus it is generally chosen to be zero).(s)
G2(s)
Y.
[Hint: Follow an "initial signal" from input to output.4. Show that
Y0(s)
_
G2(s)
Y1(s)
I + G. Efficient techniques for doing this are as yet not known. The reader can see from the expressions developed in these exercises that researchers can attempt to determine the transfer function from the given input and the observed output.7. Optimality Principles. One such system is the eye.. Show that
Y(s)
Y.(s)
Figure 4. and there is an additional signal which represents the effects
of fluctuations in the environment of the system.6
Applications
221
15.
1. Such feedback systems are represented diagrammatically as in Figure 4. 4t5 + 3t sin t
2.
Ye(s)
G.]
17.(s)
-
1
1 + G. (s)
Y. h.
Many biological systems are modeled as feedback systems or as regulators. find the Laplace transform of the given function.
.(t) cos t
'Many illustrations of this sort can be found in Rosen.7
16.(s)G2(s)
is the closed-loop transfer function.
x2)y" . especially in the process of solving some of the classical partial differential equations in mathematical physics. we review some of the properties of power series that will be used in this chapter.xy' + ply = 0 Gauss's hypergeometric equation: x(1 . We concentrate here on series solutions of second-order
linear differential equations with variable coefficients because of the importance
of these equations in applications and because the method of series solutions has been well developed for such equations.x)y" + [c .2xy' + 2py = 0 xy' + (1 . Before we explain how to obtain series solutions for the above as well as other differential equations with variable coefficients.(a + b + 1)x]y' Hermite's equation: y" .x2)y" .
.1
INTRODUCTION
In this chapter we present an effective method for solving many second-order
linear differential equations with variable coefficients by means of infinite series.x)y' + py = 0 Laguerre's equation: Legendre's equation: (1 .
We shall refer to this as the method of series solutions. each of these differential equations involves parameters whose description is associated with the problem that led to their formulation and the "separation constant" involved in the process of solving partial differential equations by separation of variables. All these equations can be solved by the method of series solutions.p2)y = 0 Chebyshev's equation: (1 . Second-order linear differential equations appear frequently in applied mathematics.
Airy's equation: y" .CHAPTER 5
Series Solutions of Second-Order Linear Equations
5. Following are some of the most important second-order linear differential equations with variable coefficients which occur in applications. It is common to refer to these equations by the name that appears to the left of the differential equation. The method of series solutions can also be used to compute formal solutions and to approximate solutions of other linear and nonlinear differential equations with constant or variable coefficients.try' + n(n + 1)y = 0
aby = 0
With the exception of Airy's equation.xy = 0 Bessel's equation: x2y" + xy' + (x2 .
a a2. See Exercises 3.x0 I < R. series (1) converges only at its center x = x0.2 REVIEW OF POWER SERIES
A series of the form
ao + a. . It is within this interval that all operations that we will perform on series. Eqs. for . in our attempt to find series solutions of differential equations.. At the endpoints of the interval
(4)._oa"(x .x0)" +
is called a power series in powers of (x . and the point x0 is called the center of the power series.x0)"
exists. We also say that (1) is a power series about the point x0. In this case the value of the limit is called the sum of the series at the point x. is called the interval of convergence of series (1)..5. the series is said to diverge at the point x.. To determine the behavior at these points. (For series in a form other than (1). such as Er_ox2n `/2".x0)2 +
+ a"(x . Finally. for x = -R + x0 and x = R + x0. if 0 < R < -.(x . we set x = ..
"-o (1)
The numbers a0. or the entire real line if
R = -.x0 I > R. To do this we compute the radius of convergence of the power series. The interval (4). If R =
it
converges for all x. .
(3)
provided that the limit in (2) or (3) exists. a".R + xo < x < R + x0.2
Review of Power Series
225
5. For such series the radius of convergence is determined by using the ratio test of calculus. (4) and diverges for I x ..x0)" converges at a specific point x.)
If R = 0. (2) and (3) are not applicable in general.
.R + x0 in
(1) and check the resulting series by one of the known tests. are legal. that is. This term is denoted by R and is given by the formula
R=
or
1
lm
(2 )
R = lim
4" a".x0) and is denoted by
i a"(x .. the series converges in the interval I x . that is. Given a power series (1). then repeat the procedure for x = R + x0. if
N
lim I a"(x. are called the coefficients of the power
series. and 24. 9.x0)".. If this limit does not exist. We say that a power series E. the series may converge or diverge. . . it is important to find all points x for which the
series converges.x0) + a2(x .
(5)
The function f(x) defined by power series (5) is continuous and has derivatives of all orders.1'
Thus. x . have the same radius of
convergence R as the initial series (5). Furthermore.1)"
(c)
Solution
(a) Here a" = n" and from formula (2). I
Illm 1 .
f (x)
na"(x .xo)
f(x) = j n(n . then for every x in the interval of convergence .xo)"
"-o
for.x0)".. Finally.
and so on. for x < 0 or x > 2. the derivatives f (x).xo )n
n-2
2. For I x .. that is. x .1)a"(x . The series diverges for j x . for x = 0 or x = 2. that is. It is more convenient to use formula (3) in this case.1 ! > 1. (b) Here a" _ (. one can see directly that the series becomes
j (-1)"(± 1)" = j ( -0 -0
1/n!
I
1)".
1
lim V-1 I
). . f"(x). . x .1 j < 1.x0 I < R. series (c) converges for all x.
If R is the radius of convergence of a power series In_oa"(x .5
Series Solutions of Second-Order Linear Equations
EXAMPLE 1
Determine the radius of convergence of each of the following
(b)
"_o
power series:
(a)
"-
n"x"
(-1)"(x . That is. f "(x).xo I < R.
and both of these series diverge.1)" and from formula (2).. these series for f'(x). that is. (c) Here a" = 1/n!. Then
R=lim
1/(n + 1)!1
=limn+1)=x. series (b) converges for all points x in the interval . of the function f(x) can be found by differentiating series (5) term by term. the sum of the series exists and defines a function
f(x) = I a"(x ..
Thus.
1
1
E
R
1
limn
Hence series (a) converges only for x = 0 and diverges for any other x.1 < 1 or 0 < x < 2.1 1 = 1.
. -1 < x .
a"(x .x0)"-' _
and
(n + 1)a". Thus.x0)" + i b"(x . x . The additional restriction for power series is to perform the operation
within the interval of convergence of all series involved. if a series is identically equal to zero. then
a" = b"
for n = 0.x0)" = i (a" + b")(x . na"(x .
(a) "-o
a"(x . subtract.xo)" provided that we increase by k the n under the summation symbol. in addition to taking derivatives of power series. (6) says that we can decrease by k the n in the general term a"(x .
.
In particular. in practice. and vice versa. (b). 1.xa)"+
"-o
(d) If i a"(x .)"
j 3(n + -0
-2
j 3(n + 2)a"-ix".i b"(x .(x .xo I < R..x0)k .x0)".
The reader should note that these changes in appearance of the summations are very much like making a simple substitution in a definite integral. For example.
(6)
which holds for every integer k. In words.o ".x0)" . we have to combine series whose general terms are not of the same
power. These operations are performed in a fashion that resembles very much the corresponding operations with polynomials.xo)"-k.x0)" _
n-k
a.x0)"
e-o
as"(x . 2.o
(b) i a"(x .b")(x . all the coefficients of the series must be zero.
(c) a(x . The basic idea behind the change of index is incorporated in the following identity:
"-0
i a"(x .2
Review of Power Series
227
In the process of finding power-series solutions of differential equations.xo)
".. in the first illustration the index n in the summation on the left is replaced by the
. However. multiply.x0)" =
b"(x . . we have to add. The operations in (a).
and equate two or more power series.x0)"
-0
-0
(a" .-.x(.xo)"
for all x in some interval . The simplest way to prove (6) is to write out the two series term by term.(x .5. For example. and (d) were performed in one step because the
general terms of the series involved were of the same power. In such cases we make an appropriate change in the index of summation of the series which does not change the sum of the series but makes the general terms of the same power.
we define the concept of "analytic function. while the function
x2-5x+7
x(x2 . . the Taylor-series expansion of any polynomial function has only a finite number of nonzero terms. a "rational function..-05 . determine the radius of convergence of the power
series. f")(xo) = n!a"
Hence.228
5 Series Solutions of Soeond-Order Uneer Equations
new index j. at the point xo we obtain f(xo) = ao. and cos x are analytic everywhere.xo)"
"-0
(7)
with a positive radius of convergence. 2..
1. and in general for n = 0. the function 3x2 . A function f is called analytic at a point x0 if it can be written as a power
series
f(x) = j a"(x . as we can see from their Taylor-series expansions. a" = f"1(xo)ln! and the power series in (7) is the Taylor-series expansion
f(x) =
"-o
n x (x . the quotient of two polynomials. 3." that is. a function f is analytic at a point x0 if its Taylor-series expansion (8) about x0 exists and has a positive radius of convergence. 1. Thus. Also. Finally.. f (x).
Let us give some examples of analytic functions. Within its interval of convergence the power series (7) can be differentiated
term by term. For example. The resulting summation is that on the right
with the index being called n again instead of j.
"-o
3"x"
3.. is an analytic function at every point where the denominator
is different from zero. . and so
it converges everywhere. f (xo) = a f'(xo) = 2a2.1)"
"-o
2. Indeed.7x + 6 is analytic
everywhere.
4'
_0 n!
S.
sin x. and -3. Also.
EXERCISES
In Exercises 1 through 6. 7
(2n + 3)
. . .
(
(2n)!
I)" xm
x"
X. Every polynomial is an
analytic function about any point x0. where j + 1 = n. . f'(x). .9)
is analytic about every point except x = 0.xo)"
(8)
of the function f at the point x0. Evaluating f(x).
(x + 1)" _0 n + 1
S'
. since the derivatives of order higher than n of a polynomial of degree n are equal to zero. the functions e'. j 3"(x ." which will be widely used in this chapter.
The points at which the denominator vanishes are singular points of the differential equation.(x)/a2(x) and a0(x)/a2(x) are analytic at every point except where the denominator vanishes.
DEFINITION 2
A point x0 is called a regular singular point of the differential equation (1) if it is a singular point [if at least one of the functions in (2) is not analytic at x0] and the two functions (x .
In Exercises 9 through 15 the student is asked to verify that all singular points of the differential equations of Table 5.(x).
DEFINITION 1
A point x0 is called an ordinary point of the differential equation (1) if the two functions
U(x)
a2(x)
and
ao(x)
a.x2)y" + (2x + 1)y' + x2(x + 1)y = 0. the rational functions a. the form of the solutions will depend very much on the kind of point that x0 is with respect to the differential equation. then x0 is called an irregular singular point of the differential equation (1).1 see Table 5.230
5
Series Solutions of Second-Order Linear Equations
In the subsequent sections we will seek series solutions of the differential equation (1) in powers of (x . then x0 is called a singular point of the differential equation (1). As we will see. If at least one of the functions in (2) is not analytic at the point x0.
(4)
.x0) a2(x)
and
(x .1. a. the coefficients a2(x). where x0 is a real number. regular singular points. If at least one of the functions in (3) is not analytic at the point x0. After cancelling common factors. which gives their ordinary and singular points in the finite real line.
EXAMPLE I Locate the ordinary points.x0)2 Q0(s)
(3)
are analytic at the point x0. and all other real numbers are ordinary points. In connection with the theory of series solutions it is important to classify the singular points of a differential equation into two categories according to the
following definition.x0).(x)
(2)
are analytic at the point x0. A point x0 can be either an ordinary point or a singular point. and ao(x) are polynomials.
In most differential equations of the form (1) that occur in applications. With reference to the differential equations mentioned in Section 5. and irregular singular points of the differential equation
(x` .1 are regular singular points. according to the following definition.
Our aim in the remaining sections of this chapter is to obtain series solutions about ordinary points and near regular singular points. Finally.5.
Since both of these expressions are analytic at x = 1. . hence we conclude that the point x0 = 0 is an irregular singular point for the differential equation (4).x2
x2(x . the two functions in (3) become (x + 1)
2x+1 _ 2x+1
X. for xo = . and -1 is an ordinary point of the differential equation (4)..1. To see which of the singular points 0. 1. = -1 is a regular singular point for
the differential equation (4). For xo = 1.1)
and
(x+1)Zx2(x+1)-(x+1)2
z .1)(x + 1) ' a2(x)
x-1
It follows from (5) that every real number except 0.1
Solution
Here
a2(x) = x° . The study of solutions near irregular singular points is difficult and beyond our scope. 1 None 0
±1
Table 5. the two functions in (3) become
(x-1)
+X =x2+1
and
(x-1)Zxzx
X)=x-1.3 Ordinary Points and Singular Points
231
Differential equation
Airy Bessel Chebyshev Gauss Hermite Laguerre Legendre
Ordinary points
All points All points except xp = 0 All points except x° = 1 All points except x° = 0. and -1 is a regular singular point and which is an irregular singular point for the differential equation (4). we conclude that the point x.
(5)
'
a. For xo = 0. 1. the two functions in (3) become
2x+1 _
2x+1
xx°-x2
x(x-1)(x+1)
and
x2
x2(x+1)-
x2
x'-x2
x-1
The first of these expressions is not analytic at x = 0. we need to examine the two functions in (3). 1 All points All points except xp = 0 All points except x0 = ± 1
Singular points
None
0 ±1
0.x2
1
x' .x2
x-1
and since both of them are analytic at x = -1 (their denominators do not vanish
at x = -1).
ao(x) = x2(x + 1).(x)2x+1
a2(x)
2x+1
ao(x)x2(x+1)=
x' . we conclude that the point xo = 1 is a regular singular point for the differential equation (4).
and so
al(x) = 2z + 1.x2
x2(x .x2.
.
The following theorem describes the form of any solution of the differential equation (1). x0 is an ordinary point of the differential equation (1) when the functions a. if R. of the series (6) can be obtained in terms of ao and a.. and R2 are the radii of convergence of the series (4) and (5).(x) _
a2(x)
A "(x
. The coefficients a" for n = 2.3.5.
(7)
. and R2. and aa(x) are polynomials in x. then as = y. in particular the form of the unique solution of the IVP (1}-(3). Finally. Our task here is to compute (or approximate) this unique solution. a.xo I < R
.
(6)
with positive radius of convergence. Theorem 1 of Section 2. = y. then the general solution of the differential equation has a power-series expansion about x0. which requires that we find the solution of the differential equation (1) that satisfies two given initial conditions of the form
Y(xu) = Yo
and
(2)
Y'(x0) = Yi. (Solutions about an Ordinary Point) If x. and R2.4
Power-Swiss Solutions About an Ordinary Point
233
in some interval about any ordinary point x0.xo)".
(3)
Let us recall that if the coefficients a2(x).
( 4)
and
ao(x) a2(x)
B"(x
for I x
2
(5)
with positive radii of convergence R. where R is the smallest of R. then the radius of convergence of (6) is at least equal to the minimum of R. More precisely. then a point x(.x0 I < R. and therefore.1)y' + 2y = 0
about the ordinary point x0 = 1.x0 I < R. The point xo is usually dictated by
the specific problem at hand.x0) " .x0) "
f or I x
.2(x .(x). by direct substitution of (6) into the differential equation (1) and by equating coefficients of the same power. if (6) is the solution of the IVP (1)-(3).. by the existence theorem. 3. a"(x .
EXAMPLE 1
Find the general solution of the differential equation
y".. The functions (4) and (5) are in particular continuous in the interval I x . and a.(x)la2(x) and ao(x)la2(x) have power-series expansions of the form
a. is an ordinary point of the differential equation (1). and R2.xo I < R . the IVP (1)-(3) has a unique solution throughout the interval I x ..
THEOREM 1
Y(x) = i. is an ordinary point of the differential equation (1) when a2(x0) # 0.. In general..
(x)
and
so(x)
a2(x)
a.2(x -1)y'. we obtain
y' = >na"(x-1)"-'=
and
y" = j n(n . n-2
1)n-2
2(x . Thus.1)". In fact. .1)
and
ao(x) = 2 a2(x)
and so R.
a'-(X)
a2(x)
. For the sake of maintaining easy bookkeeping. will remain unspecified.5 Series Solutions of Second-Order Linear Equations
Solution By Theorem 1 the general solution of Eq. while every other coefficient a a ..1)y' _ -2(x . The coefficients of series (8) will be found by direct substitution of the series into the differential equation. we write the terms y". As we can see from Eq. = R2 _ -.
Since (8) is the general solution of the second-order equation (7). (7). Writing the expressions in columns is very handy in preparing the series in the summation process. we need the radii of convergence R.
Y(x) =
"-o
a"(x .
-0 -0
The sum of the terms on the left-hand sides is zero. y' should be multiplied by . the radius of convergence of the series (8) is also
equal to -. and 2y of the differential equation in a column as follows:
Y' = 2 n(n .. the sum of the three series on the righthand side must be set equal to zero. solution (8) will converge for all x. will be expressed in terms of ao and a. and y" into the differential equation (7). because y is a solution of the differential equation (7).1) and y by 2.1)"-`
2na"(x . it should contain two arbitrary constants.
(8)
with a positive radius of convergence. the coefficients ao and a.1)a"(x "_0
na"(x-1)"
1)"-2 = j n(n . It is easier to add the
.1)a"(. (7) has a power-series
expansion about xa = 1. That is.2(x . .2(x . Hence.1). y'. . Hence..1)"
2y=2F. and ao(x) = 2. Differentiating series (8) term by term.(x) _ -2(x .(x)
Here a.1)
na"(x . To find a lower bound for the radius of convergence of the series (8).a"(x-1)"= i2a"(x-1)". a. and R2 of the power-series expansions of the functions
a.(x) = 1.1)a"(x n-2
We are now ready to substitute y.
..3
6!
=
0..2 to be calculated once a" is known...5.3 ao
2
23. Hence...On
2na"(x ..7. a3....1)"...... (9) and the recurrence formula (10)...... (12) we find
a3=0......2(x .5...-5 a` = -
ao
a = 0...
.. we rewrite the three series above in the
following suitable and equivalent form:
y" = j (n + 2)(n + 1)a".
a3
2 22 a4 =43a2= -43ao= -4iao
2..
(10)
The condition (10) is called a recurrence formula because it allows a"..7
23...
ab
6..2 ..4
Power-Serles Solutions About an Ordinary Point
three series term by term if the general terms are of the same power in each series and that the lower index n under the summation symbol is the same in the three series.5. we obtain
0 = (2a2 + 2ao) + j [(n + 2)(n + 1)a"....8...1)"
2y= 2a"(x-1)"=2ao+ j2a"(x-1)".1)y'
(n+2)(n+1)a".. from (9) we have
a2 = .3
6 5 .1)'
"-o
=2a2+
.. all its coefficients must be zero.2na" + 2a" = 0
for
n = 1. With this in mind.......4.... Using Eq.2...3 ......2na" + 2a"](x .....a0.3
22.4 ...
_
as
2-5 8..3 ao = 8!
ao
.1)
a"
for
(n + 2)(n + 1)
(12)
From Eq.3
ae = ... .. In fact.2(x.2(x .
2a2+2ao=0
and
(9)
(n + 2)(n + 1)an... of the power series in terms of the coefficients ao and a..
(11)
and from (10) we obtain
_
a"'2
2(n ..... -0 n-I
Adding the left-hand sides and the right-hand sides of these three equations. That is..
The right-hand side of this equation is a power series that is identically equal to zero..6. we can express the coefficients a2..
Observe that
a'(x)=
a2 (x)
and
-
1
= ..y' + xy = 0
Y(0) = 1
y'(0) = 1. the general solution of the differential equation (7) is
y(x) = ao + a. ... Therefore..
(13) (14) (15)
Solution
Since the initial conditions are given at the point 0.(x . .(x .1)' EXAMPLE 2
Solve the IVP
(1 .1)2 + a. Hence. we must compute the radii of convergence of the powerseries expansions of the functions a.x)y" .
aL. and so the point xo = 0 is an ordinary point..1) + a2(x .
n-o
1-x
x
IxI <1
a . (7).2.. (22/4!)(x . the IVP (13)-(15) has a unique solution of the form:
y(x) = i a"x".]. The only singular point of the differential equation (13) is x = 1.1)2 are two linearly independent solutions of Eq.
and
a.=0
.236
5
Series Solutions of Second-Ordor Linear Equations
Thus. By direct substitution of (16) into (13) and equating coefficients.
"=o
(16)
If we want at this time to find a lower estimate of the radius of convergence of the power series (16). the general solution involves the two arbitrary
constants a.Ix".(x)
Thus. the functions x .(x . = 0.
n = 1... 3.1)4 + a6(x .
REMARK 1
As we expected.
Hence. and ao.1)6 +
=a.(x)la2(x) and ao(x)/a2(x).1 and 1 ...(x-1)+ao[1-(x-1)2-2
(x-1)`-263(x-1)6-. the reader can verify that
2a2-a. we are interested
in a solution of the IVP (13)-(15) about the point xo = 0. series (16) converges at least for j x I < 1.
n = 2.= -
(2n)!
ao.
.2
z
a". .'.oa"x" of the IVP
EXAMPLE 3
y"-2x'y'+8y=0
y(o) = 0
y'(0) = 1. However.2x"
(21)
. and from (15) we find that
a1= 1.1 =
n = 2. In
REMARK 2
general.convergence of the solution (17) is equal to -.. Of course.
(17)
By Theorem I the radius of convergence of the power-series solution (17) is at least equal to 1.a. the solution of the IVP (13)-(15) is
y(x)= ia"x"=
++1
G--x"=e.1)a"_1x"
(22)
.
Y'(x) = i na"x"-'
na"x"-'
y"(x)=In
"-o
n . However.
1
a2
1
3
1
21
31
n1
Thus..1)a"x"-' _
(n + 2)(n + 1)a". we compute enough coefficients a" of the power-series solution to obtain
a "good approximation" to the solution..I . In fact.5.2x'y'
2na"x` . In Examples 1 and 2 we were able to compute all the coefficients a" of the power-series solution.1)a"x"-' = 2 n(n
"-2
y"
n(n .. this is a luxury that is not always possible. it can be larger. we always have a recurrence formula that we can use to compute as many coefficients of the power-series solution as we please.2(n .3.4
Power-Series Solutions About an Ordinary Point
237
and
n a.
(n + 1)n
From the initial condition (14) we obtain ao = 1.. the radius of. Compute the first five coefficients of the power-series solution y(x) = E.
(25) are very important in many branches of applied mathematics. However.x2)y" . is Legendre's equation.4. we can compute as many
coefficients as time permits.
APPLICATIONS 5.2(n . = 6.2(n 8a" = 0
for
n = 2. a3 = -.
Legendre's Equation
The differential equation
(1 . + 8a2.)x are the contributions of the series
(21) and (23) for n = 0 and n = 1..
"-z
where the terms (2a2 + 8a(.) and (6a.1
The method of power-series solutions about an ordinary point provides an effective tool for obtaining the solutions of some differential equations that occur in applications.. + 8ajx". Finally.3. Legendre's equation appears in the study of the potential equation in spherical coordinates.X + a2X2 + a. = 1.298
5 Series Solutions of Second-Order Unear Equations
8y = i 8ar' -0
0 = (2a2 + 8ao) + (6a3 + 8a)x
(23)
+ j [(n + 2)(n + 1)an+2 .
(25)
where p is a constant. a2 = -4ao.X3 + a4X4 +
=x-4x7+6x4+
In this example. + 8a. The solutions of Eq. there is no known closed-form expression for the general coefficients a" for all n. = 0 and so a. we find 12x4 2a.. Then a2 = 0 and a3 = -. from the recurrence formula (24) for n = 2.a and
(n + 2)(n + 1)an+2 .2xy' + p(p + I)y = 0.
(24)
From the initial conditions we obtain ao = 0 and a. For example. In fact.. using the recurrence formula.l)a"_. Hence. Thus.
y(x) = ao + a. the potential equation
z z
y2
z
axe + a
+ az .0'
..
for example.p(p + 1) =
p(p + 1)(1 + x2 + x° +
1 . When p is a nonnegative integer. we obtain
Legendre's equation (25). The form of any solution of the differential equation (25) about the point xs = 0 is
Y(x) _
n-0
a"x". and
aa(x) = p(p + 1). we find
de2 +cot0d +p(p+1)0=0.$ of the form V = rPO. the point xs = 0 is an ordinary point for the differential equation (25).x2. in quantum mechanics in the study of the hydrogen atom. these polynomial solutions are called Legendre's polynomials.
x <1.4
Power-Series Solutions About an Ordinary Point
transformed to spherical polar coordinates
x = r sin 0 cos 4).
(26)
To find a lower bound for the radius of convergence of the solution (26).
Using the change of variables x = cos 0 and replacing 0 by y. a.x2
"_o
= -2z(1 + x2 + x° +
)
=-2x2nx <1
and
a (x) a2(x)
. They appear.F -so 0 a4)2
If we are interested in a solution that is independent of.(x)la2(x) and ao(x)/a2(x). where 0 is a function of 0 only.x2
_
"-0
p(p + 1)xx2'. one of the solutions of Eq. Legendre's polynomials are widely used in applications. We have
a`(x)
a2(x)
2x
.. Here a2(x) = 1 .
becomes
a2V
y = r sin 0 sin 4). (25) about the ordinary point x0 = 0 is a polynomial. we need to compute the radii of convergence of the Taylor-series expansions about zero of the functions a. When suitably normalized (as we shall explain below).
.5.(x) = -2x.
1 a2v 2aV 1 a2V cot 08V are + r ar + r2 802 + r2 a0 + . Since a2(0) = 1 * 0.
z = r cos 0. = 0. Now we proceed to obtain two linearly independent solutions of Legendre's
equation about the point x.
.. (X) = 1 +
(2n)!
(38)
and
Y2(x) = x +
"-I
(P2 . the point x. Hermite's polynomials are also useful in probability and statistics in obtaining the Gram-Charlier series expansions.. = 0 is an ordinary point of the differential equation (40) and any solution is of the form
Y(x) = i a"x"
-0
(41)
and converges for all x. expansions in terms of Hermite's polynomials. T. T2(x)..4 Power-Series Solutions About an Ordinary Point
243
Hence.(x).
-0
or
"-0
F.. a"x" = F. that is.. if the constant p is a nonnegative integer..-0
a2"+.1)2] x2"+.2) . (36) and (37) that if p is a nonnegative integer.1)(P .(2n .2xy' + 2PY = 0. . .(x). these polynomial solutions are called
Hermite's polynomials. (40) has a polynomial solution about the point
xo = 0.2n + 2)
(2n)!
ao
(42)
a2n+
'
(2n + 1)!
The general solution of the differential equation (40) is
y(X) = F. 2. (2n + 1)!
(39)
It is clear from Eqs.32) . x.(2n . respectively... we obtain a polynomial solution called a Chebyshev polynomial and denoted by T .5.
+ j (. [p2 . . the reader can verify that for n = 1 . For example.. Eq. When suitably normalized.3) . two linearly independent solutions of Chebyshev's equation are p2 (p' .2n + 2) x2.
]
.. and T3(x). is Hermite's equation. (p . and 4x3 .2n + 1)
(2n + 1)!
+.. 2x2 .
(40)
Equation
where p is a constant.
and
2 P(P .2) (2n)!p . By direct substitution of (41) into (40) and equating coefficients.1. As we shall see.I)" 2"(P . one of the solutions is a polynomial of degree n. Clearly.
a. (p2 .22) .12)(P2 .
The differential equation
Hermits's
y" .]
+ a' [.3x are
the Chebyshev polynomials Ta(x).2)2] x2
Y. If we multiply this polynomial by 2"-'. the polynomials 1. (p . The Hermite polynomials are very important in quantum mechanics in the investigation of acceptable solutions of the SchrOdinger equation for a harmonic oscillator.xzn+
y(x) = aol 1 +
(-1)" 2 P(P ..
A deleted interval about xo is a set of the form 0 < I x . Compute a recurrence formula that can be used to
evaluate approximately the current 1(t) in the circuit.xo I < R for some positive number R.
In Exercises 25 through 27.1).1).(1-x2)y'-xy'+4y=0
In Exercises 34 through 36.(1-x2)y"-xy' +y=0
33.
31. 29.x2)y" . The radius of convergence of the power-series solution of Exercise 18 is at least equal to 1.x0 I < R. C = 0. compute the Legendre polynomial corresponding to Legendre's equation. compute the Hermite polynomial corresponding to Hermite's equation.
25.2xy' + 6y = 0 37. and V(t) = 0.
5. (1 -x2)y"-2xy'+2y=0.05
farad. compute the Chebyshev polynomial corresponding to Chebyshev's equation. y" . Verify the recurrence formulas (36) and (37). (1 -x')y"-2xy'+6y=0.y"-2xy'+4y=0
36.5
SERIES SOLUTIONS ABOUT A REGULAR SINGULAR POINT
In this section we show how to solve any second-order linear differential equation with variable coefficients of the form
a2(x)y" + a.
.x2)y" .2xy' + 12y = 0.246
5
Series Solutions of SsoondOrder Linear Equations
23. 30. (1 .
In Exercises 31 through 33. This set consists of the interval I x .11.
27. Verify the recurrence formulas (33). 24.
34. from which we delete its center x0 (see Figure 5. (1 .
26. R = (60 + 20t) ohms. The power-series solution of Exercise 15 converges for all x in the interval
-1<x<1. In an RLC-series circuit (see the electric circuit application in Section 2.(x)y' + ao(x)y = 0
(1)
in a small deleted interval about a regular singular point x0. y"-2xy'+2y=0
35. Verify the recurrence formulas (42) and (43).xy' + 9y = 0
32. assume that L = 20 henries.
28.
However.x0 )2 ao(x) =
a2(x)
"0
B" (x
for I x .x0 < R 2. and X2 be the two roots of the indicial equation
X2+(A0-1)X+B0=0.
3)
with positive radii of convergence R. Before we state a theorem that describes the form of the two linearly independent solutions of the differential equation (1) near a regular singular point.1
Let us recall that when the point x0 is a regular singular point of the differential
equation (1).x0)"
. then the functions
(x . Then the quadratic equation
x2+(A0-1)X+B0=0
is called the indicial equation of (1) at x0.x0) a'(x)
a2(x)
and
(x .x 0 )"
for I x
. its solutions. one of the solutions of Eq. (1) is of the form
Y'(x) = I x . Then. and R2.x0) a 'x) _ a2(x)
A "( x
.
"-0
(5)
. Let X. where R is the smallest of R.
(2)
(x . in general.x0 I < R. Since the point x0 is a singular point of the differential equation (1).x0 I'" i a"(x . ? X2 in case that both roots are real numbers.x0)".x0)2 ao(x)
a2(x)
have power-series expansions of the form
(x
and
( .
THEOREM 1 (Solutions near a Regular Singular Point)
Assume that x0 is a regular singular point of the differential equation (1) and assume that the expansions (2) and (3) hold. are not defined at x0. the differential equation (1) has two linearly independent solutions in the deleted interval 0 < I x . and assume that the expansions (2) and (3) hold. and R2.
(4)
indexed in such a way that k.x0 I < R.5. Our problem in this section is to compute (or approximate) these two solutions near every regular singular point.5
Series Solutions About a Regular Singular Point
247
x0-R
xo
x0+R
Figure 5.
DEFINITION 1
Assume that x0 is a regular singular point of the differential equation (1). we need the following definition.
X. b"(x . then
y2(x) = CY.(x) In I x . and the method of finding such solutions of differential equations is customarily called the method of Frobenius.xo j"2 F. and is valid in the deleted interval 0 < I x .
CASE 2 If X. . (See Exercise 29.
"-o
(8)
with bo = 1.Order Linear Equations
with ao = 1.X2 # integer. a.(x) = x . or (8).248
5
Series Solutions of Socond.x2. A second solution can also be computed by using the method of reduction of order described in Section 2. the point xo = 0 is a singular point of the differential equation (9).
Solution Here a2(x) = 2x2. as the case may be. the coefficients of the series solutions above can be obtained by direct substitution of the solution into the differential equation and equating coefficients. Since
(x-x)a.(x)=xx-x2=l-lx
a2(x)
and
2x2
2
2
(x -
xo)2 a2(x) = x2 2x2 = -
1
. (5) are called Frobenius series. First we compute the solution (5).1. = A2 + (positive integer).)
As in the case of ordinary points.y = 0
(9)
near the point xo = 0.xo I + I x . I + I x . Series of the form of Eq.
-0
(6)
with bo = 1. Since a2(0) _ 0.xo)".xo I < R is found as follows.xo)". A second linearly independent solution y2(x) of Eq.xo J
2
b"(x .(x) In I x . and 3 of Theorem 1. The constant C is sometimes equal to zero.x2)Y' . (7). (1) in the deleted interval 0 < I x .xo 112 i b"(x . where R = min (R R2).
CASE 3
If A. A second solution can be computed from (6).8. and ao(x) _ .xo)".
EXAMPLE 1
Compute the general solution of the differential equation
2x2y" + (x . then
Y2(x) = y. 2. then
Y2(x) = I x .
"=o
(7)
with bo = 1. The following three examples correspond to Cases 1.
CASE I
If X. = X2.xo I < R.
However. ?'2. Since X.2. . the two roots of the indicial equation do not differ by an integer.2. it follows that the power series in (10)
converges for all x.. and therefore the indicial equation of'the differential equation (9) at the regular singular point 0 is X2 + (2 . We have
Y(x) = x
y'(x) _
a"x" _
(n + 1)a"x"
a"x"-
and
y'(x) = i n(n + 1)a"x"-'
2x2y" =
2n(n + 1)a"x"-'
"_o
xy' _
"_o
(n + 1)a"x"-'
-x2y' _
.
"-a
(10)
with ao = 1.na. that is.1)X . It is defined (by Theorem 1) in the deleted interval 0 < I x I < oo. that is. + (n + 1)a .2 = 0. one solution of the differential equation (9) is of the form
y.ao)x +
.1 =0.
2X2-X. We shall now compute the coefficients of the solutions (10) and (11) by direct substitution into the differential equation (9) and by equating coefficients of the same power of x.
By Theorem 1.. Since R.2 and 1.
The roots of the indicial equation are .(n + 1)a"x' 2 =
-na"-ix"-
-y =
"-o
-a"x"-'
(2n(n + 1)a. Here AO = 2 and Bo = .X2 = 2. that is..5 Series Solutions About a Regular Singular Point
249
are analytic functions (with radius of convergence equal to -).5. and so a second linearly independent solution y2(x) is of the form (Case 1)
Y2(x) = I x I
b"x".
X. and we must index them so
that X..
"-o
(11)
with bo = 1. = 1
and
a2 = . for
x < 0 or x > 0. the point
xo = 0 is a regular singular point of the differential equation (9). We first compute the coefficients a of the solution (10). the solution (11) is not defined at x = 0. .a Jx""
0 = (ao .(x) = x j a"x". = R2 = w.
252
5 Series Solutions of Second Order Unser Equations
Thus. we see that for x > 0 or x < 0.
(13')
Combining (13) arid (13'). we look for a solution in the form
Y2(t) = I t
I -"I i b"t". we obtain (the calculations are omitted)
Y2(t) = t-'ae-`a. (9). = 1 we have already found a solution. (9) becomes
2t2y-(-t-t2)y-y=0. To do this. Its roots are again
X2--i
We only have to find the solution corresponding to X2 = -2 because for
X..+
(x/2)" n!
x"
n!+
= x
1/2
or
Y2(x) = x-v2e'
(13)
Now we must compute the solution (11) when x < 0.a
x <0.
"-o
Computing the coefficients by direct substitution of y2(t) into the differential equation (9'). The indicial equation of Eq.. and so
Y2(x) =
(-x)-"ae. we have t > 0 in Eq.
x Y2(x)x a(1+2
= x -la
x"
2"n!
x2
22
21+. (9').
t > 0. b"t".
X.=1
and
(91)
Since x < 0 in Eq. As before. we have Y2(x) = I x I-"evz
(14)
.
Butt = -x. we use the transformation x = -t in the differential equation (9). Eq. we have
Y2(t) = t-ia 7. (9') is the same as for Eq. (9).
"=o
Because t > 0 here. By the chain rule and using dots for derivatives with respect to t. we obtain
dydy dt
Y
dx
dt
dx
-Y
and
Y"d(-y)dt
Thus.
we obtain for x > 0 or x < 0. the general solution ofEq.x)y. (9) is
Ax)=c. Combining (15) and (16). we obtain x = t + 1.(t)=t-(12+t)_
a2(t)
and
.
Solution
Since computations about the point zero are simpler. y' = y.5
Series Solutions About a Regular Singular Point
253
Thus. Here AO = -I and Bo = 1.
y(x) = I x I" j a"x". then
y(x) = (-x)"
a"x"
(16)
is also a solution for x < 0. + y = 0
(17)
near the point xo = 1. Since
t = x . Eq. and y" = y. a"x" "-o
(15)
is a solution of Eq.
REMARK 1
It can be shown that if
y(x) = x"
. This was observed in the previous example in Eqs.=]\2= 1. (13) and (13').(x2 . (17) becomes
t2y-(t2+t)9+y=0. Thus.2 l2
-1-t
a2(t) = t2. the indicial equation is
X2-2X+1 =0
and
1`.
The point to = 0 is a singular point of Eq. Therefore. (1) for x > 0.I and find the general solution of the resulting equation near 0. (18).1. (2n+3)]+c2 X1 1aen
where c. It will be very convenient to use the following remark in the sequel. . and because
(18)
ta.
..
"-o
EXAMPLE 2 Compute the general solution of the differential equation
(x 1)2y. and c2 are arbitrary constants.xI1+5 7. we set
t = x . = 1'
it follows that to = 0 is a regular singular point.5.
1
(28)
Here we assume that the temperature is independent of the height of the cylinder.
Temperature Distribution in a Cylinder
If we know the temperature distribution in a cylinder at time t = 0.
u(r. 0. then it is proved in physics that the temperature u = u(r. The quantity k in Eq.=k u. The important method of separation of variables in solving partial differential equations is also mentioned in the process. we assume that Eq. we obtain
R"OT+rR'OT+ RO"T=kROT'.1
The method of Frobenius is a powerful technique for obtaining solutions of
certain differential equations which occur in applications. (28) has a solution u(r. called Bessel functions. (27) we show briefly how Bessel's equation appears in a specific application.
/
I
7!
`
X.. That is. As we will see below. 0. The solutions of Bessel's equation. 6) at any time t satisfies the following partial differential equation (in polar coordinates):
1
1
u. t) = R(r)0(0)T(t). -.
Bessel's
Equation
The differential equation
x2Y
+
xy'
+
(x: . the materials used in making the cylinder.258
5 Series Solutions of Second Order Linear Equations
Taking bo = 1 and equating coefficients to zero. 0. in general.4. we obtain
Therefore. t) at the point (r. a function of 0. + tX x 2 4
/
+
APPLICATIONS 5.p2)Y = 0. and a function of t. are very important in applied mathematics and especially in mathematical physics. Substituting (29) in (28) and suppressing the arguments r.
(27)
where p is a constant. +
Y:(x)=12(x2+4+40. and t.
(30)
. (28) is a constant that depends on the thermal conductivity and. 0. t) which is a product of a function of r. According to this method. For a discussion of this method see Section 11. the function R will satisfy Bessel's equation (27).
I
In x
I+I. 0.
(29)
where the functions R. We now employ the method of separation of variables to solve the partial differential equation (28). Before we study the solutions of Eq. is Bessel's equation of order p. and Tare to be determined.5.
. while the right side is a function of 8 alone. We now return to find solutions of Eq.
a. both sides of Eq.y'
dx
dr
r
and
R" =
d (Xy) dx dx dr
. we assume for simplicity that p ? 0. The only way that this can happen is if both sides of Eq. Hence.5. (35). We have
a.p2. say . (35). Thus. (31) are equal to a constant.X)
z
=1
and
x2 a°(X) :
p + x2.(x) = x. we can write Eq. (30) in such a way that the variables separate. To find the function R(r) of the
solution (29).
a.
(30) as follows:
rZR" R'r2T'_ R+rR kT
0
(31)
The left side of Eq. In fact.
Since a2(0) = 0. while the right side is a function of t.(x) = x2 . say p2.IT'
kT
33)
Now the left side of Eq. (31) is a function independent of 0. (33) are equal to a constant. If we make the transformation x = lAr and set y(x) = R(r).my = x2 y
r
Substituting R' and R" in Eq. (27) near the point xo = 0. T' + X2kT = 0
and
(34)
r2R" + rR' + (A2r2 . we find Bessel's equation (27). Then we obtain the equations
0" + p20 = 0
and
(32)
R" R'_r2T'_ p: r2R+rR kT
or
R" R +
1R'_E! rR r2
.A2. we must also solve Eq.
(36)
. the point xo = 0 is a singular point. Although p could be a complex number.
(35)
The functions 0(0) and T(t) of the solution (29) are easily found by solving the simple differential equations (32) and (34). (33) is a function of r alone. we obtain
R' = dR dx _ X Y = .(x) = x2.5
Series Solutions About a Regular Singular Point
259
The crucial step in the method of separation of variables is to be able to rewrite
Eq. But
4.p2 )R = 0.
(50) we see that if p is a nonnegative integer k.5
Series Solutions of Second-Order Unear Equations
The solution (42) with ao = 1/2-PI'(-p + 1) is also a Bessel function of the first kind but of order -p and is denoted by J_.. a second linearly independent solution of Bessel's equation is of the form (39) or (40).x. the differential equation has a solution of the form
y. In this case the solution (51) is a polynomial of degree k.(p-n+1)
(n. Since X. when 2p is not an integer. 1 . For example. the reader can verify that
a"P(P-1). = 1.x3 are
the Laguerre polynomials Lo(x).. respectively. Clearly. 2 . 1.
. one of the solutions of the differential
equation (49).0(x) and is called a Laguerre
polynomial.2 = 0. respectively. As we will see below.)2
Thus. The point x0 = 0 is a regular singular point of the differential equation (49) with indicial equation k2 = 0. The Laguerre polynomials are useful in the quantum mechanics of the hydrogen atom. When suitably normalized.k + 1.0n!1'(n-p+1) 2
x_
(-1)"
(X)2--p..
J
°()..(P-n+1)
(n!)2
X. if the constant p is a nonnegative integer. the functions J°(x) and J_P(x) are linearly independent solutions of Bessel's equation. Setting p = 0 in Eq. and 6 . Furthermore.(x)=1+
P(P-1).. these polynomial solutions are called Laguerre polynomials. (27) with an appropriate choice of the constant bo are known as Bessel functions of the second kind.(x) _
"-o
a"x"
for
-<x <
forn=1.x)y' + py = 0. one solution of Laguerre's equation is
Y.. the coefficients a" vanish for n >. This polynomial multiplied by k! is denoted by L. (49) where p is a constant.
(47
)
Thus. is Laguerre's equation.4x + x2.(x). (46) and observing that f(n + 1) = n!.(x).. and L.18x + 9x2 .(x). From Eq.2. and can be obtained by direct substitution of the appropriate series into the differential equation. Such solutions of Eq. near the point x0= 0.
Laguerre's Equation
The differential equation
xy" + (1 . we obtain
(2)2..
(51)
and converges for all x. is a polynomial..
Jo(x) = _ ((n. When p = 0 or when 2p is a positive integer.)l -
(48)
The function J0(x) is a solution of Bessel's equation (27) for p = 0.
(50)
Choosing ao = 1. L2(x). JP(x) is always a solution of Bessel's equation (27). L.
2.aby = 0. 1.
(53)
Hence.5 Swiss Solutions About a Regular Singular Point
The differential equation
Gauss's
x(1 . . . b. 1. (52) has two linearly independent solutions of the form
Y#) =
a"x"
and
y2(x) = I x 11-` F. 2. c..
Hypsrpeometric (52) Equation
where a. Since c is not an integer. For example. Thus. x). and c are constants.
x I < 1.k. b"x". is Gauss's hypergeometric equation or simply the hypergeometric equation.
. Equation (52) is of great theoretical and practical importance because many second-order linear differential equations are reducible to it and because many important special functions are closely related to its solutions.x)y" + [c .(a + b + 1)x]y' . Eq. 3.al =e`
F(-a.5.b. It is important to note that many functions
can be obtained from the hypergeometric functions for various values of the constants a. we obtain the solution
Y1(x) = 1 + ! b_x +
a(a +21c(b(b
1)1)x2 + .b.
c(c+1). b.
The two roots of the indicial equation are 0 and 1 ..
(54)
The series solution (54) is called the hypergeomeiric series and is denoted by F(a.
F(a. the two roots do not differ by an integer..b.c.c.
The reader can verify that
a"
n
= 1. (c+n-1)
n!
(55)
and converges in the interval I x I < 1. x) is a polynomial of degree k fork a nonnegative integer
xF(1. -x) = In (1 + x). b... choosing a° = 1.011)=1
limFla. by Theorem 1. and c. Let us assume that c is not equal to an integer. b. Thus. The indicial equation of (52) near the regular singular point x° = 0 is
X2+(c-1)A=0.. b. -x) = (1 + x)°
F(k + 1.
1 I +
b"(x .\2-2X+ 1 = 0.1)". The differential equation
x2y" + xy' + (x2 .5.
Y2(x) = Y1(x) In I x .is)Y = 0
near the point x0 = 0 involves a logarithm. The differential equation
x(1 -x)y"+(3-3x)y'-y=0
has two linearly independent solutions near xo = 0 of the form
Y1(x) _
a"x". Find
the two linearly independent solutions and in the process verify that C = 0 in formula (8). One of the solutions of the differential equation x2y" + xy' + (x2 .1)y"+(2-x)y'+y0
has two linearly independent solutions near x = 1 of the form
Y1(x) = i a"(x .
.(X2 + x)y' + y = 0
at xo = 0 is
.
29.()3 + x2 + x)y' + (4x + 1)y = 0
at xo = 0 is
K2-2K+ 1 = 0.
In this case. The indicial equation of the differential equation
x2y" . The differential equation
(x. 24.y' +4x'y=0.5
Series Solutions About a Regular Singular Point
265
23.1)". the roots of the indicial equation differ by an integer. The indicial equation of the differential equation
x2y" .9)Y = 0
has two linearly independent solutions near x = I of the form
27.
Y2(x) = Y1(x) In I x I + I b"x".
26.
28. Show that the point x0 = 0 is a regular singular point for the differential equation
xy" .
25.
'
'See also Amer.5. Math.7 are as described by Theorem 1. The above can be used as a motivation for series solutions about a regular singular point.268
6 Series Solutions of SecwW-Order Unear Equations
as found in Section 2. Section 5. Monthly 67(1960): 278-79.
.
as we shall see in the
examples and exercises below. where the coefficients a2(x). a. The conditions (3) which are given at the end
points (or boundary points) of the interval a s x <_ b are called boundary
conditions.3). 1
INTRODUCTION AND SOLUTION OF BOUNDARY VALUE PROBLEMS
In this chapter we present a brief introduction to boundary value problems for linear second-order differential equations. In most IVPs the independent variable x of the differential equation usually represents time. First.(x)Y' + ao(x)Y = f(x). the initial conditions. Our presentation will bring out some essential features of the basic theory by means of simple yet instructive examples.
. (2) The differential equation (1). Consider the second-order linear differential equation
(1) a2(x)Y" + a. The differential equation (1). For example. the striking difference between an IVP and a BVP is that
while the IVP (1) and (2) has exactly one solution (as we know from Theorem 1 of Section 2. together with the boundary conditions (3). we find the general solution of the differential equation and then we use
the boundary conditions to determine the arbitrary constants in the general solution. The procedure for solving a BVP is similar to the procedure used in an IVP.
y(a) = A
and
y(b) = B. together with the initial conditions (2). none. or infinitely many solutions.(x).
(3)
where A and B are two constants. The form of the boundary
conditions at the end points a and b may vary widely. However. The following examples illustrate this point. a BVP may have one. we usually want to find a solution y(x) of the differential equation (1) that satisfies a condition at each end point of the interval a <_ x < b. xo being the initial time and yo and y. There is a large body of theoretical material that has been developed for boundary value problems.CHAPTER 6
Boundary Value Problems
6. ao(x) and the function f(x) are continuous in
some interval a s x s b with a2(x) * 0 in this interval. when the independent variable x represents a space variable (as in the diffusion applications in Section 2.1). constitutes a boundary value problem (BVP). Until now we were
mainly concerned with finding a solution y(x) of the differential equation (1) which at some point x = xo in the interval a s x s b satisfied two given initial conditions
and Y(xo) = Yo Y'(xo) = y1. constitutes an initial value problem (IVP). However.
They can contain combinations of y and its derivatives at the points a and b. the BVP (13)-(15) has the unique solution
y(x) = 12 sin 2x.2' c.c. c.
y(O) = 2-:> c. We have y'(x) _ -2c.1
Introduction and Solution of Boundary Value Problems
271
EXAMPLE 3
Solve the BVP
y"+y=x
y(0) = 2. + 7r = IT . + 6c2 = 3. no solution. cos x + c2 sin x + x.
for
0:5 xsir
(10)
y(ir) = a . the BVP (10)-(11) has
infinitely many solutions:
y(x) = 2 cos x + c2 sin x + x. In this case.2y'(0) = -2. cos 2x + c2 sin 2x.2y'(0) = -2
and
c. or infinitely many solutions.. We give a simple example. = 2 while c2 remains arbitrary.
(11)
Solution The general solution of the differential equation (10) is
y(x) = c.
Now. Solving this system we obtain c.6.
Solution The general solution of the differential equation (13) is y(x) = c.2. sin 2x + 2c2 cos 2x. it is very important to know under what conditions a BVP has a unique solution.
Y(IT) t. = 2
and
y(7r) = IT . and using the boundary conditions.
(12)
The boundary conditions at the end points a and b need not always be of the type considered in the previous examples.
Thus.
Of course.
EXAMPLE 4
Solve the BVP
y"+4y=0
for
0sxsrr
(13)
(14) (15)
y(0) . . = 2. Therefore.2' . c. = 0 and c2 = . we find that
Y(0) .4c2 = -2
Y(7r) + 3Y'(n) = 3 #.3y'(7r) = 3. in
. This problem.
Let y.y. is very difficult and beyond our scope. Then the general solution of the differential equation (1) is given by
Y(x) = c1Y1(x) + c2Y2(x) + Y. and c2 of the general solution should satisfy the following linear system of algebraic equations:
c1Y1(a) + c2y2(a) + Yp(a) = A
c.y.(x) be a particular solution of the differential equation (1). if the determinant in (19) is equal to zero. we have the following. We recall (Appendix A) that a linear system of the form
a21x + a.(x)
(16)
In view of the boundary conditions (3). the following statements are valid:
(a) If
Y1(a)
Y2(a)
Y1(b)
y2(b)
. the constants c.(b) + c2y2(b) + y. and let y.(b)
Now it is clear that the BVP (1) and (3) has as many solutions as the system (17).
THEOREM 1
Let y.272
6 Boundary Value Problems
general. We only develop here a simple but interesting result about the BVP consisting of the differential equation (1) and the simple boundary conditions (3).(a)
(17)
c1Y1(b) + c2Y2(b) = B .(x) and y.2 y = b2 (18)
has exactly one solution if the determinant of the coefficients satisfies all a12
a21
a22
# 0.(x) and y2(x) be two linearly independent solutions of the homogeneous differential equation corresponding to the differential equation (1) and let y. Summarizing the facts above. Then.(x) be two linearly independent solutions of the homogeneous differential equation corresponding to the differential equation (1).
(19)
On the other hand.y.(b) = B1
or
c1Y1(a) + czy2(a) = A . the system (18) has no solution when
a b1
a21
+0
b2
(20)
and has infinitely many solutions when the determinant in (20) is equal to zero.(x) be a particular solution of the differential equation (1).
y.6.(b) y2(b)
1
0
=0
-1
0
cos a sin rr
and
y.y.(x) = x.(a)
y2(a)
0
y.(a) A .(a) y2(b)I
y.rr
-2
APPLICATIONS 6. and y. because Icos0 sin0 ly.(b)
2 -0
rr .(x) = cos x.y.1
Introduction and Solution of Boundary Value Problems
273
the BVP (1) and (3) has one and only one solution in the interval a < x <.(b) y2(b)
cos 0
sin 0
sin -w 2
1
1
0
1
cos
IT
=1#0.(b) B .y.1
Suppose that a gas diffuses into a liquid in a long. the BVP (1) and (3) has no
solution or infinitely many solutions in the interval a s x s b.1.
(22)
Let us apply Theorem 1 to the BVPs in Examples 1. The BVP (4)-(5) has a unique
solution because y.(b)
cos 0 2-0 cos.rr 1 . narrow pipe (Figure 6.
Finally. Suppose that this process takes place for such a long period of time that the
concentration y(x) of gas in the pipe depends only on the distance x from some
S
Diffusion
Figure 6. and 3.(b)
B .1r
1
#0. y2(x) = sin x. the BVP (10)-(11) has infinitely many solutions.
-1
cos.(a) A .(a) y.y.a
1
2
= 0. 2.1
. where y.1).yo(b)
is * 0 or equal to zero.(a) y2(a)
y. (b) If the determinant in (21) is equal to zero.
0
2
The BVP (8)-(9) has no solution.b.r
cos 0 sin 0
1
2
-1 1 .(b)
y2(b)
cosy sin a
cos 0
-1 0
=0
and
y. respectively.(b) B .(a) y.(a)
y.2 . depending on
whether the determinant
yl(a) A . because
y.
Air) = -1.
(26)
EXAMPLE 5
Solve the BVP (25)-(26) when f(x) = fo is constant.1.274
6 Boundary Value Problems
initial point 0 (and is independent of time). as we proved in Section 2. between two fixed points x = 0 and x = L on the x axis.
(25)
Since the ends of the string are kept fixed at the points x = 0 and x = L of the x axis.
But
and
u(0)=0'c. + c2x .
EXERCISES
In Exercises 1 through 6. A taut string.
Hence. y(x) satisfies the BVP
yY(0) = A.=0
u(L)=0=> c.
Deflection of a Taut String
y=0
y(1) = 0. A transverse (perpendicular to the x axis) load f(x) is applied to the string and produces a deflection u(x). Then it can be shown that the deflection u(x) satisfies the differential equation
.
u(L) = 0.+c2L-2T L2=0. is stretched.ZTx2. Then. y(x) = cos 3x is a solution of the BVP
y"+9y=0
Y(0) = 1. c2 = f0L/2T.
1. and the BVP (25)-(26) has the unique solution
u(x) = 2T x .
. u(x) must also satisfy the boundary conditions
u(0) = 0. c.
Solution The general solution of the differential equation (25) is
u(x) = c.sin 3x is a solution of the BVP
y"+9y=0
Y(0) = 1.Tu" = f(x).
(23) (24)
in Exercise 16 the reader is asked to solve the BVP (23)-(24). 2. lying on the xy plane. answer true or false.1. Y('n) _ I.27x2 = 2Tx(L .x). = 0. under a tension T. y(x) = cos 3x .
respectively. The BVP (1)-(2) is also called an eigenvalue problem (EVP). These values of x are called the
eigenvalues of the BVP (1)-(2).
6. However.
2 EIGENVALUES AND EIGENFUNCTIONS
From the point of view of applications. Solve the BVP
1 :5x!52 xzy" .
(1)
where p(x) (> 0) and q(x) are continuous in some interval a 5 x 5 b.
15x52
y(1) = A. Solve the BVP (25)-(26) when
f (x) = x.276
6 Boundary Value Problems
14. y(2) . Show that the BVP
x2y"-3xy'+3y=1nx.
17.Y'(1) = 2. Solve the BVP consisting of the differential equation (23) and the boundary conditions (24). the important issue in this type of BVP is to determine values of x for which
the BVP (1)-(2) has a nontrivial solution. A (one-dimensional and steady-state) diffusion process in a medium bounded by the planes x = 0 and x = I which are maintained at constant
concentrations c.
18. Solve this BVP and show that the concentration y(x) changes linearly from c.
L = 1.Y(b) + R2Y'(b) = 0.2y'(2) = 4y(1) .
16. there is a very important class of BVPs in which the differential equation is of the form
(p(x)y')' + (q(x) + X)y = 0. a is a real parameter.
. and the two boundary conditions at the end points a and b are of the form
o1Y(a) + a2Y'(a) = 0
P. y(2) = B has a unique solution for all values of A and B. especially problems in mathematical
physics.
15.
(2)
The BVP (1)-(2) is always satisfied by the trivial solution y(x) = 0. The corresponding nontrivial solutions are called eigenfunctions. and c2.3xy' + 3y = 0. is governed by the BVP
y"(x) = 0
Y(0) = c Y(I) = c2 (provided that the diffusion coefficient D is constant).
T = 2. to c2 through the medium.
+ c2 = 0
and
(6)
y(rr) = 0' c. the solution (8) is identically zero. and its general solution is
CASE 1
y(x) = c. = c2 =
0.2
Elg nvalues and Eigenfunctions
277
In this section we shall compute the eigenvalues and eigenfunctions of some simple but representative EVPs and state some of their properties.
(7)
Solving the system of simultaneous equations for c. from Eq.6. (6) we obtain c2 = -c and inserting this value into Eq.e*') = 0.k2. or positive. Then
c2 = 0 and the solution (5) is identically zero. and c2.e-"x + c2e". We have
y(0) = 0 ' c.
EXAMPLE 1
Compute the eigenvalues and eigenfunctions of the EVP
y" + Ky = 0
Y(0) = 0.(e-" . which implies that a = 0 is not an
.
Solution Our problem here is to find all the values of the real parameter a for which the EVP (3)-(4) has a nontrivial solution and to find the corresponding nontrivial solutions. + c2x. Hence.
Using the boundary conditions (4). This implies that the EVP (3)-(4) does not have negative eigenvalues.
K < 0 Set k = . The characteristic roots of the differential equation (3) are
\. and
its general solution is
y(x) = c. we see that
(8)
y(0) = 0
and
c. = 0. and so c.key = 0. = 0
' c2rr = 0
c2 = 0. the form of the solutions of the differential equation (3)
depends on whether X is negative. we find that c. we find c. a-"° # e". + c2rr = 0
Thus. and c2. Then the differential equation (3) becomes y" .
y(rr) = 0
eigenvalue. zero. (7).
c. where k > 0.e-*' + c2e*' = 0. Since k > 0.rr
(3)
(4)
y(rr) = 0. we use the boundary conditions (4) to determine the constants c.
CASE 2 \= 0 In this case the differential equation (3) becomes y" = 0.
for
0 s x <. So in solving the eigenvalue problem (3)-(4) we have to examine three cases. In fact.
(5)
Next. Remember that we want to find solutions that are not identically equal to zero.
. Now k = k2. Indeed.. = 0 and c2 sin krr = 0.278
6 Boundary Value Problems
X > 0 Set X = k2. and k = n. and therefore the eigenvalues are
given by
X = n2
for
n = 1. . up to a nonzero constant multiple (that is. Recall once more that we try to compute
nontrivial solutions of the differential equation (3)...
(14)
(15)
Solution
The characteristic roots of the differential equation (14) are . zero. and its general solution is
CASE 3
y(x) = c. . 2.
for 0sx<1
y'(1) = 0.
(10)
Equation (10) gives all the eigenvalues of the EVP (3) -(4).k is negative. Thus. In this case... .. 2. the eigenvalues and eigenfunctions of the EVP (3)-(4) are given by
n = 1.
. Then the differential equation (3) becomes y" + k2y = 0. To find the eigenfunctions corresponding to these eigenvalues. we use (9) with c. c. ..
We now employ the boundary conditions
(9)
y(o)=0=> c. Thus.. cos kir + c2 sin kzr = 0 ' c2 sin kIr = 0. must be zero.. = 0. Then y = c2 sin nx. k = n for n = 1. c2)..
(13)
where the subscript n is attached to indicate the correspondence between
EXAMPLE 2
Determine the eigenvalues and eigenfunctions of the EVP
y"+(-4+ \)y=0
y'(0) = 0. c. . 2.
(12)
y" = sin nx
eigenvalues and eigenfunctions. hence..=0
and
y(7r) = 0 c.
for n= 1. or positive. Hence. c2 sin k7r = 0 if sin kzr = 0 or ka = nrr for n = 1. 2... 2. the eigenfunction of the EVP (3)-(4) that corresponds to the eigenvalue X = n2 is given by
y = sin nx
k" = n2
and
for for
n = 1.. that is. .
(11)
Hence.. the form of the solutions of the differential equation (14) depends on whether the quantity 4 .. 2. where k > 0. cos kx + c2 sin kx. c2 * 0. a nontrivial solution is possible only if c2 can be chosen different from
zero.
To this end we first define the concepts of inner product and orthogonal functions.
(22)
where series (22) converges uniformly in the interval a <_ x s b. The inner product of the functions f and g is denoted by (f..
When (f.. Let us verify (21). . -
+00
(20)
and a corresponding sequence of eigenfunctions
Y Y2.
In Example 1 it is clear that property (20) is satisfied. (16) and (17) for n = 0 give us o = 4 and yo = 1.R
x s b.. we conclude that the eigenvalues and eigenfunctions of the EVP (14)-(15) are given by the equations a = 4 + n2..
(18)
y = cos nax
n =0...
(19)
Finally.1...
. g) and is defined by
(f.
..2. we state a theorem that summarizes some important properties of the eigenvalues and eigenfunctions of the EVP (1)-(2).. In the EVP (3)-(4) we found that [see Eq.280
8 Boundary Value Problems
If we observe that Eqs. 2. 1.
THEOREM 1
The EVP (1)-(2) possesses an infinite sequence of eigenvalues
a a2.
.
. we say that the functions f and g are orthogonal.
X...
DEFINITION 1
Let f and g be two continuous functions defined in the interval a s x <_ b.
(21)
Furthermore. the eigenvalue and eigenfunction found in Case 2. any function f(x) that is twice continuously differentiable and satisfies the boundary conditions (2) has the following eigenfunction expansion:
f(x) _
(f. g) = J f(x)g(x)dx.
if nxm
if
(Y >Ym) =
1. with the property that in the interval a
10. ..n2
and
for for
n = 0.
n = m.
.Yoy.
that is. g) = 0. (13)]
y = sin nx
for n= 1. 2...
(13)
.
ym) =
1o
'r . Churchill.
(wave equation). For example.. then
f( x) _
2
(f(x)..
Equation (24) appears in the study of heat conduction and many other diffusion processes (for a derivation of this equation and more discussion of its solution and some applications of it. (New York: McGrawHill. Assume that the bar is oriented so that the x axis coincides with the axis of the bar. consider an insulated uniform bar of length 1. 2nd ed. sin nx) sin nx =
2
(ff(x) sin nx dx sin nx. for Theorem 1 we have to replace the eigenfunction (13) of the EVP (3)-(4) by
/2112 ` J
y"=(2)112
sinnx
for n= 1..)
With respect to these eigenfunctions. In particular. V.
(13.+ uYY = 0
(Laplace's equation).
(26)
Let us demonstrate this in the case of Eq. that is. we have to multiply each eigenfunction in (13) by (2/7r)".1
Eigenvalue problems occur frequently in problems of mathematical physics.
(y.c=u = 0
and
u. 1963).
sin nx sin mx dx =
if if
nm
n = m. y. by the method of separation of variables. t) denote the temperature in
'See R.2. Fourier Series and Boundary Value Problems. = «:u_
(heat equation).2 Eigenvalues and Eigenfunctions
281
Computing the inner product (y.6. f(O) _
f(irr) = 0.
(23)
and this series converges uniformly in the interval 0 <_ x s rr. they appear in the process of solving. which is acceptable since the eigenfunctions are defined up to a nonzero constant multiple. see Section 11.") we obtain that
r
0.7). (24).
(24)
(25)
U . Let u(x.. The series (23)
is known as the Fourier sine series' of the function f(x) in the interval 0 s
xsa. initial-boundary value problems which involve one of the following partial differential equations:
u..
APPLICATION 6. 2
In order that Eq.. f" exists and
is continuous) and satisfies the boundary conditions (4). property (22) indicates that if f(x) is a function that is twice continuously differentiable (in other words. (21) be satisfied. Thus.2.
The Heat Equation
.
we need to know some initial and boundary conditions.5. That is. that is. The
constant a2 is called the thermal diffusivity and depends only on the material of which the bar is made. Then u satisfies Eq. that is. t) = X(x)T(t).
(27)
and that the ends of the bar are kept at zero temperature. (30) is a function of x alone.
Using the boundary conditions (28).
(28)
Using the method of separation of variables (see Section 5. and (28) of the form
u(x.
We have
(29)
u.t) = X(0)T(t)
which imply that
and
0 = u(1. (30) are equal to some constant X. t) = X(1)T(t).
.4).
and Eq.
0 <. we look for a solution u(x. and its right-hand side is a function oft alone. 0) = f(x). (24) becomes
XT' = a2X"T
or
X" X
1 T' a2 T
(30)
The left-hand side of Eq. The only way that this can happen is if both sides of Eq. we find
(32)
0 = u(0. = XT' and u = X"T. t) of the initial-boundary value problem (24).
u(0. (27).
(33)
X(0) = X(l) = 0. (24) for 0 < x < 1.
X"
X -X
and
1 T'
a2T
or
X"-AX=O
and
(31)
' . For example.282
6 Boundary Value Problems
the bar at the point x at time t.
u(x.a2AT = 0. assume that the initial temperature f(x) in the bar is a known function of x.x s 1.1 or Section 11. t) = u(1. t) in the bar. To determine the flow of heat u(x. t) = 0.
(36)
In view of the equations (29).
(35)
For the sake of completeness. (34).y=0for0:s xs2
y(0) = 0. .. each of the functions
u. y"+\y=0for 0sxs1
Y'(0) = 0.
(39)
which means that the constants c must be the coefficients of the Fourier sine series of the function f(x) in the interval 0 s x s 1... and (36).rtz
a=
and
/'
n=1. . is given by
T (t) = exp(-a2n'ar2l-2t). (32).. Y(2) = 0
4. the function X(x) of the solution (29) satisfies the EVP consisting of the differential equation (31) and the boundary conditions (33). 2.. it can be shown that any linear combination
t)
(38)
of the functions in (37) is a solution of (24) and (28) provided that the series (38) converges and can be differentiated term by term a sufficient number of times with respect to t and x. in such a way that the series (38) satisfies the initial condition (27). . That is..6. y(2) = 0
2. and (28). Furthermore. . n7rx .
n = 1..
1
. y" + Jay = 0for0 <_x s ar
Y(0) = Y'(0). c. (35).
EXERCISES
In Exercises I through 8. We leave it to the student to show that the eigenvalues and eigenfunctions of the EVP (31).2
Eigenvalues and Eigenfunctions
283
Thus.7. Y('rr) = Y'('T)
. Y(I) = 0
3. up to a constant multiple. t) = exp(-a2n'Tr2l-2t)sinnx
(37)
is a solution of the heat equation (24) and satisfies the boundary conditions (28). Y" + Xy = 0for -2sx<_2
Y(-2) = 0. . . y"+1.(x.
1.. n = 1 .
n = 1..2. compute the eigenvalues and eigenfunctions of the
EVP. 2. All that we have to do now is to choose the constants c. Since A has one of the values given by Eq. (33) are given by
nz. The procedure just outlined is carried out in detail in Section 11..
(34)
XX(x) = sinnIx. A -1 0<x<-1.. (27)..
n = 1. the solution of Eq. 2.. 2. we now indicate what remains to be done to find the solution u(x. . t) to the initial boundary value problem (24).
. is not what is meant by numerical solutions of differential equations. sin x + c2 cos x as a formula for its solutions.(3n-2)x3"
(3n)!
+c2x+
2. More specifically. although it does point out one aspect of the problem.y = 0 has y = c. For instance. Of course. the formula for y may itself impose computational difficulties. such computational difficulties can be easily overcome if the accuracy desired is within the capability of the device used.1
INTRODUCTION
Up to now our treatment of differential equations has centered on the determination of solutions of the differential equation.. sin x.xy = 0 above. the accuracy desired in the numerical value of y may cause some computational problems. Furthermore. It is natural to adopt the attitude that these
formulas supply all the information about the solution that we need.. In this same context there are many differential equations for which it is impossible
. this
terminology applies to methods associated with the determination of approximate numerical values of the solution when it is generally impossible to obtain a formula
for the solution.4 7. or cos x. matters are not always that simple. For example.CHAPTER 7
Numerical Solutions of Differential Equations
7. and the differential equation y" .e' + c2e-x as a formula for its solutions.8 .1
1
as a formula for its solutions.. Unfortunately. with the present-day proliferation of pocket calculators and the widespread availability of digital computers.xy = 0 has
y=c. This is the
case in the description of the solution of y" . Rather.
Our description of determining the numerical value of the solution of a differential equation. we simply substitute the given value of x into the appropriate formula and compute the corresponding value of y. The reader perhaps recalls from calculus that many functions do not have antiderivatives. Even in those cases where the solution is represented by such relatively simple functions as e`. 1+
L
1.(3n-1)
(3n+1)!
x3". the differential equation y" . the differential
equation y" + y = 0 has y = c. our methods have provided us with formulas for the solution.5. and hence it is impossible to represent the indefinite integral of these functions by a formula in terms of elementary functions. if we want the value of y corresponding to a specific value of x in the cases above.
(1) (2)
To obtain a feeling for the goal of numerical methods. Suppose that we
want to know the coordinates of the point (x y). let us imagine that the curve in Figure 7. The Analysis of Numerical Methods (New York: John Wiley & Sons.
. is it still possible to determine y. Any numerical method involves approximations of one sort or another. any reasonable attempt
at answering such questions here would take us far beyond our goal of presenting a brief introduction to numerical methods. we touch upon only a few of the elementary methods.? The answer to this question is the basic goal of any numerical method. 1966). y). Consequently. B. knowing x.
y
(x0.7. We note that all differential equations can be written in this form or as a system of differential equations of this form. questions arise concerning the accuracy of the results. Keller. Since our treatment is admittedly introductory and brief.1 represents the solution to the IVP (1)-(2). In the case that y is given
explicitly by a formula involving x.1
If y is not known in terms of x and the solution curve of Figure 7.Yo)
17
Figure 7. Isaatson and H. the interested reader is referred to E.1 exists only in theory (from the existence-uniqueness theorem). we assume that the differential equation in question is of the form y' = f(x. There are at present a large number of numerical methods for solving differential equations. how much error is involved in the approximation? Although we present a few
definitions and results concerning the error involved.y)
Y(xo) = Yo. In such cases we "solve" the differential equation by applying certain numerical methods. we concentrate on initial value problems. Thus we investigate the IVP
y' = f(x. it is simply a matter of knowing x.1
Introduction
287
to obtain a formula for the solution. to avoid cumbersome treatment of integration constants.
'For some answers to questions about the accuracy of the methods. Furthermore. from one point of view there is no loss of generality in making this assumption. Hence. In other words.'
In our subsequent discussion.
. where Ox = x.. 3. We can then approximate y.2). Now dy = (dy/dx)dx = f(x.y. 2. The Euler method can be summarized by the following formulas of the coordinates of the points (x.):
X1 = x. + repeated to obtain additional points. y. one considers x. Naturally. Thus. = xo + Ox and y.x0). + h.
i =
(1)
(2)
s. y2 = y. with x = b and y.
to approximate y2 for some given x2 (> x.288
7
Numerical Solutions of Differential Equations
We will say that we have solved a differential equation numerically on the interval [xo. Given the IVP (1)-(2) of Section 7.)Ox = y. is given by the Euler method to be yo + f(xo..
These changes are denoted by Ax and Ay. . The Euler method is to approximate Ay by the differ-
ential dy.). In this One can then use x.yo)Ax = yo +
f (xo.. by approximating Ay.x0 and Ay = y.1 and a specific value of x. The
spacing is normally taken to be uniform (in which case the spacing is often called
the "mesh width" or simply "mesh") and is denoted by h. we can consider that in transferring from the initial
point (xo. we can write x. Equivalently. . then h is given by the formula when plotted. +
y
1.2 EULER METHOD
Referring to Figure 7. n i = 1. n.. say
x then y. one can calculate y. directly from the information given.). . and hence
Ax to be known exactly..xo)/n..
7. 3.
x
Rgure 7. = yo + Ay. if this
approximation is to be reasonably accurate. + f(x.yo.f(x. This process can then be case. then Ox should be quite small (see Figure 7. . 2.b] when we have obtained a set of points {(xo._.y)dx.. If it is desired to obtain n points in addition to the initial point.
yj = y.y)dx .2
. . Of course. The sequence of
close to the solution curve. an approximation of the solution when x has the (x. yo)(x..
value x. .2.yo). hopefully lie very (b . Usually. this process is carried out in a specified interval with a prescribed spacing between the x coordinates. y0) to the new point (x y) we have induced changes in the coordinates.
The results are displayed in Table 7. and y.3
This solution The "solution" to the IVP is then the sequence of points is usually displayed in tabular form.7. is that the smaller h is. of course. the smaller the value of h will be and so the more accurate our results will be.2
Euler Method
Yo
Yt
Yx
Ys
h
Rgure 7.0).0.
. We choose n = 5. (0. (1. the choice is at our disposal. h = (1 . Here and in subsequent applications.0.. (There are problems of accuracy associated with round-
off also.2.
Solution We were not told how many points to obtain.008). (See Figure 7.0). (0.0)/5 = 0.2.3.6.0. but again we skirt the issue since it would take us beyond our
treatment.1. the more calculations we require.4. The price to be paid. (0. therefore. the collection of points ((0.040013).242857)} is our numerical solution (by the Euler method with n = 5) of the IVP (3)-(4) in the interval 0 s x s 1. (0.112333). we present the intermediate calculations to assist the reader in following the details.8. Even though we are only interested in x.x s 1. Thus. it is to be considered that if one were to plot the points of our numerical solution and connect these points with straight lines.)
EXAMPLE 1
Use the Euler method to solve numerically the IVP
YI = x2 + ya
Y(O) = 0
(3) (4)
on the interval 0 s x s 1. Where necessary we rounded numbers
off to six decimal places. The more points we ask for.) Thus. although it could be displayed geometrically by plotting each of the points. the resulting polygonal curve would be an approximation to the graph of the solution on the interval 0 <.0.
y). y) I and y = min (a. One fairly straightforward procedure to suggest a possible best solution for
a given computer is to write a program that repeats the sequence of steps
illustrated in the flowchart several times. until the computed results [that is.y). to approximate a true solution is called truncation or discretization error. We shall do this in the remaining sections.y). A program for Example 1 would use the initial values x = 0.K. y = 0. The error resulting from the use of a numerical method. This implies that the final digit in any number represented in the computer may be in error due to round-off. many programmers find this practical rule very useful for a first guess. As n increases. y). This is due to the limitation of all computers that can carry only a specified number of digits. the accumulated round-off error can more than offset the gains made in improving the accuracy of the method. we decrease the value of h = Ox and generally improve the accuracy of the method. In particular.R ax
'The authors wish to thank Thomas M. we state the following theorem without proof.
THEOREM 1. xo s x s xo+a. (x..2 A flowchart of the steps involved is shown in Figure 7. yo-b s y s y0+b}. called round-off error. tends to become more significant as h gets smaller.
. However.
Suppose that the functions f(x.1 can be carried out with the aid of a computer or programmable calculator.
With regard to the error in the Euler approximation. Of course.4.2
Euler Method
291
The calculations presented in Table 7. increasing n (decreasing the step size) reduces the truncation error. of and of are defined and continuous on
the rectangle R = {(x. While this is no guarantee of the accuracy of the results. Generally.
(=. if we increase the value of n.y). such as the Euler method. Green for the development of the flowcharts and the
computer and calculator exercises.7. y )] no longer differ in the number of decimal places required.. Once a program has been written it is a simple matter to produce a solution with any other set of initial values. ay
(.R ay
maxIaf+fafIsL. Let K and L represent two positive numbers
(2. one way to obtain a better approximation of the solution is to use
another. each time doubling the size of n. more accurate method with fewer steps. another type of error. p-). and n = 5. Numerical analysts have spent and continue to spend much time searching for the proper balance between reducing the truncation error (increasing n) and the round-off error (decreasing
n). b = 1. Let M =
max I f(x..R
such that
maxi -I<.
) by the Euler approximation y.
14.2.125. Compare with the exact results. use the Euler method with h = 0.1 and (b) h = 0.5. The
. then
E.7. is printed and not the other points Adjust the program so that only (x.1 and (b) h = 0.2.
7.. . Use the
Euler method with h = 0. 0.1. Write a computer or calculator program to repeat Exercise 3 using (a)
h = 0. use the Euler method with h = 0. with the results of Exercise 5. y'=3x2-2y
y(0) = 0
6.
EXERCISES
i = 1. 4.
where a = L(e'K .85)._. y(jo) = 0.01 and compare. allowing h to decrease until there is no change in the first three decimal places of the approximations for y(1). + hf(x. Write a computer or calculator program to repeat Exercise 5 using (a) h = 0. Write a computer or calculator program to repeat Exercise 1 using (a) h = 0.
y' = siny + x
y(0) = 0
5. y'=a-y-y+1
Y(0) = 1
3. Solve Exercise 6 exactly and compare the actual set of points (x.25 to approximate y(0.
8.2 to solve each of the IVPs on 0 s x s 1.). on the interval 0 < x s 1.
n
In all problems. 5 with the approximate set obtained in Exercise 6. Compare your result with the actual value. In Exercises 1 through 6. and in other cases it is not possible (or perhaps not feasible) to obtain the complete series. 2. represents the error made in the ith step by approximating the actual solution
y(x.01 and compare with the results of Exercise 1.2. 3..1 and (b) h = 0.. maintain four-decimal-place accuracy. on the interval for i < n.
1.sah. Write a computer or calculator program to repeat Exercise 12 with
h = 0.y(0) = 0.1)12K.
11.3
Taylor-Series Method
293
If E.0625.
2. In some cases it is possible to generate the complete series._. .01 and compare with the results of Exercise 3. y'=y2-2x2+1
y(0) = 0
y'=2xy'
y(0) = 1
7.
13. y.
12. i = 1._ y.. Repeat Exercise 12 with h = 0..). 0. y'=x-2y
y(0) = 3
4.25. 9. = y. Given the IVP y' = 3xy2. 0..3 TAYLOR-SERIES METHOD In Section 2. 10.9 and in Chapter 5 we demonstrated how the Taylor series can be used to generate series representations of solutions of a differential equation. Solve the IVP y' = x + y.
Given the IVP
y' = f(x.. + h.
+-xo
r-rO
. In this language the Euler method is a Taylor approximation of order 1.m
Y-YO
= dx(f(x.. For example.. Yo) + fr(xo. Yo)
Y (xo)
dx
(Y (x)) ==. .b].
Y("nxo)
= Y(xo) + Y'(xo)h +
L o) h2 + Yi o) h3
2!
+ . = x(.Yo)h + _
3!
h3 +
The Euler method can be thought of as an approximation of the Taylor series
by retaining only the first two terms of this series..f (xo.Yo). if the Taylor series is approximated by
Y(xo) + Y'(xo)h + . we have
Y(x1) = Y(xo + h) =
o
h
(x .
then the differential equation (1) and the initial condition (2) can be used to determine each y()(xo). Yo). .xo)".y)}^-o that lie close to the solution curve on the interval [xo..
0) h2 +
= Yo + f(xo.
.y) of any order exist at (xo. .. In general. 3.Y))
x-YO
r-YO
of + afdy
ax
ay dx
= f (xo. Yo) .y). We recall that the Taylor-series representation of the solution y about xo is
Y(x) = i
Setting x = x.
x .. . and consequently y. +
h'. k = 2..
Y(xo) = Yo. : 5 x : 5
(1) (2)
we again wish to obtain a sequence of points {(x.
Y'(xo) = y'(x)
= f (xo.7 Numerical Solutions of Differential Equations
Taylor series can be used in another way to obtain an approximation to the
value of the solution of an IVP at specific values of x.
we say that the approximation is a Taylor approximation of order 1.
If we assume that all partial derivatives of f(x.
) = Y(x. (0. (0. = y(x. by the Taylor approximation
of order 2 with h = 0. + h) .331469)).Y. +
2.4.
f(x. From the differential equation (3).-. 0. the more accurate we anticipate the results to be.
For the Taylor approximation of order I it can be shown that the error is proportional to h'.Y. The larger 1 is. our numerical solution of the IVP (3)-(4)..
Y2 = Y(x2) = Y(x.0).-. (0. +
(x2
y.
Solution We choose h = 0. is the set
1(0.2.0).2.
In general.) + Y'(x.)h
h2
+ (2x. Thus. one must specify the order I of the approximation. we obtain
Y' = x2 + y2
y" = 2x + 2yy' = 2x + 2y(x2 + y2)
=2x+2x2y+2y3. 0.-. Thus. + 2x?_. we have
. on the interval 0 s x s 1. and Y"(x.
We present the results in Table 7._. We illustrate the method by repeating Example 1 of Section 7. In any application of the Taylor method.Y(x.2. Once again the computations involved increase with increasing 1.Y.) =
EXAMPLE 1 Use the Taylor approximation of order 2 to obtain a numerical solution of the IVP
Y' = x2 + Y2
(3)
Y(0) = 0
(4)
on 0:x<_1.
+
{Mx.) = y Y'(x.3
Taylor-Series Method
Subsequent derivatives can be obtained in like fashion. incorporating the intermediate calculations. 0.2. (1.-.0.-.2..016).)h2
y.
where Y(x.) = so on.064154).-.'-. + h)
h
and x2. Thus we anticipate that the Taylor approximation of order
.8. Additional points are obtained in the same way. (0.6. Knowing we can then calculate y2 in a similar manner.).)
2
Y.161911).7.)h +
=y.
y) +fy(x. The forms presented are those
commonly called the Runge-Kutta methods.
CALCULATE
4.5.y +f(x. Some exercises involve IVPs that can be solved exactly. the more laborious the calculations. Namely. use the Taylor approximation of order 2 with h = 0. the higher the order. Naturally. y)f(x.2 to Produce Flowchart for Taylor-Series
Approximation of Order 2.Y)1 h2/2
Figure 7.
y'=x2+y
Y(0) = 0
2.
1.7.3
Taylor-Sari" Method
297
2 is more accurate than the Taylor approximation of order 1. Of course a higherorder Taylor approximation would be even more accurate. Changes in Flowchart Presented in Section 7.
y' = siny + x
y(0) = 0
5.5.
y'=x2+er
y(0) = 0
3.1
.Y)f(x.Y)h + Lf (x.
y'=x -y
y(0) = 1
4.
EXERCISES
In all problems.2 to solve the IVPs on
0sxs1. maintain four-decimal-place accuracy. Special forms of "higher-order
methods" are presented in the next section.
f(x. y)l h2/2
ASSIGN
xt-x+h
y.
y'=xy
y(0) = 1
S.
y' = 3x + 2y y(0) = . Y)+fy(x. The reader will be asked to compare the various methods of this chapter in Review Exercises. We have only to make minor modifications in our flowchart for the Euler method to produce a flowchart for the Taylor-series method. we change steps 4 and 5 as in Figure 7. In Exercises 1 through 8.Y)h [fx(x.Y) f(x. and therefore the numerical results of each method can be compared to the actual results as well as comparisons with the other methods.
12. yo)h + V . yo)} 6
Based on our knowledge of the Taylor series (or convergent series in general) from calculus. Repeat Exercise 6 with (a) h = 0. Use the Taylor approximation of order 2 with h = 0.01.1 and (b) h = 0.2) of this exercise
is to be compared with y(0. [Hint: Calculate y" and y"' directly from the differential equation.
y' = x2y y(O) = 2
8. yo)) 2
The Taylor approximation of order 3 would have the additional term
{fa(xo.y(xo) + f(xo. yo) + 2f:r(xo. yo) f(x(. For instance. y(0.
7. Compare with the results of Exercise 9 and with the exact results.4 RUNGE-KUTTA METHODS
In the description of the Taylor approximation of order 2. Repeat Exercise 10 with h = 0. yo) + fy(xo. Repeat Exercise 9 with h = 0.] 11.1 and compare with the corresponding
results of Exercise 10 and with the exact results (see Exercise 11). Repeat Exercise 1 with (a) h = 0.
10.(x(. y0)12
+ Mx.1 and compare with the corresponding results of Exercise 9 and with the exact values.
14.1 and (b) h = 0.) + f (xo.
In Exercises 13 through 15 write a computer or calculator program for the Taylor approximation of order 2.298
7
Numerical Solutions of Differential Equations
7..2 to solve the IVP
y'=x+y
y(o) = 0
on 0 ss x :s 1.2) of Exercise 9 and with the exact value of
y(0. y(. we know that the accuracy of our approximation improves with
.1
9. Also compare with the Eule method (Exercise 12 in Section 7. Repeat Exercise 3 with (a) h = 0. yo)(flx0.
13..2 to solve Exercise
9.2). Use the Taylor approximation of order 3 with h = 0.2). we wrote
y(xo + h) . yo)l2f(xo.01. 15. Compare with the exact results.1 and (b) h = 0.01. yo) f(xo.' yo)f(xo.
y' = e'' sin y y(0) = 0. yo) + [f(xo.
This process is repeated until we have the desired number of points. from formulas (1)-(4) by setting i = 0. (0. and with x.
RRunge first developed the approximation of order 3.6.+h.
The Runge-Kutta approximation of order 3 is defined by the following collection of formulas:
k. we obtain y2 from formulas (1)-(4) by setting i = 1.=x2+y2
y(0) = 0
(5) (6)
on o:5xs 1.2 to
solve the IVP
y. it is convenient to display our solution (see Table 7.)
(1)
k2 = hf(x.072509). For a development of these formulas and an analysis of their accuracy.+2k2-k.). Once again we show the intermediate calculations.
(3)
(4)
The scheme then is to obtain y. (1. S. the description of the higherorder terms gets more and more complicated and the associated calculations more profuse. 0. Numerical Analysis (New York: McGraw-Hill. y. (0. + 6(k.2.4
Runpe-Kutta Methods
the number of terms retained. = y. 1957). 0. 0). + 4k2 + k. and Kutta generalized and refined the approximations of orders 3 and 4.k(2)
k3=hf(x.
. 0. The basic idea of the Runge-Kutta methods' is to preserve the
order of a Taylor approximation (in the sense of the error involved) while
eliminating the necessity of calculating the various partial derivatives off that are involved.2 of the IVP (5)-(6) on 0 s x s 1 is given by the set
Solution
{(0. Kunz. On the other hand. (0.021371). the reader is referred to K. = hf(xny.7. The alternative proposed by these methods involves evaluating the function f at certain judicious points rather than evaluating the specific partial derivatives. 0. Our numerical solution using the Runge-Kutta approximation of order 3 with h = 0. + 12 h.)
y .174273).350721)}. We present here the Runge-Kutta formulas of order 3 and
order 4. With this value of y.4. 0.3). + 1.002667).
As before. in the hope that they help the reader to follow the development.y. (0.8.
EXAMPLE I
Use the Runge-Kutta approximation of order 3 with h = 0.
. use the Runge-Kutta approximation of order 3 with h = 0.072451).)
12 = hf(x.
(11)
We employ formulas (7)-(11) in exactly the same manner as the Runge-Kutta formulas of order 3.7. + 212)
14=hf(x.174090).
It can be shown that the error involved in the Runge-Kutta approximation of order k(k = 3. 0. 0. 0. In Exercises 1 through
8. 0. maintain four-decimal-place accuracy. = hf(xi. it has the feature that the
calculations depend on knowledge of the function f only and not on derivatives off as in the case of the Taylor method. (0.-. we demonstrate our results in a table (see Table . = hf(x. (0. 0. + Ih.
EXAMPLE 2
Solve the IVP
y'=x2+y2
Y(0) = 0
(12)
(13)
and 0 <_ x <_ 1. Use the Runge-Kutta approximation of order 4 with h = 0. We illustrate with an example.4
Runge-Kutte Methods
301
The Runge-Kutta approximation of order 4 is defined by the following formulas:
1.2 to solve the
IVPson 0:5x<_1.
Steps 4 and 5 of our flowchart given in Section 7. (1.
EXERCISES
In all problems. 4) is proportional to h*. 0).
. The Runge-Kutta method is highly accurate (small truncation error) even
when n is relatively small (small round-off error).2.2 would be changed to those
in Figure 7.)
y.
{(0. + ih. For these reasons it is a widely used method.=y.+1.) (7)
(8) (9)
(10)
1. y.6.4).
Solution
As in the previous example. y.+212+21.+h. (0.002667). Our solution of the IVP (12)-(13) on 0 :s x <_ 1 using the Runge-Kutta approximation of order 4 is the set
(0.8.350257)}.y.021360).6 in order to produce a flowchart for the Runge-Kutta method of order 4.4. y.+14).2.+}(1. Thus the Runge-Kutta method is more accurate than the Euler method and although it has the same order of accuracy as the Taylor approximation of order k. + i 1.
c.. "The Long Jump Miracle of Mexico City.308
7
Numerical Solutions of Differential Equations
7.=Mg+D. R.3c.itics attributed the length of the jump to the thinness of the air at Mexico City. and the magnitude of D is given experimentally as kpV2. the center of mass G of the long jumper describes a path that one can study as a projectile moving through a resisting medium. Stoker. r = (v0 + co)t . The results are rather cumbersome. g is the acceleration vector due to gravity. V is the velocity vector of G at any instant. vo is the initial velocity of the wave. is an unknown function that is related to the
depth y of the river by the approximation c.ca)/ir. The direction of D is opposite to that of V. t = x.(4) . In his article Brearley attempts to disclaim these critics. 'M. so numerical methods might be more appropriate as a means of solution (see Exercises 1 through 4). where p is air density. Hence. Beamon of the United States set a world rewrd of 29 feet 212 inches in the long jump.(V .
. Furthermore. S represents the slope of the river bed. This improved the existing world mark by an astounding 2 feet 78 incl. B represents the width of the river (any cross section of the river perpendicular to the direction of the river flow is assumed to be a rectangle of constant width B but variable depth y). however.
(1)
In this equation g represents the acceleration due to gravity (g = 32. J.
(2)
where M is the mass of the long jumper.es. where U is a unit vector in the
direction of V. and D is the air drag on the long jumper. by co = (gyo)"2 (note that co has units of velocity). Equation (1) is a Bernoulli equation and hence could be solved by the method associated with that type of differential equation (see Exercise 19). and k is constant for
fixed body posture. and t is time.16 ft/sec2). Stoker. We present here Brearley's model only and recommend that the reader consult the article for Brearley's complete argument.' derives the
following differential equation:
(vo + co)'-dc (-l
d
. Brearley. D = -pkV2U.x. Water Waves (New York: Interscience. Magazine 45 (1972).6 APPLICATIONS
Flood Waves
In his study of the behavior of flood waves in rivers. + cgS
vo
3co
gBgB2c2I = 0. x is a space variable whose axis is
parallel to the direction of flow of the river." Math. N.
A novel and entertaining application of differential equations is presented An Outstanding Long Jump in an article by Brearley. V = I V 1.°
During the 1968 O'jmpic Games in Mexico City. co is related to the initial depth y.
'J. The equation of motion in vector form is
M
dt. 1957). After takeoff. Many c.
Brearley gives the
values
uo = 9. can more conveniently be solved numerically as indicated in Section 7.
(4)
We also have the initial conditions u(0) = u0.15 m/sec
(5) (6)
k = 0.6
Applications
YJ
mg
Figure 7.7
The path of G is considered relative to a two-dimensional coordinate system
(x.5.7).182 m'
p = 0. treated as a system. we shall modify the problem by making an assumption that Brearley did not make. S = 0.pku(u' + v')"
M
(3)
dt
= -Mg .7. Beamon).4 ft/mile. However.984 kg/m'
(at Mexico City)
(7)
(8)
M = 80 kg
(approximate mass of R.pkv(u' + v')"`. At time t after takeoff.)
EXERCISES
For the differential equation (1) applied to the Pawcatuck River in Rhode Island. y) with the origin located at the position of G at t = 0 (takeoff). yo = 15 ft.
take the following values: B = 500 ft. v(0) = vo. Equations (3) and (4). Equation (2) can be written as the following system du M = . (See Exercises 5 through 8. u and v represent the velocity components of G parallel to the
x and y axes.19
. vo = 2.
(9)
Equations (3) and (4) are difficult to solve in their present form and hence can be studied numerically.45 m/sec
vo = 4. respectively (see Figure 7.
2 by:
5.
6. With this information and c. A = 3 x 10-4. the Taylor approximation of order 2. the Euler method. the Taylor approximation of order 2. The rate of loss of heat from the mass is given by the differential equation
M
J= -A(T' . the Euler method.pku[u2 + . Observe that the
expression involved in the solution is not convenient for obtaining numerical values for T. solve Eq.)
(11)
Assuming that M = 100. = 10.1.
13. so we obtain
M du
= . = T(0) = 30. (7). the Runge-Kutta method of order 4.310
7
Numerical Solutions of Differential Equations
mph. (3) we assume that v = . solve Eq. T.3. (11) numerically on 0 :s t s 1 with h = 0. 2. 3. the Taylor approximation of order 2.
11.
Heat Loss Consider a mass M of water of specific heat unity that is mixed so well that its temperature T is the same throughout. the Runge-Kutta method of order 3. the Runge-Kutta method of order 4.
14. Gas Ionization A gas is to be ionized in such a way that the number of
electrons per unit volume equals the number of positive ions per unit vol-
. represents the temperature of the medium surrounding the mass.
8. the Runge-Kutta method of order 3.T. Solve Eq. and T.2 by:
9.2 by: 1.15)2]" 2.(u . the Runge-Kutta method of order 4.15). solve the differential equation (10)
numerically on the interval 0 s t s 1 with h = 0. T. the Euler method.
(10)
which is a differential equation involving only the unknown function u. (1) numerically on the interval 0 s k s 1 using h = 0. (11) by these techniques.(u .(0) = 22.6 mph. The differential equation (11) can be solved by separation of variables and
partial fractions. the Runge-Kutta method of order 3. Using the values given in (5). 4. In Eq. and co = 15.
10. 12. and (9).1. (8).
7.
Positive ions and electrons recombine to form neutral molecules at a rate equal to kn2. and for t > 0.
17.2 cos a)"
/
(14)
where 0 is the angle that the pendulum makes with the vertical. 1974).
16. I is the length of the pendulum. We assume that the gas is initially (t = 0) un-ionized. compare the numerical results of Exercise 14 with the actual results. 0 are polar coordinates. If we assume that g/1 = 2 and at = 7r/6.2 using the Taylor approximation of order 2.
.u
It
v
16
For simplicity we take GM/h' = 1. (16) numerically on 1 s u s 2 with h = 0.A. h is twice the area swept out by the radius vector in unit
time. Physics In the study of the simple pendulum (see Chapter 8). C. solve Eq. u = r-'. and 0(0) = eo = a. (14) numerically on 0 s t s 1 with h = 0. M is the mass of the sun. Solve Eq. and e is a small parameter.2 by the Runge-Kutta method of order
3. Lin and L. the equation of the orbit is assumed in the form r = r(e). E = 0.01 and v(1) = 1. g is the acceleration due to gravity. we can write
d'u/de' = v(dv/du) and the differential equation (15) takes the form
dv _ (GM/h')(1 + Fu') . where the constant k is called the constant of recombination. Using the values of A and k given in Exercise 14.6 Applications
311
time. G is a universal constant. Astronomy If one takes into account the general theory of relativity when studying planetary orbits.2 by Euler's method. the following differential equation occurs:
de = .kn'
(12)
n(0) = 0. solve the IVP (12)-(13) numerically on 0 :5 t s 1 with h = 0. This leads to the
1VP
dodt=A
. the following differential equation occurs':
d'u
GM
(1
(15)
In this differential equation r.7.000 and k = 5 x 10-6. Mathematics Applied to Deterministic Problems in the Natural Sciences (New York: Macmillan.J 1(2 cos e . Segel.
'C. If we set v = du/de.
15. Solve the IVP (12)-(13) exactly by separation of variables.
(13)
If we take A = 100. A (a constant) ions per unit volume are produced.
2.. Solve the initial value problem (C)-(D) on the interval 1st :s 2 by the Euler method with h = 0.6
Applications
313
6.8) + 1]
x sin x '
dx
(F)
Al) = 2
(G)
'This is Exercise 3 in Philip M. show that
y
43x
11.2 if it is given that v = 700 cm/sec when x = 0.y(t)) is a solution of the initial value problem (C). show that this equation can be
dv__1 /
dx
R Ix + v(1 + 13v2)
rRly
Taking R = 900/(K/R).2. Morse and K.2. 9. 883. 7.2. and R = 2 10-5. or 9 to determine whether or not the relationship in Exercise 10 is valid for all t 1. Theoretical Acoustics (New York:
McGraw-Hill.7. KIR = 100. Setting v =
(E)
rewritten in the form
d and vdx = drx in Eq. p.5):
dy
(4 -y)[(x cos x) In (2y . 8.
10.
. If (x(t). (E).
Exercises 13 through 15 relate to the following initial value problem (see Example 2. Use the results of either Exercise 6. Uno Ingard. Section 1. an approximate equation for free oscillations of such a resonator is of the form
.d'tx 4-P-li +pldtl2J+Kx=O
12.
If we account' for the nonlinear dissipative effects in the neck of a Helmholtz cavity resonator. 1968). 7. solve this equation on the interval 0 <_ x s 1 by the Runge-Kutta approximation of order 3 with At = 0.-(D).. Solve the initial value problem (C)-(D) on the interval 1 s t s 2 by the Taylor approximation of order 2 with h = 0. Reprinted by permission of McGraw-Hill Book Company.
8. Solve the initial value problem (C)-(D) on the interval 1 :s t s 2 by the Runge-Kutta approximation of order 4 with h = 0. Solve the initial value problem (C)-(D) on the interval 1 s t <_ 2 by the
Runge-Kutta approximation of order 3 with It = 0.
Solve the initial value problem (F)-(G) on the interval 1 s x s 2 by the Runge-Kutta approximation of order 3 with h = 0. Solve the initial value problem (F)-(G) on the interval 1 s x s 2 by the
Taylor approximation of order 2 with.
.h = 0.
15.
14.2 and compare the
approximate results with the exact results.314
7
Numerical Solutions of Dlferential Equations
13. Solve the initial value problem (F)-(G) on the interval 1 s x s 2 by the
Euler method with h = 0.2 and compare the approximate results with the exact results.2 and compare the approximate results with the exact results.
the situation is not as hopeless as it seems.
(1)
subject to the initial conditions x1(to) = 4
xx(to) = xi. the coefficients and initial conditions of dif-
ferential equations are often determined approximately. x2)
X2 = f2(t. Fortunately.
(2)
. However. The same is true for nonlinear systems. only a few types of nonlinear differential equations (for example. that is. It is often possible to answer some of the questions above by utilizing the form of the differential equation and properties of its coefficients without the luxury of knowing the exact solution.2 EXISTENCE AND UNIQUENESS THEOREMS
Here we state an existence and uniqueness theorem for systems of two differential equations of the first order with two unknown functions of the form
Xt = fA(t.1
INTRODUCTION
Nonlinear differential equations and systems of nonlinear differential equations occur frequently in applications. It is therefore very
important to know what effect "small" changes in the coefficients of differential equations and their initial conditions have on their exact solutions. x2). We are often primarily interested in certain properties of the solutions and not the solutions themselves. Also. For example. x1. exact) can be solved explicitly. x2. homogeneous. Our aim in this chapter is to present some elementary qualitative results for nonlinear differential equations and systems.
8. to approximate them by linear differential eq rations. The questions raised above belong to that branch of differential equations known as qualitative theory.CHAPTER 8
Nonlinear Differential Equations and Systems
8. are the solutions bounded for all time? Are they periodic? Does the limit of the solution exist as t --+ +o? A very effective technique in studying nonlinear differential equations and systems is to "linearize" them. separable.
subject to the initial conditions
Y(0°) = yo.(t. it can be shown that f satisfies a Lipschitz condition. 9(to) = Y. and L2 are called
Lipschitz constants. The constants L. Y Y2)I < L. (8) with x° = y° and x112 = y'. x1. -
(3)
(4)
(5)
In fact. . x1.
. x2). x1. x) be continuous and satisfy a Lipschitz condition (with respect to x. and x2) in the region
THEOREM 1
= I
t . I + L2 Ix2 . x x2) be continuous and satisfy a Lipschitz condition (with respect to x.(to) = y°.f(t. x2) be a function defined in a region 9R of the three-dimensional Euclidean space R3. x1. Ix. x2). and x2) in the region 9t defined by Eq.x° I < B
(8)
Ix2-x0IsC
Then the initial value problem (1)-(2) has a unique solution defined in some interval a < t < b about the point to. x x2) and f2(t.
the differential equation (3) is equivalent to the system
Xl = x2
(6)
x2 = f(t.t°I s A
(t. if we set
y = x1
and
y = x2. Then the PVP (3)-(4) has a unique solution defined in some interval a < t < b about the point to. and df/8x2 exist and are continuous. and x2) in 91 if there exist constants L. We say that f(t. Y.
and the initial conditions (4) are equivalent to
x2(to) = y' (7) x. x2) satisfies a Lipschitz condition (with respect to x. When the partial derivatives of/ax.Y2I for all points (t.316
8 Nonlinear DHfrential Equations and Systems
The same theorem applies to second-order differential equations of the form
y = f(t.
(Existence and Uniqueness) Let each of the functions f.
COROLLARY 1
Let the function f(t. y2) in 9t. x1. x x2):
I x1 . y).
DEFINITION 1
Let f(t. (t.y. and L2 such that
If(t. Before we state the existence and uniqueness theorem we explain what is meant by a Lipschitz condition. y1. x x2) .
EXERCISES
1. y') does Corollary 1 imply that the IVP
Y(ro) = Y. With the hypotheses of Theorem I it can be shown that the solutions of system (1) depend continuously on the initial conditions (2). Y).
Y=y+yia
Y(10) = y°
0<_x2<1. z)
(2)
.3 Solutions and Trajectories of Autonomous Systems
317
Theorem 1 and its corollary can be easily extended to systems of n differential equations with n unknown functions and to differential equations of order n. For what points (t°. Suppose that we make an error of magnitude 10-3 in measuring the initial conditions A and B in the IVP
y+y=0
y(O) = A
y(O) = B.
respectively. Second-order differential equations of the form
z = F(x.3 SOLUTIONS AND TRAJECTORIES OF AUTONOMOUS SYSTEMS A system of differential equations of the form
x = f(x. Therefore. Show that the function f(t.
3sx. because we can never measure the initial conditions exactly. This result is very important in physical applications.n + x2 does not satisfy a Lipschitz
condition (with respect to x. Y = g(x.
0:5 x2s 1?
-1 <t<1.
has a unique solution?
3.51. y°. x x) = x. "small" changes in the initial conditions (2) will result in small changes in the solutions of system (1). Y).
How about in the region
0sx.8. and x2) in the region
-1 sts1. (1) where the functions f and g are not time-dependent is called an autonomous system.
2. 54.
What is the largest error we make in evaluating the solution y(r)?
8.
Let us emphasize again that a solution of system (1) is a curve in the three-dimensional space t. then we obtain explicitly the trajectories. Then. defined in some interval a < r < b which contains the point t0. (6). if (xo. And. the parameter t from the two equations in (3). In fact.< t < oo. x. Then by Theorem 1 of Section 8. the solution (3) defines a curve in the three-dimensional space t. assume that in some region D. yo) is any point in 9t and if to is any real number. if possible. f(x. there exists a unique solution
x = x(t). y(t)) trace out a curve in the xy plane called the trajectory or orbit of the solution (3). x. the points (cos t. The trajectory of this solution is again the unit circle
shows. Y = y(t) (3) of system (1).
the points (x(t). y)
The solutions of Eq. However.
(5)
This solution is a helix in the three-dimensional space t. therefore. . as t varies
x' + y' = 1. Even
. we obtain the first-order differential equation
dy = dx
dy dt dt dx
g(x. Y) x = Y. using the two equations in (1) and the chain rule. y. we can get a lot of information about the trajectories of a system even when we cannot find the solutions explicitly. For the sake of existence and uniqueness of solutions. different solutio is of a system may have the same trajectory. y = sin t. while an explicit solution of the system is impossible. Thus. in some cases we can find the trajectories explicitly. If we regard t as a parameter. If we can solve Eq. To see this. y. For example. x.cos t. Another solution of system (5) is x = sin t. the xy plane is usually called the phase plane.
y. sin t) trace out the unit circle x2 + y' = 1 an infinite number of times.
Y(to) = Yo
(4)
Clearly. while a trajectory is a curve in the phase plane that is described parametrically by a solution of system (1). This was
actually done in the example above.318
8
Nonlinear DUtsrential Equations and Systems
can be also written as an autonomous system by setting
y = F(x.oo < t < oo is a solution of the system
= -y. the trajectory of this solution is the circle x' + y' = 1 in the xy plane. (6) give the trajectories of system (1) through the points of D.
y=X. In the study of physical systems. With the assumptions on f and g made above. yo) E R. y)
(6)
f(x. y) is called a phase of the system and. it can be shown that system (1) has exactly one trajectory through any point (xo. y = . y) 0. the pair (x. x = cos t. As this
in the interval -. If we know a solution (3) of system (1). then as t varies in the interval a < t < b. we assume that each of the functions f and g is continuous and satisfies a Lipschitz condition (with respect to x and y) in some region 9t of the xy plane. and satisfying the initial conditions
x(to) = xo. we can find the corresponding trajectory by elimi-
nating.2.
we can use Eq. Similarly. (6) to compute the slope of any trajectory through a point of D. if x represents the position and y the velocity of a particle moving according to the differential equation (2).
Y = F(x. (12) for c = 0. Hence. and hence
(9) x(t) = x0. are of special interest. yo) in the phase plane. (0. then x(t) = x0. 0) is also a trajectory that is contained in Eq. (6). yo) is a critical point of system (1). y(t) =
yo is a solution for all t. For such points neither Eq. As we mentioned above. Thus.
(11)
Solution The critical points of (10) are determined by the two equations
-y = 0. (0.
(12)
For y = 0 we see from (10) that x = 0.
(7) is valid. 0) is the only critical point of (10). y) * 0 in some region S.
EXAMPLE 1 Find the critical points and the trajectories of the following system of differential equations:
x= -y. yo) in the xy plane (the phase plane). it follows that x = xo is a position of equilibrium of the motion. (12). we can utilize the first-order differential equation dx f(x.
(10)
In particular. we see that critical points are identified with positions of equilibrium. x = 0. y) (7)
dy g(x.the trajectories through the points of S. if g(x. (6) nor Eq. if (x0. y0) is a critical point of (8). y)
(8)
If (x0. gives all the trajectories of system (10)
. y(t) = 0 is a solution of (8) for all t.8. where both functions f and g are zero. Thus.
initial conditions
y=x. Such points are called critical points (or equilibrium points) of system
(1). (2) can be easily put in the form of system (1)
by setting z = y. which is a one-parameter
family of circles centered at the origin. The trajectory corresponding to this solution is the single point (x0. Eq. With such an interpretation. Then
x = y. The points (x0. Using Eq. y) to comput.
y(O) = \. Eq. we see that the trajectories of (10) for y * 0 are the solution curves of the
differential equation
-= -x' dx y
y#0. it follows that yo = 0.3
Solutions and Trajectories of Autonomous Systems
319
if we are unable to find explicit solutions. Thus.
The general solution of this separable differential equation is
xz+y'=c'. draw the trajectory corresponding to the solution of (10) with x(0) = 1. Clearly.
it =x.1
(see Figure 8. x(0) = 1.320
8 Nonlinear Differential Equations and Systems
Rgure 8.y=Y
2. The arrows on the trajectories indicate the direction of increasing t and can be found from system (10). compute the trajectory of the solution of the IVP. the first equation in (10) implies that x < 0.
In Exercises 5 through 10.Y.1).y= -x
x(0) = 1.
x(0) = 1. x = x.y=x
5.5 contains many worked-out examples in which we find the shape of all trajectories of linear systems. Indicate the direction of increasing t. Indicate the direction of increasing t.
x(0) = 1.
4. Section 8. 1 = -x. y = 2y. find the critical points and the trajectories of the
system.z=y.z=y. This means that x(t) decreases (and so it moves in the counterclockwise direction). (12) we see that the trajectory of the solution of the IVP (10)-(11) is the circle x2 + y2 = 4.
1. For example. centered at the origin with radius equal to 2.
y(0) = 0
Y(0) = 0
8. y = -x .
.
EXERCISES
In Exercises 1 through 4. x = x. From Eq. y = x + y.
y(0) = -1
Y(0) = 1
6.z= -x. y = -2y.
2y
3. when y > 0. 7. x = x.
yo) is a critical point of (1). A.
DEFINITION 1
The critical point (xo.
11.
y=cx+dy
(b) Show that the system has a line of critical points if ad .
(1)
We recall that a point (xo. More precisely. yo) = 0. Y(t) ° Yo (2) is a solution of (1) for all t. (a) Show that the point (0.
y = g(x.be * 0. yo) (or the constant solution (2)) of system (1) is called stable if for every positive number E there corresponds a positive number 8 such that. yo) is a critical point (or an equilibrium point) of system (1) if f(xo.xo)2 + [y(t) . y = -sinx
13.
. is called stable.x=y. 19.be = 0.y= .4 STABILITY OF CRITICAL POINTS OF AUTONOMOUS SYSTEMS
Consider again the autonomous system
z = f(x. x(0)= -1. x = -x2 + y2. x . the solution (2). z + x = 0 16.x=x-xy. yo) = 0 and g(xo.
8.x0)2 + [Y(0) . every solution (x(t). we give the following definitions. transform the differential equation into an equivalent
system (by setting x = y) and compute the equation of the trajectories.4
Stability of Critical Points of Autonomous Systems
321
9.8.y.x= -x+y.Yo)' < 8
exists and satisfies
(3)
NO . y(o) = -1 In Exercises 11 through 14.x + x' = 0
In Exercises 15 through 18.y= -4sinx
15. find the critical points and the equations of the trajectories of the solutions of the system. i = -y. x . A. yo). Since the derivative of a constant is zero.Yo)2 < e
(4)
for allt>-0. it follows that if the point (xo. y(t)) of (1) which at t = 0 satisfies
1x(0) . y = try
12. z + sin x = 0
14. x = y.x0.y+xy
17. or the critical point (xo. r = x.y= -x .
if and only if ad . If this is true. y(0)=1
x(o) = 1. then the pair of constant functions x(t) . In many situations it is important to know whether every solution of (1) that starts sufficiently close to the solution (2) at t = 0 will remain close to (2) for all future time t > 0.x = 0 18.
10. 0) is the only critical point of the system
X=ax+by.
(5)
(6)
DEFINITION 3
A critical point that is not stable is called unstable. respectively. sin t . stability means that a small change in the initial conditions causes only a small effect on the solution. yo) [or the constant solution (2)] is called asymptotically stable if it is stable and if in addition there exists a positive number So such that. y(t)) of (1) which at t = 0 satisfies [x(0) . The concepts of stable. asymptotically stable.
lim Y(t) = y0.c2 cos t.Yo]2 < au exists for all t ' 0 and satisfies
lim x(t) = x0.Yo)
x
(a)
(b)
(c)
Figure 8.
is stable. Choose 8 = E. Roughly speaking. Is it asymptotically stable?
Y=x
(7)
Solution We shall apply Definition 1.322
8
Nonlinear Differential Equations and Systems
DEFINITION 2
A critical point (x0. Every solution of (7) is of the form
x(t) = c. and (c).
. while instability means that a small change in the initial conditions has a large effect on the solution. every
solution (x(t).x0]2 + [Y(0) . 0) of the system
x= -y.2(a). Let e > 0 be given.
(x0. and unstable critical point are illustrated in Figure 8. cost + c2 sin t y(t) = c. asymptotic stability means that the effect of a small change tends to die out.2
EXAMPLE I
Show that the critical point (0. (b).
Y.
0) of system (11) depends almost entirely on the roots of Eq.
The three examples considered were all linear. both roots of Eq. In fact. In Example 1. using Theorem 1(c) it follows that the critical point (0. c. using Theorem 1(b) we see that the critical point
(0. 0) is a critical point of system (1). Its roots
are ±i. 0) of system (11) is unstable if one (or both) of the roots of Eq. yo) is a critical point of the system. Since a transformation of the form X = x : xo. [See Exercise 19(a) in Section 8. 0)
of system (11) is easy to study.
When the autonomous system (1) is linear with constant coefficients.
THEOREM 1
(a) The critical point (0. 0) of system (10) is unstable. b. Since they are real and negative. We shall assume that ad . (12) are real and negative or have negative real parts. and d = 0) X2 + 1 = 0. and only if. and only if. 0) is an unstable critical point of the system (10). 0) is the only critical point of (11). the characteristic equation is (here a = -1. Since one of the
roots is real and positive. c = 0.324
8 Nonlinear Differential Equations and Systems
which cannot be true for all t ? 0. In Example 2. the solutions of system (11) are of the form
x=Ae". b = -1. b = 0. Since they have zero real parts. b = 4. (c) The critical point (0. Hence.(a + d)X + ad . Y = y . (0. 0) of system (11) is asymptotically stable if.3. Its roots are X. 0) as a critical point. (12) is real and positive or if at least one of the roots has a positive real part. Its roots are 1. and
d = -1) a2 + 2X + 1 = 0. (b) The critical point (0. it follows that the critical point (0.. the characteristic equation is (here a = -3. = X2 = -1 (a double root). we have the following theorem.3. c = 1. using Theorem 1(a).be = 0. and d are constants. both roots of Eq. c = -2.
(11)
where a.
In Example 3. An effective technique in studying the auton-
.be f 0. (12) are real and negative or have nonpositive real parts. 0) of system (9) is asymptotically stable. the character-
istic equation is (here a = 0. = 1 and a2 = -1.
(12)
where X is a root of the characteristic equation
X 2 . it is not surprising that the stability character of the critical point (0. that is. and d = 3) X2 . Therefore.
y=Be". we assume. Then the
point (0. (12).
The stability character of the critical point (0.] As we explained in Section 3.1 = 0. Consider again the autonomous system (1) and assume that (x0.yo transforms the autonomous system (1) into an equivalent system with (0. without loss of generality. that (0. 0) of system (7) is stable. when
X=ax+by.
y=cx+dy. we can obtain the solution explicitly. 0) of system (11) is stable if.
and ad . 0) is just a
stable point of system (11).
Solution
Here a = -I.
This theorem offers no conclusion about system (13) when (0. and F(0.
(14)
[Roughly speaking. 0) of system (11) is unstable.Y2). b = 1. the conditions (14) are satisfied.0
G(x.0.4
Stability of Critical Points of Autonomous Systems
325
omous system (1) near the critical point (0. 0). by Theorem 2(a).
F(x. Hence. 0) of the nonlinear system (13) is unstable if the critical point (0. the functions F and G are continuous.
Clearly. 0) is to approximate it by a linear system of the form (11). the point (0.8. 0) of the nonlinear system (13) is asymptotically stable
if the critical point (0. 0) is also an asymptotically stable critical point of the nonlinear system (15). The linearized system is
x= -x+y. 0) = 0.lim
i-.be # 0 and F(0.] Then the following result holds.] Assume. the condition (14) means that the linear system (11) is a good approximation of system (13). furthermore. y) = (x2 + . The result that we are about to state is of this nature. Y) ryl
(13)
. In many cases one can prove that if the approximation is "good". c = 0. [Hence (0. with ad . y = cx+dY +Gx y).be =2 # 0. Assume that system (1) is of the form
i = ax + by + F(x. d = -2. have continuous first partial derivatives. 0) = G(0. y)
x2 + y2
. (b) The critical point (0.
THEOREM 2 (a) The critical point (0. 0) = G(0. 0) is an
asymptotically stable critical point of (16).
(16)
The characteristic equation of system (16) is Xz + 3X + 2 = 0. = -1 and k= = -2.
EXAMPLE 4 Show that the critical point (0. y)
(. 0) = 0. Since they are both negative. 0) is a critical point of (13). that near the origin (0. the point (0. the linear system (11) has solutions which themselves are "good" approximations to the solutions of system (1). 0) of the "linearized" system (11) is asymptotically stable. G(x.
y= -2y. 0) of the nonlinear system
z= -x+y+ (x2+y2)
(15)
Y= -2y-(xz+yz)aa
is asymptotically stable.(x2 + y2)'a. y) = .
. and that
lim F(x. Its roots are
k.
Assume that the system
5. 0) is an unstable critical point of (18).y=5x-y
8. x=x .y=x+3y 7.
6. or unstable.y= -x
4.y= -x 2. y = r sin 0.x . The linearized system is
z= -3x+4y. 0) is also an unstable critical point of the nonlinear system (17). and d = 3 with ad .x=y.sin20)
x2 + y2
and
x and y in polar coordinates: x = r cos 0. 0) of each of
the systems in Exercises 1 through 3 is stable. determine whether the critical point (0.x= -3x+5y.y= . G(x. or unstable.be = -1 4
0.x=3x-2y.y= -x+y
x=4x-6y. Its roots are as = 1 and \2 = -1.y
3.
EXERCISES
By using the definition. 0) of the
system is stable. 0) of this system is asymptotically stable and therefore both populations are headed for extinction.y2.
Solution
(17)
Here a = -3.y= -2x+3y 9. 0 is equivalent to r -+ 0) F(x.
1. asymptotically stable.0 and r
r2 cos 0 sin 0
= r cos 20
+0
as r . 0) of the nonlinear system
x=-3x+4y+x2_y2
-2x+3y-xy
is unstable.
(18)
The characteristic equation of system (18) is k2 . determine whether the critical point (0. F(x. where x is the desirable population
and y is a parasite.z=5x-6y.c= -2. z= -3x+4y.y= -2y.y.
y= -2x+3y. y) = x2 . the point (0. Since one of them is positive.y=6x-7y 10. the conditions (14) are satisfied. We express
y --). Then (x . By Theorem 2(b). 0) = 0.
. the point (0.x=x-y.x= -x+y.x= -y. y) _ r2(cos2 0 . Show that the critical point (0. y)
x+y2
r
=
Hence. with F(0. asymptotically stable.326
8 Nonlinear Differential Equations and Systems
EXAMPLE 5 Show that the critical point (0. 0) = G(0.1 = 0.b =4.0
G(x.y=4x-y
11.
y=8x-lOy
represents two competing populations. y) = -xy. x= -x+y.
In Exercises 4 through 10.
0) in each of the following cases. 0) is the only critical point of system (1).5
Phase Portraits of Autonomous Systems
327
12.
13.
Y = y . by means of examples.x=5x-6y+xy. whether it is asymptotically stable or unstable. show that the point (0. Show that the critical point (1. and investigate the stability character of the resulting system at the critical point (0.x=
-x-x2+y2. p2>4q.p2=4gandp>0 23. p2>4gandq<0
22.]
In Exercises 13 through 17.y= -y+xy
17.
d). In this section we first present. We shall assume that (0. (See Exercise 19.7y + 1.
where X is an eigenvalue of the matrix
(c
y = Be'. 1) of the system
x = 5x .y2)5
15. x= y+x2-xy. y = 0.x=x-xy.y + (x2 . 0) is a critical point for the given system and determine.3.y= -y+2xy
Transform the differential equation t + pi + qx = 0 into an equivalent system by setting x = y.
which appears in Turing's theory of morphogenesis is asymptotically stable. the solutions of system (1) are of the form
x = Ae". andp < 0
20. p2 > 4q.
y = 6x . and this is equivalent to assuming that
ad .y= -2x+3y+y2
16. As we explained in Section 3.2y + (X2 + y2)2.
cx+dy
(1)
near the critical point x = 0. all possible phase portraits of the linear autonomous system
x=ax+by. 0) by means of X = x .p2<4gandp<0
19.8. x = 3x .5 PHASE PORTRAITS OF AUTONOMOUS SYSTEMS
The picture of all trajectories of a system is called the phase portrait of the
system.) At the end of this section we state a theorem about the phase portraits of some nonlinear systems. p2<4gandp>0
25.be * 0.p2=4gandp<0 24.y=6x-7y-xy
14.
[Hint: Transform the critical point to (0.1.p=0andq>0
8.andp>0 21.6y + 1. if possible.q>0. q > 0.1.
. y = 4x .
18.3. Section 8.
<0
CASE 2
or
A. < 0) Draw the phase portrait of the linear system
x= -2x+y.=k+il.
Equal roots.328
8
Nonlinear Differential Equations and Systems
that is. and A2 of Eq.. that is. (2). ] There are five different cases that must be studied separately. (2).+ad-bc=0.
It is sufficient to illustrate these cases by means of specific examples.>X.2=k-il
with
k<0
or
k>0.
CASE 4
d-b
1.
X2 = -il
with
I * 0.
CASE 5
i.
y=x-2y.
(2)
The phase portrait of system (1) depends almost entirely on the roots \.
Real roots with opposite sign.
1`1 = il.
Pure imaginary roots. that is. that is.>0.
(3)
.
CASE 1
Real and distinct roots of the same sign. that is. This is true because any two systems falling into one and the same of these cases (and corresponding subcase) can be transformed into each other by means of a linear change of variables. [The assumption ad .<0<A2. = A2 = A
matrix
with
A<0
or
A > 0.
A2<A.
In this case we obtain two distinct portraits.
EXAMPLE 1
(A2 < \. that is.
A. depending on whether or not the
ac
is equal to zero.
>.be * 0 implies that A = 0 is not a root of Eq.)
Complex conjugate roots but not pure imaginary.
CASE 3
.\. k is a root of the characteristic equation
A2-(a+d)>.
3. we have the solutions x= -3% y = -c2e-3'
(6)
For c. all the solutions (5) have the common trajectory. whose trajectory is the origin (0. respectively. Similarly.
and when c.e-'. When c. except the pair y = -x > 0 and y = -x < 0. = c2 = 0. The general solution of system (3) is x = cle-. from (4). y = 0. First. all the trajectories of system (3). (4)
where c. When c. 0) as t approaches infinity. all the solutions (5) have the same trajectory.e-' . _ -1 and K2 = .c2e-u
x
c.5
Phase Portraits of Autonomous Systems
329
Solution Here the characteristic roots are X. we have y c.
Thus. # 0 and c2 # 0. = 0 and c2 # 0. y = x < 0.e ' + c2e
'
c. and c2 are arbitrary constants. In fact. Furthermore.e _ c2e-3.e-'. From (6) with c2 > 0 and c2 < 0.. we can still obtain a good picture of the phase portrait of the system as follows.8.3
. + 0 and c2 = 0. for c. + c2e-3. we have the solution x = 0.3. y = c. These four trajectories are half-lines shown in Figure 8. We want to find the trajectories of all the solutions given by (4) for all different values of the constants c. + c2e ''
-*1
asr ->m. Figure 8. x = c.c2e-3' c . To obtain the other trajectories explicitly. and c2. 0) in Figure 8. The arrows on the half-lines indicate the direction of motion on the trajectories as t increases. < 0. we must eliminate t from the two equations in (4) and investigate the resulting curves for all nonzero values of the constants c. it is clear that every trajectory of system (3) approaches the origin (0. we find the solutions (5) Y = c. and c2. for c. we also find the trajectories y = -x < 0 and y = -x > 0. approach the origin tangent to the line y = x. > 0. y = x > 0.3 shows a few
Figure 8. When this approach is complicated (as in this example). all these trajectories approach the origin tangent to the line y = X.3.
we have the solutions
x = c2e. °°.
2c.330
8
Nonlinear DiNerential Equations and Systems
trajectories of the phase portrait of system (3). In the case of an unstable node.
y=2x-2y. The arrows on these trajectories point toward the origin because both x and y approach zero as t .e -t + c2eh
!2ex. solution of (7) is
I and x2 = 2. > 0. and c2 are arbitrary constants.e ' + c2
1 ->2
ast->°°. > X2 > 0. The arrows on these trajectories point away from the origin.
Hence.
. On the other hand. approach infinity as t -.°D.e-'. as in Exercise 1.
while for c2 < 0. the trajectory is the half-line y = j x < 0 (see Figure 8. we obtain the solutions
x = c. # 0 and c2 # 0.+ C2eL.
which implies that all trajectories are asymptotic to the line y = 2x as t tends to -°°. all trajectories are asymptotic to the line y = j x as t tends to 00.°°.4).z
asr-> -°°. y = !c2e'. The arrows on the trajectories indicate the direction of increasing t. This type of critical point is called an unstable node.
When X. the trajectory is the half-line y = 2x < 0 (see Figure 8. + c2e
31 .4 shows a few trajectories of the phase portrait of system (7). except that the direction of the arrows on the trajectories is reversed and the trajectories approach the origin as t .. we have
y x
2c.e-'. < 0. When c.
EXAMPLE 2
(X.e-' + c.
while for c.
For c. y = 0.e-" + k2
c. y = 2c. The general
Zc2e". the phase portrait is exactly the same. < 0 < X2) Draw the phase portrait of the linear system
s=3x-2y. This type of critical point is called a stable
node.4).e. since in this case both x and y tend to ±°° as t-> °°. Figure 8.
(9)
and when c. all the solutions. the trajectory of the solutions (9) is the half-line y = 2x > 0. This observation follows from (9). except the solution x = 0. as can be seen from (10).
For c2 > 0. + 12c2?
=
c.
(8)
where c.
-_ X
2c. * 0 and c2 = 0. = 0 and c2 * 0.e
+
(7)
Solution Here the characteristic roots are X.
x = c. the trajectory of the solutions (10) is the half-line y = x > 0.e
+ c2e2'
2c.
y = 2c. (10) For c.e-' + k2e2i
c.
EXAMPLE 3A
(X. = X2 = k < 0 and the matrix (a
C
X
d b
AI
is zero)
Draw the phase portrait of the linear system
x = -2x. = X2 = -2 < 0.y = c2x with slope c2/c. (Thus.e 2`. (12) where c.
.8.5
Phase Portraits of Autonomous Systems
331
Y .. The trajectories of the solutions (12) are the half-lines c.Ix<
Figure 8.4
This type of critical point is called a saddle point. The phase portrait of an unstable node is as in Figure 8. Since one of the roots of the characteristic equation is positive.5.
y = -2y. As we see from Figure 8. and c2 are arbitrary constants. all slopes are possible.4. the rest tend away from it. This type of critical point is called a stable node when X < 0 and an unstable node when x > 0. it follows that a saddle point is an unstable point of the system.5 except that the direction of the arrows on the trajectories is reversed.
(11)
Solution Here the characteristic roots are X. and the general solution of the system is
x = c. y = c2e 2'. only two trajectories approach the origin.) The phase portrait of system (11) has the form described in Figure 8.
(13)
Solution Here the characteristic roots are a.5
EXAMPLE 38
a < 0 and the matrix (t2_ '\ d b \ i is not zero)
Draw the phase portrait of the system
x = -x. For c. and the general solution of the system is
x = c. When c. = 0.
(16)
Figure 8. the positive and negative y axes are the trajectories of the solutions (15) (see Figure 8.te" + c. c.
y = -x .Y. From (14) we see that all trajectories approach the origin as t --> w.e-'. _ -1 < 0. = K. obtaining
y=fi=x+xln c.6).
y = -c.
y = c.e
(15)
Clearly. and c.332
8
Nonlinear Dl ferential Equations and Systems
Figure 8. we find the solutions
x = 0. * 0. are arbitrary constants. # 0.
(14)
where c.6
.e". we can eliminate t between the two equations in (14).
8.5 Phsso Portraits of Autonomous Systems
Equation (16) gives explicitly the trajectories of the solutions (14) for c, P, 0. All these trajectories approach the origin tangent to the y axis (see Figure 8.6). This type of critical point is called a stable node when A < 0 and an unstable node when k > 0. The phase portrait of the unstable node is as in Figure 8.6 except that the arrows on the trajectories are reversed.
EXAMPLE 4
(a, = k + il, A2 = k - it with k < 0) Draw the phase
portrait of the linear system
x= -x+y
9= -x - y.
Solution Here the characteristic roots are a, _ -1 + i and k2 The general solution of the system is
(17)
x = c,e-'cost + c2e-'sin t
y = - c,e -'sin t + c2e -'cost.
It seems complicated to find the trajectories through this form of the solution. For this reason let us introduce polar coordinates in (17) by setting
x = rcos4),
Then
y = rsin4).
9 = r sin 4) + (r cos 4,)4.
z = r cos 4) - (r sin 4)4
and
Substituting these values of x, y, x, and y into (17) and solving for r and $, we obtain the equivalent system
r= -r,
The general solution of (18) is
4)= -1. 4)_ -t + B.
(18)
r=Ae-',
(19)
The trajectories are therefore logarithmic spirals approaching the origin as t tends to infinity. The phase portrait of system (17) has the form described in Figure 8.7. This type of critical point is called a stable focus when k < 0 and an
unstable focus when k > 0 (see Figure 8.8). As is expected, in the unstable focus the arrows have opposite direction.
EXAMPLE 5
(A, = il, x2 = -il with I # 0) Draw the phase portrait of the
linear system
x=y,
y= -x.
(20)
Solution Here the characteristic roots are A, = i and K2 = -i, and the general solution of the system is
x = c, cost + c2 sin t,
y = -c, sin t + c2 cos t,
(21)
334
8
Nonlinear Diffsr.ntiel Equations and Systems
Figure 8.7
Figure 8.8
8.5
Phase Portraits of Autonomous Systems
335
yl
I'
7--
Figure 8.9
where c, and c2 are arbitrary constants. From (21) we see that x2 + y2 = c; + cZ. Therefore, the trajectories of the system are circles centered at the origin with radius c; + c2. The phase portrait of system (20) has the form
described in Figure 8.9. This type of critical point is called a center. The direction
of the arrows in Figure 8.9 is found from system (20). From the first equation in (20) we see that i > 0 when y > 0; that is, x increased when y is positive. This implies that the motion along the trajectories is in the clockwise direction.
Pt 9y
Asymptotically Stable
Focus
o
A
Unstable Focus
rs2-4p-0
J
Unstable
Nodes
0. cd
A sYSt
able
yo
b'cNode
'ON`
ally
Nodes
Figure 8.10 Summary of Portraits
336
8 Nonlinear Differential Equations and Systems
Summary In Figure 8.10 we summarize the nature and stability of the critical point (0, 0) of the system
z=ax+by,
cx+dy,
when ad - be * 0. If A, and A2 are the characteristic roots, then X, + A2 =
s2 - 4p. In the sp plane we have graphed the parabola s2 - 4p = 0. Since we have assumed that ad - be # 0, the entire s axis is excluded from the
Finally, we state a theorem about the phase portrait, near a critical point (0, 0), of a nonlinear system of the form
x= ax + by + F(x, y)
(22)
y = cx + dy + G(x, y),
where a, b, c, d, and the functions F and G satisfy the conditions mentioned before Theorem 2 in Section 8.4. Let \, and A2 be the two roots of the characteristic equation (2) of the linearized system (1). Under these hypotheses the
following result holds.
THEOREM 1
The critical point (0, 0) of the nonlinear system (22) is (a) a node if a, and A2 are real, distinct, and of the same sign; (b) a saddle if A, and A2 are real and of opposite sign; (c) a focus if X, and A2 are complex conjugates but not pure imaginary; (d) a focus or center if 1A, and A2 are pure imaginary.
In Example 5 of Section 8.4 we verified that the functions F(x, y) = x2 - y2 and G(x, y) = -xy satisfy the hypotheses of Theorem 1. Then from Theorem 1 and the Examples 1, 2, and 4, we conclude that the critical point (0, 0) of the
systems
In Exercises 1 through 10, determine whether the critical point (0, 0) is a node, saddle point, focus, or center, and draw the phase portrait.
1.x=x,y=3y 3. x=x,y= -y 5.x=2x,y=2y
2.z= -x,y= -3y 4. x= -x,y=y S. z= -x,y= -y
8.6
Applications
337
7.x= -x,y=x-y
9. x= -2x+y,y= -x-2y
S. i=x,y= -x+y
10.1= -y,y=x.
In Exercises 11 through 14, determine whether the critical point (0, 0) is a node, saddle point, focus, or center, and graph a few trajectories.
11.x= -x+y,y= -x -y
13.x=4x+6y,y= -7x-9y
12.x= -x+y,y= -x-3y 14.x=2x-y,y= -x+2y
In Exercises 15 through 19, determine whether the critical point (0, 0) is a node, a saddle, or a focus.
15.x= -2x+y-x2+2y2,y=3x+2y+x2y2 16.x= -x+x2,y= -3y+xy
17.x= -x+xy,y=y+(x2+y2)2 18.x=2x+y2,y=3y-x2
19.x=x-xy,y=-y+xy
Transform the differential equation z + pi + qx = 0 into an equivalent system by setting x = y and investigate the type of the phase portrait of the resulting system near the critical point (0, 0) in each of the following cases:
8.6 APPLICATIONS
In this section we present a few applications of nonlinear differential equations and systems. In comparison to mathematical models described by linear equations and systems, nonlinear equations and systems describe situations that are closer to reality in many cases.
338
8
Nonlinear Differential Equations and Systems
Biology Interacting Populations
Consider two species and assume that one of them, called prey, has an
abundant supply of food while the other species, called predator, feeds exclusively on the first. The mathematical study of such an ecosystem was initiated independently by Lotka and Volterra in the mid 1920s. Let us denote by x(t) the number of prey and by y(t) the number of predators at time t. If the two species were in isolation from each other, they would vary at a rate proportional to their number present,
z = ax
and
y = -cy.
(1)
In Eq. (1), a > 0 because the prey population has an abundant supply of food and therefore increases, while - c < 0 because the predator population has no food and hence decreases. However, we have assumed that the two species interact in such a way that the predator population eats the prey. It is reasonable to assume that the number of fatal encounters per unit time is proportional to x and y, and therefore to xy.
Then the prey will decrease while the predators will increase at rates proportional to xy. That is, the two interacting populations satisfy the nonlinear system
z=ax - bxy,
where b and d are positive constants.
y= -cy+dxy,
(2)
Although system (2) is nonlinear and there is no known way to solve it
explicitly, it is nevertheless possible, using the qualitative theory of such systems, to obtain many properties of its solutions, and in turn to make useful predictions
about the behavior of the two species.
Solving the system of simultaneous equations ax - bxy = 0 and - cy + dxy = 0, we find the critical points (0, 0) and (c/d, a/b) of system (2). Thus, system (2) has the two equilibrium solutions x(t) = 0, y(t) = 0 and x(t) = c/d, y(t) = a/b. Of course, only the second of these is of interest in this application. Let us compute the trajectories of the solutions of (2). Clearly, x(t) = 0, y(t) = y(0)e- is a solution of (2) with trajectory the positive y axis. Also, y(t) = 0, x(t) = x(0)e°' is another solution of (2), with trajectory the positive x axis. Because of the uniqueness of solutions, it follows that every solution of (2) which at t = 0 starts in the first quadrant cannot cross the x or the y axis, and therefore should remain in the first quadrant forever. The remaining trajectories of the solutions of (2) satisfy the differential equation
dy=-cy+dxy_(-c+dx)y
dx
ax - bxy
(a - by)x
which is a separable differential equation. Separating the variables x and y, we obtain
a - byd-c+dxdx y
y x
or
v'
-b)dy=
l \-c+d)dx. \ x /
(3)
8.6
Applications
339
Integrating both sides of (3), we obtain the general solution of (3),
a in y - by = -c In x + dx + k,
where k is an arbitrary constant. Equation (4) can be rewritten as follows:
(4)
lny'+Inx`=by+dx+k
Iny°x`=by+dx+k
eXc = eby.dx,k
x' _
e, ;
7
K,
(5)
where K is ek. If we allow K to be equal to zero, Eq. (5) gives all the trajectories of system (2). It can be shown that for each K > 0, the trajectory (5) is a closed curve, and
therefore each solution (x(t), y(t)) of (2) with initial value (x(0), y(0)) in the
first quadrant is a periodic function of time t (see Figure 8.11). If T is the period
of a solution (x(t), y(t)), that is, if (x(t + T), y(t + T)) = (x(t), y(t)) for all
t >- 0, then the average values of the populations x(t) and y(t) are, by definition, given by the integrals
X=
1
r
n
T
x(t)dt,
y= T
1
T
o
y(t)dt.
Surprisingly, these average values can be computed directly for system (2) with-
out knowing explicitly the solution and its period. In fact, from the second equation in (2), we obtain
y= -c+dx.
y
Integrating both sides from 0 to T, we find that
in y(T) - In y(0) = - cT + d
J
Y
T x(t)dt.
x
Flgure 8.11
340
8 Nonlinear Differential Equations and Systems
Since y(T) = y(O), it follows that r - cT + d x(t)dt = 0
J0
or
r
T
x(t)dt = d
o
Therefore,
(6)
Similarly, from the first equation in (2), we obtain
- =a - by.
X
Integrating both sides from 0 to T and using the fact that x(T) = x(0), we find
_a
y b.
(7)
From (6) and (7) we can make the interesting prediction that the average sizes of two populations x(t) and y(t) which interact according to the mathematical model described by system (2) will be exactly their equilibrium values x = c/d
and y = a/b. We can utilize this observation to make another interesting prediction. In
addition to the hypotheses about the prey and predator populations which lead to the mathematical model described by system (2), assume that the prey population x(t) is harvested in "moderate amounts." Then both the prey and predator populations will decrease at rates, say, ex(t) and Ey(t), respectively. In this case, system (2) should be replaced by the system
z=ax - bxy - Ex,
or
y= - cy+dxy - Ey
y = -(c + E)y + dxy.
(8)
x = (a - E)x - bxy,
Applying equations (6) and (7) to system (8), we conclude, with surprise, that if the harvesting of the preys is such that a > e, the average size of the prey population will be
x=
C + E
(6' )
d
which is somewhat higher than before there was any harvesting. On the other hand, the average size of the predator population will be
_ a - E
y
b
(7')
'
which is somewhat smaller than before there was any harvesting.'
'For more details on this and other related applications, the reader is referred to U. D'Ancona, The Struggle for Existence (Leiden: Brill, 1954).
8.6
Applications
341
Nonlinear A simple pendulum consists of a bob B of mass m at the end of a very light Mechanics and rigid rod of length L, pivoted at the top so that the system can swing in a vertical plane (see Figure 8.12). We pull the bob to one side and release it from The Motion of a rest at time t = 0. Let 0(t), in radians, be the angular displacement of the rod Simple Pendulum from its equilibrium position OA at time t. The angle 0(t) is taken to be positive when the bob is to the right of the equilibrium and negative when it is to the left. We want to study fi(t) as the bob swings back and forth along the circular arc CC'. From this information we know already that
00(0) = 0,
and
6(0) = 0,
(9)
where 8 is the initial angular displacement of the rod and 6(0) = 0 because the bob was released from rest. There are two forces acting upon the bob B at any
time r. One is its weight - mg, where m is the mass of the bob and g is the
acceleration of gravity. The other force is the tension T from the rod.
The force -mg is resolved into the two components -mg cos 0 and
- mg sin 0, as shown in Figure 8.12. The force - mg cos 0 balances the tension T in the rod, while the force - mg sin 0 moves the bob along the circular arc BA. By Newton's second law of motion, we obtain
z
m d-2 = - mg sin 0, 7
(10)
where s is the length of the arc AB, and d's/dt' is the acceleration along the arc. Since L is the length of the rod, we have s = LO and therefore
d's d'0 dt'=Ldt''
Using Eq. (11) in (10) and simplifying the resulting equation, we obtain
(11)
9+Lsin9=0.
(12)
The nonlinear differential equation (12), together with the initial conditions (9), describes completely the motion (in a vacuum) of the simple pendulum.
0
C
I
I.
\
T
/
c
B
eneR
I
A LL -.Or
-mg sino
Figure 8.12
0). The critical points of system (13) are the solutions of the simultaneous equations
y=0
and
. vertically upward.13). 0). we conclude that when x = 0. (2'n. The arrows on the trajectories point in the
clockwise direction because.. and when x = ar.. as we see from the first equation in (15).w2 sin x = 0. the differential equation (12) is equivalent to the system
X = y. 0)..
Since sin x = 0 for x = 0. (-2ar.12). 21r. The characteristic equation of (15) is X2 + w2 = 0 with characteristic roots
X = ± wi (pure imaginary). the critical points are
(14) (0. we study the critical point (0... Hence. If we now set 0 = x and 0 = y.. 3ar. all of them are located on the x axis. 0)..w2x.. . 0). . Since x = 0 is the angular
displacement of the rod from the equilibrium position OA (see Figure 8. the critical point (0. the pendulum is pointing vertically 7the pendulum is pointing downward. 0) and (ar. (n. 0). Clearly. .. -ar. ±2ar. Since
sinx=x z = y. 0) only.
(13)
In what follows we present a phase-plane analysis of system (13).13
.
+ 3!5!
x3
x5
we can approximate (13) by the linear system
(15) y = .
y = -w2 sin x... . . 3 r . . 0) is stable and a center for system (15) (see Figure 8.342
8
Nonlinear Differential Equations and Systems
Let us denote for simplicity the constant g1L by w2.ar. Because of the periodicity of the function sin x it suffices to study the nature of the critical points (0.tar. x increases
Figure 8. First. ( . :t 7r.
. n = 0.. Then (16) becomes
v = Y.. 0) by setting v = x . it can be shown that it is a center (the simple pendulum is a conservative system. System (13) also has a saddle point at each of the
critical points ((2n + 1)a. 0) is also a saddle point and hence an unstable equilibrium point of the linearized system (16) (see Figure 8. 0). we conclude that (a. 0). Thus. ±1. the linearized system is
X=Y. System (13) also has a center at each of the critical points (27rn.n in (16).
Rgure 8.8.
Next we study the critical point (a.r)+(x-a)'(x-. 0) is either a center or a focus point for the nonlinear system (13). 0) is a saddle point and therefore an unstable equilibrium point for system (13). 0) of (17) is a saddle point and hence an unstable equilibrium point. We can transform the critical point (7r. however. .w2 = 0 with characteristic roots
± w. ±2. 0) for n = ± 1. (. .. ±2. The trajectories in this
case are the heavy loops shown in Figure 8.14.5 (whose hypotheses are satisfied here). Since the roots are real with opposite sign. The Taylor-series expansion of sin x about x = a is given by
sin x=
-(x-..14
. it follows that the critical point (0.. ±2.
Y=w2(x-Ir)
(16)
and has a critical point at (w.. from Theorem 1(b) of Section 8. 0) to (0.)5
3! 5!
Hence.r. and hence there is no gain or loss of energy as in the case of a focus). In this case.5.13). it follows that (0.
Y = w2V. 0) for n = ±1.
(17)
The characteristic equation of (17) is Xz . . The closed trajectories
in Figure 8..8
Applications
343
when y is positive. By Theorem 1(d) of Section 8. 0). Now.14 correspond to the oscillatory motions of the pendulum about the
stable points (21rn..
In fact. The trajectories in this case are the wavy curves on the top and bottom of Figure
CASE 2
8.14.
yo I > 2w In this case the pendulum makes complete revolutions.. it is convenient to express c in terms of initial conditions.w2
y2 + 2w2(1 . and they are given by
x Y = ±2w cos 2.14 correspond to the following three cases.
(18)
where c is the integration constant.344
8 Nonlinear Differential Equations and Systems
The equations of the trajectories of the solutions of system (13) can be found explicitly. we obtain the family of trajectories
2 'y2-w2cosx=c. The equations of these trajectories can be found from (19) if we substitute yo by 4w2.
. from (13) we see that the trajectories satisfy the separable differential equation
dy dx
sin x
y
Separating the variables x and y and integrating.14. = 2 aresin Zw < tr..
(19)
The three types of curves shown in Figure 8..cosx) = yo
y2 + 4w2 sing 2 = YO. Assume that y = yo when x = 0. Then from (18) we find that
2 c=2Yo-w 2
1Y2 . For the sake of describing the phase portrait.
CASE 1
x = 0) is attained when y = 0 and
In this case the pendulum oscillates between the extreme angles ±xm. The trajectories in this case are the closed curves shown in Figure 8.w2cosx = 1Y0 . I yo I < 2w Then the maximum value of the angle x (remember that
xm.
CASE 3
1 yo I
= 2w In this case the trajectories are the heavy loops that
separate the closed and wavy trajectories described in the previous cases.
0) and determine whether (0. and e2 are nonnegative constants.
y = . or focus. Assume now that the prey has a limited supply of food available. The differential equation
6+k6+w2sin0=0.] Find the critical points of the system (negative populations are not permissible).
describes the motion of a simple pendulum under the influence of a frictional
force proportional to the angular velocity 6.
where a. ± 1. In the special case
z=3x-xy-2x2.
k>0.e2y2.
3. In each case. we assumed that the prey population has an abundant supply of food. [Fore.
where x and y are measured in hundreds of organisms. Biology In the predator-prey interaction that we discussed in the text.
In each of the following cases.Exercises
EXERCISES
1.bxy . 0) when µ < 2 and when µ > 2. y = Y + a/b.1)z + x = 0. 0) is a node. or focus.e. 0) of the linearized system. saddle. 0) for n = 0. are the only critical points of the system. we obtain system (2) in the biology application. ± 2. determine whether (0. study the stability of each critical point and determine whether it is a node. Show that (0.
y= -y+2xy-y2. study the stability of the system at (0. (a) k < 2w (b) k = 2w (c) k > 2w
4.]
2. Electric Circuits The differential equation
k + µ(x2 . b.cy + dxy .
saddle. Linearize system (2) in the biology application in the neighborhood of the critical point (cld. a/b) and study the nature and stability of the critical point (0. 0) is the only critical point of
the system. and d are positive constants and e. Transform the differential equation into an equivalent system by setting i = y. = e2 = 0. saddle. or focus.
. Transform the differential equation into an equivalent system by setting x = 0 and y = 6.
µ>0
is called the van der Pol equation and governs certain electric circuits that contain vacuum tubes.x2. In this case it is reasonable to assume that the predator-prey interaction is described by a system of the form
x = ax . Study the stability of the system at the point (0. [Hint: Set x = X + c/d... Show that (na.. 0) is a node. c.
x=x.4x-y+xy
12.y=4x-y
15.3x + 2y. or center.y
7.2y
14. saddle point. Find the critical points of the system
y=x+2y.10y
17. x = .
3. or unstable.(x2 + y2)
In Exercises 13 through 16. y = x . 0) of the
system is stable. y = 5x . Find the critical points and the equations of the trajectories of the solutions of the system
y= -x2+2y. 2.x= -3x+2y+x2-y2.
. x = 4x . X = 4x .
y(0) = 0
In particular.
6.y
16.
13.y=5x.
y=5x-Joy
-(x2+y2)v2. 0) is a node. Find the critical points and the equations of the trajectories of the solutions of the system
x=x-4y.2.9y + (x2 + y2)3. determine whether the critical point (0.8y.
y(0) = 0
has a unique solution defined in some interval about the point to = 0.
y=2x-y. determine whether the critical point (0.x=4x-9y.
-2x+y. y = 4x .9y.
x=2x2-xy.y= -x
9. x = 2x . and graph a few trajectories. x=3x+8y. y = x . 0) of the system
x=4x-9y+ (x2+y2)2.8 Nonlinear Differential Equations and Systems
REVIEW EXERCISES
1. 0) is the only critical point of the system
x=2x2-xy.y= -x . Determine the nature of the critical point (0.Joy .
and indicate the direction of increasing t.
5. x=3x+8y-xy. Show that (0.2y
8.
4.
In Exercises 6 through 12. x = 2x . y = 5x .8y.x= -3x+2y.9= -x -y
11. asymptotically stable.l0y
y+xy
10. draw the trajectory corresponding to the solution with initial conditions
x(0) = 1. focus. x= 3x+8y. Show that the IVP
i=2x2-y+t
y=x+2y2-e
X(O) = 1.
16. = . Determine' the stability of the linear time-invariant network shown in Figure 8.2-2 in Behrouz Peikari. Inc. N.Q and x2 = L)
Figure 8. ® 1974.2-3. Let the initial current through the inductor be io.15
19.. Consider' the linear time-invariant RLC network shown in Figure 8. 1974). 'This is Example 11. Inc.: Prentice-Hall. p.)
Figure 8. = 1 and x2 = Q. (The state variables are x. Reprinted by permission of Prentice-Hall. Reprinted by permission of Prentice-Hall. (The state variables are x. Determine the stability of this
network.
.Exercises
347
18. 0 1974.16
This is Example 11. 455. Fundamentals of Network Analysis and Synthesis (Englewood Cliffs.J. ibid.15. Assume that the initial voltage on the capacitor is va.
2... is an equation of the form
F (k. .l.l ..3Yk.CHAPTER 9
Difference Equations
9.
(1)
where F is a given function. As we will see in this chapter.
3Yk. the theory and solutions of difference equations in many ways parallel the theory and solutions of differential equations. ... ..
Yk..
Difference equations are the discrete analogs of differential equations.) = 0
. 2. 2. . 1. Yk.1 + 6yk = 0
Yk+3 . 1. 1. or derivatives. Yk. 1. 2.lyk = V. and k = 0.. . which
involve instantaneous rates of changes. . Yk. n is some positive integer. they appear as mathematical models in situations where the variable takes or is assumed to take only a discrete set of values..8Yk = 0
Yk.
. For example.. .4yk = -2k + 5
yk-1 .
DEFINITION 2
A solution of the difference equation
F(k.
(1)
is a sequence Yk which satisfies (1) for k = 0..
DEFINITION 1
A difference equation over the set of k-values 0. Yk. The following are examples of difference equations over the set of k-values
0.l..1
INTRODUCTION AND DEFINITIONS
Difference equations are equations that involve discrete changes or differences of the unknown function. Yk.2yk.. k =0..
.1.. the sequence
Yk = 2*. of the unknown function. This is in contrast to differential equations.n) = 0.2 + 6Yk.
.2 + 5Yk.1 . .2.
For example. (2). it is a linear difference equation with variable coefficients. 2. 1. . (2) is called homogeneous.k)Yk = 1
kyk.n + an-1(k)Yk..1 = 3k ...2 + 2Yk. (3). + a.n-1 + .3 .
DEFINITION 3
A difference equation over the set of k-values 0. is said to be linear if
it can be written in the form
(2) an(k)Yk.1
Introduction and Definitions
349
is a solution of the difference equation
Yk.1 . When f is not identically zero.. .
On the other
Yk-2 .3yk = 0
(3) (4) (5)
(6) (7)
Yk. 2. (3) and (6) are homogeneous. and (7) are linear difference equations with constant coefficients.. a0.. 2. Eq.1 + yk =
k2
Yk. Eq. (4). ao are constants. we speak of Eq. 1.2yk = 0.2 2k = 0
for
k = 0. (5) and (6) have variable coefficients.... The following are examples of linear difference equations over the set of k-
values 0.2 2k = 2 21 . (2) as
a linear difference equation with constant coefficients. none of the following difference equations can be written in the form of Eq. (5). Eqs..yk. 1. (2) is called nonhomogeneous. and (7) are nonhomogeneous.1+ao(k)Y0 = f(k).
Yk.
If all the coefficients an.2 + Yk = 0 Yk. together with the function f. . when the function f is identically zero.. .Y2k = 0
Yk Yk.
5yk. Eqs. . otherwise.l .2Yk = 2k -1 ... but Eqs.2
1
2+y2 + Yk.(k)Yk. .. 2.an. but Eqs.9.. 1.2 = 1 k"
Such equations are called nonlinear. For example.
In fact. where n is some positive integer and the coefficients an.. ..t .4Yk. an . . As in the case of ordinary differential equations.4 + yk = k
hand. (4).(1 .3 . are given functions of k defined for k = 0..
..
. Then
Nk+. 4. Equation (8) can be written in the
Nk+. (3) is of order 1. In this case. . Eq... in order to define the order of the
linear difference equation (2). and optics. this is the case in some insect populations.. 1 . (6) has no singular point and its order is 3. For example. 1.
APPLICATIONS 9.Nk = aNk... Eq.(k) and ao(k) are nonzero for all k = 0.
Biology
The Malthusian law of population growth (see the biology application of Section 1.350
9
Difference Equations
For reasons that will become clear in subsequent sections (see Remark 1 in Section 9.1. 3.
(9)
.. (5) has no singular points over the set of k-values 2. and Fibonacci sequences will be given in subsequent sections.. For example. As we have already seen in Chapter 7. . . of a power series solution y(x) = E%o a x" in terms of the coefficients ao and a. (5). We say that k = 1 is a singular point for Eq. But Eq.. if N(t) denotes the size of the population at time t. Eq..3). Eq. where one generation dies out before the next generation hatches.1) assumes that the rate of change of a population is proportional to the population present.
In this section we see how difference equations appear as mathematical models
describing realistic situations in biology.. from one generation to the next.. then N(t) = kN(t).1.(k) vanishes when k = 0. 3. For some populations. its order over this set is 1 . Eqs. which enabled us to compute the coefficients a a3.(1+a)Nk = 0. probability theory. Here N(t) is assumed to be a differentiable function of time. and Eq. Eq. . (2) is said to be of order n.2 and Remark 1 in Section 9.. (6). (5) and (6) may not have any singular points. (5) is not of order 1 since the coefficient ao(k) vanishes when k = 1.1
Difference equations appear as mathematical models in situations where the variable under study takes or is assumed to take only a discrete set of values. 2. Also. Also over the set of k-values 1 . where k is the constant of proportionality. (6) is not of order 3 because the leading coefficient a. we must make the assumption that the coefficients
a. The basic idea there is to approximate the differential equation by a difference equation. (7) is of order 4. however. We should remark here that over a set of k-values different from 0. k = 0 is a singular point. 2. electric circuits. That is. . is proportional to the size of the former generation. Eq. 2. . it is more realistic to assume that the size N of the population is a step function. A simple model for such a population will be to assume that the increase in size. (4) is of order 2. . Difference equations were also encountered in Chapter 5 as recurrence formulas. .. difference equations are useful in connection with the numerical solution of differential equations. Further applications on these topics and applications to business. For example. Let N denote the size of the population of the kth generation. For Eq.
form
(8)
where a is the constant of proportionality.
. will give the value of Vk for k = 1 .
(10)
which is the mathematical formulation of the assumption that the size of the population of the (k + 2)nd generation depends (linearly) on the population of the previous two generations. A ray PQ passing through L will be refracted by the lens in the direction QR. which can be found as follows: From the center 0 of the lens.
Figure 9..1
Consider the converging thin lens L shown in Figure 9.9. I = V/R. 2. together with the boundary conditions Vo equal to a given constant and V. we have
Ik.1 . 1 . ...2-13Vk. n satisfies a linear homogeneous difference equation of order 2 which. = 0. Applying this law at the junction point corresponding to the voltage Vk..2 + pNk. .. Each resistance in the horizontal branch is equal to R and in the vertical branches equal to 6R. In some cases it may be more realistic to assume that
Nk. .1
Introduction and Definitions
351
which is a linear homogeneous difference equation of order 1.2 +
Using Ohm's law.Vk. = 0.I . . + qNk = 0. the above equation can be replaced by
VI . Its solution is
(see Section 9. according to Kirchhoff's current law. . n-1. Assume that Va is a given Electric Circuits voltage.4)
Nk = NO(1 +a)k.
where No is the initial size of the population. 2.2.
Consider the electric circuit shown in Figure 9.1+6Vk=0.Vk. the sum of the currents flowing into a junction point is equal to the sum of the currents flowing away from the junction point. draw a line parallel
Optics: Rays Guided by Lenses
.2 + R
V11 .1. Equation (9) assumes that the size of the population depends on the population in the previous generation... We want to find the voltage V k for k = 1 .. In fact. We will show that the voltage V k for k = 0. and the shaded region indicates the ground where the voltage is zero. 2..1
R
= Vk.0
6R
(11)
Equation (11) can be simplified and reduced to
6Vk. n -1.l = 4. V.
352
9
Difference Equations
I
x
F
L
Figure 9. of the ray PQo through the system of lenses. The two lines meet at the point R. =
mk f Yk =
.4.mkf). of the line segment QkQk. Now consider a system of identical converging thin lenses. we want to find the y-coordinate yk of the point Qk..2 . Thus the coordinates of the point A are (f. it follows that the slope mk+. satisfies the linear difference equation
fYk.yk). as we are about to prove. draw a line parallel to the
lens. Since Qk has coordinates (O. through which the refracted ray passes..
(12)
where f is the focal length of each lens and d the distance between lenses.1 Yk + mk.-.3
.(2f-d)Yk. is that the displacement Yk of the ray. and its equation is y = mx. each of focal length f and spaced d units apart on the x-axis.3.. as it traverses the lenses. with respect to the optical axis x.) Then the line OA also has slope mk. Let us denote by mk the slope of the segment Q.Q2 .
(13)
QO
P
Qk
Qk + I
I Qn
ex
4+d+
f
Lo
Lk
Lk + l
Ln
Figure 9.l + fYk = 0. The result. is given by
mk. We want to study the path QQQ. (See Figure 9.. That is.Q.2
to the ray PQ. as shown in Figure 9. from the focal point F of the lens..
. the proof of the corresponding theorem for difference equations is almost trivial. Eq. y. of the solution
yk are uniquely determined.
In fact.. (3) and using the values of y1 and y2. . 2. we should continue by mathematical induction as follows: Assume that the terms yo.9.l + gkyk = rk.
yo = A. 2. On the other hand.. (3) becomes
Y2 + poy1 + g0yo = ro.
. (1) by a2(k) and setting
a1(k) a2(k)
. . Using the initial conditions (4) we find
Y2=ro-p0B-q0 A.l + gkYk = rk. 2.1.1 and the result follows.
f(k) k a2(k) . we find
y_2 = -P Y.. yl = B..
.2 + al(k)yk.. 1. For a rigorous proof that every term yk of the solution is uniquely determined.2 + PkYk.
.1 + aa(k)Yk = f(k)... from Eq.
The proof of the existence and uniqueness theorem for differential equations (see Appendix D) is so involved that it is almost never proved at the elementary level. and qk # 0 for all k = 0. .
ao(k)
a2(k) = qk. .
Hence the value of y2 is uniquely determined in terms of known quantities. has exactly one solution. 2.. 2. Since Eq. setting k = 1 in Eq..r
for k=0. We must prove that yn+2 is also uniquely determined. the coefficients a2(k) and ao(k) are nonzero for all k = 0.. . we find that y3 is uniquely determined. 1. A solution of Eq. and rk are given sequences of numbers. (3) with k = n. (1) is assumed to be of order 2..
where pk. B are given constants. y 1 . qk. . qk.. and so forth. Next. .
where Pk. 1. Dividing both sides of Eq.. a ao and the function f are given functions of k defined
for k = 0.
THEOREM 1 EXISTENCE AND UNIQUENESS
The initial value problem
Yk. and A... .
(2)
we get
Yk-2 + PkYk.
Proof
For k = 0.Pk. . 1.
.2 . . where the coefficients a2. (2) for
all k = 0. 1. (2) is a sequence yk which satisfies Eq. and rk are given sequences of numbers with qk + 0 for all k = 0.2
Existence and Uniqueness of Solutions
355
Consider the second-order linear difference equation
(1) a2(k)Yk.....
not both zero. .)
DEFINITION 1
Two sequences ak and bk. such that
Aak+Bbk=0
for
k=0. 3.
(11)
called the Fibonacci sequence. (See Definition 1 of Section 2. Every new pair of rabbits does exactly the same.=y2=1.358
9 Difference Equations
Fibonacci Sequence
In the sequence
1. For example.
(12) (13)
y.2.
9 .
.3.. k = 0. How many pairs will there be after k months if we begin with a pair of newborn rabbits?
17. k = 0..5. if yk denotes the kth term of (11)..
W+ (1+
L\
2
J
In Exercise 17..1. the sequences
ak = 3k +
12
5
and
bk = 5k + 4. have many fascinating properties and appear mysteriously in many diverse situations. 2.. + yk. are said to be linearly dependent if
there exist constants A and B.. 1.. Show that the unique solution of the IVP (12)-(13) is given by (14).8.. the reader is asked to show that the unique solution of the IVP (12)-(13) is given by
f1-2 51k1
(14)
Yk
)J
The numbers in the Fibonacci sequence. 1. They were discovered in the famous Fibonacci rabbit problem. 2... 2. Therefore. .3 LINEAR INDEPENDENCE AND THE GENERAL SOLUTION
The following definition of linear dependence and independence of sequences is similar to the corresponding definition for functions..2 = Yk.. The Fibonacci Rabbit Problem A pair of rabbits gives birth to a new pair of rabbits every month after the pair is two months old. each term after the second is the sum of the two preceding terms. then the Fibonacci sequence is the unique solution of the IVP
Yk.2. k = 1.
(1)
Otherwise the sequences are said to be linearly independent.2. Liber abaci (1202). called the Fibonacci numbers..'
16.
(2)
The rabbit problem first appeared in Leonardo "Fibonacci" da Pisa.13...1.
. not both zero. 2.. . k = 0.3
Linear Independence and the General Solution
359
are linearly dependent because (1) is satisfied with A = 5 and B = -3.bk) _
ak ak.9.. 1.
(4)
Setting k = 0 and k = 1 in (4). 2.2(2 3k)
= 2 k 3 k = 6' # 0
for all k = 0. k = 0. 1. the Casoratian plays for difference equations the role that the Wronskian plays for differential equations. 1.1
3k.
DEFINITION 2
The Casoratian of two sequences ak and bk. .. is denoted by
C(ak.
.
3'
C(2k. This contradicts the hypothesis and establishes our claim that the two sequences in (3) are linearly independent. 3k) = 2k. Before we present a simple test for linear dependence and independence of solutions of linear homogeneous difference equations. 2.bk) and is defined to be the determinant
C(ak..(5k + 4)
\\\\\\13k+ 5?/
=+3k+ S )(5k
=0
and
j2k
5
for all k = 0.
//
\\
3k+
5
12
5k +4
12
3(k+1)+
5(k +1)+41
9). . which means there exist constants A and B. On the other hand.
A2k + B31 = 0
for
k = 0. the sequences
ak = 2k
and
bk = 3k... That is.' bk bk
(6)
As we will see. 2. .
2A + 3B = 0. Otherwise.. 1. 2. 1. they are linearly dependent...2`3k = 3(2k3k) .
(3)
are linearly independent. we find
A+B=0
and
(5)
But the only solution of system (5) is A = B = 0.. such that (1) holds. 1
= 2k3`' . We have
C{3k+ 5 5k+41
.. Let us compute the Casoratians of the sequences in (2) and (3). we need the following definition..
0. yk')
0
if k=0 if k#0
k = 1." + By. ..
Otherwise the Casoratian is identically zero.
Consider the homogeneous system
Ayo" + Byo) = 0
Ay.. 0.
and
Yk2' : 0.Yk2)) # 0
for
k = 0. 2. B). . 1. Their Casoratian is
11
C(yk ). 0.3
Linear Independence and the General Solution
361
considered over the set of k-values 0. 15. 1. by (11).. are linearly independent if and only if their
Casoratian
(2) CUM +Yk) _
-
A
Ykl. 0. that the system has a nontrivial solution (A.. . 1. on the basis of Definition 1. yk" and y(k2). of the linear homogeneous difference equation
Yk.1
is different from zero.. 1.
. .2.
On the other hand..
." and yk2) are linearly independent.
Yo
Y(V)
Yo
y(2)
= 0. 3. 2. 2. from Appendix A.2) = 0
(12)
(13)
Since.1
Yk2'
Yk2.. 2. and..3. it follows. . 6. although
Yk2'=2yk`)
for
THEOREM 2 Two solutions. ..
are solutions.
Proof Assume that the solutions y. the determinant of the coefficients of the system is zero.. 0. In fact. We
should prove that
C(y(k`).. in particular.
(10)
where qk # 0 for k = 0.9. . 0. the solutions ykl) and yk2) are linearly independent. 1... 2. . 30.2 + Pk Yk«1 + qk Yk = 0.
yk 1
:
1.
This contradicts our assumption and the proof is complete.. Yk2)) # 0. The case B * 0 is treated in a similar fashion. Otherwise they are linearly dependent which.2 .362
9
Difference Equations
Multiplying both sides of (12) by q0 and both sides of (13) by po and adding the results. we have
y( I) = . 1.
Since A and B are not both zero. As in the case of differential equations. This contradiction proves the claim that
C(Yk'>. using the fact that yk'> and yk2> are solutions of (10).
(16)
Assume A * 0.Yk `2)) = yo> cn ..
(15)
One can show by induction that
Aykl) +Byk2) = 0
for
k = 0. See Exercises 9 and 10. such that
Ayk'> + Byk2> = 0
for
k = 0.
Conversely.2> = 0.
yk'>
B A Yk2>>
and so
C(yu> .
. The proofs are straightforward. by Definition 1.
(14)
Now.. 21 . it follows that the solutions y(I) and yk2 are linearly dependent. means there are numbers A and B. we find
go(AYo ) + Byo2>) + Po(AYi'> + BY(2)) = 0
or
A(Po Yi'> + qo Yo >) + B(P0 y.
We must prove that yp"and yk2> are linearly independent...y. not both zero. Then. from
(16).. 1.(Po Yi'> + qo )Q0o'>)
and
22> = -(po yti2) + qo y( o')). we have the following theorem for the general solution of a linear homogeneous and a linear nonhomogeneous difference equation of order 2.
and (14) becomes
Ay(I) + By.. yk2)) # 0.2>ycu
k
k Yk-l
0.2> + qo yo >) = 0. assume that
C(Yk".
the
general solution of Eq.366
9
Difference Equations
9. .2-8Yk. Here the idea is to look for a solution of the form
Yk =
.
. They are linearly independent because their Casoratian
3'
3k
5k
5k"
is never zero for k = 0.
EXAMPLE 2
Find the general solution of the difference equation
3Yk.3. Substituting (2) into Eq. We present the main features of the method by means of examples..1
.1 + 15yk=0.4 HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS
In this section we show how to find the general solution of a linear homogeneous difference equation with constant coefficients.
Solution
(1)
With differential equations. are Jk. (1).. 1. Hence.15\k = 0. .
Ik2-8X+ 15=0. = 3 and A2 = 5.2Yk = 0
Solution Characteristic equation: 3X2 . called the characteristic roots.. Hence
Ykl) = 3k and
yk2) = 5k
are two solutions of Eq. The roots of the characteristic equation (3).2 = 0. and c2 are arbitrary constants.5). (1) is
Yk = c13k + c25k.k..5Yk.
where c. we looked for a solution of the form y(x) = e'.
(3)
Equation (3) is called the characteristic equation of the difference equation (1).
EXAMPLE I
Find the general solution of the linear homogeneous equation
Yk. 2. . (1). by Theorem 3 of Section 9. we find
Xk+2
-
8Xk.2 .
(2)
where X # 0.
and dividing by lkk.
.l+9yk=0. Eq. in addition to (5).4 Homogeneous Equations with Constant Coefficients
367
Characteristic roots:
5_
25+245±7
6
6
2
={-1/3'
General solution: yk = c.(.
Hence. a. we find
x2-6k+9=0. the product kyk'> gives always another solution which.
Characteristic roots: K _ -1 ± i. but it does not have two linearly independent solutions of the form Kk. in the case of equal roots. However.)
Yk = c. = 1\2 = 3. (4) is
EXAMPLE 4
Find the general solution of the equation
Yk_2 + 2yk.1 . and so Eq.
REMARK 1
(8)
We will show here how to write the general solution (8) of Eq.
But here the characteristic roots are equal. yk2)
(6)
are linearly independent.
(9)
.1 + i)k + C2(.
that K must be a root of the characteristic equation
(4)
Solution Looking for a solution of the form yk = K".
(7)
Solution Characteristic equation: K2 + 2K + 2 = 0. together with yk". the general solution of Eq. consider the difference equation
a2 Yk-2
+ alYk l + a0 Yk = 0.9. as in Example 1. are linearly independent. (4) has the solution
Yk2' = k3k.3* + c2k3k. (Show it.i)k.
and yk'). (7) as a linear combination of two real and linearly independent solutions.2k + c21 -
EXAMPLE 3
Find the general solution of
Yk. That is. (4) has the solution y°' = 3k (5) (and all its multiples).2-6yk. In general.
General solution:
Yk = C. + 2yk = 0.
ib=re °
Thus.
Taking complex conjugates. for
economy in notation. (12) and (13).388
9
Difhrence Equations
with real coefficients. Then they must be complex conjugate. (9).i sin k9)
= rk(c. -ice) sin k9
= C. + c2 and C.
. using Euler's identity. you are asked to show directly that the sequences
yk') = rk cos k9
and
Y") = rk sin k9
(14)
are two linearly independent solutions of Eq.
(12)
and the argument 9 satisfies the equations
cos 8 = a.
where C. That is. That is. In polar form.rk cos k8 + c2rk sin k9. with
r
-r<9sn. in the form
a + ib = r(cos 9 + i sin 9). where a and b are real numbers. In summary.rkeik° +
c2rke-k°
= c.+c2) cos ke + rk(ic.
-1 + i = r(cos 0 + i sin 0). we also have
a .
r
sin 9 = b. = c. (7). and assume that the roots of the characteristic equation
a2X2 + a.
where r and 9 are determined by Eqs. Now the crucial step is to write the complex number a + ib in complex polar form.k + ao = 0
(10)
are complex numbers. the roots of Eq. we find
a + ib = reB.ice are arbitrary constants which. we can still denote by c. (9) is Yk = c.
(13)
From (11). the general solution of Eq.(reiB)k + C2(re-ie)k
= c. the general solution of the difference equation (9) is
Yk = c.
where the modulus r of a + ib is given by
(11)
r=
a2 -+b'.
Let us apply the above results to Eq. In Exercise 38. and c2. (10) are of the form a ± ib.(a+ib)k + C2(a-ib)k = c. The characteristic roots are -1 ± i.rk cos k9 + C2rk sin k9.rk(cos kO + i sin k9) + c2rk(cos k9 . . = ic.
3).. and y.3 with C.. (12). Find the voltage Vk for k = 1 . However. 1..1. q * 0
to approach zero as k --' x.. Given that yk = I is a solution of the equation
(2k + 1)Yk-2 .2. we can use a transformation to obtain an equation of lower order.. 2. 1. .000 students..1. what will the enrollment be after 10 years?
44.1. if we know a nontrivial solution of the homogeneous equation.]
47.. 2.
46. n . = 2.
Use Eq.
find the general solution.n . [Hint:
Uk-.372
9 Difference Equations
are necessary and sufficient for all solutions of the difference equation
Yk-2 + PYk-1 + qyk = 0. But there exists no general method for solving linear equations with variable coefficients of order higher than one. where qk * 0 for k = 0.. .Yk)
XkXk-..
. In the electric circuit application (see Section 9.
-Y=
Xk
C(Xk.1..4(k + 1)yk-. Show that the transformation
Yk = Xkuk
produces a first-order linear difference equation for uk. 2.1). y. . In the optics application [see Eq. Reduction of Order First-order linear difference equations with variable coefficients can always be solved explicitly (see Exercise 20. Assume that Xk is a solution and xk * 0 for k = 0.
42. If the initial size of the colony is 1000. .1) are all equal to R. Section 9. assume that d = 4f.
43. .1].. (9) of Section 9.J'k-. Section 9. Assume that the resistances in the electric circuit application (see Section
9.Uk =
Yk-1
Xk-. If the college currently has 20. Investigate the displacement yk of the ray as it traverses the lenses. = 1.. find the voltage Vk for
k= 1. what will its size be after 12 hours?
45. + gkyk = 0. It has been estimated that the student enrollment in a college undergoes a 5 percent relative decrease in size per year. as with differential equations. = 1.1.
41. A certain colony of bacteria undergoes a 5 percent relative increase in size per hour... Consider the second-order linear equation
yk-2 + P. + (2k + 3)yk = 0.
cos yx. and yk is a particular solution of the nonhomogeneous equation (1). As in the case of differential equations.5. a and ap of Eq. pk. there are two methods which can be used to find a particular solution of Eq. where a is a positive integer or zero. where y is a nonzero constant. k°.1. cos yk.5. one can find the form of a particular solution of a difference equation
. 5. sin 8k.5 Nonhomogeneous Equations with Constant Coefficients
373
9. That is. cos yk. where S is a nonzero constant. (1).
The sequences k°. From Theorem 4 of Section 9. A (finite) product of two or more sequences of types 1-4. The method of undetermined coefficients requires that the coefficients a2. (1) is the sum of the general solution of Eq.
(3)
where yk is the general solution of Eq. es`. and to equations
with variable coefficients only if we know the general solution of the corresponding homogeneous equation. 4. sin 8x listed under types 1-4 in Section 2. In Section 9.5 NONHOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS
Consider the linear nonhomogeneous difference equation with constant coefficients
a2Yk. however. (2) and a particular solution of Eq. (1) be constants and that the nonhomogeneous term rk be of a special form. yk is the general solution of the homogeneous equation (2).2 + a1yk. where 3 is a nonzero constant. 3. sin 8k correspond to the functions x. Hence Y. It should be remarked.
yk = Yk + YPk . the method of undetermined coefficients usually requires less labor. will be completely known if we show how to obtain a particular solution of Eq. 2. (1). (1).) On the other hand. (1) is a linear combination of sequences of the following types:
1.4 we explained how to obtain yk. (See Section 9.
(2)
where a2ao # 0.1
UNDETERMINED COEFFICIENTS
This method applies successfully when the nonhomogeneous term rk in Eq. we know that the general
solution of Eq. the method of variation of parameters imposes no restriction on rk. that when
we have a choice.11. With this correspondence in mind.3.l + a0yk = 0. it applies to equations with constant coefficients always. + aOYk = rk
(1)
and the corresponding homogeneous equation
a2yk. pk.
9..9.2 + alyk. (1): the method of undetermined coefficients and the method of variation of parameters.
(5).5
3k.
Characteristic equation: K2 .8Yk. (5) since the solution of Eq.35
812A2k+3B(k+1)1 + 15(A21 +Bk3k) = 3 25 . The rule is to multiply by k".8yk.
EXAMPLE I
Find a particular solution of the nonhomogeneous difference
Yk. (4).5 3k
11
3A2k .5 3'
3A = 3
and and
-6B = -5
B=5/6. where n is the smallest positive integer such that k"31 is not a solution of Eq. (4) and equating coefficients of similar terms.8K + 15 = 0.2k-5.
Characteristic roots: K.11 for a particular solution of a differential equation.11. = 3. We now have
[A22+B(k+2)35-2]_ 81A2 k'`+B(k+1)35"1+ 15(A2k+Bk3k) = 3.. = c.11 and treating terms of the form ¢k much in the way we treated ea` in Section 2.
Yk = A21 + Bk3k
(9)
is a particular solution of Eq. Using notation similar to that of Section 2. The undetermined coefficients A and B will be determined by substituting (9) into Eq. first we find the general solution of the corresponding homogeneous equation Yk-2 . Therefore. (5). + 15yk = 3 2k .
equation
(4)
Solution
As with differential equations. we have
3
2k*{2k}
3k -> {3k} -a {k
35}.I + 15yk = 0.
-5
(7) (8)
where in (8) we multiplied 3k by k because 3k is a solution of Eq.2 .
A=1
.3' + c25k. (5) can influence the form of a particular solution of Eq. (4).374
9
Difference Equations
in a manner similar to that used in Section 2.
(6)
Here rk is a linear combination of two sequences of type 2.6B3k = 3 25 . Homogeneous solution: yk. 752 = 5.
(4).
.
EXAMPLE 3
(11)
Substituting (11) into Eq.5 Nonhomogsnsous Equations with Constant Coef dents
375
Hence.2 . we
Find a particular solution of the equation
4Yk.
Yk = 2k + 5 k3k
is a particular solution of Eq.
(10)
Homogeneous solution yk = c.
(i)"
Treating cos (krr/2) in the way cos (x(TT/2) was treated in Section 2.
Substituting (13) into Eq.
(12)
Homogeneous solution: y= c.11. (We multiplied by k because 1 is a solution of the corresponding homogeneous equation. k}. y4=2k3-6k2+4k. (10) and equating coefficients of similar terms. we find A = . k.5.) Hence. and C = 4.
Yk = . As in Example 1. B = 0. sin 7r 1. Therefore. B = -6. k2. 1) --* {k. the rule is to multiply all terms in the bracket by the lowest positive integral power of k so that no term in the last bracket is a solution of the homogeneous equation.5 cos 2 .or higher-order difference equations are found in a similar fashion.
find A = 2. Treating k2 in the same way x2 was treated in Section 2.Yk = 12k2.
3
(13)
kit
Particular solutions of lower.
Hence. Therefore.
EXAMPLE 2
Find a particular solution of
Yk. 1 + c2( -1)k.
+ c221 .
Solution
yk=Ak'+Bk2+Ck.11.
yk=Acosk2 +Bsink2. we have
3 cos
(-
--
cos 2r.9. we have 12k' {k2.2-Yk=3cos
Solution
kir
2. (12).
as we just proved. Dividing both sides of Eq. we find 1 (n + 1)2
1
k2
k-.
where c. (30) is
Yk=Y2+Yk=c.08) 500 = $40 to the account. (29) is
Yk = c.1. Thus. and c2 are arbitrary constants.
yk =
=o
(2n + 3)
1
1
2n+3)
(n + 1)2
-o(
k2 . we find that
rk=2k+3. Next we will find a particular solution
of Eq.(n + 1)2
(n +2)'-(n + 1)2
_
-a
(n + 2)2 [k2-(n+1)2]=k. the unique solution of the IVP (30)-(31) is
Yk=2+
k2
4k3 -3k2-k
6
4k3 -k
6
Business:
9. the general solution of Eq.2..+c2k2+4k'-3k2-k
6
Using the initial conditions (30). (29) because they are solutions. using formula (27).
Now. At the end of the first year the bank will add (0.P b l ems m compound interest often lead to simple difference equations..
1 + c2k2. = 0 and c2 = 1/2. (30). We must use variation of
parameters because the coefficients of the equation are variables. (30) by the leading coefficient (2k + 1). (29) and (30). at the end of the second year
._k(k+1)(2k+1)=4k'-3k2-k
6 6
where we used the fact that k-'
-o
k(k + 1)(2k + 1)
6
Hence.5. The two sequences in (28) are linearly independent solutions of Eq... to Fmance
First let us recall the meaning of simple and compound interest. and their Casoratian
is
I
1
k'
(k+l)z1=(k+1)2-k2=2k+1#0
for
k=0. we find c.
Thus. Suppose we invest $500 principal at 8 percent simple interest.380
9
Difference Equations
Eqs.3 APPLICATIONS
Mathematics of . Here we have no choice of method. the general solution of Eq.
after 2 years the account will be
worth $585.02)8 = 500(1.. denotes the value of the account at the end of the kth period.02)8 = $585. k = 4 2 = 8 periods. we find
A8 = 500(1 + 0. bringing the total to 510 + 10. and so forth. Continuing this process.9. and iAk is the interest during the next period. .
or
Ak_.02) (510) = $10.5 Nonhomogeneous Equations with Constant Coefficients
381
it will add another $40. On the other hand.
The solution of this IVP is given by (32).02) 500 = $10 to
the account. or
annuities. together with the interest.. the bank will add (0. k = 0.20. with interest compounded quar-
terly. at the end of the second quarter the bank will add (0. What will the amount be after 2 years?
Solution P = 500. = Ak + iAk. This interest is now compounded with the principal and from now
on will also earn interest.
EXAMPLE 7 I invested $500 at 8 percent annual interest compounded quarterly. Interest rates are usually quoted as annual rates.
Proof
(32)
Since A.(1+i)Ak = 0. at the interest rate of i per period. bringing the total to $580.
Theorem 3 gives the formula for periodic payments of a fixed amount.02. from (32).
.
Theorem 2 is a basic model for compound interest. we earn interest only from the principal investment of $500. . Hence.82. it follows that
Ak_. Show that the value A4 of the account at the end of the kth period will be
Ak = P(1+i)k. That is.20 to the account. 2. i = 0. so 8 percent interest compounded quarterly really means 2 percent per quarter. that is.
THEOREM 2
Assume that the amount P is deposited in a savings account and. 1. (See Example
7). is kept there for k time periods. compared to $580 in the case of simple interest. suppose we invest $500 at 8 percent compound interest.82. At the end of the first quarter (each quarter is 3 months). Hence.
Also
A = P.20 = $520. and the interest is 4 = 2 percent
per quarter.
q #-0. xk+I = 4xk. find the general solution of the system by the method of elimination (see Exercise 56). xk+.
57.(all + a22)xk+l + (a11a22 .
In Exercises 57 through 60.2k
7Yk + 1
Yk+l = 5Xk . Linear Systems with Constant Coefficients: Method of Elimination systems with constant coefficients of the form
xk+l = allxk + a12Yk + rk
Yk+1 = a21xk + a22Yk + tk
can be solved by the method of elimination.
1 -q>0.2yk '.Yk . Write the first equation in the
form
xk+2 = allxkfl + al2Yk+l + rk+I+
then eliminate y by using the second equation and then the first (exactly as in the case of differential equations in Section 3.
.1
61.
Linear
56. xk+1 = -3xk + 4yk
Yk+1 = -2Xk + 3Yk
59. = 3xk .a22rk + rk+1
After we find xk we compute yk from the first or second equation.
58.(k
(k+2)3-k'
+ 1)' k'Yk+1
+(k+2)'-(k+1)3
(k + 1)' .2).k'
Yk _ -
k+2)'-(k+1)'. The equilibrium value E is stable if every solution of the equation approaches E as k-k -.Yk
Yk+1 = 2xk + Yk
xk+l =5Xk-6yk+ 1
Yk+l .
if yk = E is a solution. Show that. Show that the value r
E_
I+P+q
and
is a stable equilibrium of the above difference equation if and only if the
following conditions are satisfied:
q-p+1>0.6Xk
-
60. we obtain the second-order linear equation with constant coefficients
Xk+2 .q+p+1>0. after we eliminate y.386
9
Difference Equations
Then use the method of variation of parameters to find a particular solution of the equation
Yk+2 . Equilibrium and Stability A constant E is called an equilibrium value for the difference equation Yk+2 + PYk+I + qyk = r.al2a2l)xk al2tk .
Yo= -2
16. assume that each resistance in the horizontal branch is equal to 1 fl and each resistance in the vertical branch is equal to 211. find the voltage V5. solve the IVP. the winner wins one dollar. Y1 = 0. There are no ties. That is.I = -6xk . Given that yk = 1 is a solution of the homogeneous equation
4(k+l)
Yk+2 --
2k+3 2k+1 Yk+' + 2k+1Yk = ' 2k+3
2k+
solve the IVP
4(k+1)
Yk+2 . where p + q = 1. what will its size be after 5
days?
17. Probability: The Gambler's Ruin Two people. Principle of Superposition Show that if yk11 is a solution of
Yk+2 + PkYk+1 + gkyk . Xk+1
= 5Xk + 6yk + I
3 3
Yk. play a game (cards. tennis. and the outcomes of the successive plays are
.
In Exercises 14 and 15. On each play of the game.
13. Use the principle of superposition (see Exercise 11) to find a particular solution of the equation
Yk+2 .1).rk1
and yk21 is a solution of
Yk+2 + PkYk+l + gkYk =
rk21.4Yk
Yk+l = 2rk + 3Yk
15. and the probability that B wins
is q.7Yk -1
Xo = 1.9
Difference Equations
10. chess) in which their skills or chances of winning are rated as p to q. If the initial size of the colony is 1000.
18.2k+1 Yo = 1.
Xk+1 = -3Xk . A colony of bacteria undergoes a 10 percent relative decrease in size per day. A and B.Y2=2
11.4 ' Y.y1 =1. If V0 = 10 volts and
V10 = 0 volts.1. Yk+i + Yk = 3
yo=5. YO = -1
XO=z.1 (see also Figure 9.8Yk+1 + 15yk = 3 ' 2k . the probability that A wins is p.
then aykl" + by12j is a solution of
J
Yk+2 + PkYk+1 + gkYk = ark) + br'kl
12. In the electric circuit application of Section 9.
14.
yl + Y3 +
+ y2k..
'This is Exercise 2. 246. Aseltine.4-2 in J. Y1 + Y2 + . Transform Method in Linear System Analysis (New York:
McGraw-Hill Book Co. New Jersey. 1979). and if they continue to play until one of them is ruined.700 at 9 percent compounded monthly. Seely and A.D. What is the monthly payment?
In Exercises 20 through 23. Mary has a 25-year mortgage on her house. + (Yk+1 + 1 23.
. 83. plot the ray for the distance of five lenses.5 Nonhomoganaous Equations with Constant Coafflciants
389
independent.` assume that all resistances except R( on the end are of the same value R.1 = Y2k
22 Y2 +Y4+
+Y2k-Y2k+1 . Inc.J. p. Poularikas.
'See Section 16-1 in J. Inc. p.
'This is Exercise 11-9.A.. yI . Find the necessary value of r. An optical system2 of identical converging thin lenses is constructed with d = 5 cm and f = 16 cm.2 in S.. Suppose an investor' wishes to double her money in six years by loaning it out at a rate of r compounded yearly. p. If A starts with a dollars and B with b dollars.Y2 + Y3 . If the ray enters the first lens at height 2 cm and at an angle of 352.
20. 1973). Englewood Cliffs..: Prentice-Hall.Y4 + 24. Reprinted by permission of Prentice-Hall. Cadzow.9.1
1`k+l
Yk = . + Yk = Yk+2 . Show that the current i in the nth loop satisfies the difference equation
and solve for i.Y. Electromagnetics: Classical and Modern Theory and Applications (New York: Marcel Dekker.I
21. She borrowed $25. N. 1958). N. Discrete-Time Systems: An Introduction with Interdisciplinary Applications (Englewood Cliffs. Used with the permission of McGraw-Hill Book Company. The borrower is to pay the loan
back in one lump payment at the completion of s4 years.A.
(_ 1)k+l
25..
26. In the network shown in the figure. what is the probability that A
will ruin B by winning all of B's money?
19. 533. Reprinted by permission of Marcel Dekker. verify the summation property of Fibonacci numbers.
2
PERIODICITY AND ORTHOGONALITY OF SINES AND COSINES
In this section we study the periodic character and orthogonality properties of the functions
cos n j x
and
sin n x
for
n = 1. for n = 1.CHAPTER 10
Fourier Series
10. . Fourier discovered an ingenious method for computing the coefficients a and b of (2) and made systematic use of such series in connection with his work on heat conduction in 1807 and 1811.2. they were considered to be nonsense by many of his contemporaries. Series like the ones which appear in the right-hand sides of (1) and (2) are called trigonometric series or Fourier series in honor of the French -scientist J. l > 0. in such a way that a given function f can be expressed as a trigonometric series of the form
f(x) _
(1)
This. we still have to show how to choose constants b. and the more general problem of expressing a given function f as a series of the form
f(x) =
+
(2)
will be the subject matter of this chapter. .. the theory of Fourier series has been developed on a rigorous basis and has become an indispensable tool in many areas of scientific work. thus leading him to the false conclusion that any arbitrary function can be expressed as a series of the form (2).
.
2
10. Consequently. This is not true. . when Fourier published his results. B. 3..1. . 2. as we will see in Section 10.4.1
INTRODUCTION
In connection with the solution of the heat equation in Section 6. Fourier. 3. However. . Fourier's work lacked mathematical rigor. 2.
(1)
which are the building blocks of Fourier series. Since then.
are
periods of sin x and cos x. 47r.10..
which means that 2T.. .. 61T. and the motion of a pendulum are examples of periodic functions.. Periodic functions have many periods. sound waves.. The Smallest positive number T for which Eq. The functions sin x and cos x are simple examples of periodic
functions with period 27r. p > 0. .. and in general (see Theorem 1) the fundamental period of each function in (1) is 21/n.
where 1 is a positive number.. is called the fundamental period of f For example.2
Periodicity and Orthogonality of Sines and Cosines
391
Let us recall that a function f is called periodic with period T > 0 if for all x in the domain of the function
f(x+T) = f(x)
(2)
Geometrically.
.1
Periodic functions appear in a variety of real life situations. 3. are also periods of f. this means that the graph off repeats itself in successive intervals
of length T. For example. if it exists. the fundamental period of sin x and cos x is 2a. More generally.. no matter how small. are periodic with fundamental period T = 21/n.
Y
i4TiT--o
£--c
0
l1
1 23
-1 0
Figure 10. 3T.and
sin .1. Alternating currents. 2.1(a) has period 2 and that in Figure 10. A constant function is also periodic with period any positive number. a constant function has no fundamental period.. it follows from (2) that if f has period T. then
f(x) = f(x+ 7) = f(x+2T) = f(x+3T) =
. . This is because any positive number. (2) holds.
n = 1. are periodic with fundamental period 27r/p.
THEOREM 1
The functions cos px and sin px. On the other hand. In particular. the functions nax nTrx
cos .. 27r.. is a period.1(b) has period 1. The function in Figure 10. the vibrations of a spring. Other examples of periodic functions are the functions in (1) with period 21 (see Remark 1) and the functions shown graphically in Figure 10.
The proof for sin px is similar. More precisely. 2.. Assume that T is a period of f(x) = cos px.x s 1. after expanding the cosine
of the sum px + pT. Then the statement f(x+ T) = f(x) for all x is equivalent to cos (px+pT) = cos px or.x s 1.
THEOREM 2
The functions
cosnIX
and
sinnx
for
n = 1. 2. each of the functions listed in (1) is periodic with
fundamental period 27r/(n7r/1) = 21/n.2...
(3)
But (3) is true for all x if and only if cos pT = 1 and sin pT = 0.
J
cos n j x cos k l x dx
= 111
sinnrxsinklxdx={0
if if
n# k. with n = 1. Next we turn to the question of orthogonality of the functions in (1). for all x. This means that any two distinct
functions from (1) are orthogonal in -1 <. we have
Theorem 2. Thus T = 2n7r/p. n = 1 . (For a review of odd and even functions. see Section 10.. Recall the definition of orthogonal functions given in Section 6.392
10
Fourier Series
Proof We give the proof for the function cos px. cos px cos pT .
n. (recall that p and T are positive). Two functions f and g defined and continuous in an interval a < x S b are said to be orthogonal on
REMARK 1
a<-x!5 bif
Jbf(x)g(x)dx
= 0. That is.. pT = 2mr.. . .5. 3.1>0
satisfy the following orthogonality properties in the interval -1 <.)
.
The following result establishes the fact that the functions listed in (1) are
mutually orthogonal in the interval -1 S x S 1. and so the fundamental period (the least positive) of cos px is 27r/p. it follows that each one has also period 21. .2.
J
r cos
n!
xx sin
k l x dx = 0
for all
Proof We can (and do) immediately establish (6) by using the fact that the integrand is an odd function and so its integral over the interval -1 S x S l
(which is symmetric with respect to the origin) is zero. Clearly.
Since the functions in (1) have (fundamental) period 21/n..3.
n
(4)
(5) (6)
J
1
#k.. k.sin px sin pT = cos px.
10.2
Periodicity and Orthogonality of Sines and Cosines
Next we prove (4). For n = k we have, using the identity cos'x
I + cos 2x
2
-cos-j-dx=Jrl(cosnlxl2
JJ cos-
ax
2n1 xl
= I r' zl 1 + cos
dx =
IX
+
I
sin
2nI xl
Now for n * k we find, using the identity
cos x
cos y = Z [cos (x+y) + cos (x-y)),
I
/ cos
n 1 x cos
k x dx = J
(n+k)irrx
1
cos
(n + k)rrx
I
+'(n-k)arsin
+ cos (n - j )mix] dx
J
_
,
l
(n-k),rr
I
I '(n+k)Trsin
0.
The The proof of (5) is similar to that of (4). In the case n = k we need the identity sin' x = (1 - cos 2x) /2, and in the case n * k we need the identity
sin x sin y = 2 [cos (x -y) - cos (x +y)).
17. Assume that the functions f and g are defined for all x and that they are periodic with common period T. Show that for any constants a and b, the functions of + bg and fg are also periodic with period T.
18. Assume that the function f is defined for all x and is periodic with period
T. Show that if f is integrable in the interval 0 <- x s T, then for any
constant c,
1 T f (x ) dx =
0
I T f (x) dx.
J
19. Find a necessary and sufficient condition for all solutions of the differential equation y" + py = 0, with p constant to be periodic. What is the period?
20. Find a necessary and sufficient condition for all solutions of the system with constant coefficients
x= ax + by
ycx+dy
to be periodic.
1 0.3 FOURIER SERIES
Let us begin by assuming that a given function f, defined in the interval
10.3
Fourier Series
395
-1 <- x s 1 and outside of this interval by f(x + 21) = f(x), so that f has period 21, can be expressed as a trigonometric series of the form
11
and sin (nirx/1) for n = 1, 2, 3.... in the interval -I s x <_ 1. (i) To compute the coefficients a. for n = 1 , 2, 3, ... , multiply both sides
of (1) by cos (kirx/1), with k a positive integer, then integrate from -I to /. For the moment we assume that the integrals exist and that it is legal to integrate the series term by term. Then, using (4) and (6) from Section 10.2, we find
1'
That is, ap is twice the average value of the function f over the interval -1 <_ x s 1. Note that the value of ao can be obtained from formula (2) for n = 0. Of course, if the constant ao in (1) were not divided by 2, we would need a separate formula for a.. As it is, all a are given by a single formula, namely,
a, _ !
f (x) cos
nn x
dx,
n = 0, 1, 2, ....
(4)
396
10
Ssries
(iii) Finally, to compute b, for n = 1, 2, 3, ... , multiply both sides of (1)
by sin (k7rx/1), with k a positive integer; then integrate from -I to 1. Using (6) and (5) from Section 10.2, we find
r'
f(x) sin
J r
krrx
1
dx = 2
a0
'
sin
k7rx
1 dx
0
+
[a. J cos
'
nirx
sin
kirx
dx + b
r`
I
JJJ
sin
nrrx
sin
kirx
dx
bkl.
0
0 if n#k
I if n = k
Thus, replacing k by n, we find
b,, = 1
f (x) sin n j x dx,
n = 1, 2, 3.... .
(5)
When the coefficients a and b, of (1) are given by the formulas (4) and (5) above, then the right-hand side of (1) is called the Fourier series of the function f over the interval of definition of the function. The formulas (4) and (5) are known as the Euler-Fourier formulas, and the numbers a, and b are called the Fourier coefficients of f. We will write
f (X) - 2 +
a
cos n
x + b. sin n xl
(6)
to indicate that the right-hand side of (6) is the Fourier series of the function
I.
Before we present any examples, the following remarks are in order. So far we proved that, if the right-hand side of (1) converges and
REMARK 1
has sum f(x), if f is integrable in the interval -l s x s 1, and if the term by
term integrations could be justified, then the coefficients a. and b in (1) must
be given by the formulas (4) and (5) respectively. On the other hand, if a
function f is given and if we formally write down its Fourier series, there is no guarantee that the series converges. Even if the Fourier series converges, there is no guarantee that its sum is equal to f(x). The convergence of the Fourier series and how its sum is related to f(x) will be investigated in the next section.
REMARK 2 To compute the Fourier coefficients a and b,,, we only need the values of f in the interval -I <- x s 1 and the assumption that f is integrable there. It is a fact, however, that an integral is not affected by changing the
values of the integrand at a finite number of points. In particular we can compute
the Fourier coefficients a, and b, if f is integrable in -1 s x < 1, although the function may not be defined, or may be discontinuous at a finite number of
10.3
Fourier Seri"
397
points in that interval. Of course the interval does not have to be closed; it may be open or closed at one end and open at the other.
REMARK 3 When the series in (1) converges for all x, its sum must be a periodic function of period 21. This is because every term of the series is periodic with
period 21. For this reason Fourier series is an indispensable tool for the study of periodic phenomena. Assume that a function f is not periodic and is only defined in the interval -1 -5 x < l (or -I < x <- 1, or -1 < x < 1). We can write its Fourier series in -1 s x < 1. We also have the choice of extending f outside of this interval as a periodic function with period 21. The periodic extension of ff. F, agrees with f in the interval -1 s x < 1. Therefore, a function f, defined in -1 s x < 1, and its periodic extension, which is defined for all x, have identical Fourier series. (See Example 4.) Finally we should mention that if f is defined
in a closed interval -1 s x s / and if f(-1) # f(l), then f cannot be extended periodically. In such a case we can either ignore (as we do in this book) or
modify the values off at ±1 and proceed with the periodic extension.
REMARK 4 When f is periodic with period 21, the Fourier coefficients off can be determined from formulas (4) and (5) or, equivalently, from
a=
and
.v
n = 0, 1, 2, ...
(4)
c+Z
b _ 1 fc
f(x) sin n x dx,
j
n = 1, 2, 3, ... ,
(5')
where c is any real number. This follows immediately from Exercise 18 of Section
Assume that a function f is defined in the interval -1 s x < 1 and outside this interval by f(x + 21) = f(x), so that f has period 21. In Section 10.3 we defined the Fourier series of f,
f(x) - 2 +
formulas
(a,, cos
n!
xx + b,, sin
n xl
(1)
where the Fourier coefficients a and b of f are given by the Euler-Fourier
f(x)cosn,xdx,n=0,1,2....
and
(2)
1('
nnx
1 dx,n=1,2,3,....
(3)
When Fourier announced his famous theorem to the Paris Academy in 1807, he claimed that any function f could be represented by a series of the form
f(x) = 5 +
.
a cosnjx + b. sin nx)
,
(4)
DEFINITION 1
A function f is said to be piecewise continuous on an interval I if I can be
subdivided into a finite number of subintervals.
f(c . in each of which f is continuous and has finite left.c from the left. and to the
average (f(x -) + f(x + )1/2 of the left. In this section we state conditions which are sufficient to insure that the Fourier series converges lot all x and furthermore that the sum of the series is equal to the value f(x) at each point where f is continuous. As we will see in Theorem 1.2. Clearly. nevertheless. W. then
f (x) if x is point of continuity of f
i x is point of if discontinuity
nax+b. These conditions. generally satisfied in practice. The notation f(c -) denotes the limit of f(x) as x ..:
. for example.and right-hand limits at each point x where f is discontinuous.c from the right. necessary and sufficient conditions for (4) to hold have not been discovered.402
10
Fourier Series
where the coefficients a and b are given by (2) and (3). Such discontinuities (where the left. there is a huge class of functions for which (4) fails at the points of discontinuities of the functions.
The hypotheses of the above theorem' are known by the name Dirichlet
conditions. Hence. Then the Fourier series off converges to the value f(x) at each point x where f is continuous. if f satisfies the Dirichlet conditions. are. That is. then
f(c-) = f(c+) = f(c)
THEOREM 1
(5)
Assume that f is a periodic function with period 21 and such that f and f' are piecewise continuous on the interval -1 s x s 1. Examples are also known of functions whose Fourier series diverge at "almost" every point.sinn"xl=f(x-)+f(x+) gg+ i (acos-7._
!
/
(6)
2
Addison-Wesley Publishing Co. Mass. a piecewise continuous function on an interval I has a finite number of discontinuities on I. Kaplan.and right-hand limits exist but are unequal) are called jump discontinuities. If f is continuous at c. Fourier was wrong in asserting that (4) is true without any restrictions on the function f. although not the most general sufficient conditions known today. Sufficient conditions for (4) to be true were given by Dirichlet in 1829. 1973). An example of a piecewise continuous function is shown graphically in Figure 10. Advanced Calculus (Reading.
Similarly we write f(c+) to denote the limit of f(x) as x . However.and right-hand limits.) = lim f(x + h).
'For a proof of Theorem 1 see.
if f is continuous everywhere and satisfies the Dirichlet conditions. On the other hand. we will continue using the symbol .10. false at the point where f is discontinuous. the sum of the Fourier series of fat I is
F(I -) + F(1+)
2
. and the left-hand side of (6') reduces to f(x). if F denotes the periodic extension of f. 3. Then the periodic extension of f.to indicate that the right-hand side is the Fourier series of the function to the left. if x is a point of discontinuity of f. the sum of the Fourier series of f at each of the endpoints ±l is Z(f(l-) + f(-1+)]. Furthermore. if x is a point of continuity. Using (5) we can write (6) in the form
fix ) + fix +)
2
as
2
+ i (a. from (6'). The periodic extension off can also be utilized in finding the sum of the Fourier series of fat the endpoints ± 1.
.2
where a and b are given by (2) and (3). (6') agrees with (6). satisfies the Dirichlet conditions. then (4) is true for all x. we have f(x -) = f(x +) = f(x). which agrees with f on I. In fact. in
general. The Remarks 2. It follows from (6) that (4) is. In fact. provided that f and f ' are piecewise continuous on I. The conclusion of Theorem 1 is also true for functions f which are only defined
on an interval I with endpoints -1 and +1.4 Convergence of Fourier Series
403
Figure 10. the Fourier series off and its periodic extension are identical. then. Unless (4) is true for all x.j(1-) + J(-I+)
2
and at -1 is
F(-I-) + F(-1+) f(l-) + f(-1+)
2
2
Thus. and 4 of the last section are relevant to this section as
well. cos mrx + b sin narx
(6')
which is true for all x.
It is identical to f everywhere except at the discontinuities of f..and right-hand limits at
these points exist and are finite. the Fourier series converges to the average of the
left. ±Tr.. ±2ar. From Theorem 1. ±2Tr.. sin nx). and x = ±a. . the graph of the function to which the Fourier series off converges is now completely known.3
. At each of the points 0. See Figure 10.
Solution The function f.] The graph off is sketched in Figure 10. the only points in the interval -Tr s x s IT where for
f' is not continuous are x = 0. satisfies the Dirichlet conditions (the hypotheses of Theorem 1) with 1 = Tr.
-it<x<ar
1.
±2rr.. . [Note: f'(x) = 0 in -Tr < x < 0 and
0 < x < Tr... ±Tr.* cos nx dx = 0. the Fourier series of
f converges to f(x) at each point except 0..
a_r
0
f (x) cos nx dx = . where the value of the Fourier series is zero. In fact. where f is not even defined. ±n. whose graph is known as a square wave of period 21r and amplitude 1. the left.3.10
Fourler Series
EXAMPLE I
Find the Fourier series of the function
AX) __
1.
Sketch for a few periods the graph of the function to which the series converges...4.
0<x<Tr '
f(x + 27r) = f(x).
The function f is continuous everywhere except at the points 0.
C
n=0. Therefore.
Next we compute the Fourier series of f.... . . Here I = Tr and
f (x) with
1)
+ i (a cos nx + b. which in this case is 0.
Y
1
I
-2rr
[-rr
0
rrl
2rrl
x
Figure 10.1.2.-
* cos nx dx + .and right-hand limit.
6 shows the graph of the function to which the Fourier series of
f converges. 3. ±2a.2 +
with
(a cos nx + b sin nx).10
Fourier Series
Y
Figure 10. where f is discontinuous. the
Fourier series converges to the value IT.3 with c = 0 and i = -rr.rr) = g(x). (See Figure 10.
Figure 10.
g(x+2.6.. . . 2. Theorem 1 applies and Figure 10.5
The function in this example is different from the function
g(x) = x.. it is advisable to use formulas (4) and (5') of Section 10. The Fourier series off is
f(x) .
2Tr.
In the interval 0 s x < 2Tr.
n = 1 .
-a s x < rr. we compute the Fourier series of f.) At all other points the graphs off and the function to which its Fourier series converges are identical.8
.. Therefore. ." x cos nx dx =
1r
n=0
10. Since the interval 0_ x < 27r is not symmetric with respect to the origin. . At the points 0. the functions f and f are piecewise continuous with jump discontinuities only at the points 0 and 21r. which is the average value at the jumps.
a =1
Tr Jo
f(x) cos nx dx = 1 f. Next. ±4rr.
f(x+21r) = f(x)
is equal to f(x) everywhere.f(x + 2) = f(x)
satisfies the hypotheses of Theorem 1. 0 s x < 27r. The function
f(x) _. . The Fourier series of the function
f(x) = I x 1.f(x+21r) = f(x)
does not involve any sine terms. -Tr <.1 s x s I.
25.f(x+2) = f(x)
converges to f(x) everywhere. The Fourier series of the function
f(x) = x2.1 <.
27.coslx n
(2)
.10.f(x + 2) = f(x)
is continuous everywhere.5
FOURIER SINE AND FOURIER COSINE SERIES
As we saw in Section 6. The Fourier series of the function
f(x) = x.. The Fourier series of the function
f(x) = x2. The function
f(x) = V.
29.x < Tr. -1 :5 x<1. sometimes it is necessary to express a given function f as a Fourier series of the form
f(x) _
b sin n jx
(1)
In other cases it is necessary to express f as a series of the form
f(x) = 2 +
a.f(x+27r) = f(x)
does not involve any sine term.2. -Tr s x s 7r.
28.x s 7r. f(x+2Tr) = f(x)
converges to f (x) everywhere.5
Fourier Sine and Fourier Cosine Series
24. The Fourier series of the function
f(x) = x2.
10. -Tr <. .1 in connection with the solution of the heat equation.x s 1.
26.
30.
5x . (v) even x odd = odd. (iv) odd x odd = even. x2.
DEFINITION 1
A function f. even and odd functions have the following properties:
(i) even + even = even.7
. (iii) even x even = even. x. then in the interval 0 < x < I we have the choice to represent f as a Fourier sine series or a Fourier cosine series. and those in Figure 10. is called even
if f(x) = f(-x) for each x in the domain off and odd if f(x) = -f(-x) for
each x in the domain off.. whose domain is symmetric with respect to the origin. (The others are proved in a similar fashion.x2 + x4 cos 2x are even and sin ax. Set F = fg. Before we establish the above claim. (v). 1. With respect to the operations of addition and multiplication. we will review the concepts of odd and
even functions and see how.
which proves that F is odd.8 are odd.x2 sin 4x are odd. The functions whose graphs are sketched in Figure 10.410
10
Fourl. In this section we will show that if a function f is defined in the interval 0 < x < 1. With respect to integration. For example. Geometrically speaking. the functions cos ax. a Fourier series of the form (2) is called a Fourier cosine series. for such functions. I x 1. Then
F(-x) = f(-x)g(-x) = f(x) (-g(x)) = -f(x)g(x) = -F(x). Sedss
A Fourier series of the form (1) is called a Fourier sine series. even and odd functions have the following useful properties:
Ir
r
(even) dx = 2 J r (even) dx
o
(3)
Y
Y
x
x
Figure 10. 3 . for example. a function is even if its graph is symmetric with respect to the y-axis and odd if its graph is symmetric with respect to the origin.7 are even.
(ii) odd + odd = odd.) Assume f is even and g is odd. the labor of computing the
Fourier coefficients is reduced. and if f and f' are piecewise continuous there.
Let us prove.
i b. Therefore.
f(-x). 3. and it converges to the average
g(x-) + g(x+)
2
AX-) + f(x+)
2
at each point in 0 < x < I where f is discontinuous. -1<x<0
0 < x < l
h(x + 21) = h(x).
n jx
dx.
mrx
(10)
Hence. the Fourier series of an odd function is reduced to
f(x) -. Then. it has a Fourier cosine series which converges to g(x) = f(x) at each point x in the interval 0 < x < 1 where f is continuous. the integrand in (6) is odd and in (7) is even. = 0.4 are satisfied for the functions g and h.. we find
a. g.
If a function f is defined only in the interval 0 < x < 1.
. we define its even
periodic extension by
g(x) = If(x). since h is odd. are given by (8). 2. . 1. sin
where the coefficients b are given by (10). Similarly. for even or odd functions.. it has a Fourier sine series which converges to h(x) = f(x) at each point x in 0 < x < 1 where f is continuous.. the hypotheses of
Theorem 1 of Section 10.
. Since g is even.
(9)
Odd Functions When f is odd..
It should be remarked that. .
and
n = 0. Furthermore. and to the average
h(x-) + h(x+)
f(x-) + f(x+)
. the Fourier series of an even function is reduced to
f (x) -. -1<x<0'
f(x).2 +
where the coefficients a. 2.
(13)
Note that the functions f. then g and g' and also h and
h' are piecewise continuous on -1 < x < 1. if f and f are piecewise continuous on 0 < x < 1.412
10 FounerSeries
Hence.
b_
u
f(x) sin
n = 1. and h agree in the interval 0 < x < 1. from (3) and (4).
a cos n j x. the formulas for the Fourier coefficients use the values of the function in the interval 0 < x < I only.
0 < x < 1
g(x + 21) = g(x)
(12)
and its odd periodic extension by
h(x) _ {
l-f(-x).
/n=0
n ir' I cos 2 Hencc. From (14). the Fourier cosine series off is n2x f(x) 2 + a cos
with
a.414
10
Fourier Series
EXAMPLE 2
Sketch the even and the odd periodic extension of the function
x... and ±1.1) + n sin
2
J cos
I=
. = fa
1
f(x)cosn2xdx
= t xcosn2xdx
2. and the odd periodic extension is sketched in Figure 10.9
.
-6
-5
-4 -3
O
-2 -1
o
2
O
Z7--G
4
5
o
6
iX
I
3
Figure 10.9.4 is applicable. .
) AX)
0<x<1
0.
The even periodic extension of f.10.1 < x < 2
Solution The even periodic extension off is sketched in Figure 10. 3.q + j [n4 (cos Z . is shown graphically in Figure 10.
The only discontinuities of g are at the points 0. . ±2.
Solution
I = 2.
EXAMPLE 3
Compute the Fourier cosine and the Fourier sine series of the f(x)
function
Ix. From this graph we see that Theorem 1 of Section 10.9. referred to here as g.
Y
. 0<x<1 0..
1 + n-Tr sin 2
n = 1. . 1<x<2-
Sketch the graph of the function to which the Fourier cosine series converges and the graph of the function to which the Fourier sine series converges. ±6. ±4.
f(x) -. 2.
.12.9 and from Theorem I of Section 10.10..11
. t 3.11 (which is easily constructed from Figure 10.i b sin
with
nax
nlrx 6 = f (x) sin 2 dx =
((2
x sin
u
nirx
2
4 dx = n-':-s'na 2
2
n7t
cos
nir
2
.. From (15). ±5. we
conclude that the Fourier series converges to 0 at the points t 0. with I = 2.10
t3. t 4..4 is applicable. From Figure 10.10. is shown graphically in Figure 10. Next we compute the Fourier sine series of f... t 2.. ... and converges to z at the points ± 1. ± 1. The graph of the function to which the series converges is shown in Figure 10. .
AX)
nn 4 sin ni-Z 2-
n rr cos 2) sin 2 . . The only discontinuities of h are at the points 0..
JU
Hence. we have
AX) .9).
2
nlr
n rrx
The odd periodic extension of f. The graph of the function to which the series converges is sketched in Figure 10. referred to here ass h. t 2.4. At every other point x the
series converges to g(x). From this graph we see that Theorem 1 of Section 10.5
Fourier Sine and Fourier Cosine Series
415
Figure 10. ± 3.
y
Figure 10. ± 2..
p. 1977). Inc. 'This is Example 6.
0<x<a. 195. Electrical Network Science (Englewood Cliffs.
H
F-1
-W
I
F-1
Fl'
t
III--TH
'This is Exercise c6.J. Use the result in Exercise 13 to prove that
1 1 1
n=
15.
. Find the Fourier series' of the half-wave-rectified cosine shown in the figure.4 in ibid. Kerr. B.: PrenticeHall.
Compute the Fourier cosine series of the function
f(x)=1-x.. New Jersey.. Find the Fourier series expansions of the pulse train shown in the figure. New Jersey. Englewood Cliffs. N. Reprinted by permission of Prentice-Hall. Inc. Show that.. and a repetition period of T seconds.
0<x<-rr. p.
Compute the Fourier sine series of the function
fx(x) = 1 +x. Each pulse has a height H. Englewood Cliffs.
18. Reprinted by permission of Prentice-Hall. 335.
n-'
(-ln
2
cosnIrx=2Ix14
14.
1(r)
17. a duration of W seconds.420
10
Fourier Soria
13.
16.4 in R. for -1 s x < 1.
In the case of ordinary differential equations.
.6 through 11. and the wave equation.4. and 11. The general solution of this ordinary differential equation contains one arbitrary constant. a partial differential equation seems to differ from an ordinary differential equation only in that there are more independent variables. the initial value fixes the value of the arbitrary constant in the general solution. That is to say. Uniqueness of
solution (in other words. The general theory of partial differential equations is beyond the scope of this book. If one considers a first-order linear partial differential equation (to be defined in Section 11. the results are very different. In contrast.2). the potential (Laplace) equation. In this chapter we present an elementary treatment of partial differential equations.
At first glance. although the study of partial differential equations frequently utilizes known facts about ordinary differential equations.CHAPTER 11
An Introduction To Partial Differential Equations
11. The reader interested in the more practical aspects of this subject should concentrate on Sections 11. Let us assume that the independent variables are x and y and that the unknown function is u. For instance.4). Our treatment will focus on some simple cases of first-order equations and some special cases of second-order equations that occur frequently in applications. no simple condition serves to isolate (uniquely determine) a
specific surface from this collection. for partial differential equations the unknown function depends on two or more independent variables. The general solution can be interpreted geometrically as a collection of three-dimensional surfaces. the unknown function depends on a single independent variable. hence this general solution can be interpreted geometrically as a set of two-dimensional curves. Example I illustrates this fact. However.
Unfortunately.1
INTRODUCTION
Partial differential equations are equations that involve partial derivatives of an unknown function. isolating a specific curve in the set of two-dimensional curves) is accomplished by specifying a two-dimensional point (an initial value y = yo when x = x0) that the general solution must contain. 11.3. each coming
about by assigning a different value to the arbitrary constant. one of the simplest types of ordinary differential equation is the
first-order linear equation (see Section 1. In this latter category we concentrate on the classical equations of mathematical physics-
the heat equation. and we make no attempt to develop it here. in general it is quite different.10.
y)"-'(-1).Y) .f'(x . where n is any positive integer.y) => ax ` [f'(x . on the other hand. suppose that f(x .
Y
The solution in part (a) demonstrates that we cannot expect to obtain unique-
ness of solution by specifying a single point that the general solution must contain.
au
ay
= n(x .y)"=> au = n(x . In particular.
au
ax ay
(a) u = f(x . In spite of the above differences. The solution in part (b). then there is an infinity of solution surfaces (corresponding to the values of n) which contain this curve.
au
(b)
u = (x .
au
au ax+a =n(x-y)^'-n(x-y)^'=0. there is an infinity of solution surfaces (corresponding
to the values of c) that contain this point.+ -au= 0.y).Y)](1).y) ° c(x .2 we will discuss a few of them. In Section 11. for partial differential equations. Example 1 illustrates that. u = 0 when y = x).y)^. where c is an arbitrary constant. where f is any function having a continuous derivative.
au
ay
au
= [f'(x .
Thus.422
11
An Introduction to Partial Differential Equations
EXAMPLE 1
Show that the following functions are solutions of the differential
equation
.Y) = 0.
.y). Appropriate conditions that produce uniqueness of solutions usually depend on the form of the partial differential equation. but rather we will discuss particular situations.
Solution
(a) u = f(x . One point cannot uniquely determine the arbitrary function f. there are some similarities and analogies between ordinary differential equations and partial differential equations.y)"-'(1). We will not attempt to investigate all of the ramifications of this issue. uniqueness of solutions is not (in general) accomplished by simply specifying a point or a curve that the general solution must contain. if it is specified that the solution surface contain the curve y = x in the xy-plane (in other words. More specifically.
au
ax
+ ay = f'(x . demonstrates that we cannot expect (in general) to obtain uniqueness of solution by specifying a particular curve that the general solution must contain.
and
(b) u = (x . u = 0 when x = y = 0). If we specify that the solution surface contain the origin (in other words.Y)](-1).
Thus.
.. denoted by x. Give an argument to support
the statement that there is an infinity of solutions which contain the point (0. u. u.. and (b) u = (x + y)^. ) = 0.2
DEFINITIONS AND GENERAL COMMENTS
To simplify the notation.
_ au
u
az
u ... that is.
is.2
DaflnItions and General Comments
EXERCISES
1.. u. With these constraints we can define a partial differential equation to be a relation of the form
F(x..... 0.
dq 3.y). u u. ur = u. = u.
. u... F is a linear combination of the unknown function and its derivatives. and no more than three independent variables..
u. u. and if g possesses a continuous
derivative. denoted by u. z..
We always assume that the unknown function u is "sufficiently well behaved" so that all necessary partial derivatives exist and corresponding mixed partial derivatives are equal.. is a solution of the partial differential equation du + 2
= 0. and so on. If g is any function such that g(0) = 0.x in the
11. where f is any function possessing a continuous
derivative.. where n is a positive integer are solutions
of the partial differential equation
au ax
ay Y
= 0. (1) we have used the subscript notation for partial differentiation. u. we restrict our attention to the case of one unknown function. u". where f is any function possessing a continuous derivative. u. 0). 2.. = u. u. for example.. we define the partial differential equation (1) to be linear if F is linear as a function of the variables u. ...... Give an argument to support the statement that
there is an infinity of solutions which contain the curve y = . show that u = g(3x + 2y) is a solution of the partial differential equation 2 ax xy-plane. Equation (1) is said to be quasilinear if F is linear as a function of the highest-order derivatives.. Show that (a) u = f(x + y).y
. .
au
3
au
aY = 0.. y. Show that u = f(2x . .
(1)
In Eq.
and soon... u.. that
_
a2u
axeY
a"'
_
a3u
ax3 . y. u-.
Just as in the case of an ordinary differential equation. u u. we define the order of the partial differential equation (1) to be the order of the partial derivative of highest order appearing in the equation. z. Furthermore.
.11.
and u is unknown. Thus. y. reduces Eq. (8) is called a partial
differential equation with constant coefficients.
+ a. y.(x. (5). y. + a. Y. which.y) was a solution of u + u. (8) to an identity.(x.3zu. Eq. and (5) are nonhomogeneous. with continuous first. is not a constant. Y. is a constant.
7u. By a solution of Eq. Note that Eqs.
EXAMPLE 1
Find a solution u = u(x. y.424
11
An Introduction to Partial Differential Equations
The following are examples of partial differential equations:
u... + a.. are known.. y. (8).
If f(x. Equation (6) is quasilinear. otherwise it is called nonhomogeneous (or inhomogeneous).. y) of the partial differential equation
u.-2x2-3z
u = x2
(2)
(3)
5xyu. (3). z)u.(x. z) = 0. z)u + a2(x. (8) we mean a continuous function u. Thus our most general partial differential equation can be written in the form
a. we will limit our discussion to linear partial differential equations of order one or two.. when substituted in (8). Eq. in Example 1 of Section 11.1. (8) it is understood that the function f and the coefficients a. and Eqs. (3).and second-order partial derivatives. the partial differential equation (8) is called homogeneous. z)u = f(x. Y. = x + Y. + 2u = 0. Y. (4) has
variable coefficients.
(9)
.. With very few exceptions. + a. y. y.. + 3u = 2xey. + a9(x.. z)u. Equation (6) is nonlinear because it is not of the form of Eq. If every one of the coefficients a. z)u.(x. z)u
+ a. and the rest are linear. z). (2). and (7) have constant coefficients.(x.. z)u.+u. z)u. + a6(x. + a. z.=3u. y. z)u.
Equations (2). we demonstrated by direct substitution that
u = f(x . Y. (8) is called a partial differential equation with variable coefficients. of the independent variables x. z)u. = 0.o(x. + 8u. and Eq..
(8)
In Eq. If at least one of the a. (4)
and (7) are homogeneous. .
(4)
(5)
(6)
(7)
Equations (2) and (5) are first order and the rest are second order.(x.
(12)
Solution First we integrate partially with respect to y (treating x and z as constants) to obtain
ux = yz + xy + f. since
afa(y)
= 0. z).(x. z) + g(y. z). with c replaced by f(y). Next we integrate partially with respect to x (treating y and z as constants) to obtain
u = xyz + gx2y + Jfi(s.2
Definitions and General Comments
425
Solution We begin by integrating Eq.
. is an arbitrary function of the variables y and z.(s. would still be a solution of the partial differential
equation (9).(y. Since we seek the most general form possible for the solution. we need only substitute this expression for u in Eq.
(11)
where f is an arbitrary function of y. we integrate with respect to x. z) =f2(y. is a solution of Eq. (10). When verifying this. and g is an arbitrary function of y and z. z)ds.). (9).=z+x. treating the variable y as if it were a constant) to obtain
u=zx2+xy+c. This example illustrates another strong contrast between partial differential equations and ordinary differential equations in that solution (11) contains an arbitrary function rather than an arbitrary constant. z).
EXAMPLE 2
Find a solution u = u(x. where f. we write
u = 2x2 + xy + f(y). z) = JYf. y. is an arbitrary function of the variables x and z. (10) is not the most general
result possible unless we emphasize that c is to be replaced by an arbitrary
function of y. (10).
(13)
where f is an arbitrary function of x and z.
u = xyz + 2x2y + f(x. such as f(y).
(10)
We note that the "constant of integration" is denoted by c.z)ds + f. If we set
f(x. In order to verify that u. then u as given by Eq. as given in Eq. z).
then our solution takes the form
g(y.
where f. f and g are to have continuous first and second partial derivatives with respect to their arguments.11. We conclude that Eq. (9) "partially with respect to x" (in other
words. notice that even if c were not a constant but a function of the variable y. (9). z) of the partial differential equation
u.
and determine whether it has constant or variable coefficients. and the general solution [given as solution (a)] contains the arbitrary function f.. (12) [for the particular choices f(x. determine whether it is homogeneous or nonhomogeneous.-u.. + 2 x2y + z cos x + ye. state its order.
u=2x=+xy+e'
is a particular solution of Eq... the following is a reasonable claim for the partial differential equations to be treated in this text. u. (Not necessarily the same
set of (s .. Each specific assignment of the arbitrary function(s) in the general solution gives rise to a particular solution of the corresponding partial differential equation.u.. u. u.5u. z) = ye].. assume that u is a function of the two independent variables x and y.
Even though it is difficult (if not impossible) to make all-inclusive general statements about partial differential equations.1) variables.. and
u = xy. u. u.. .+u. (9) [for the particular choice f(y) = e']. In Example 1 of Section 11..
In Exercises 11 through 20.= cosy + ex
u..
1. u = 0
18.
2. The general solution to a linear partial differential equation of order n.) Thus.u...u.1. n = 2. = 0
.
3. = 0 17. involves n arbitrary
functions. u..
EXERCISES
For each of the linear partial differential equations in Exercises 1 through 10.y. n = 1.. each of which depends on (s . which depends on the single variable x .=cosz 10... In Example 2.+17u=0
5.. u. u. u.-3zu+u. which depends on the single variable y.-u.
11.
is a particular solution of Eq. and the general solution contains the two arbitrary functions f and g.=27y+z2 S.. Integrate each equation to obtain the general solution. u. n = 1..+5xu.3u..1) variables applies to each arbitrary function.=sinx-siny
16. s = 2.=0
13. f depending on the variables x and z. and g depending on the variables y and z.. for an unknown function depending on s independent variables.+2zu = 3u. in Example 1.=3x2+4y
15.11
An Introduction to Partial Differential Equations
As in the case of ordinary differential equations. 3u . Thus. s = 2...=0 8. ..y + 4u. = 0
S.=0
14.u. + u . (9) and (12) respectively. s = 3.y2
4. = 3u
7. we call the solutions (11) and (13) the general solution of Eqs.
12..= 3x' . and the general solution contains the arbitrary function f. z) = z cos x
and g(y. u..
The shearing stress and the normal stress in an elastic body are obtainable from Airy's stress function. and z. u.
21.. Verify that the general solution of Exercise 23 conforms to the claim about the form of the general solution made at the end of this section. (8). z)u + .11..2.]. = sect y
In Exercises 21 through 30. u.
. y.. (8) in the abbreviated form
A[u] = f. Our partial differential equation has the appearance of Eq.. u.... is a solution of the partial differential equation 0..x
23. 32... = 0
25. where 4.._. and u2.. + a9(x.(x.u. z)u.=0
24. u =y+ 3x
28. Verify that the general solution of Exercise 25 conforms to the claim about the form of the general solution made at the end of this section. = sec'. + c2u2] = c.
A[u] = a. u.
22.
DEFINITION 1
An operator A is called a linear operator if A[c. 33. u. Thus. = x2 + z
27. Section 11.3 The Principle of Superposition
427
19.. y. Integrate each equation to obtain the general solution. we write Eq. + c2u2] make sense) choice of functions u. u. y.A[u. assume that u is a function of the three independent variables x. + 24.]. =y 29... In some instances the methods will be further illustrated
in subsequent sections. +
Classify this equation utilizing all definitions of this section that are appropriate.. (8).u.(x. + a. c2 and for every permissable (in other words.
sec' x
31. z)u. u. 4. Verify that the general solution of Exercise 21 conforms to the claim about the form of the general solution made at the end of this section.=2
30. Elasticity.. u.3
THE PRINCIPLE OF SUPERPOSITION
In this section and the next two sections we outline some general ideas and methods of solution. and A[c..] + c2A[u2] for every choice of the constants c.. A[u.. (8) we write... = 0
26. for the operator of Eq. = cos y + e'
20. and the manner in which the operator
A "operates" on the function u (denoted by A[u]) is defined by the left-hand side of Eq.
11. For simplicity.
(1)
We speak of the symbol A as an operator.. u. 4. y. u.
34. A[u.
. if u. A[u. is also a solution of A[u] = 0. and up is a
particular solution of A[u] = f. the sum of a solution of the homogeneous equation and a particular solution is also a solution.u. . the equation
A[u] = 0 is called the associated homogeneous equation... = f(x. . u. = f where
.. solutions of the equations A[u. .. and that u satisfies the partial differential equation
EXAMPLE 1
uxy = z + X. then u = c.. z). + up is the general solution of A[u] = f.. x...... be any constants. u... is a solution of A[u] = f.. . z) + g(y... then u = u. are any
constants. The same approach is sometimes useful in trying to solve a
nonhomogeneous linear partial differential equation. where f and g are arbitrary func-
tions.. As in the case of ordinary differential equations. = 0 is given by u..A[u. y.
Let f f2..
Principle of Superposition
are.]
c1f +c2l2 T
+cJ....] = f A[u2] = f2.. . (ii) u. . c.. + c2u2 +
A we have
A[u] = A[c. + c. + c2u2 +
+ cmum]
= c. is a solution of A[u] = 0 and if up is a particular solution of A[u] = f. that is. that is. then u = u. A very important property of linear partial differential equations is contained in the following principle..u.. c.
These two consequences are similar to the manner in which we generated the general solution for a nonhomogeneous linear ordinary differential equation (see Chapter 2). c.. = xyz is a particular solution of the equation u. {A'' [u1] + c2A[u2] + .. If A is a linear operator and if u u2.. then E . Two important consequences of the principle of superposition are as follows:
(i) If u u2.u. (ii) If u.. any linear combination of solutions is a solution. ... is a solution of the equation = f. ..u.. respectively. is the general solution of the homogeneous equation A[u] = 0.
and the proof is complete.] + c.
Suppose that u is a function depending on the variables. ... and z.
(2)
Show that (i) the general solution of the associated homogeneous equation u.428
11
An Introduction to Partial Differential Equations
DEFINITION 2
Given the nonhomogeneous partial differential equation (1). . . are solutions of A[u] = 0 and c c2. f be any functions and let c c2... but the methods of solution are not as concrete or systematic for partial differential equations as for ordinary differential equations... + u...
(the potential or Laplace equation). where the functions X.
. or three-dimensional. The basic ideas and manipulations involved in the method are illustrated in the following examples. two-.8.11.y = X(x)Y"(y)
Substitution of these results in Eq.
Solution If
u = X(x)Y(y).7. the one-dimensional heat equation is discussed in Section 11. (1) leads to
X°(x)Y(Y) .
then
u.4
Separation of Variables
433
Remark The classical equations of mathematical physics are
u. each of which depends on exactly
one of the independent variables.6.
0 = A [u]
In each case the equation is called one-. depending on whether the Laplacian operator is one-. = aO [u]
(the heat equation). and the twqdimensional Laplace equation is discussed in Section 11.r = X"(x)Y(Y). For example.X(x)Y"(Y) = 0. or three-dimensional. = X'xx)Y(y). The one-dimensional wave equation is discussed in Section 11. two-. where the functions X. For the partial differential equation
we would try to write the solution in the form u = X(x)Y(y). u. u.. we would try to write the
solution of the partial differential equation
u = X(x)Y(y)Z(z). In this method we try to write the solution as a product of functions. = X(x)Y'(y).
u. and Z are to be determined. Y. U. = c2O [u]
(the wave equation)..4 SEPARATION OF VARIABLES
A frequently used method for finding solutions to linear homogeneous partial differential equations is known as separation of variables.
EXAMPLE 1
For the partial differential equation
(1)
find a solution in the form u = X(x)Y(y). Y are to be determined.
11.
X Y(Y) = 0.
C3 cos
x=0
y + C 4 sinVy. In fact.
C3C'Y + c4e "
A>0
Y(Y)=
C3+c4Y.ru + c2e -. changes in x should not affect the expression The net conclusion X(x) is that in order for (2) to be an equality. the expressions L (X) and Y "(Y) must X(x) Y(Y) be constants.
A<0 . if (2) is to be an equality.5 to yield.
X"(x) .
.e .
X(x)
A=0
+ c2 sin N/ -\x. they must be the same constant.
X"(x) Similarly. we note that changes in y will not X(x)
have any effect on the expression X( ).rte
X> O
C1+C2X.
.
el cos VC
A<0.
c. If the constant is denoted by A."(X)
X(x)
or
Y"(y)
Y(Y)
=0
X"(x)
X(x)
Since
Y"(Y) Y(y)
(2)
X"(x) does not contain the variable y.XX(x) = 0
and
3)
Y"(Y) .11
An Introduction to Partial Differential Equations
Divide this latter equation by u = X(x)Y(y) (assuming that u # 0) to obtain
X. Thus. we can write
X(x)
X(x)
and
Y(Y) =
Y(Y)
a. (4) Equations (3) and (4) are ordinary differential equations with constant coefficients and can be solved by the methods of Section 2. it must
happen that changes in the variable y do not affect the expression
Y
either.
Thus.
A>0
0
X(x)Y(y)
(c
c.5X(x)Y(y)Z'(z) = 0.10).
Without further information we have no way of knowing the value of.
Solution
If
u = X(x)Y(y)Z(z)
then
u. hence we cannot specify the form of the solution.\.
I (c3 cos Vy + c4 sin V y).x
A < 0. -5u. = X'(x)Y(y)Z(z).=0. say N.
e
A + c4 e Y
" 1.
U
(ce
J
x + c2) (c3 e -.4
Separation of VarlabMs
435
Thus.-2u.
find a solution in the form u = X(x)Y(y)Z(z).ru
crr) (c3 CJ').0
3X'(x) 2Y'(y) + SZ'(z) (5) Z(z) ' Y(y) X(x) Using the same type of argument as in Example 1.
Dividing by u = X(x)Y(y)Z(z) (assuming u # 0).2X(x)Y'(y)Z(z) .
Substitution into the partial differential equation yields
3X'(x)Y(y)Z(z) .
EXAMPLE 2
For the partial differential equation
3u.
and
u. these conditions usually dictate the value of A and the form of the solution (see Sections 11. we have
3X'(x) X(x)
or
2Y'(y)
Y(y)
5Z'(z)
Z(z) .6-11. = X(x)Y(y)Z'(z). In many practical problems there are other conditions that the solution must satisfy. (5) can be an equality is that both sides of the equation equal a constant.
uy = X(x)Y'(y)Z(z). Thus 3X'(x) (6) X(x)
and
-
2Y'(y)
Y(y) +
5Z'(z)
Z(z) -
(7)
. we conclude that the only way that Eq.11. cos N/--lx + cI sin
->.
e(")') (C2e(w2)Y) (C3el`". Therefore a solution to the partial differential
is
u = (c.=0
2. that is. Equation (7) can be rewritten as 5Z'(z) 2Y'(y) (8) Z(z) .
EXAMPLE 3 Show that the variables "do not separate" for the partial differential equation
Solution We try a solution in the form u = X(x)Y(y).51t) =
ke)b3)x-(W2)!-[(" -")15]t
In Examples 1 and 2.
Y(Y) -
Once more we argue that in order for (8) to be an equality.-3u. U.=0
.+3u. Example 3 is a case in point.436
11
An Introduction to Partial Differential Equations
Equation (6) has as general solution X(x) = c. that every linear homogeneous partial differential equation can be solved by the method of separation of variables. assume a solution in the form u = X(x)Y(y).el"3j`. Show that the equation "separates. U_ = X"(x)Y(Y) Substitution of these results into the partial differential equation leads to
X'(x)Y'(y) + X"(x)Y(y) + X(x)Y(y) = 0. however.")'si'. the partial differential equations have constant
coefficients." and find the differential equations that X and
Y must satisfy. (9) is Y(y) = ce("`2)Y. and the solution to Eq. Then ut = X'(x)Y(Y). therefore we conclude that the method does not work for this partial differential equation. The solution to Eq."
EXERCISES
In Exercises 1 through 22.. u. = X'(x)Y'(y). but the method can also be applied to equations with variable
coefficients. the variables "do not separate.
Z(z)
(10)
where µ is a constant. It is not possible to algebraically manipulate this latter equation to a form P(x) = Q(y).el('.").
1. we must have 2Y'(y)
Y(y)
= µ>
(9)
and
XSZ(z)=µ. It is not the case. There are many equations for which the method does not apply.4u. (10) is Z(z) = c.
)"
lu. (a) Set u = X(x)Y(y) and determine the differential equations for X and
Y.. Methods of Mathematical Physics. There are other techniques that may also be called separation of variables. the velocity potential satisfies the equation
(MIwhere M(>1) is a constant known as the Mach number of the flow.11. To illustrate..
66.(u. Fy. Set u = X(x)Y(y)Z(z)T(t) and determine the differential equations for X. .
c
I
where the constant c represents the velocity of sound in the medium.. + u) = 0 x+y
occurs in the one-dimensional isentropic flow of a compressible fluid.. Z.4 Separation of Variables
439
62. a is a constant and represents the velocity of sound in the medium.
63.
64.
. Set u = X(x)T(t) and determine the differential equations for X and T. y. Section 11. z. and T. Isentropic Fluid Flow The second-order linear partial differential equation
a u. 67.
65. + a2U = U.. p.y. where gradient F is the vector [F.. Acoustics The nonlinear partial differential equation
(u. = azu
occurs in the study of the propagation of sound in a medium.
where a2 is a constant.+ U =
. Courant and D. 459.. 2 (New York: lnterscience
Publishers. Acoustics In the study of the transmission of sound through a moving fluid. Given
u. Supersonic Fluid Flow In the study of the supersonic flow of an ideal compressible fluid past an obstacle. Set u = X(x)Y(y) and determine the differential equations for X and Y. one considers the velocity potential u(x. F. Y.]. t). vol. Solve these differential equations..3)
U + U. (The term potential is usually used in physics to describe a quantity whose gradient furnishes a field of force. Set u = X(x)T(t) and determine the differential
equations for X and T. Separation of variables for partial differential equations normally relates to the method described in this section. let us assume a solution of the equation
'R.' a is a constant which depends on the fluid.u. In this case the gradient of u yields the velocity of the flow. 1962). Hilbert.
(b) Solve the differential equations of part (a).) It can be shown that u satisfies the three-dimensional wave equation (see the remark following Exercise 50.
6
THE HOMOGENEOUS ONE-DIMENSIONAL WAVE EQUATION: SEPARATION OF VARIABLES
Consider that we have a string that is perfectly flexible (that is to say. that is.(x).. The distance along the string will be denoted by s and as usual
ds =
(dx)2 + (dy)2. A6 are known constants.. . The string is
then given an initial displacement and/or an initial velocity parallel to the y-axis. if we set F.
In Problem 1.1. the
length of the string is approximately unchanged (since s . one or more of the boundary conditions. y. thus setting it in motion.. y). and F(x. y) = 0. .10 we consider special cases of Problem 1 with F(x. (4) and (5) are called the initial conditions. f. x = 0 and x = 1.
11 . f.6-11..11.9 and 11. f. Consequently. we have the situation illustrated in Figure 11. If no other forces are acting
on the string. and Eqs. For a given problem. Equations (2) and (3) are called the boundary conditions.
Thus. f2(y). The main prerequisite for utilizing the method we discuss is that the method of separation of variables be applicable to the partial differential equation.. y).8 The Homogeneous One-Dimensional Wave Equation: Separation of Variables
441
where a. we do provide the necessary ingredients for the solution of many problems that fall into the category of Problem 1. where T represents the (constant) force due to tension. and f. for specific cases. each equal to zero. or both.
as
ax
+ dye
ax
. a6(x. While we do not solve the general case of Problem 1.
If we assume that the displacement. y) are assumed to be known functions and A. the string
is capable of transmitting tension but will not transmit bending or shearing
forces) and that its mass per unit length is a constant. f.8 we demonstrate.x). then approximately
as
= 1. we obtain the associated homogeneous problem. y) * 0. Is there a function u of two variables x and y that satisfies the partial differential equation (1) and each of the conditions (2)-(5)? Such a problem is referred to as an initial-boundary value problem (I-BVP). f2. how to solve Problem 1 with F(x. or one or more of the initial conditions may be missing. the tension in the string is approximately constant. f3(x).
may be infinite.. I or m.(x. is small enough so that (ax)2 is a very
small quantity (in comparison to 1). In Sections 11.
. The string is to be stretched
and attached to two fixed points on the x-axis.(y). . In Sections 11. .
azy
P
:
at. the vertical component of the force T is
T sin 02.1 is given by pds. T as .tan 9z = y'(x + Ax).
Dividing by . y) is given by
. R.
.
Ax + R. Expanding y'(x + Ax) in a Taylor's series. however. Since the displacement is considered to be small. we have by Newton's Second Law )of Motion
pdxaa
=
Tas+ Tay1x+R+l-Tax)
= T ds Ox + R.
If p represents the mass per unit length.-O Ox
lim 1 R=0.x and taking the limit as Ax tends to zero. which is approximately pOx. = T ax2
. Thus. has the property
. ax ax ax as
At the point (x + .Tay= -Tay as = -TaY. y + Ay). namely.x.T sin 0.11 An Introduction to Partial Differential Equations
The vertical component of the force T at the point (x.
az where the remainder. the vertical component
is
z
T
+T
ax. for motion in the vertical direction. sin °2 . we obtain the equation of motion of the vertical displacement of the string. then the mass of the portion of string in Figure 11.
A typical initial-boundary value problem for the one-dimensional wave equation is the following.=0.8 and 11. 0 < x < 1
(2)
(3) (4)
u(0. Substitution of (7) into Eq. the initial-boundary value problem (2)-(6) has a unique solution. 0 < x < l
u. 0<x<1. Eq. Note that the wave equation is an example of a hyperbolic partial differential equation. t 0.(x.7 respectively.
(1)
where c2 = Tip is a constant according to our assumptions. we rewrite the equation of motion in the form
u . Conditions (5) and (6) can be interpreted as indicating that the ends of the string are attached ("tied") to the x-axis for all time.
T"
c2X"
T
X = X. Consequently we have two separate problems:
T" _
T .
t > 0. are discussed in Sections 11.
0 < x < 1.
Thus. (1) is frequently referred to as the equation of the vibrating string. T are unknown functions to be determined. To this end we assume a solution of Eq.
(5) (6)
It can be shown that if f and g satisfy the Dirichlet conditions (see Theorem 1. (2) yields
XT"-c2X"T=0. (2) to be of the form
u(x. A standard approach to solving this initial-boundary value problem is to use the separation of variables method of Section 11. For obvious reasons. The other two. t) = X(x)T(t).t`
(8)
. t>0
u(x. 0) = f(x).4 and Fourier series (Chapter 10).. (1) the one-dimensional wave equation. t) = 0. -czu. Condition (3) represents the initial position of the string and condition (4) is the initial velocity.
u. t z 0 u(l.6 The Homogeneous One-Dimensional Wave Equation: Separation of Variables
Conforming now to our convention of denotirig the unknown function by u.4).
with X a constant.11. The wave equation is one of three partial differential equations known as the classical equations of mathematical physics. It is also customary to call Eq. t) = 0. the potential (Laplace) equation and the heat equation. 0) = g(x).
(7)
where X.c=u = 0. Section 10.
What about conditions (3) and (4)? Let us investigate condition (3) when u(x. 3. 2 . restricts g to be of the form B sin nlx .. too. 3.
u(x. where A is a constant.. c=
X(0) = X(1) = 0.
n
=1 .
(9)
The boundary condition (5) demands that X(0)T(t) = 0 for all t ? 0. then.X X=0. (10). therefore. t) _ X. ..
!Z
H ence
T(t) = a.
. The
function X.. to yield the eigenvalues
z:
n jz
.. (2) that satisfies conditions (5) and (6).. X(0) = 0. If f is of this form. 2..444
and
11 An Introduction to Partial Differential Equations
c2X
X
= a. Similarly.
(12)
where a and b. 2.. where B is a constant..0)=f(x)=>
(sinnIx)s
=f(x)
(13)
The only way that (13) can be satisfied is that f(x) be restricted to be of the
form A sin nn
that
xx
. thus.(x)T (t). condition (4) demands
narx
!
Hare
sin
(1 b = g(x)
(14)
This.
This problem can be solved by the methods of Section 6. we consider an alternate approach.2..
n = 1.).(x)T (t) is a solution of Eq.
Conditions (13) and (14) place too great a restriction on the permissible forms for f and g.. is to be a solution of the eigenvalue problem
X". (8) takes the form n2arZ
T" +
.3. We conclude that for each specific value of it (n = 1. are the integration constants in the general solution.. Eq..
(10)
and the corresponding eigenfunctions
1
. the function X.
n=1. 3 .. with it not specified but otherwise considered fixed.
T=0
(11)
With K given by Eq. the boundary condition (6) indicates that X(1) = 0..2. cos
nc
t + b sin
n cc t.
condition (4) will be satisfied provided that
b"Isinnx = g(x)
In other words. We will not investigate this question here but rather emphasize the method of solution. (2) for each value of n (n = 1. u(x. (2). X (x)T (t) is a solution of Eq. t) satisfies conditions (5) and (6). 2.
(>O'
u(x.(x.). t) = 0. t) = 0. (2).x). u(x.
u(a. given by (11) and T given by (12).
U11 .
(17)
We conclude that the solution of the initial-boundary value problem (2)-(6) is
given by
u(x t)
where a" and b" are given by Eqs.11. (16) and (17) respectively. and since Eq. as is easily verified.
t ? 0. a" is given by (see Section 10.
t ? 0.5)
-1
a" =
1
Of(x)sinnjxdx._. t) = i X (x)T(t). 0 < x < IT. 0) = x(ir .
(16)
Likewise. Naturally there is the question of whether or not this infinite series converges. to be a solution of Eq. Consequently. b" is given by
b"
nac o 8(x) sin n x dx.
(18)
EXAMPLE 1
Solve the following initial-boundary value problem.4u_ = 0. 3.
.
0 < x < n. it seems reasonable to expect that I.
u..
u(0.
0 < x < IT. 0) = 0. that the Fourier sine series for f(x) in the interval 0 s x s I be nrx
a" sin
.6 The Homogeneous One-Dimensional Wave Equation
445
Since X"(x)T (t) is a solution of Eq. We consider
u(x.
(15)
with X.
that is. .. t) will satisfy condition (3) provided that
a" sin
mrx
= f(x). (2) is a linear partial differential equation.
The method introduced in this section is applicable to many problems that come under the classification of Problem 1, Section 11.5. Other types of equations are presented in Sections 11.7, 11.8, 11.9, and 11.10. We conclude this section with another example involving the wave equation but with different boundary conditions.
EXAMPLE 3
Solve the initial-boundary value problem
0<x<1, t>0
u(x, 0) = f(x),
u,(x, 0) = g(x),
0 < x < 1, 0 < x < 1,
(19)
(20) (21)
ux(0,t) = 0,
t > 0,
t > 0.
(22) (23)
uY, t) = 0,
Solution Conditions (22) and (23) are different than those imposed for the
problem (2)-(6); therefore, if we assume a solution in the form u(x, t)
X(x)T(t), X must satisfy the eigenvalue problem
X"-iX=0,
X'(0) = X'(1) = 0.
11
An Introduction to Partial Differential Equations
Except for a slight change in symbolism, this eigenvalue problem is that of Example 2, Section 6.2. Therefore, we have
=-n12
and
z
,
n=1,2,3,...
nirx
,
n=1,2,3,....
Thus we can repeat the development of the solution as we did in the beginning of this section. The solution is
u(x, r) _
where
(a. cos
nrrc
i + b sin
nrrc
t cos
n rrx
1
,
a=
and
2
u
f(x) cos nn xx dx
b"
-
ug(x)cosnrxxdx.
(25)
That is, a and (ill") b are the coefficients of the Fourier cosine series of the
functions f and g respectively.
EXERCISES
In Exercises 1 through 14, solve the initial-boundary value problem (2)-(6) for the conditions given.
31. Suppose that u is a solution of the initial-boundary value problem
0<x<1, t>0,
0 < x < 1, 0 < x < 1,
t > 0,
(26)
(27)
(28)
(29)
(30)
u(1, t) = B, t > 0, where A and B are constants. Show that if
v(z,t)=u(x,t)+X111A-IB,
then v is a solution of the initial-boundary value problem
(31)
0<x<1, t>0,
v(x, 0) = Ax) + I 1X A - l B, Il
\\
/
0 < x < 1,
v,(x, 0) = g(x),
0<x<
t > 0,
v(0, t) = 0,
v(1,t)=0,
1>0.
450
11
An Introduction to Partial Differential Equations
In Exercises 32 through 45, solve the initial-boundary value problem (26)-(30) of Exercise 31 for the conditions given. [Hint: Find v, then determine u from Eq. (31)] 32. A = 3, B = 0 and the conditions of Exercise 2.
33. A = 0, B = 3 and the conditions of Exercise 1. 34. A = -3, B = 2 and the conditions of Exercise 4. 35. A = 2, B = 2 and the conditions of Exercise 3.
36. A = 0, B = -2 and the conditions of Exercise 6. 37. A = 10, B = it and the conditions of Exercise 5. 38. A = 8a, B = it and the conditions of Exercise 8. 39. A = 7, B = 2 and the conditions of Exercise 7. 40. A = 4, B = 0 and the conditions of Exercise 10. 41. A = 0, B = 5 and the conditions of Exercise 9. 42. A = 9, B = 5 and the conditions of Exercise 12. 43. A = 0, B = 8 and the conditions of Exercise 11. 44. A = 13, B = -3 and the conditions of Exercise 14. 45. A = 6, B = 0 and the conditions of Exercise 13.
46. Give a physical interpretation of the initial conditions of Example 1 (assume that the problem relates to a "string").
47. Give a physical interpretation to the initial conditions of Example 2 (consider that the problem relates to a "string").
48. A tightly stretched string 3 feet long weighs 0.9 lb. and is under a constant tension of 10 lb. The string is initially straight and is set into motion by imparting to each of its points an initial velocity of 1 ft/sec. (a) Find the
displacement u as a function of x and t. (b) Find an expression for the
displacement of the midpoint one minute after the motion has begun. 49. Wave Equation with Damping If there is a damping force present, for example, air resistance, the equation of the vibrating string becomes
u + 2au, - c2u = 0,
where a is a positive constant known as the damping factor. (a) Set u(x, t) = e-°'v(x, t) and show that v satisfies
(e) Describc u u2 as "waves" and determine their speed (see Exercise
50. A taut string of length lm and c = 1 is subjected to air resistance damping
for which a = I (see Exercise 49). Using the method of separation of
variables and Fourier series, find the displacement as a function of t and x if the initial displacement is zero and the initial velocity is I m/sec.
51. Torsional Vibration in Shafts A shaft (rod) of circular cross section has its axis along the x-axis, the ends coinciding with x = 0 and x = 1. The shaft
is subjected to a twisting action and then released. 0(x, t) denotes the
angular displacement undergone by the mass in the circular cross section located at the position x and time t (that is, the mass of a very thin disc located there). It can be shown that 0 satisfies the one-dimensional wave equation with c = G/p, where G (a constant) is known as the shear modulus of the shaft and p (a constant) is the density of the shaft. Furthermore, it is known in the theory of elasticity that 0, = T/Gµ, where r is the twisting moment (torque) and µ is the polar moment of inertia. Two types of end conditions are common. They are a fixed end (for which 0 = 0) and a free end (for which 0, = 0, since T = 0). Set up and solve the initial-boundary value problem for the angular dis-
placement of a rod that is fixed at the end x = 0, free at the end x = I
(take I = 1), whose initial velocity (0,) is zero, and whose initial displacement
is given by 3x. Take c = 1.
52. Repeat Exercise 51 by considering all the information to be the same except that the end at x = 0 is free instead of fixed, and the end at x = 1 is fixed instead of free.
53. Plucked String When the initial displacement of the string is of the form
mx,
0:5 X!5 X
f(x) =
=0 (1-x), xosx:Sl
- xo
one can say that the string has been "pinched" at the point x = x0, lifted
(in case m > 0) to the height mxo, and then released. This action is described as plucking the string. Find the displacement of the string that is plucked
at its midpoint and released from rest. Take c = 1 = 1, m = 1.
Remark
It is reasonable to think of a guitar string as being plucked, since a guitar pick or a person's fingernail can be thought of as acting at a point
on the string. The same would be true for a harp string except that the plucking very often occurs at two or more positions on the string. On the other hand, piano strings are set into motion when struck by a hammer,
452
11 An Introduction to Partial Differential Equation
which does not act at a certain point on the string, but rather on a segment of the string. In this case, it is reasonable to think of the string as being in an initial horizontal position; the hammer imparts an initial velocity to the
portion of the string it strikes, and the rest of the string has zero initial
velocity.
54. Piano String Find the displacement of a 2-foot piano string that is struck by a 2-inch hammer having an initial velocity of 1 ft/sec, if it is known that the center of the hammer strikes the center of the string. Take c = 1. [Hint: See the preceding remark.] 55. A 2-inch-wide acorn travelling at 3 intsec. strikes a taut spider's web consisting of a single horizontal thread 6 inches long. Find the displacement of the web if it is known that the center of the acorn strikes the web at a
point 2 inches from its left end. Take c = 1. [Hint: See the preceding
remark.]
56. Harmonica The solution given in Eq. (18) can be written in the form u(x, t) _
where
u"(x, t),
u"(x, t) =
and
n `x
r
,
1
A" = (an + b,)12, h(x) = sin
and
g(t) = cos I n 7 c t
(Recall from trigonometry that A cos at + B sin at = C cos (at - R), with C = (A2 + B2)11 and cos P = A/C.) By itself, u" is a possible motion of the string and is called the nth normal (or natural) mode of vibration or the nth harmonic; u, is called the fundamental mode or the fundamental harmonic. If we consider x to be fixed, then u" is a simple harmonic motion of period
_
P.
21
nc
and frequency
nc v"=u.
v" is called the nth natural frequency, and v, is called the fundamental frequency of vibration. If u" = 0, then v" is taken to be zero. Find the first four harmonics and the first four natural frequencies for the string of Example 1 of this section.
57. A clothesline 10 feet in length is to be considered as a taut flexible string. A boy strikes the clothesline with a paddle that is one foot wide, so that the paddle is travelling with speed 3 ft/sec at the time of contact. The point of contact of the center of the paddle is at the point 3.5 feet from the left
11.6 The Homogeneous One-Dimensional Wave Equation
end of the clothesline. Find the displacement of the clothesline. Take
c = 1.
58. The transverse displacement, u(x, t), of a uniform beam satisfies the partial differential equation u,, + c'u_ = 0, where c2 = ElIpS, with E a constant (Young's modulus), I a constant (moment of inertia of the cross section area), p a constant (the density), and S a constant (the area of cross section).'
An end of the beam is said to be hinged if u = 0 and u = 0 at that end. Set up and solve the initial-boundary value problem for the transverse
displacement of a beam of length I that is hinged at both ends, having initial displacement f(x) and initial velocity zero. (Hint: Use the method of sdparation of variables and Fourier series.]
59. If the beam of Exercise 58 is subjected to axial forces, F(t), applied at its ends, the transverse displacement, u(x, t), satisfies' the partial differential equation
Elu,,,, - F(t)u + pSu = 0.
Assume that the beam is of length I and has hinged ends. This equation does not "separate," and we cannot use the method of this section. Set
u(x, t) _
(a) Show that u satisfies the boundary conditions. (b) Determine a differential equation that T .(t) must satisfy if u is to be a solution of the partial differential equation. 60. The Telegraph Equation If u(x, t) and i(x, t) represent the voltage and current, respectively, in a cable where t denotes time and x denotes the position in the cable measured from a fixed initial position, the governing equations' are
Cu,+Gu+i,=0 Li,+Ri+u,=0.
The constants C, G, L, and R represent electrostatic capacity per unit length, leakage conductance per unit length, inductance per unit length,
and resistance per unit length, respectively.
(a) Eliminate i from the above system to show that u satisfies the telegraph equation
LCu,, + (LG + RC)u, + RGu = u,,.
(34)
[Hint: Differentiate the first equation with respect to t and the second with respect to x.] (b) Eliminate u from the above system and show that i satisfies Eq. (34).
'S. Timoshenko, Vibration Problems in Engineering (Princeton, N.J.: Van Nostrand, 1928), p. 221. 'Ibid., p. 374.
61. Travelling Waves Divide by LC and rewrite the telegraph equation, Eq. (34), in the form
u + (a + (3)u, + apu = C'u,,,
with c' = 1/LC, a = GIC, p= RIL.
(a) Set u(x, t) = e-1121°'a''v(x, t), and show that if u is a solution of the telegraph equation, then v satisfies
(b) If a = p, that is, GL = RC, then v satisfies the one-dimensional wave equation; although we do not have initial-boundary conditions, we can still discuss the solution. Using the method of Exercise 40, Section 11.4, show that v has the form
v(x, t) = f(x + ct) + g(x - ct).
The expression f(x + ct) can be thought of as a "wave"8 travelling to the left with speed c, and g(x - ct) as a "wave" travelling to the right
with speed c. Thus the solution u of part (a) can be described as a voltage (or current) subjected to damping that is travelling in both directions in the cable. Thus, for appropriate properties of the cable (a = p), signals can be transmitted along the cable in a "relatively"
undistorted form yet damped in time.
62. Maxwell's Equation for the Electromagnetic Field Intensity in a Homogeneous
Medium The partial differential equation
4o(E), (E) _ 1A[E] -
(35)
is one of Maxwell's equations that occur in electrodynamics.' E is a vector representing the electromagnetic field intensity, and a (conductivity), a (die-
lectric constant), and µ (permeability) are constants associated with the medium; 8 is a constant associated with the conversion of units. The onedimensional version of Eq. (35) can be written in the form (32) with c= a = 2ao/e. Repeat the procedure outlined in Exercise 49. In this application the factor e-°' is called the attenuation factor.
63. Determine the tension in a string of length 100 cm. and density 1.5 gram per meter, so that the fundamental frequency of the string is 256 cycles per second (256 cps is middle C on the musical scale).
'See, for example, J. M. Pearson, A Theory of Waves (Boston: Allyn and Bacon, 1966), p. 2. 'Pearson, Theory of Waves, p. 29.
11.7 The One-Dimensional Nest Equation
11.7
THE ONE-DIMENSIONAL HEAT EQUATION
In the investigation of the flow of heat in a conducting body, the following three laws have been deduced from experimentation.
LAW 1 Heat will flow from a region of higher temperature to a region of lower temperature.
LAW 2 The amount of heat in the body is proportional to the temperature of the body and to the mass of the body.
Heat flows across an area at a rate proportional to the area and to the temperature gradient (that is, the rate of change of temperature with respect to distance where the distance is taken perpendicular to the area).
LAW 3
We consider a rod of length I and constant cross sectional area A. The rod is assumed to be made of material that conducts heat uniformly. The lateral surface of the rod is insulated so that the streamlines of heat are straight lines perpendicular to the cross sectional area A. The x-axis is taken parallel to and in the same direction as the flow of heat. The point x = 0 is at one end of the rod and the point x = I is at the other end. p denotes the density of the material, and c (a constant) denotes the specific heat of the material. (Specific heat is the amount of heat energy required to raise a unit mass of the material one unit of temperature change.) u(x, t) denotes the temperature at time t (> 0) in a cross sectional area A, x units from the end x = 0. Consider a small portion of the rod of thickness Ax that is between x and x + Ox. The amount of heat in this portion is, by Law 2, cpAOxu. Thus, the time rate of change of this quantity of
heat is cpAAx
au
. Thus,
cpAOx dl = (rate into this portion) - (rate out of this portion).
From Law 3 we have
rate in = - kA
au
ax
x,
rate out = - kA ax
x + 'IX,
where the minus sign is a consequence of Law (1) and our assumption regarding the orientation of the x-axis. The constant of proportionality k is known as the thermal conductivity. Thus,
cpAAx a!
kA 2X
x
+ kA - x + Ox ax
i
.
(2)
(3) (4)
(5)
u(x.4)..
For k as given in (9).3.
.. t > 0..2. 0) = f(x).
EXAMPLE 1
Solve10 the initial-boundary value problem
0<x<1.
u(0.
(9)
(10)
X(x)=sinnjx. t>0. respectively.
X(0) = 0. 0 < x < 1.e Thus..
t > 0.456
or
11
An Introduction to Partial Differential Equations
[au
au at
aul
ax x
Ax
k axlx+Ax
cp
Taking the limit as Ax tends to zero. u(l. we find that X is to be a solution of the eigenvalue problem
X"-aX=O. 3. Using the method of separation of variables. t) _
X (x)T"(t)
1OIt can be shown that if f satisfies the Dirichict conditions (Theorem 1. n=1. Note that the heat equation is a
parabolic partial differential equation.
n27r2a
(8)
l2
.
The eigenvalues and eigenfunctions of problem (6)-(7) are. we obtain the one-dimensional heat equation
(1)
where a = klcp is known as the diffusivity. .
Solution We note that conditions (3) and (4) indicate that the ends of the rod are in contact with a heat reservoir of constant temperature zero. t) = X(x)T(t).
and T is to satisfy the differential equation
(6) (7)
T'-XT=O.. .
n = 1. We seek a solution in the form u(x. x(l) = 0. problem (2)-(5) has a unique solution. t) = 0.
u(x. Section 10. 2. (8) is T(t) = c. the general solution of Eq. t) = 0. (5) gives the initial distribution of temperature.
]
28. . :)u.
u(x. 32. If the temperature is a function of the radius only. the rod is receiving heat (thus. u). For this reason we have restricted our attention to the one-dimensional heat
equation.3. Section 11. [Hint: Set u = e-' v(x.). however. The following special cases are of interest. t). u(x.au_ = q(x.1) = 0. if r is negative. t) = 0. t. t > 0. L. t. a heat source exists). however.9 for a specific example. p. u(0. u) is the rate of production of heat energy per unit volume per unit time.
0<x<1. u) = r(x.
0 < x < 1. show that the twodimensional heat equation can be written in the form
U.x). and 32 through 36).
(18)
[Hint: See Exercise 50.
(19)
where h(x.
"P. =
a
(ru. t) = Au(0. McGregor. This situation corresponds to heating caused by an electric current through the rod and gives rise to a nonhomogeneous problem.
26. Berg and 1.7 The One-Dimensional Heat Equation
Consequently one can consider heat diffusion problems in two and three
dimensions also. the rod is giving off heat (thus. then the equation" of heat flow is
u. This case can be thought of as heating of the rod caused by a chemical reaction that is proportional to the local temperature. See Exercise 24. If r is positive.(0. 1 = a = 1. t. Automatic Heat Control The initial-boundary value problem
u. t>0
t > 0.=au_. t) = u(1. These functions are considerably more difficult to deal with than the sine and cosine functions heretofore encountered. a heat loss
exists).
.1
27. t. 1966). (a) q(x.and threedimensional problems that lead to relatively simple solutions (see Exercises 26. 0) = x(1 .
(19) with q = -au. Section 11. There are. Heat Equation: Source Terms If internal sources of heat are present in the rod and if the rate of production of heat is the same throughout any given cross section of area. 0) = f(x).
u.
u(l. t). the analysis is
complicated by the fact that the corresponding eigenvalue problems involve Bessel and Legendre functions. (b) q(x. the initial-boundary
value problem for v is solvable by the methods of this section. u) = F(x.11. Elementary Partial Differential Equations (San Francisco: HoldenDay. a(> 0) a constant.
Set up and solve the initial-boundary value problem consisting of Eq. t). u) = cpq(x. W. Some of these higher-dimensional problems can be handled
in precisely the same manner as in this section. some special types of two. t.
2<x<1 . with the hot rod in the middle. (b) Show that there are no real eigenvalues in the case A < . The initial distribution of temperature in the rod is f(x)..). The rods are placed end to end. the other has constant temperature 0°C throughout. 0<x< 2
U. Three rods of the same material. and whose right end is equipped with an automatic heat control which keeps the temperature at this end proportional to the temperature at the left end. 31. no heat is lost through this end. If u(r. from this common face twenty minutes after contact is made. =
a
r
(r2U.1.
1
where A is a constant. Two rods of the same type of material.
33. show that the heat equation in spherical coordinates for a temperature distribution that depends only on the radius r and the time t is
u. find to the nearest degree the temperature at a point on the common face and at points 5 cm.). =
r (ru) . a = 1. Solve the initial-boundary value problem (2)-(5) when
/Tx) = 1
IA.02.
(b) Show that
(r 2u. long. long. The outer faces are maintained at 0°C. Just prior to contact. t) denotes the
. (a) If .
30. one of the rods has constant temperature 100°C.I < A < 1. [Hint: See Exercise 29. are each 10 cm.3 and the Remark which
follows that exercise. Temperature Distribution in a Sphere The surface of a homogeneous solid sphere of radius R is maintained at the constant temperature 0°C and has
an initial temperature distribution given by g(r).
. (a) Find the temperature distribution as a function of x and t after contact is made. The rods each have constant temperature.] (b) If a = 0.
29. hence ux = 0).
32. (a) Referring to Exercise 50(c) of Section 11. two at 0°C and one at 100°C. each 10 cm.2 for the material.11
An Introduction to Partial Differential Equations
where A is a constant (A # 1). describes the temperature in a rod whose left end (x = 0) is thermally insulated (that is. Find the temperature at a point on the middle (center) cross section of the "new" rod if the ends of the "new" rod are maintained at 0°C. are placed face to face in perfect contact. solve this initial-boundary value problem.
then u(r.r2. t). t) satisfies the problem (see Exercise 32)
u. u(r.
0 <.
t > 0.5j. is made of material for which a = 0. A solid sphere of radius 10 cm. we determine that the gain of heat in the horizontal direction for this portion is
kDAy
ax ix + Oz
au ax au
and in the vertical direction it is
kD. therefore. 0<r<R.
Y+Ay ayY
kDt
or
1
Thus. a constant. (a) Set up and solve the
corresponding initial-boundary value problem for v. Repeat Exercise 33 for g(r) = A. t) = ru(r. the total gain of heat for this portion of the sheet is
alx+ax
au
ax Ix + AX
Ox
au I ax
ay du
di
ax
au
AY
kDOxOy
+ ay I Y +
aY
Y
AY
. Using the same arguments as in Section 11.8 THE POTENTIAL (LAPLACE) EQUATION
We begin this section with a consideration of heat flow in two dimensions. and consider the flow of heat in the differential portion of the sheet depicted in Figure 11.2. Initially the temperature is 100°C throughout the sphere.
It is reasonable to assume that u is bounded at r = 0. we would have v(0. 35. and the sphere
is cooled by keeping the surface of the sphere at 0°C. 34.7. if we set
v(r.8 The Potential (Laplace) Equation
461
temperature in the sphere as a function of the radius r and the time t only.2. 0) = g(r).1 1. Refer
to Section 11. Find the temperature
(to the nearest degree) at the center of the sphere 20 minutes after the
cooling begins..
The conducting material is in the shape of a rectangular sheet of constant
thickness D.7 for the appropriate definitions and the experimental laws associated with heat flow. [Hint: See Exercise 33. :>0
u(R. t) = 0..]
11. = a(ru). Repeat Exercise 33 for g(r) = RZ . (b) Find u. t) = 0.r < R. 36.
there are many other physical applications in which the potential equation occurs.
(1)
Equation (1) is called the two-dimensional potential (Laplace) equation. a common situation is that u is to satisfy Eq. Note that the potential equation is an elliptic partial differential equation.2
By Law 2 of Section 11. for R.
. this rate of gain of heat is also given by
cpDOxAy at . It is the geometry of R and the nature of the boundary conditions
that dictate whether or not it is easy (or even possible) to solve the Dirichlet
problem. and that the values of u are known on the
boundaries of this region. the temperature u is a function of x and y only
(that is. say R. a typical problem associated with the potential
equation is a boundary value problem.
where a = k/pc.7. of the xy plane. + uyy = 0. Hence. commonly referred to as Dirichlet's
problem. = 0).
If at a future point in time. In this case and for that of two-dimensional heat flow (as well as others). Consequently. (1) in a fixed bounded
region..462
11
An Introduction to Partial Differential Equatlona
(x. One application is that u represents the potential of a two-dimensional electrostatic field.x and Ay tend to zero. u is independent of time and u. one says that steady state conditions
have been achieved. the partial differential equation for the
steady state temperature for two-dimensional heat flow is
u.
du du ax x + Ax Ox
_ du
ax
au
+ -ay
cPDt1r0Y at = kDOxY
Y + Ay
Ay
du ay
y
Simplifying and taking the limit as both . we obtain the two-dimensional heat equation
u.y)
Figure 11. Thus. Y)
(x+Ax. = a(uu + ayy). In addition to the steady state temperature for two-dimensional heat flow.
(5) and T . By Theorem 1 of Section 6. it is sometimes possible to introduce a change of variables that makes the boundary conditions have this form (see Section 11.. (10) is given by
c = T JO) = J f(x)4.(X) = -. As in Example 1.
EXAMPLE 2
Solve the initial-boundary value problem
0<x<1.
and we seek a solution of problem (12)-(16) in the form
u(x.
t > 0. (5) and (6).7. 0) = fjx).2.(t) is given by Eqs.6. t) = 0 is solvable and that the
"boundary terms" in the integration by parts process [see Eq. 1>0
u(x.
0 < x < 1. t)
T
(x) is given by Eq. we set +.
0 < x < 1.8.10). the solution of the initial-boundary value problem (1)-(4) is given by
u(x. we note that
TJO) = fu(x.
Consequently.(x.
u(l. t) = 0 in Eq.nsous Partial Diftarsntial Equations: Mathod I
471
From Eqs. 40 = J u(x.
0
(11)
Finally.11. The method of solution outlined in Example 1 is useful for many problems of the type encountered in Sections 11.6. The latter requirement will be met if the boundary conditions for the original problem are u(0.
(14) (15) (16)
u(0. xx
.
(12)
(13)
u. (12). t) = 0.9 Nonhomog. t) = 0. and 11. (8)) either vanish
or are known as a function of t.(1) are to be determined.(x)dx. t) = i
.-1
where the functions T. It is important that the
associated problem obtained by setting F(x.
(17)
. 0) = g(x).
Solution If we set F(x. (10) and (11). 11.
If this is not the case.. we can write
T JO = (u. we then have the initial-boundary value problem of Section 11./2-/l sin
n.
t > 0. t) = 0. the constant c in Eq. t) = 0 and u(1.
EXAMPLE 1
Find the solution of the initial-boundary value problem
u. Substitution of (1) into Problem 1 yields
PROBLEM 1'
A[v] = F(x.9 or by one of the methods in Sections 11. 0) = f3(x) . t) = 0.
t > 0. 0).11.
11. t> 0.
u(l.(x. since the only criterion is that Problem 1' be a solvable problem. 0) = f3(x). The string is
initially at rest and is subjected to the external force 7r2x.(1.
10
IOl GENEOUS PARTIAL DIFFERENTIAL EQUATIONS:
METHOD
We write the partial differential equation (1) in Problem 1 of Section 11. y) .A2w. y) = f.(0. in other words. y)
A3v(x. Once w is chosen and v is calculated. Find the displace-
ment of the string as a function of the time i and the distance x.
A.(x.A4w.
t > 0. the desired solution u is obtained from Eq. y). or 11. Y) .w(0.8. Y) + A4v.(x) .
0 < x < I. A string is stretched between the points (0.10 Nonhomogeneous Partial Differential Equations: Method II
475
displacement of the string as a function of the time t and the distance x. y) + w(x.
.5 in the form A[u] = F(x.A. Take
c = 1. y)
A3v(1.(y) . y) is an unknown function to be determined. y) + A2v. (1).w(x. y) = v(x. Set
u(x.6. 0).(1. 11.
u(0.
(1)
where v(x.A. 0) and (1. u = w + v. y). and w(x.A[w]. y) is to be a new unknown function. For a given problem we may have considerable flexibility in the choice of w. (1).au_ = F. 0) = f. 0) A6v. t).
u(x.(0. t) so that Problem 1' reduces to a problem that is solvable either by the method of Section 11.(x. y) = f2(y) .A3µ'(1.7. .v(0. t) .A6w. where A is a linear operator. The idea is to choose a specific w(x.. 26. t) = f. 0 <x<1. Take c = 1.
t) = u(1. Section 11. for the beam satisfies Eq.(x. 0) = 0.
u.12) (the time t is measured in hours.
. y) _ . and midnight corresponds to t = 24). [Hint: The original rectangle is .11
An Introduction to Partial Differential Equations
14.
)2
puN. For convenience of the eigenvalue problem. 0) = 0. Consider that the surface of the earth is warmed by the sun so that the temperature.6). = f(x.
t > 0.
u(x. t) = t3/3!. Suppose that a circular shaft (refer to Exercise 51.
u + a2u = uu. If a beam of uniform cross section has its axis coincident with the z-axis. Solve the new boundary value problem as a first step. If the initial temperature
distribution in the earth's crust is f(x) = 28 IX 1 1J . set up and solve the
initial-boundary value problem for the temperature in the earth's crust when o = 1 = 1.a/2 s y s a/2.t)+11-xP
\\\
1
3!
. Set up and solve the
initial-boundary value problem for the angular displacement.
u(0.x <_ 7/2.
17. Take c = 1.
0<x<1. Assume that the shaft undergoes torsional vibrations due to a periodic rotation at
the end x = /(take l = 1) of the form f(t) = cos t. where a.
0 < x < 1.6) has the end x = 0 fixed and is initially at rest in its equilibrium position.)
16. t = y + a/2. Find the stress function of a square
beam of side a units. t). set s = x + a/2. u.a/2 <. and note that Poisson's equation becomes
u + u = -2. then linear elasticity theory shows that the stress function. (4) with Ax. u(x. Section 11. Assume a section of the earth's crust to be a rod with one end at the surface of the earth (considered to be at x = 0) and the other end (considered to
be at x = 1) inside the earth at such a depth that the temperature at that end is fixed (taken to be 0°C). 1 AM corresponds to
t = 1. 1)
=v(x.
t > 0. y). (Assume that the effect of time units in hours rather than seconds has already been accounted for and that no further modifications are necessary. Y) (4)
is known as the (two-dimensional) Poisson equation. at the surface is given by 28 cos i2 (t .]
15. p are constants.2 and u = 0 on the boundary of the area of intersection
of the beam with the xy-plane. Set
u(x. Torsion of a Beam The two-dimensional nonhomogeneous Laplace equation uu + uy. The following initial-boundary value problem corresponds to a special case of the telegraph equation (Exercise 60. t>0
0 <X < 1.
5. and f(r) = 100(101 . Integrate the equation to obtain the general solution.5y2u. < r < r2. and we wish to in-
vestigate the loss of heat through the sides of the pipe. 18..r). The pipe can be considered a hollow cylinder of inner radius r outer radius r2. the temperature in the pipe satisfies the initial value problem u.=z'+2xy-t
6. Loss of Heat through the Sides of a Pipe Suppose we have a cylindrical pipe filled with a hot fluid (of constant temperature). < r < r2. t > 0
u(r2..r)
produces a nonhomogeneous equation (for v) with homogeneous initialboundary conditions (the solution of which involves Bessel functions).10 Nonhomogeneous Partial Differential Equations: Method 11
481
and determine an equation for the eigenvalues of the initial-boundary value problem for v(x.. u.2(u.)2 + 2u2u.
r. indicate whether it is homogeneous or nonhomogeneous and whether it has constant
coefficients or variable coefficients. assume that u is a function of four variables x.. t) = v(r. Show that u = x' + cos (x + 3y) is a particular solution of 9u . t) + 100(101 . t). uu + 5u. u.
.
u(r. B = 0°C.. =
54x. = a
A. u. therefore.11.
1.
t>0. if it is linear. + 3u = 0 2. = 0
In Exercises 5 and 6. state the order of the partial differential equation. y.
r. u . 2xy [Hint: set v = u...]
7. For the case A = 100°C. 8.
where A is the temperature of the fluid. t) = B.. and. + u...
r2 = 101 cm. state whether it is linear or quasilinear... and f(r) is the initial distribution of temperature in the pipe.. r. B is the temperature of the air (or medium) surrounding the pipe. + uY = 5x + 3y. u-. Show that u = 3xy + x' is a particular solution of u. = 3xy 3. . take a = 1 and show that the
substitution
u(r.u..
REVIEW EXERCISES
In each of Exercises 1 through 4. 0) = f(r).. z and t.. t>0. = 100 cm. | 677.169 | 1 |
Algebra is where we substitute a letter (called a pronumeral) to make equations (and even maybe solve them). Algebra isn't hard - if you know the basics. Thi...
Brendon C's insight:
This quick video will explain what algebra is and how it can be used. Have a look at this video before next class, it is a fantastic starting point and grounding to the fundamentals of the next content area, algebra | 677.169 | 1 |
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0.28 MB | 120 pages
PRODUCT DESCRIPTION
Get up and MOVE with this review that will get students out of their seats and engaged with the material. Students work independently on three sets of problems that are linked to make it easy to check correct answers. As students move around the room working problems and searching for answers, they will circle back to their starting place when they have done them correctly. Included answer keys make it easy to see which problem students got off track on | 677.169 | 1 |
CCSS 6.EE.9 Test on Functions, Tables, and Rules
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0.6 MB | 7 pages
PRODUCT DESCRIPTION
Aligned to Common Core State Standareds 6.EE.9
This test covers functions, tables, and rules, including using a pattern of figures to create a table and find the function rule. The test is divided into the following sections:
Questions 1-4: Students are given incomplete function tables and the function rules (in the form of equations). They are to use the rule to complete the table.
Questions 5-8: Students are given incomplete function tables, without the function rules. They are to write the function rule (as an equation) and use it to complete the table.
Questions 9-11: These are multiple choice questions that require students to read a word problem based on a real life scenario, use a given function table, and answer questions related to the function rules and later terms (i.e. Using the rule, how much will Sarah earn in 10 months?)
Questions 12-14: Students are given a pattern of figures and are required to complete a function table, write the function rule as an equation, and find a later term (nth term).
Questions 15-16: This is a constructed response question. Students read a description of a real life situation and are required to find the function rule, complete a table, and find a later term. However, the function rule will be a "two-step" equation; there will be a constant. Below is the real life situation:
"Josh is trying to earn money for a field trip for his class. Mrs. Johnson agreed to give him a one-time donation of $22. For every touchdown Josh scores, Coach Andrews said he will give him $8."
Bonuses: There are two bonus questions included. One requires students to analyze a function table and write the function rule. It will be a two-step rule. The other requires the student to complete a function table when given a two-step function rule. This requires substitution.
Please note that each question is given a point value, such that the entire test is worth 100 points. Point values are stated in each section of the test. This makes for easy grading!
The answer key is provided, as well!
Please check out the downloadable preview for a sample of what the test will look like | 677.169 | 1 |
MATH 112 is the designed to give prospective elementary education majors the depth of understanding necessary to teach mathematics in the elementary classroom. Topics will include representations of and operations on the natural numbers, integers, rational numbers and real numbers, and properties of those operations. Emphasis will be on communication, connections to other parts of mathematics, problem solving, representations, and reasoning and proof.
Upon successful completion of MATH 112, the student should be able to:
Demonstrate various representations of Natural numbers and Integers.
Define the operations on Natural numbers and Integers.
Identify, describe, and demonstrate the proper use of the properties of operations on Natural numbers and Integers.
Demonstrate various representations of Rational and Real numbers.
Define the operations on Rational and Real numbers .
Identify, describe, and demonstrate the proper use of the properties of operations on Rational and Real numbers. | 677.169 | 1 |
pinner says: Today's project in College Algebra was making the color coded #logarithmic definition patterns "cheat sheets." It worked like a charm! Almost the whole class was engaged and understood how to convert from one to the other! ~Bon
What are logarithms and why are they useful? Get the basics on these critical mathematical functions -- and discover why smart use of logarithms can determine whether your eyes turn red at the swimming pool this summer. Lesson by Steve Kelly, animation by TED-Ed. | 677.169 | 1 |
Publisher's Description
Here's real help for math students. From abacus to zero property of multiplication, this handy reference guide for students contains more than five hundred common mathematical terms. Written in simple language and illustrated with hundreds of helpful photographs and drawings, Math Dictionary takes the mystery out of math.
Author Bio
Eula Ewing Monroe is a former classroom teacher. She lives in Provo, Utah, where she teaches mathematics education at Brigham Young | 677.169 | 1 |
Exploring Mathematics is intended for a one or two-term course in mathematics for college students majoring in the social sciences, English, history, music, art, education, humanities, philosophy, criminal justice, government, or any of the other majors inMore...
This textbook is intended for a one or two-term course in mathematics for college students majoring in the social sciences, English, history, music, art, education, humanities, philosophy, criminal justice, government, or any of the other majors in liberal arts. Most colleges and universities require liberal arts students to fulfill a math or quantitative component. The liberal arts mathematics course with an algebra prerequsite is a popular selection for these students. Our 9-chapter textbook offers modern applications of mathematics in the liberal arts as well as the aesthetic features of this very rich facet of history and ongoing advancement of human society.Johnson believes that both the use of the concept of function as a central theme and the inclusion of several chapters seldom seen in other texts of this genre considerably broadens the appeal of this text. It provides a sample of the next level of mathematics for which a college student who has passed intermediate algebra should be able to master. The book attempts to answer the questions, "How does mathematics help us to better our society and understand the world around us?" and "What are some of the unifying ideas of mathematics?" The central theme helps to impress upon the student the feeling that mathematics is more than a disconnected potpourri of rules and tricks. Although it would be inappropriate to force a functional connection in every single section, the theme is used whenever possible to provide conceptual bridges between chapters. Developing the concept of a function augments the presentation of many topics in every chapter.The Text's Objectives:The author chose the topics based on meeting the specific NCTM curriculum standards to:1. Strengthen estimation and computational skills.2. Utilize algebraic concepts.3. Emphasize problem-solving and reasoning.4. Emphasize pattern and relationship recognition.5. Highlight importance of units in measurement.6. Highlight importance of the notion of a mathematical function.7. Display mathematical connections to other disciplines.Key Features:Writing StyleAbundance of ExamplesNonstandard MaterialVocabulary ListChapter Review TestsHistorical FiguresFlexibility | 677.169 | 1 |
Availability: Available to Backorder, No Due Date for Supply, Not for Xmas
Thinking about Ordinary Differential Equations by Robert E. O'Malley, Jr.
Book Description
Ordinary differential equations - the building blocks of mathematical modelling - are also key elements of disciplines as diverse as engineering and economics. While mastery of these equations is essential, adhering to any one method of solving them is not: this book stresses alternative examples and analyses by means of which the student can build an understanding of a number of approaches to finding solutions and understanding their behaviour. This book offers not only an applied perspective for the student learning to solve differential equations, but also the challenge to apply these analytical tools in the context of singular perturbations, which arises in many areas of application. An important resource for the advanced undergradute, this book would be equally useful for the beginning graduate student investigating further approaches to these essential equations | 677.169 | 1 |
A treatise on spherical trigonometry; with applications to
Teachers and Home schoolers use the math worksheets on this website to measure the children's mastery of basic math skills, give extra practice, homework practice, and save precious planning time. The following general relations between a ratio and its reciprocal should be noted: (a) When the ratio is increasing its reciprocal is decreasing, and vice versa. (b) When a rauo IS a maximum us reciprocal will be a mtntmum, and vice versa Consequently Since the maximum value of the Sine and COSine In the first quadrant I" Unity, the minimum value of the cosecant and secant must be Unity If a number IS very large, ItS reciprocal IS very small.
Pages: 58
Publisher: RareBooksClub.com (March 6, 2012)
ISBN: 1130543153
Bevel Gear Tables
Algebra 1 holt work book answers, math solver online, japanese abacus lcm, check for nagative in javascript, introductory algebra fourth edition chapter 9 help Modern We have presented Mathematics in an interesting and easy-to-understand manner. Mathematical ideas have been explained in the simplest possible way. Here you will have plenty of math help and lots of fun while learning and teaching math step-by-step. Keeping in mind the mental level of a child, every effort has been made to introduce new concepts in a simple language, so that the child or student understands them easily Elements of geometry and trigonometry Collected Papers of G. H. Hardy, Volume III; including joint papers with J. E. Littlewood and others. Trigonometric Series; Mean Values of Power Series.
When in addition there is a CE key to clear the last entry, the ONtC key usually clears the whole machine including the memories Seven place logarithmic tables of numbers and trigonometrical functions. The sine and cosine functions are fundamental to the theory of periodic functions such as those that describe sound and light waves Trigonometry: An Analytic Approach. After studying the problems I missed with Zoom Math's Trace feature as my guide I was able to pull my math section of the ACT up to a 31. I now have a 30 composite score and a full ride to the University of Alabama and you guys are 100% responsible!" - Andrew T., Sacramento, CA "We received my son's MCT scores from 10th grade (end of last year) The Way Out: Kick-starting Capitalism to Save Our Economic Ass. It may even reveal a new lucky number or, if you're on a roll, tons of lucky numbers! An electronic laboratory for exploring probability. Flip a coin, spin a spinner, roll a die from one to thousands of times, and examine results in a table, bar chart, pie chart, or innovative line graph. Challenge your intuition about random events: you may be surprised Logarithmic and Trigonometric Tables!
Algebra with Trigonometry for College Students
The Elements of Plane and Spherical Trigonometry; With the Construction and Use of Tables of Logarithms, Both of Numbers, and for Angles
Trigonometry: development and use
All of these books are complete courses, with plenty of college and high school classrooms, 28 half-hour video programs high level extra-curricular math concepts (primes, topology, chaos, etc) The use of graphs as a way of recording the data comes from Neugebauer's book The Exact Sciences in Antiquity Plane Geometry.... A common mathematical function known as factorial is represented by an exclamation point following a positive integer number. The following are all examples of factorials: Explain in your own words what this "factorial" function represents The Elements Of Trigonometry.... MATH 15300 covers applications of integration, an introduction to infinite sequences and series and Taylor expansions, and an introduction to multivariable calculus including functions of several real variables, partial derivatives, gradients, and the total derivative, and integration of functions of several variables Trigonometry: for schools and colleges. Chapter 12, containing 66 Sanskrit verses, was divided into two sections: "basic operations" (including cube roots, fractions, ratio and proportion, and barter) and "practical mathematics" (including mixture, mathematical series, plane figures, stacking bricks, sawing of timber, and piling of grain). [62] In the latter section, he stated his famous theorem on the diagonals of a cyclic quadrilateral: [62] Brahmagupta's theorem: If a cyclic quadrilateral has diagonals that are perpendicular to each other, then the perpendicular line drawn from the point of intersection of the diagonals to any side of the quadrilateral always bisects the opposite side Algebra and Trigonometry 5th (Fifth) Edition bySullivan. Pre-algebra with pizzazz! 224, SUARE ROOT, saxon math7/6grade answers to set 81, teach permutations powerpoint, solving addition and subtraction expressions, interactive positive and negative numbers. Cube root to fraction, adding variables with exponents, how to change decimals to fractions on ti 89, free ebooks of math for primary student, free 7th grade math worksheet printouts. 6th grade chapter 5 review for fractions, how do I solve for variables withe a caculator, diference in solving equations by graph,substitution and by elimination, Different Math Trivia, simplifying square root of 50 x to the 3rd, adding decimal piunts worksheets Combo: College Algebra with Trigonometry with Student Solutions Manual. | 677.169 | 1 |
Accelerated Algebra
Sunday, December 4, 2016
Algebra is a level of mathematics typically taught in high school which focuses on working with equations. While this course covers topics similar to that of a typical high school algebra class, we focus on developing students problem solving and critical thinking skills through real world applications and complex problems. We require students to develop a deeper understanding of algebraic concepts than that of a typical classroom by helping students to understand how and why rather than providing them with formulae. | 677.169 | 1 |
The fundamentals of probability are integrated into diverse math courses taught in high school and college. Now your can master introductory probability quickly and easily with Video Aided Instruction's Probability. The ultimate resource for high school students, college students, and adult learners, this set covers the standard probability topics taught in math classes and is jam-packed with practice questions and strategies for tackling even the most confusing problems.
Authors: Biagini, Francesca, Campanino, Massimo Provides an introduction to elementary probability and to Bayesian statistics using de Finetti's subjectivist approach Introduces as fundamental the concept of random numbers Includes numerous exercises covering most topics treated in the theoretical part of the book
Editors: Podolskij, M., Stelzer, R., Thorbjørnsen, S., Veraart, A.E.D. (Eds.) Honors the numerous scientific achievements of Ole E. Barndorff-Nielsen on the occasion of his 80th birthday Presents fascinating current research results regarding the theory of probability, statistics, as well as their applications Includes contributions by international experts | 677.169 | 1 |
ISBN 13: 9780007194636
Active Revision - KS3 Maths
One of the hardest things about revision is planning what to revise and when. Active Revision helps students to structure their revision effectively.
Key concepts are revised through clear worked examples in the accompanying full colour book. Interactive tests on the CD-Rom check students' understanding. An electronic revision planner and timetable helps students to plan their revision. Active Revision will provide students with a unique insight into where to concentrate their revision time and energies.
An exam practice section at the end of the book provides an opportunity for students to put all the skills into practice and shows them how to improve their grades. With examiners' advice and fully worked answers, this adds yet another dimension to this revision package.
For permission to use this software on more than one computer simultaneously, please purchase the relevant network licence.
"synopsis" may belong to another edition of this title.
From the Publisher:
Collins Revision Guides: A revision guide to suit every age, every stage and every pocket. New stunning colour books make learning fun, unique learning models will boost grades, clear concise information makes revision less daunting with all the facts for study and help presented in one portable book you simply can't be without. Written by GCSE Examiners and National Test Senior Moderators. The complete KS3 Do Brilliantly series; English, Science and Maths.
From the Back Cover:
• Written by the examiners using the latest – real – National Test papers • Complete KS3 Test practice paper (Reading and Writing, Shakespeare) • Model answers written at all levels
• Tutorials from Test Markers on improving Test scores and boosting performance • 'What's the question looking for' section provides advice on how to work out what skill is really being tested
Book Description Collins, 2005. Paperback. Book Condition: Good. Complete with CD40584
Book Description Collins 20/02194636
Book Description Collins 20/02194636
Book Description Collins 20/0219463663444
Book Description Collins, 2005. Paperback. Book Condition: Very Good. EXCELLENT value for money and ready for dispatch. Delivery normally within 3/4 days. Our reputation is built on our Speedy Delivery Service and our Customer Service Team. Bookseller Inventory # mon0003101123 | 677.169 | 1 |
The Common Core places increased emphasis on student understanding of function behavior, especially when it relates to interpreting graphs, tables, and symbolic representations. There are five separate high school standards that directly address interpreting the behavior of polynomial functions (HSA.APR.B.3, HSA.SSE.B.3, HSF.IF.B4, HSF.IF.C.7, HSF.IF.C.8). Can the students recognize roots, intercepts, and end behaviors on a graph? Can they determine those same values from a factored polynomial function? Can they manipulate a polynomial in standard form so that it is factored?
This Polynomial Graph Matching Activity is an entry point for this important content. Ten polynomial functions are presented in their factored form and compared to graphs in a one-to-one relationship. This allows students to apply what they know about end behavior, and then make and test predictions about intercepts. From this foundation they can begin to generalize their understanding and hopefully see the benefit of manipulating polynomial expressions from their standard form to a factored form.
Supporting Standards with Polynomials
Standards Alignment: Supporting Standards with Polynomials
Graphing Polynomials - Roots and the Fundamental Theorem of Algebra
Graphing Polynomials - Roots and the Fundamental Theorem of Algebra
Unit 4: Polynomial Theorems and Graphs
Lesson 10 of 15
Objective: SWBAT Produce a sketch of the graph of a higher order polynomial function using what they have learned about end-behavior, extreme values and roots of polynomial functions.
While I circulate around the room to check homework with the homework rubric at the start of today's lesson, students answer some TI NSpire Navigator questions about end behavior: Warm-up Polynomial End Behavior is a set of five questions that assess students' understanding of end behavior and the formal notation used to express it.
After the warm-up, my students work on Polynomial Graph Matching. I arrange students in pairs for this activity, choosing partners based on ability. For each partnership, I'd like to have one student who is strong with graphing and one who is strong in algebraic manipulation. Both skills are required for success in this activity.
Polynomial Graph Matching is a set of 20 cards with algebraic and graphical representations of polynomial functions. I included only algebraic functions in factored form to make it easier for my students to connect the graphs to the functions. The first time I used this activity, I printed out the cards on card stock and cut apart the cards. I have a set of storage drawers in my closet labeled with each unit I teach, so after I make them the first time I collect then and put them in the appropriate drawer for the next time I teach the unit. The last page of Polynomial Graph Matching is a record sheet. I print this out on plain paper and provide one sheet to each pair.
Up to this point, we have not yet explicitly discussed the relationship between the zeros of a polynomial and the x-intercepts of its graph so the matching activity will require students to think through and discuss this relationship [MP3]. While my students work I will use the 3 Cup System to provide support where needed without being "too helpful" [MP1]. I make written notes on which students seem to understand end behavior and which students are able to connect the linear factors with the roots of the equation. When students have matched up the nine pairs, they fill in the record sheet and turn it in to me. I remind them to leave the matched cards in front of them for our discussion.
By lining up all the record sheets in front of me, I can quickly see which of the graphs were hard and easy for my students. We go through each of the matches, discussing how we know which algebraic function pairs with each graph [MP2, MP3].
Resources
After reviewing the correct pairs for the Polynomial Graph Matching Activity, we discuss how the process of finding the x-intercepts of a polynomial relates to the Fundamental Theorem of Algebra and the Remainder Theorem. It's important to me that my students to not think of Algebra 2 as a long list of unrelated topics, so I am careful to give them time to connect new knowledge to what we have already learned. I will structure this discussion so that students can make sense of the following concepts:
The Remainder Theorem says that f(a)=0 if and only if f(x) divided by x-a has a remainder of zero. This makes sense because plugging in the x-intercept into the function gives and output of zero. All the x-intercepts come from setting the linear factors equal to zero.
The Fundamental Theorem of Algebra talks about the number of complex roots. The degree of the polynomial is sometimes bigger than the number of x-intercepts, so the complex roots must not be x-intercepts.
The Fundamental Theorem of Algebra says that a polynomial of degree n has n complex roots provided repeated roots are counted separately. The "counted separately" refers to roots where the graph touches and then turns around rather than crossing through.
This last point leads to a discussion of multiplicity, which will be a new concept for my students. Using examples of factored polynomial functions like
f(x)=(x-2)2(x-3)
f(x)=(x-2)3(x-5)4
After these final polynomial graph concepts have been explored, we take notes on the most efficient way of producing a good sketch of a polynomial graph. I advise my students to use a process something like this:
Think about end behavior and maybe draw in some small arrows to remind themselves
Calculate the y intercept and plot it on the graph
Factor the polynomial function completely in the real number system (if it isn't presented that way) to make it easy to determine the x-intercepts. Plot these on the graph, noting whether the graph will touch or cross at each intercept.
Connect the intercepts in such a way that the graph turns out to have the expected end behavior.
Resources (1)
Resources
As an exit ticket, I ask students to produce a sketch of two polynomial functions using the strategy presented in the notes. Exit Ticket-Graphs of Polynomials has two functions for students to graph:
One factorable quartic function whose factors all have a multiplicity of one
One factorable cubic function with one double root and a single root
I want students to reflect on the differences between even and odd functions.
For homework, student will work on 16 questions related to the graphs of polynomial functions in WS Graphing Polynomials. These exercises focus on my students' ability to connect the algebraic form of a polynomial function to the graph of the function [MP2]. | 677.169 | 1 |
This course provides an instruction in linear systems, linear programming, matrices, set theory, permutation and combinations, symbolic logic and switching networks, discrete probability and probability distributions, Markov chains and theory of games, and arithmetic and geometric progression. It satisfies some of the mathematics requirements for students of Business, Management, biological and social sciences, computer science and technology and information systems.
Learning Outcomes: Upon successful completion of this course, the student should be able to:
- Identify linear and quadratic functions, equations, and the slope of a line.
- Conduct a break even analysis.
- Solve and manipulate systems of linear equations, using matrices and matrix operations.
- Perform linear programming operations.
- Distinguish between permutations and combinations, conditional probability and independence.
- Define sample spaces, events and probability.
- Produce a bar graph, pie chart, and graph. Draw a histogram.
- Explain measures of central tendency and dispersion.
- Recognize various probability distributions and use them to solve problems.
- Use decision-making theories.
- Recognize and apply the properties of Markov chains.
Grading System: Letters
Passing Grade: D (50%)
Percentage of Individual Work: 100
Textbooks:
Textbooks are subject to change. Please contact the bookstore at your local campus for current book lists. | 677.169 | 1 |
Synopsis
Success in Maths for the Caribbean by Gerry Rose, Althea Foster
The Workbook that accompanies Success in Maths for the Caribbean (Workbook 2), is designed to stimulate students' intrest and to increase their confidence in mathematics along with providing additional exercises that can be used for homework or further classwork.
About the Author
The authors are well-known names in mathematics education. Althea Foster serves on the CXC Examining Committee and is co-author of the highly successful Mathematics for Caribbean Schools course. Gerald Rose is Research Fellow in the School of Education, UWI Cave Hill, and has over 20 years' experience of teacher training for mathematics education throughout the English-speaking Caribbean. | 677.169 | 1 |
[EAN: 9780070612068], Gebraucht, guter Zustand, [PU: McGraw-Hill Science/Engineering/], MATHEMATICS CALCULUS GEOMETRY | 677.169 | 1 |
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