text stringlengths 6 976k | token_count float64 677 677 | cluster_id int64 1 1 |
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Functional equations and functional identities with the range of functions. Using CAS and algebraically. Illustrated with past exam questions and typical examples. Notes pages to speed up answering MC questions.
Power point presentation on how to use the compound interest formula when compounding periods vary. Typical exam scenarios considered. Detailed instructions how to use Finance solver as a built-in financial package on TI Nspire .
Graphing sine and cosine curves with dilations, reflections and translations. Determining period, amplitude, coordinates of endpoints and axis intercepts. Useful strategies how to sketch the graphs by hand and using technology. Writing equations of a given graph. Stating which transformations were applied.
A set of tests on exponential & logarithmic algebra, trigonometry, functions, transformations, quadratics, polynomials, probability, counting methods, calculus, sequences and series. Each topic test has two papers, with and without GDC. A zipped file with 10 tests, each having two papers.
Domain and range, even and odd functions, implied domain, piecewise functions, applications. A zipped file containing a power point presentation, student calculator handout, tns file. What CAS features are available for functions.
A class activity to introduce solving simultaneous linear equations with a parameter. Geometrical interpretation. Using a slider on TI Nspire to illustrate the possible cases. Followed by solving with matrices. | 677.169 | 1 |
Product Description:
The hallmark of this text has been the authors' clear, careful, and concise presentation of linear algebra so that students can fully understand how the mathematics works. The text balances theory with examples, applications, and geometric intuition.Learning Tools CD-ROM will be automatically packaged free with every new text purchased from Houghton Mifflin.Section 3.4, now named Introduction to Eigenvalues, has been broken into two separate sections to provide more emphasis on the early introduction of eigenvalues. The new Section 3.5, Applications of Determinants, covers the Adjoint of a Matrix; Cramer' s Rule; and the Area, Volume, and Equations of Lines and Planes.All real data in exercises and examples have been updated to reflect current statistics and information.This edition features more Writing Exercises to reinforce critical-thinking skills and additional multi-part True/False Questions in the end-of-section and chapter review exercise sets to encourage students to think about mathematics from different perspectives.Additional exercises involving larger matrices have been added to the exercise sets where appropriate. These exercises will be linked to the data sets found on the web site and the Learning Tools CD-ROM.Eduspace is Houghton Mifflin' s online learning tool. Powered by Blackboard, Eduspace is a customizable, powerful and interactive platform that provides instructors with text-specific online courses and content. The Larson Elementary Linear Algebra course features algorithmic exercises and test bank content in question pools.
REVIEWS for Elementary Linear Algebra | 677.169 | 1 |
Mathematics A Simple Tool for Ge demonstrates mathematics using geological examples to illustrate mathematical ideas. This approach emphasizes the relevance of mathematics to geology, and gives examples of mathematical concepts in a context familiar to the reader.
This work demonstrates mathematics using geological examples to illustrate mathematical ideas. This approach emphasizes the relevance of mathematics to geology, and gives examples of mathematical concepts in a context familiar to the | 677.169 | 1 |
Technical Analysis and Applications with MATLAB, 1st Edition
This text combines technical and engineering mathematical concepts at a basic level using MATLAB® for support and analysis. Once math concepts are introduced and understood using conventional techniques, MATLAB® is then used as the primary tool for performing mathematical analysis. Featuring practical technical examples and problems, the text is designed for math courses within an engineering technology or engineering program or any courses where MATLAB is used as a supporting tool. The text provides a review of differential and integral calculus with an emphasis on applications to technical problems. | 677.169 | 1 |
This
course is designed to prepare students to achieve at the Proficient or Advanced
level on the Algebra 1 Keystone Exam.Students will utilize a variety of resources to provide supplemental
instruction for items tested on the Algebra 1 Keystone Exam.Instruction will be based on topics from the
core mathematics curriculum that are aligned with the eligible content that is
assessed on the Keystone through inquiry-based learning in real world
scenarios.Students may exit the course
when they achieve a minimal score of Proficient on the Exam or complete a
project-based assessment approved by the Pennsylvania Department of Education.
ESSENTIALS OF ALGEBRA 1 (Part 2) (IMPACT)
No. 3333 quadratic,
radical, and rational functions.As
students learn about each family of functions, they will learn to represent
them in multiple ways.They will also
learn to model real-life situations using functions in order to solve problems
arising from those situations.
ESSENTIALS OF ALGEBRA 1 (Part 2)
No. 3301
quadratic, radical, and rational functions.As students learn about each family of functions, they will learn to
represent them in multiple ways.They
will also learn to model real-life situations using functions in order to solve
problems arising from those situations.
ACADEMIC ALGEBRA 1
No. 3101
Full Year/Full Time
Grade 9
Credit 1.0
Academic
Algebra 1 is the first formally structured course of the Academic
sequence.The content is organized
around the families of functions, with special emphasis on linear and quadratic
functions, along with representing functions in multiple ways through inquiry-based
learning in real world situations. In addition to its Algebra content, the
course offers lessons on probability and data analysis as well as numerous
examples and exercises involving mathematical connections to Geometry. Algebra
1 provides instruction and practice on standardized test questions in a variety
of formats including multiple-choice, short response, and extended
response.
HONORS ALGEBRA 2
No. 3202
Full Year/Full Time Honors Wt.
Grades 8,9,10
Credit 1.0
(listed in Grade 10 section for description)
ACADEMIC ALGEBRA 2
No. 3103
Full Year/Full Time
Grades 9,10,11,12
Credit 1.0
(listed in Grade 10 section for description)
HONORS GEOMETRY
No. 3201
Full Year/Full Time Honors Wt.
Grades 7,8,9,10
Credit 1.0
This
is a rigorous course for students who had Advanced Algebra 1 in grades 6, 7, or
8. This is the second year of an Honors transformations, perimeter, area, circumference, surface
area, and volume to solve real-world problems. In addition to the Geometry
content, this course includes numerous examples and exercises involving Algebra
and trigonometry. Honors Geometry provides inquiry-based learning and practice
on standardized test questions in a variety of formats including multiple
choice, short response, and extended response. Technology support will be used
for both learning Geometry and for preparing for standardized tests.
ACADEMIC GEOMETRY
No. 3102
Full Year/Full Time
Grades 9, 10
Credit 1.0
This
is the second course of the Academic perimeter, area, circumference, surface area, and volume to
solve real-world problems. In addition to the Geometry content, this course
includes numerous examples and exercises involving Algebra and trigonometry.
Academic Geometry provides inquiry-based learning and practice on standardized
test questions in a variety of formats including multiple-choice, short
response, and extended response. Technology support will be used for both
learning Geometry and for preparing for standardized tests.
HONORS PRE in Grade 11 section for description)
COMPUTER SCIENCE A
No. 3523
Full Year/Full Time
Grades 9,10,11,12
Credit 1.0
(listed at end of Math section for description)
COMPUTER SCIENCE B
No. 3524
Full Year/Full Time
Grades 9,10,11,12
Credit 1.0
(listed at end of Math section for description for description)
ESSENTIALS OF GEOMETRY (IMPACT)
No. 3334ESSENTIALS OF GEOMETRY
No. 3302HONORS GEOMETRY
No. 3201
Full Year/Full Time Honors Wt.
Grades 7,8,9,10
Credit 1.0
(listed in Grade 9 section for description)
ACADEMIC GEOMETRY
No. 3102
Full Year/Full Time
Grades 9,10
Credit 1.0
(listed in Grade 9 section for description)
HONORS ALGEBRA 2
No. 3202
Full Year/Full Time Honors Wt.
Grades 8,9,10
Credit 1.0
This is a rigorous course for students who had Honors Geometry in grades 7, 8, or 9. This is the third year of an honors mathematics Honors Algebra 2 includes topics on probability, data analysis, Geometry, and Trigonometry. Honors Algebra 2 provides instruction and practice on standardized test questions in a variety of formats including multiple-choice, short response, and extended response. Technology support for both learning Algebra 2 and for preparing for standardized tests is available at
The Advanced/Honors Mathematics courses are intended to be more challenging than Academic courses and are designed to provide multiple opportunities for students to take an increased responsibility for their own learning and achievement. These courses are designed for students who have demonstrated an advanced level of achievement in mathematics. The curriculum is distinguished by a difference in rigor and the quality of work, not merely the quantity.
ACADEMIC ALGEBRA 2
No. 3103
Full Year/Full Time
Grades 9, 10, 11, 12
Credit 1.0
This is the third year of the Academic Mathematics Academic Algebra 2 includes topics on probability, data analysis, Geometry, and Trigonometry. Academic Algebra 2 provides instruction and practice on standardized test questions in a variety of formats including multiple-choice, short response, and extended response. Technology support for both learning Algebra 2 and for preparing for standardized tests is available at
NOTE: Students with less than a 'B' in Honors Geometry will be recommended for Academic Algebra 2.
HONORS PRE- under Grade 11 section)(listed under Grade 11)
COMPUTER SCIENCE A
No. 3523
Full Year/Full Time
Grades 9,10,11,12
Credit 1.0
(listed at end of Math section)
COMPUTER SCIENCE B
No. 3523
Full Year/Full Time
Grades 9,10,11,12
Credit 1.0
(listed at end of Math section)
ESSENTIALS OF ALGEBRA 2 (PART 1)
No. 3303
Full Year/Full Time
Grades 11, 12
Credit 1.0
This course is the fourth course of the Essentials sequence. The Algebra strand of this course includes topics arranged around family of functions including linear, absolute value, polynomial, and quadratic. Other topics include data analysis and probability, discrete mathematics, and an introduction to trigonometry. Essentials of Algebra 2 (Part 1ACADEMIC ALGEBRA 3
No. 3623
Full Year/Full Time
Grade 11
Credit 1.0
This is the fourth year mathematics course for an academic student. Major emphasis includes the topics of modeling problem situations, family of functions, including linear, absolute value, quadratic, polynomial, exponential, logarithmic, radical, and rational functions. Students will also learn to model real-world situations using functions and transform the graphs of functions. In addition to its algebra content, Academic Algebra 3 includes topics on probability and counting and sequences and series. Academic Algebra 3 provides instruction and practice on standardized test questions in a variety of formats including multiple-choice, short response, and extended response.This
is the fourth year mathematics course for an academic student. Major emphasis
includes the topics of modeling problem situations, family of functions,
including linear, absolute value, quadratic, polynomial, exponential,
logarithmic, radical, rational, and circular and trigonometric functions.Students will also learn to model real-world
situations using functions and transform the graphs of functions. Academic
Pre-Calculus with Trigonometry provides inquiry-based learning and practice on
standardized test questions in a variety of formats including multiple-choice,
short response, and extended response.Completion of the course will provide a smooth transition to Foundations
of Calculus (Academic), but will NOT satisfy the prerequisite for Honors
Calculus or AP CalculusThis
is a rigorous course for the accelerated student. It requires a strong
foundation in Algebra and Geometry. Major emphasis is placed on algebraic
concepts and analysis of curves, functions, and graphing techniques. This
course also contains a study of Trigonometry from the circular and right
triangle perspective. The analysis of conic sections and other geometric curves
from a coordinate point of view is also studied. This is an Honors course which
leads to Honors Calculusor AP
Calculus.This course is required as a
prerequisite for Calculus.PERSONAL FINANCE
No. 3411
Semester/Full Time
Grades 11, 12
Credit .5
This practical course is designed to empower students to become more responsible consumers and to prepare them to be financially successful in the years ahead. The major topics of the course are as follows: foundation of financial planning; short and long-term financial goal writing; an in-depth look at the influence of today's economy; budget preparation and money management; banking and investing; consumer credit; local, state and federal taxes; car buying and financing; home mortgages; protection against identity theft; insurance basics; and the mathematics behind key financial ratios. This course is designed as an elective and is not part of any specific mathematics sequence. By developing a strong background in financial literacy, students will be ready to take control of their own personal success towards a secure future.
INTRODUCTION TO PROBABILITY AND STATISTICS
No. 3812
Full Year/Full Time
Grades 11, 12
Credit 1.0
This course develops the basic tools of probability theory and statistics. Topics studied include countingmethods using permutations and combinations, axiomatic probability, descriptivestatistics, and statistical inference. Statistical inference topics includeparameter estimation, sampling theory, and hypothesis testing. This courseprovides a smooth transition to statistics needed at the college-level. Thepossibility for college credit may be available for this course. Detailsrelated to this option and registration procedures will be discussed in class.
AP STATISTICS
No. 3014
Full Year/Full Time AP Wt.
Grades 11, 12
Credit 1.0
This course is devoted to developing the student's ability to interpret and investigate statistical data. The activities of decision-making and justifying hypotheses are of the highest importance. The course uses an activity/project oriented approach to develop the concepts.
It will be necessary for each student to have a TI-83/TI-83+ calculator. This calculator will be used to produce, analyze, and interpret data.
It is strongly recommended that the student take the AP exam upon completion of this course. The student should have a high level of maturity and interest in mathematics.
COMPUTER SCIENCE A
No. 3523
Full Year/Full Time
Grades 9,10,11, 12
Credit 1.0
(listed at end of Math section)
COMPUTER SCIENCE B
No. 3524
Full Year/Full Time
Grades 9.10,11,12
Credit 1.0
(listed at end of Math section)
AP COMPUTER SCIENCE
No. 3011
Full Year/Full Time AP Wt.
Grades 11,12
Credit 1.0
(listed at end of Math section)
ESSENTIALS OF ALGEBRA 2 (Part 1)
No. 3303
Full Year/Full Time
Grades 11,12
Credit 1.0
(listed under Grade 11)
ESSENTIALS OF ALGEBRA 2 (PART 2)
No. 3304
Full Year/Full Time
Grade 12
Credit 1.0
This is the fifth course of the Essentials sequence. The Algebra stand of this course includes topic arranged around family of functions including polynomial, radical, exponential, logarithmic, and rational. Other topics include systems of equations, an introduction to trigonometry, and conic sections. Essentials of Algebra 2 (Part 2TRIGONOMETRY WITH FUNCTIONS
No. 3624
Full Year/Full Time
Grade 12
Credit 1.0
This
course is an introduction to and application of functions. It includes
inquiry-based learning and the analysis and application of functions such as:
linear, polynomial, rational, trigonometric, and transcendental. It is the
fourth course in the Academic Mathematics sequence for students who completed
Algebra 1 at the ninth grade level. The course provides a smooth transition to
college mathematics(listed under Grade 11)
FOUNDATIONS OF CALCULUS (ACADEMIC)
No. 3105
Full Year/Full Time
Grades 11, 12
Credit 1.0
This
is a fifth-year mathematics course for the academic student.Foundations of Calculus (Academic) is an
advanced level of mathematics equivalent to a college freshman course.This course will provide a foundation in
calculus which deals with change and how the change in one quantity affects
other quantities.We will discuss many
of the functions used in calculus and review techniques from pre-calculus used
to obtain the graphs of functions, and to transform known functions into new
functions.This course will show
students how to define and calculate limits, derivatives and integrals which
are the three concepts that distinguish calculus from algebra and
trigonometry.The development of these
topics will explore the connection of these mathematical concepts and the
relationship to other subject areas.
HONORS CALCULUS
No. 3422
Full Year/Full Time Honors Wt.
Grades 10,11,12
Credit 1.0
This is a rigorous course for the accelerated student.If students have completed four years of
Honors Math with a high level of achievement, they should consider this course.
This course is the standard first course in calculus for science, engineering
and mathematics students.Challenging
for the able student, Honors Calculus meets many of the requirements for an AP
course. Students with a high level of achievement may elect to, but are not
expected to, take the AP exam for college credit.These students would need to complete
additional work outside of class to prepare for that exam. This course covers
many similar concepts as AP Calculus AB but at a slower pace.
The possibility for college credit is
available for this course, which follows the curriculum set by the University
of Pittsburgh's Analytic Geometry & Calculus 1 (Math 0220) course.If students elect to register for the College
in High School course (4 credits) they must pass an online placement exam and
pay the registration fee. However, all students will learn the same content and
take the same tests throughout the course, whether they are taking the course
for college credit or not.The only
exception is the final exam – only CHS students will take the University of
Pittsburgh Final exam.This
course in sequence with Honors Pre-Calculus will enable the student to take the
AP exam (AB) for college credit and/ or placement. Because of the rigor and
fast pace, only those students with a high level of achievement in previous
math courses and the recommendation of the Honors Pre-Calculus Math teacher
will be accepted.The course will cover elementary
functions, limits, derivatives of algebraic and transcendental functions, and
basic integration with some application to area and volume.This course differs from
course AP Calculus BC in that it is somewhat less rigorous, and because it
meets only five periods per week, involves less homework and covers less
material.
NOTE: For students taking this course in grades 10, 11, or 12, another Calculus course may have already been taken prior to this (3422) or (3022) may be taken after this. For a maximum number of credits earned in Calculus not to exceed 2.5 credits.
AP CALCULUS BC
No. 3022
Full Year/Full Time AP Wt.
Grades 10, 11, 12
Credit 1.5
This
course in sequence with Honors Pre-Calculus will enable the student to take the
AP exam (Level BC) for college credit and/or placement. Because of the rigor
and fast pace, only those students with the highest level of achievement in
previous math courses and the recommendation of their Honors Pre-Calculus Math
teacher will be accepted.The course
will cover elementary vector and parametric functions, rigorous definitions of
limits, derivatives of algebraic, transcendental, vector and parametric
functions, integration involving area, volume, trigonometric substitution and
integration by parts and by partial fractions, and sequences and series.This course differs from course AP
Calculus AB in that it meets seven periods per week, carries 1.5 credits, moves
at a faster pace, is more rigorous, and involves more homework.
NOTE: For students taking this course in grade 11 or grade 12, another Calculus course (3422, 3012) may have already been taken. For a maximum number of credits earned in Calculus not to exceed 2.5 credits.
PERSONAL FINANCE
No. 3411
Full Year/Full Time
Grade 12
Credit 1.0
(listed under Grade 11)
INTRODUCTION TO PROBABILITY AND STATISTICS
No. 3812
Full Year/Full Time
Grade 11,12
Credit 1.0
(listed under Grade 11)
AP STATISTICS
No. 3014
Full Year/Full Time AP Wt.
Grades 11,12
Credit 1.0
(listed under Grade 11)
COMPUTER SCIENCE A
No. 3523
Semester/Full time
Grades 9, 10, 11, 12
Credit .5
Computer Science A is a one semester course designed to be the student's first experience in structured programming. The student will learn to use top-down design and step-wise refinement in designing programs using an appropriate programming language. The course will concentrate on problem-solving applied to familiar topics from mathematics, science, and business. It is essential that students have a grade of 'C' or better in previous math courses. The programming language used in this course is C++.
COMPUTER SCIENCE B
No. 3524
Semester/Full Time
Grades 9.10,11,12
Credit .5
The major emphasis in this course is on extending the student's proficiency in programming methodology and understanding of algorithms and data structures. The implementation of this extension will be accomplished using an appropriate programming language. The high-level structured nature of the programming language will be utilized to develop solutions to problems by applying top-down design and modular programming methods. The topics and algorithms learned provide an excellent background for taking AP Computer Science (3011). The programming language used in this course is C++.
AP COMPUTER SCIENCE
No. 3011
Full Year/Full Time AP Wt.
Grades 11,12
Credit 1.0
Advanced Placement Computer Science is an introductory course in computer science focusing on Object Orientation. A large part of the course is built around the development of computer programs that are understandable, adaptable and when appropriate, reusable. In addition, an extensive library, packages for developing GUI (graphical user interface) applets, multiple classes, and methods make Java very suitable for the Internet. Programs are used in the development of algorithms, the development and use of fundamental data structures and real-world applications. A Case Study, large real-world program, is included as part of the AP curriculum. In addition, an understanding of the basic hardware and software components of computer systems and the responsible use of these systems are integral parts of the course. The programming language used in this course is Java.
AP COMPUTER SCIENCE PRINCIPLES
No. 3010
Full Year/Full Time AP Wt.
Grades 9,10
Credit 1.0
Advanced Placement Computer Science Principles
is an introductory computing course focusing on computational thinking
practices.The major area of studies in
this course are organized around seven big ideas: creativity, abstraction, data
and information, algorithms, programming, the Internet, and global impact of
technology. | 677.169 | 1 |
Relevant For...
Solving systems of linear equations is a common problem encountered in many disciplines. Solving such problems is so important that the techniques for solving them (substitution, elimination) are learned early on in algebra studies. This wiki will elaborate on the elementary technique of elimination and explore a few more techniques that can be obtained from linear algebra.
Contents
Row Reduction Techniques
A system of equations can be represented in a couple of different matrix forms. One way is to realize the system as the matrix multiplication of the coefficients in the system and the column vector of its variables. The square matrix is called the coefficient matrix because it consists of the coefficients of the variables in the system of equations.
An alternate representation called an augmented matrix is created by stitching the columns of matrices together and divided by a vertical bar. The coefficient matrix is placed on the left of this vertical bar, while the constants on the right hand side of each equation are placed on the right of the vertical bar.
The matrices that represent these systems can be manipulated in such a way as to provide easy to read solutions. This manipulation is called row reduction. Row reduction techniques transform the matrix into reduced row echelon form without changing the solutions to the system.
The reduced row echelon form of a matrix \(A\) (denoted \( \text{rref}(A)\)) is a matrix equal dimensions that satisfies:
The leftmost non-zero element in each row is \( 1 \). This element is known as a pivot.
Any column can have at most \( 1 \) pivot. If a column has a pivot, then the rest of the elements in the column will be \( 0 \).
For any two columns \(C_{1} \) and \(C_{2}\) that have pivots in rows \( R_{1} \) and \( R_{2}, \) respectively, if pivot in \( C_{1} \) is to the left of pivot in \( C_{2}\), then \( R_{1} \) is above \( R_{2} \). In other words, for any two pivots \( P_{1},\) and \(P_{2}\), if \( P_{2} \) is to the right of \( P_{1} \), then \( P_{2} \) is below \( P_{1}\).
Rows that consist of only zeroes are in the bottom of the matrix.
To convert any matrix to its reduced row echelon form, Gauss-Jordan Elimination is performed. There are three elementary row operations used to achieve reduced row echelon form:
The top left element is a pivot, so the rest of the elements in the first column must be 0. This can be done by subtracting the first row from the second row. Furthermore, the first row can be added to the third row to obtain the necessary 0s in the first column.
Case 1. If \(\text{rref}(A)\) is the identity matrix, then the system has a unique solution. \[\]
Case 2. If \(\text{rref}(A)\) contains a row of zeroes followed by a zero augmented value, then the system has infinitely many solutions. \[\]
Consider the following augmented matrix in reduced row echelon form. The last row reads \(\left[0 \hspace{.15cm} 0 \hspace{.15cm} 0 \hspace{.15cm} | \hspace{.15cm} 0\right] \) or \(0 = 0,\) which is not a contradiction. The other rows read \(x + 4y = -1\) and \(z = 2.\) This determines a value for \(z,\) but there are infinitely many pairs of real numbers \(x\) and \(y\) such that \(x + 4y = -1,\) so this system has an infinite number of solutions.
Case 3. If \(\text{rref}(A)\) contains a row of zeroes followed by a nonzero augmented value, then the system has no solution. \[\]
This case is similar to the first in that the bottom row contains mostly 0s. However, observe that the last row of the matrix reads \(\left[0 \hspace{.15cm} 0 \hspace{.15cm} 0 \hspace{.15cm} | \hspace{.15cm} 1\right] \), which reads as \(0 = 1.\) This is a contradiction; thus the system has no solutions.
The first row of this reduced form reads \( x = 1 \) and the second row reads \( y = 4 \). \(_\square\)
Note that this process is essentially the same as standard elimination. The main advantage of this procedure is the brevity of notation (not having to write variable names such as \(x, y,\) and \(z\) repetitively) and ease of keeping track of steps taken.
Multiplication by Inverse of Coefficient Matrix
As seen before, a system of equations can be represented by the matrix multiplication \(Ax = b.\) From here, the solution represented by the column matrix \(x\) can be obtained by left multiplying both sides of the equation by the inverse of the coefficient matrix \(A^{-1}\):
\[Ax = b \Rightarrow A^{-1}Ax = A^{-1}b \Rightarrow x = A^{-1}b. \]
One pitfall of this method arises when \(\det(A) = 0.\) Because the determinant of \(A\) is 0, it cannot have an inverse, meaning this method will fail. In fact, \(\det(A) = 0\) is equivalent to \(\text{rref}(A)\) containing a row of zeroes. As such, the system of equations that \(Ax = b\) represents will not have a unique solution. However, the system could have infinitely many solutions or no solution and a different method will be needed to discover it.
Let \(z\) and \(w\) be the complex numbers satisfying the equations above.
Compute \(|z+w|^2\).
\(\) Note: \(i^2 = -1\), and \(|z|\) is the absolute value of \(z\).
Cramer's Rule
Cramer's rule is a formula that uses determinants to provide a solution to a system of linear equations. The statement of the rule below uses only three variables, but the rule can be applied to a system of any size.
If \(\det(A)=0\) and if any one of \(\det(A_1)\), \(\det(A_2),\) or \( \det(A_3)\) is not equal to \(0\), then the system has no solution.
If \(\det(A) = 0\) and \(\det(A_1)\), \(\det(A_2),\) and \( \det(A_3)\) all equal 0, then a solution may or may not exist. If a solution does exist, then the system of equations will have infinitely many solutions.
The computational complexity of this method increases on the order of O(\(n!\)), so it is unwieldy after \(3\times 3 \) systems.
As previously stated, a system of equations can be be represented as a product of matrices like
At this point, it is worth pointing out that column reduction techniques can be used as though they were row reduction techniques while preserving the solution. So adding \(y\) times the second column and \(z\) times the third column to the first column yields | 677.169 | 1 |
Chapter 4 More on Sets Solutions to Homework Exercises (Solutions to oddnumbered exercises are in textbook)
4.1 #1, 2, 3
2
Write the converse of parts (a), (c), and (e) of Exercise 1 and prove or disprove them.
The converse of (a) is:
If A C then A B and
MAT 1033C - Martin-Gay
Intermediate Algebra
Chapter 8 (8.1, 8.2, 8.5, 8.6
MAT 1033C
Intermediate Algebra - Martin-Gay
Practice for the Final Exam
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Use a calculator to approximate the square root to 3 decimal places. Check to see th
MAT1033C Intermediate Algebra
Name _
Self-Diagnostic Test Instructions
This instrument is a self-diagnostic test which will measure your own
preparation for MAT 1033. All topics in this diagnostic are pre-requisite
knowledge that you will need to have in
MAT 1033C - Martin-Gay
Intermediate Algebra
Chapter 7 or phrases
MAT 1033C - Martin-Gay
Intermediate Algebra
Chapter 3 (3.1 - 3.5, 3.7 word
r——_"—~————-i
VCC Prealgebra Exam Page 1 of 2
VCC Erealgebra Exam General Information:
Calculators will EOT be allowed on this exam.
Formulas will NOT be provided on this exam.
Students should know the following information:
Perimeter and area of a square
MAT 1033C - Martin-Gay
Intermediate Algebra
Chapter 6 (6.1 - 6.7)
MAT 1033C - Martin-Gay
Intermediate Algebra
Chapter 4 (4.1, 4.2, 4.3, 4.4)
Practice for the Exam
Name_
Date _
Day/Time: _
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Fill in the blank with one of
MAT0024 Beginning Algebra Chapter 4 Practice Problems
C/G/J
Name_
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
BONUS QUESTION 1. Question 2 is not a bonus: Write the equation of a line through the give
MAT0024 Carson/Gillespie/Jordan
Practice Problems Chapter 3
Name_
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Write the ratio in simplest form.
1) An athlete ran 24 miles this week, including 9 miles
MAT0024 Practice Probems Chapter 2
Carson/Gillespie/Jordan
Name_
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Decide whether the given number is a solution to the equation preceding it.
59
1) 7x + 5 =
MAT 0012C Name:
Chapter 3 6i 4 Date:
Exam (v2) Practice Section:
You MUST show your work to receive full credit. Express all your answers in IOWest terms.
This exam is worth 100 points (4 points per problem).
1. Which of the following are proper fractio | 677.169 | 1 |
Category Archives: Math Analysis
Under this category student will learn about topic and concepts related to Math Analysis and Pre-calculus. The students will learn about trigonometry, basic derivatives and advanced algebraic concepts. The students will also use technology to aide in their understanding of the topics. The technology that they will use will be computers, mobile tools like tablets & phones and lastly graphing calculators such as the TI-84 class of devices.
Solve SSA Triangle (Ambiguous Case) 1 Solution In this tutorial the students will learn how to use the Law of Sines to solve a triangle given a side, side, and an angle (SSA). This is an ambiguous case that involves checking for alternate angle measurements.
Solve SSA Triangle (Ambiguous Case) No Solution In this tutorial the students will learn how to use the Law of Sines in an attempt to solve a triangle with the given information for a side, side, and an angle(SSA). The triangle in this tutorial is an example of the ambiguous case.
Complex Number to Polar Form In this tutorial the students will learn how to convert a complex number to its polar form equivalent in just 6 minutes! Complex numbers are a combination of real and imaginary numbers in the form a+bi. Polar form is the length/magnitude and angle of the vector. Virginia Standards of Learning… Read More » | 677.169 | 1 |
In this algebra instructional activity, high schoolers evaluate equations, identify the inverse of numbers, translate words into a number sentence, compare expressions, and solve radicals. This six page instructional activity contains a variety of 67 problems. | 677.169 | 1 |
...
Show More curriculums. It will help your student build confidence in his or her schoolwork and get better grades! Appropriate for students in grades 6 through 9, Practice Makes Perfect: Pre-Algebra gives your child the tools to master: Integers Rational numbers Patterns Equations Graphing Functions And | 677.169 | 1 |
Designed specifically for business, economics, or life/social sciences majors, Calculus: An Applied Approach, 8/e, motivates students while fostering understanding and mastery. The book emphasizes integrated and engaging applications that show students the real-world relevance of topics and concepts. Several pedagogical features–from algebra review to study tips–provide extra guidance and practice. | 677.169 | 1 |
Secondary and Middle School Mathematics
Overview fourth edition offers a balance of theory and practice, including a wealth of examples and descriptions of student work, classroom situations, and technology usage to assist any teacher in visualizing high-quality mathematics instruction in the middle and secondary classroom. | 677.169 | 1 |
Product Description:
<> MATLAB for Engineers, 3e, is ideal for Freshman or Introductory courses in Engineering and Computer Science.
With a hands-on approach and focus on problem solving, this introduction to the powerful MATLAB computing language is designed for students with only a basic college algebra background. Numerous examples are drawn from a range of engineering disciplines, demonstrating MATLAB's applications to a broad variety of problems.
This book is included in Prentice Hall's ESource series. ESource allows professors to select the content appropriate for their freshman/first-year engineering course. Professors can adopt the published manuals as is or use ESource's website to view and select the chapters they need, in the sequence they want. The option to add their own material or copyrighted material from other publishers also exists.
REVIEWS for MATLAB for Engineers | 677.169 | 1 |
Elementary Number Theory, Seventh Edition, is written for the one-semester undergraduate number theory course taken by math majors, secondary education majors, and computer science students. This contemporary text provides a simple account of classical number theory, set against a historical background that shows the subject's evolution from antiquity to recent research. Written in David Burton's engaging style, Elementary Number Theory reveals the attraction that has drawn leading mathematicians and amateurs alike to number theory over the course of history.
"synopsis" may belong to another edition of this title.
From the Publisher:
HISTORICAL EMPHASIS: The author has carefully placed the topics of number theory within the larger historical frame of mathematical development. Historical remarks are woven throughout the text along with theory, bringing out the point that number theory is developed piece by piece, with the work of each individual contributor built upon the research of many others. A student who is aware of how people of genius found their way slowly through the creative process may be less likely to be discouraged by his or her own difficulty with the subject. EXTENSIVE EXERCISE SETS: More than 750 problems are an integral part of the text and range in difficulty from the purely mechanical to challenging theoretical questions. The computational exercises develop basic techniques and test understanding of concepts, while those of a theoretical nature give practice in constructing proofs. NEW SECTIONS: Two new sections have been added to this edition: Section 6.4, An Application to the Calendar and Section 15.3, An Application to Factoring: Remote Coin-Flipping. Section 6.4 uses congruence theory to determine the day of the week on which a given date falls. Section 15.3 describes a number-theoretic protocol for flipping a fair coin over the telephone. ADDITIONAL TOPIC COVERAGE: Additional topic coverage in this edition includes coverage of the quadratic sieve method with the factorization algorithms of Section 15.2. The material on Pell's equation has been expanded and clarified. A treatment of polyalphabetic ciphers, focusing on Hill's cipher, now appears in the section devoted to secrecy systems. MODERN ADVANCES IN NUMBER THEORY: The resolution of certain challenging conjectures such as the confirmation of the composite nature of the Fermat number F24 is discussed in this edition, emphasizing the vitality of number theory as an area of research mathematics. In addition, certain numerical information has been updated in light of the latest finding83149025761
Book Description McGraw-Hill Education, 2010. QBINT-16
Book Description McGraw-Hill Education. Hardcover. Book Condition: New. 0073383147 International Edition Paperback OR Softcover Edition with Same Contents. Fast Delivery. Book Cover and ISBN may be different from the US Edition. Customer satisfaction guaranteed. Bookseller Inventory # INDMRK-978125902576183149
Book Description McGraw-Hill Education. Book Condition: New. 00733831478314983147 globe, including India depending upon the availability of inventory storage. Customer satisfaction guaranteed. Bookseller Inventory # INAB146694346 | 677.169 | 1 |
"Difference Equations, Second Edition" presents a practical introduction to this important field of solutions for engineering and the physical sciences. Topic coverage includes numerical analysis, numerical methods, differential equations, combinatorics and discrete modeling. A hallmark of this revision is the diverse application to many subfields of mathematics. The features include: phase plane analysis for systems of two linear equations; use of equations of variation to approximate solutions; fundamental matrices and Floquet theory for periodic systems; LaSalle invariance theorem; Additional applications: secant line method, Bison problem, juvenile-adult population model, probability theory; appendix on the use of Mathematica for analyzing difference equations; exponential generating functions; and many new examples and exercises.
"synopsis" may belong to another edition of this title.
Review:
"The first edition of this book has been the best introduction to difference equations available; the second edition improves this even further." --Martin Bohner, University of Missouri-Rolla "The authors have their finger on the current trends in difference equations. This is a well-written textbook by authors who are known as teachers and expositors." --Johnny Henderson, Auburn University
Book Description Academic Press. Hardcover. Book Condition: New. 012403330X International Edition Paperback OR Softcover Edition with Same Contents. Fast Delivery. Book Cover and ISBN may be different from the US Edition. Customer satisfaction guaranteed. Bookseller Inventory # INDSKT-9788131202324033306
Book Description Academic Press. Book Condition: New. 0124033240333062403330X
Book Description Hard123475 316453
Book Description Paperback. Book Condition: New. New Softcover International Edition, Printed in Black and White, Different ISBN, Same Content As US edition, Book Cover may be Different, in English Language. Bookseller Inventory # 11998ELS-INRS-121 | 677.169 | 1 |
...
Show More all or a portion of your course online, whether your students are in a lab setting or working from home. MyMathLab provides a rich and flexible set of course materials, featuring free-response exercises that are algorithmically generated for unlimited practice and mastery. Students can also use online tools, such as video lectures, animations, and a multimedia textbook, to independently improve their understanding and performance. Instructors can use MyMathLab's homework and test managers to select and assign online exercises correlated directly to the textbook, and they can also create and assign their own online exercises and import TestGen(R) tests for added flexibility. MyMathLab's online gradebook--designed specifically for mathematics and statistics--automatically tracks students' homework and test results and gives the instructor control over how to calculate final grades. Instructors can also add offline (paper-and-pencil) grades to the gradebook. MyStatLab, part of the MyMathLab series, includes all the great features of MyMathLab, but is customized for Prentice Hall textbooks in statistics. MyStatLab courses include essay questions that can be assigned for online tests and quizzes; Javaa applets; statistical software; and other resources designed specifically to help students succeed in statistics. MyMathLab is available to qualified adopters. For more information, visit or contact your sales representative | 677.169 | 1 |
Notes on Geometry|4th Edition
Product Description:
In recent years, geometry has played a lesser role in undergraduate courses than it has ever done. Nevertheless, it still plays a leading role in mathematics at a higher level. Its central role in the history of mathematics has never been disputed. It is important, therefore, to introduce some geometry into university syllabuses. There are several ways of doing this, it can be incorporated into existing courses that are primarily devoted to other topics, it can be taught at a first year level or it can be taught in higher level courses devoted to differential geometry or to more classical topics. These notes are intended to fill a rather obvious gap in the literature. It treats the classical topics of Euclidean, projective and hyperbolic geometry but uses the material commonly taught to undergraduates: linear algebra, group theory, metric spaces and complex analysis. The notes are based on a course whose aim was two fold, firstly, to introduce the students to some geometry and secondly to deepen their understanding of topics that they have already met. What is required from the earlier material is a familiarity with the main ideas, specific topics that are used are usually redone.
REVIEWS for Notes on Geometry | 677.169 | 1 |
Piecewise Functions Project
Compressed Zip File
Be sure that you have an application to open this file type before downloading and/or purchasing. How to unzip files.
9.59 MB | 7 pages pages
PRODUCT DESCRIPTION
Roller Coaster Engineer: Students will create a roller coaster that demonstrates their knowledge and understanding of the following skills:
• Graph a continuous function that represents a roller coaster using parent functions. (Editable for teacher to choose which parent functions)
• Write the equation for each section of the roller coaster, using a transformational approach.
• Write a piecewise-defined function to represent the entire roller coaster.
• Determine the height of the roller coaster at a specified time.
• Identify function attributes: Domain and Range, Local Minima and Maxima, and Intervals of Increase and Decrease
100% editable rubric included!
Check out the preview to see what skills are covered and more details of this fun project | 677.169 | 1 |
The
goal of the Mathematics Department is to
facilitate the construction of meaningful mathematical knowledge and skills,
with the ultimate goal of promoting the personal and professional growth of
each student.
The
curriculum is designed to meet and exceed the New Jersey Student Learning Standards for
Mathematics, as students will have opportunities to take courses such as Precalculus,
Calculus, Statistics, and Computer Science, as well as the required courses of Algebra
I, Geometry, and Algebra II. Each of these courses will emphasize conceptual
understanding as well as procedural fluency.
The
mathematics department is committed to using technology to enhance instruction
and help students engage with mathematics in the 21st century. | 677.169 | 1 |
I would recommend taking it if you like math or need it for your major. I didnt learn a lot about how the material applies to real situations which would help in my engineering major but I did learn a lot about math ideas and how to solve problems.
Course highlights:
I loved learning about dot products and cross products and vectors. I learned how to calculate and solve problems in many different ways.
Hours per week:
3-5 hours
Advice for students:
Pay attention in class and go back through the textbook to learn some more because you can't learn everything in the short amount of time of class.
Course Term:Fall 2016
Professor:Nathan Geer
Course Required?Yes
Course Tags:Background Knowledge ExpectedMany Small Assignments
Dec 01, 2016
| Would recommend.
This class was tough.
Course Overview:
I thought the teacher knew her subject well and was willing to help students who were struggling.
Course highlights:
She set up out of class reviews right before exams just so you could work out any last minute kinks in a concept you might be stuck on. I learned that often people are struggling in the same areas as you. | 677.169 | 1 |
NCERT Solutions for Class 10 Maths in PDF
Free NCERT Solutions Class 10 Maths PDF format are available to download. Buy NCERT books or download along with the answers given at the end of the book. NCERT Exemplar books provide excellent practice including higher order thinking skills (HOTS) for each chapter. So, it is advised to explore these books to make your knowledge foundation better. Some of the books are available as named Revision Book which are confined to latest CBSE Syllabus 2017, providing even better strengths for learners. Students of class 10 must go through the Sample papers and Previous Year Board Papers issued by Central Board of Secondary Education as a supplementary practice for Science and maths. Books and their solutions of other subjects as Science, Social Science and Hindi are also available to download in PDF format. If you are having any suggestion for the improvement, you are welcome. The improvement of the website and its contents are based on your suggestion and feedback. Download NCERT solutions for Science, Social Science and Hindi.
POLYNOMIALS
Zeros of a polynomial. Relationship between zeros and coefficients of quadratic polynomials. Statement and simple problems on division algorithm for polynomials with real coefficients.
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically – by substitution, by elimination and by cross multiplication method. Simple situational problems. Simple problems on equations reducible to linear equations.
QUADRATIC EQUATIONS
Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solutions of quadratic equations (only real roots) by factorization, by completing the square and by using quadratic formula. Relationship between discriminant and nature of roots. Situational problems based on quadratic equations related to day to day activities to be incorporates.
ARITHMETIC PROGRESSIONS
Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems.
1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.
3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar.
4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar.
5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar.
6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other.
7. (Prove) The ratio of the areas of two similar triangles is equal to the ratio of the squares on their corresponding sides.
8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.
9. (Prove) In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angles opposite to the first side is a right angle.
INTRODUCTION TO TRIGONOMETRY Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined); motivate the ratios whichever are defined at 0 degree and 90 degree. Values (with proofs) of the trigonometric ratios of 30, 45 and 60. Relationships between the ratios. Proof and applications of the identity sin2A + cos2A = 1. Only simple identities to be given. Trigonometric ratios of complementary angles.
HEIGHTS AND DISTANCES
Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30, 45, 60.
CIRCLES
Tangent to a circle at a point.
1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact.
2. (Prove) The lengths of tangents drawn from an external point to circle are equal.
CONSTRUCTIONS
1. Division of a line segment in a given ratio (internally).
2. Tangent to a circle from a point outside it.
3. Construction of a triangle similar to a given triangle.
AREAS RELATED TO CIRCLES, 90 and 120 only. Plane figures involving triangles, simple quadrilaterals and circle should be taken.)
SURFACE AREAS AND VOLUMES
(i) Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right circular cylinders/cones. Frustum of a cone.
(ii) Problems involving converting one type of metallic solid into another and other mixed problems. (Problems with combination of not more than two different solids be taken.) | 677.169 | 1 |
Algebra 1 Curriculum
This bundle includes notes, homework assignments, quizzes, tests, warm-ups, and 170+ games and activities to teach the units listed below. This is a complete curriculum that can be used as a stand-alone resource or used to
Algebra 2 Curriculum
This bundle contains warm-ups, notes, homework assignments, quizzes, unit tests, a midterm test, end of year review materials, and a final exam for Algebra 2. This bundle does not contain activities. The activities are sold
Geometry Curriculum
Pre-Algebra Curriculum
What is currently included in this curriculum?
So far, Units 1-8. I opted to release this early for those that wish to pre-order the Pre-Algebra units instead of purchasing them separately.
What will this curriculum
Pre-Algebra Activities Mega Bundle
What does this bundle contain?
This bundle contains 824 activities in my store related to Pre-Algebra (grades 6-8 math). Sold separately, they total $283. If more activities are added, you will get these for
Algebra 2 Activities Growing Bundle
What is currently included in this bundle?
This bundle contains 66 activities to supplement and align with my Algebra 2 Curriculum. Click on the preview to view these activities. As I continue to add more
This set of FREE binder covers and spine labels will help organize your instructional materials, and make them pretty too! Choose between color or black and white. Simply customize the unit titles to match your own. The spine labels are ideal for
Ask All Things Algebra a question. They will receive an automated email and will return to answer you as soon as possible.
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TEACHING EXPERIENCE
I have taught Math 8, Algebra, Honors Algebra, and Geometry here in Virginia!
MY TEACHING STYLE
I am a fan of structured notes that are easy to for students to read, followed by lots of fun activities to keep them engaged and practice their new skills!
I am ALWAYS looking for new ideas that present concepts to my students in a fun and exciting way. I believe that the appearance of what we give our students matters greatly. In a day where we are competing with the newest gadgets, we need to grab their attention and interests.
HONORS/AWARDS/SHINING TEACHER MOMENT
Yet to be added
MY OWN EDUCATIONAL HISTORY
I graduated from the State University of New York at Brockport in 2006 with a degree in Mathematics, Computer Science, and Education.
ADDITIONAL BIOGRAPHICAL INFORMATION
Originally from Orchard Park, NY, I relocated to Virginia with my husband, and our two young boys. I have taught Pre-Algebra, Algebra I, and Geometry to 8th graders and LOVE IT! Not only is teaching my passion, but also creating materials that students can relate to, find more appealing, and make learning more fun. | 677.169 | 1 |
The three-volume series History of the Theory of Numbers is the work of the distinguished mathematician Leonard Eugene Dickson, who taught at the University of Chicago for four decades and is celebrated for his many contributions to number theory and group theory. This second volume in the series, which is suitable for upper-level undergraduates... more...
Ideal either for classroom use or as exercises for mathematically minded individuals, this text introduces elementary valuation theory, extension of valuations, local and ordinary arithmetic fields, and global, quadratic, and cyclotomic fields. more...
This text uses the concepts usually taught in the first semester of a modern abstract algebra course to illuminate classical number theory: theorems on primitive roots, quadratic Diophantine equations, and the Fermat conjecture for exponents three and four. The text contains abundant numerical examples and a particularly helpful collection of exercises,... more...
An imaginative introduction to number theory, this unique approach employs a pair of fictional characters, Ant and Gnam. Ant leads Gnam through a variety of theories, and together, they put the theories into action?applying linear diophantine equations to football scoring, using a black-magic device to simplify problems in modular structures, and... more...
Classic two-part work now available in a single volume assumes no prior theoretical knowledge on reader's part and develops the subject fully. Volume I is a suitable first course text for advanced undergraduate and beginning graduate students. Volume II requires a much higher level of mathematical maturity, including a working knowledge of the theory... more...
An advanced monograph on Galois representation theory by one of the world's leading algebraists, this volume is directed at mathematics students who have completed a graduate course in introductory algebraic topology. Topics include abelian and nonabelian cohomology of groups, characteristic classes of forms and algebras, explicit Brauer induction... more...
An engaging treatment of an 800-year-old problem explores the occurrence of Fibonacci numbers in number theory, continued fractions, and geometry. Its entertaining style will appeal to recreational readers and students alike. more...
Unusually clear and interesting classic covers real numbers and sequences, foundations of the theory of infinite series and development of the theory (series of valuable terms, Euler's summation formula, asymptotic expansions, other topics). Includes exercises. more... | 677.169 | 1 |
Designed for the person who needs to learn algebra as a prerequisite for further study or for a refresher course before moving on, the book covers all of the basic algebra concepts such as variables, equations, quadratic equations, factoring algebraic expressions, exponents, roots, radicals, and more. It includes numerous step by step examples and practice exercises that help the reader to understand the topics in a "self-study" format, designed for those who are uncomfortable with mathematics. The companion disc includes self-correcting exercises and all the figures from the text. Instructor resources available for use in course adoptions.
FEATURES:
•Presents basic concepts in an easy to understand style, designed for those who are uncomfortable with mathematics
•Provides hundreds of step by step examples and practice exercises that help the reader to understand the topics in a "self-study" format
•Includes a companion disc with self-correcting exercises and all the figures from the text | 677.169 | 1 |
Lecture 1: general greens theorem.mov
Lecture Details :
An example of the generalization of green's theorem. For more math shorts go to
Course Description :
Contents:
General greens theorem - scalor line integrals - Find a potential function - Conservative vector field test - volume between two surfaces - Two more ways to set up a tripple integral - Sketch the solid of integration - Set up a tripple integral more general region - Limits of multivariate functions - intersection of a plane and a surface - Graphing a vector valued function - contour maps - planer form of curvature - integration of vector valued functions | 677.169 | 1 |
College Algebra DeMYSTiFieD, 2nd Edition/i>
Overview
Don't let quadratic equations make you irrational beginning with the math fundamentals then introducing you to this advanced form of algebra. As you progress, you will learn how to simplify rational expressions, divide complex numbers, and solve quadratic equations. You will understand the difference between odd and even functions and no longer be confused by the multiplicity of zeros. Detailed examples make it easy to understand the material, and end-of-chapter quizzes and a final exam help reinforce key ideas.
It's a no-brainer! You'll learn about:
The x-y coordinate plane
Lines and intercepts
The FOIL method
Functions
Nonlinear equations
Graphs of functions
Exponents and logarithms
Simple enough for a beginner, but challenging enough for an advanced student, College Algebra DeMYSTiFieD, Second Edition is your shortcut to a working knowledge of this engaging subject.
Most Helpful Customer Reviews
I purchased this book to brush up on math, as I need to take a graduate proficiency exam. This book is very good at conveying the subject matter. The one downside I have come across is that there are several mistakes in the solutions to problems which can become annoying. After a couple of headaches trying to figure out how my answer was different, I got wolfram alpha pro to check my work. I have found 2 errors that were just wrong and a couple other misleading typographical errors. Despite this annoyance the author conveys the material in a very easy to read way. Furthermore, the chapters all lead into each other which makes for an extremely streamlined experience.
Pros:
Is really easy to follow.
Effective and efficient.
Short and to the point.
Cons:
Has a handful of mistakes which has caused me much confusion. - 1 Star.
Is an ugly book, with ugly brown pages, that looks like it is for kids. -1 Star. | 677.169 | 1 |
About this item
Comments: Brand new. We distribute directly for the publisher. A successful mathematical career involves doing good mathematics, to be sure, but also requires a wide range of skills that are not normally taught in graduate school. The purpose of this book is to provide guidance to the professional mathematician in how to develop and survive in the profession. There is information on how to begin a research program, how to apply for a grant, how to get tenure, how to teach, and how to get along with one's colleagues.
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Comments: Has minor wear and/or markings. SKU:9780821846292 | 677.169 | 1 |
"Head First Algebra is a clear, easy-to-understand method to learn a subject that many people find intimidating. Because of its somewhat irreverent attitude in presenting mathematical topics for beginners, this book inspires students to learn algebra at a depth they might have otherwise thought unachievable."
— Ariana Anderson
"The way this book presents information is so conversational and intriguing it helps in the learning process. It truly feels like you're having a conversation with the author."
— Amanda Borcky
"What do punk bands need to know about algebra? How will quadratics make your listening experiences better? Crack the spine on this to find out in a fun and engaging way!"
— Cary Collett
"This has got to be the best book out there for learning basic algebra. It's genuinely entertaining." — Dawn Griffiths, author of "Head First Statistics"
"I wish I had a book like Head First Algebra when I was in high school. I love how the authors relate math concepts to real-life situations. Not only does it make learning Algebra easy, but also fun!"
— Karen Shaner
"Head First Algebra is an engaging read. The book does a fantastic job of explaining concepts and taking the reader step-by-step through solving problems. The problems were challenging and applicable to everyday life."
— Shannon Stewart, Math Teacher
"The book is driven by excellent examples from the world in which students live. No trains leaving from the same station at the same time moving in opposite directions. The authors anticipate well the questions that arise in students' minds and answer them in a timely manner. A very readable look at the topics encountered in Algebra 1."
Head First JavaTM Head First Object-Oriented Analysis and Design (OOA&D) Head First HTML with CSS and XHTML Head First Design Patterns Head First Servlets and JSP Head First EJB Head First PMP Head First SQL Head First Software Development Head First JavaScript Head First Ajax Head First Physics Head First Statistics Head First Rails Head First Web Design Head First PHP & MySQL
Beijing • Cambridge • K ln • Sebastopol • Taipei • Tokyo
Tracey Pilone M.Ed. Dan Pilone
Head First Algebra
Wouldn't it be dreamy if Algebra was useful in the real world? It's probably just a fantasy...
O'Reilly Media books may be purchased for educational, business, or sales promotional use. Online editions are also available for most titles (safari.oreilly.com). For more information, contact our corporate/institutional sales department: (800) 998-9938 or corporate@oreilly.com.
The O'Reilly logo is a registered trademark of O'Reilly Media, Inc. The Head First series designations, Head First Algebra., was aware of a trademark claim, the designations have been printed in caps or initial caps.
While every precaution has been taken in the preparation of this book, the publisher and the authors assume no responsibility for errors or omissions, or for damages resulting from the use of the information contained herein.
No variables were harmed in the making of this book.
[M]
ISBN: 978-0-596-51486-0
Vinny Pilone Nick Pilone
This book uses RepKover™, a durable and flexible lay-flat binding.
This book is dedicated to my parents and teachers for believing that I could be good at math, even when I didn't agree.
— Tracey
This book is dedicated to the amazing teachers I've had in life—starting with my parents who taught me that the most important is to keep learning.
— Dan the authors Authors of Algebra
Dan Pilone is a Software Architect for Vangent, Inc. and has led software development teams for the Naval Research Laboratory and NASA. He's taught graduate and undergraduate Software Engineering at Catholic University in Washington, D.C.
This is Dan's second Head First Book, but it still comes with some firsts: his first book outside of Computer Science and his first book co-authored by his wife (who, incidentally, is much better looking than his last co-author. Sorry, Russ.) Working with Tracey on this book changed it from being work to being family fun time. Well, not entirely, but still an amazing experience.
Dan's degree is in Computer Science with a minor in Mathematics. For anyone who needs inspiration that Algebra can be fun, fire up a good game of Halo and think about all the x's, y's, and z's that make it all possible.
Tracey Pilone would first like to thank her co-author and husband for being unwavering in his support and open enough to share the Head First world with her.
She is a freelance technical writer who supported mission planning and RF analysis software for the Navy, right before she decided to write a math book.
She spent several years before becoming a writer working as a construction manager on large commercial construction sites around Washington DC. That's where she actually used Algebra on a somewhat regular basis and saw first hand that math is what makes buildings stay up.
She has a Civil Engineering degree from Virginia Tech, holds a Professional Engineer's License, and received a Masters of Education from the University of Virginia.
Dan Pilone Tracey Pilone table of contents vii
Table of Contents (Summary)
Table of Contents (the real thing)
Your brain on Algebra. Here you are trying to learn something, while here your brain is doing you a favor by making sure the learning doesn't stick. Your brain's thinking, "Better leave room for more important things, like which wild animals to avoid and whether naked snowboarding is a bad idea." So how do you trick your brain into thinking that your life depends on knowing Algebra? | 677.169 | 1 |
Designed for non-physical science majors such as business, management, behavioral science, and social science. Topics include: relations, functions, exponential and logarithmic functions, derivatives and their applications, integrals and their applications, and functions of several variables. Prerequisite: 2.0 or better in MATH 147 or MATH& 141 or appropriate placement or permission from department lead. | 677.169 | 1 |
A Friendly Approach to Complex Analysis
Description
The book constitutes a basic, concise, yet rigorous course in complex analysis, for students who have studied calculus in one and several variables, but have not previously been exposed to complex analysis. The textbook should be particularly useful and relevant for undergraduate students in joint programmes with mathematics, as well as engineering students. The aim of the book is to cover the bare bones of the subject with minimal prerequisites. The core content of the book is the three main pillars of complex analysis: the Cauchy-Riemann equations, the Cauchy Integral Theorem, and Taylor and Laurent series expansions.Each section contains several problems, which are not purely drill exercises, but are rather meant to reinforce the fundamental concepts. Detailed solutions to all the exercises appear at the end of the book, making the book ideal also for self-study. There are many figures illustrating the text. | 677.169 | 1 |
Mathematics
At King's Academy, there is an appreciation for the intrinsic value of the study of mathematics: its power lies not within the mere manipulation of numbers, but in developing a mental discipline for approaching the solution of problems methodically and rationally. Leveraging a solid basis of mathematical knowledge and expertise in fundamental skills, the focus of the curriculum is the development of critical thinking by incorporating an ever increasing number of word problems of greater complexity as students transition from introductory to advanced courses in mathematics.
King's Academy has adopted a well-tested and proven course program that moves from Algebra I and Geometry to Algebra II to establish basic mathematical thought processes and skills in both computation and visualization. Having completed those, students have a variety of higher level math courses from which to choose, including Pre-calculus, Calculus, AP Statistics, AP Calculus AB and AP Calculus BC. Mathematics is one of the tools at the school's disposal to prepare students to meet challenging problems in their future lives.
This is a course in first-year algebra with a focus on numerical, algebraic, graphing and verbal methods of problem-solving. The algebra topics of study include equations, proportions, and inequalities in one variable, writing, solving and graphing linear equations and inequalities, solving and graphing systems of linear equations, operations involving polynomials and factoring, solving quadratic equations, fractions, exponents and data analysis. Following Algebra I, students take either Integrated Mathematics or Geometry. Course length: One year Prerequisites: Department consent
The Geometry course is designed to provide a solid foundation of basic and fundamental algebraic and geometric concepts. Upon completion of the course, students should have a firm and confident grasp of Euclidean geometry and be well prepared for further study in mathematics, namely Algebra II and beyond. Constructions, investigations, proofs and projects are used to explore the various facets of geometry. The topics include both inductive and deductive reasoning, and plane, spatial, coordinate, and transformational geometry. Course length: One year Prerequisites: Department consent
Covering all major topics of algebra and geometry, Integrated Mathematics is the foundation for all higher-level math courses. It enables students from a range of math backgrounds to tackle challenging problems with a variety of approaches and to improve their critical thinking skills. The algebra topics of study include writing, solving and graphing linear equations and inequalities, solving and graphing systems of linear equations, operations involving polynomials and factoring, solving quadratic equations, and exponents and radicals, while the geometry topics of study include the properties of lines in a plane, triangles, polygons, similar polygons and right triangles including trigonometric ratios, circles, area and volume. Course length: One year Prerequisites: Department consent
Fundamental to the study of advanced Algebra is the thorough development of the concept of functions. Course material includes an emphasis on slope as an average rate of change, introduction of inverse functions, exponential and logarithmic functions, polynomial functions, rational expressions and functions, radical expressions and functions, the introduction of imaginary numbers, right triangle trigonometry and matrices, and an overview of statistics and probability. A graphing calculator is required. Course length: One year Prerequisites: Completion of Algebra I and Geometry or Integrated Mathematics
Algebra II Honors includes conic sections, series and sequences and partial fractions along with all Algebra II topics (linear functions and systems, matrices, quadratic and polynomial functions, exponential and logarithmic functions, radical and rational functions, right triangle trigonometry and probability and statistics) with particular emphasis on challenging word problems and applications of the concepts. This course is an excellent choice for students who want to enhance and develop furthermore their critical thinking and problem-solving skills and prepare well for the Pre-Calculus Honors course the year after. A graphing calculator is required. Course length: One year Prerequisites: Completion of Algebra I and Geometry or Integrated Mathematics and department consent
Designed to supplement the material presented in Algebra II, FST completes the study of the elementary functions; linear, quadratic, exponential, logarithmic and trigonometric. Additionally, the course develops some material from finite mathematics including an introduction to probability and statistics, additional applications of trigonometry, and sequences and series. The topics cover a wide range of mathematics and are designed to significantly enhance students' ability to undertake the study of advanced statistical applications. Throughout the entire course, modeling of real phenomena is emphasized. A graphing calculator is required. Course length: One year Prerequisites: Algebra II and department consent
Pre-Calculus is not a specific, discrete study in mathematics, but rather a course that focuses upon establishing the student's knowledge and skills in preparation for undertaking more advanced math studies. While many of the topics introduced in Algebra II are revisited, they are covered in greater depth and breadth. Included are more challenging studies in functions, analysis of their domains and ranges, recognition of families of curves and their transformations, the study of conic sections, advanced trigonometry, arithmetic and geometric series, and statistics and probability. A graphing calculator is required and integral to the course as methods of solution include algebraic, numeric and graphical approaches. Course length: One year Prerequisites: Algebra II, Algebra II Honors or FST, and department consent
Pre-Calculus Honors consolidates algebra and geometry skills, and emphasizes application and synthesis of those topics as a preparation for AP Calculus. The topics include solving algebraic equations and inequalities, function operations, polynomial and rational function analysis, exponential and logarithmic functions, trigonometric functions and applications, sequences and series, and conic sections. Problems are solved numerically, graphically and algebraically, and a graphing calculator is used extensively for modeling and analyzing functions. Course length: One year Prerequisites: Algebra II or Algebra II Honors and department consent
From opinion polls and customer satisfaction surveys to drug trials, people seem to be surrounded by data everywhere. The importance of statistical literacy has been steadily increasing over the years, and data analyses often drive decision-making. Thus, students taking this course will rarely question the relevance of course content to real life. Introductory Statistics is primarily a project-based course in which students often collect and analyze their own data. They study proper collection and inference techniques to determine the significance of the data they collected. Students also learn how to build probability models by observing data and design experiments to reduce variability. A graphing calculator is required. Course length: One year
Prerequisites: Pre-Calculus or Functions, Statistics and Trigonometry and department consent
This course covers all of the first semester as well as some of the second semester topics of a college-level calculus survey course. Included are studies in limits and continuity, derivatives and integrals and selected applications of them and an introduction to differential equations. Pre-calculus topics are reviewed when appropriate to ensure contextual presentation of new material. A graphing calculator is required. Course length: One year Prerequisites: Pre-Calculus or Pre-Calculus Honors and department consent
This course follows the College Board Advanced Placement syllabus and is designed to introduce students to the major concepts and tools for collecting, analyzing and drawing conclusions from data. Students are exposed to four broad-conceptual themes: exploring data (describing patterns and departures from patterns), sampling and experimentation (planning and conducting a study), anticipating patterns (exploring random phenomena using probability and simulation) and statistical inference (estimating population parameters and testing hypotheses). A graphing calculator is required. After the completion of this course, students are expected to take the AP Exam. Course length: One year Prerequisites: Pre-Calculus, Pre-Calculus Honors, or Functions, Statistics and Trigonometry and department consent
A rigorous and challenging course comparable to courses in colleges and universities, AP Calculus AB is designed for students with excellent mathematical skills who seek college credit, college placement or both from institutions of higher learning. Based on the College Board Advanced Placement AB syllabus, the course approaches the calculus concepts (limits and continuity, derivatives and integrals and their applications) from multiple perspectives – graphically, analytically, numerically and verbally. A graphing calculator is required. After the completion of this course, students are expected to take the AP Exam. Course length: One year Prerequisites: Pre-Calculus or Pre-Calculus Honors and department consent
Designed as an extension of Calculus AB rather than an enhancement, AP Calculus BC includes, along with all Calculus AB topics, additional topics such as: integration by parts and by tables, improper integrals, Euler's Method and L'Hôpital's Rule, infinite series, parametric equations, and polar coordinates and polar graphs. A graphing calculator is required. After the completion of this course, students are expected to take the AP Exam. Course length: One year Prerequisites: AP Calculus AB and department consent
Unlike AP Calculus AB and BC in which students study calculus of a single variable, Multivariable Calculus is a rigorous college course focused on functions of two or more independent variables. The concepts studied in this course are applied in many different fields – thermodynamics, electricity and magnetism, economics, modeling fluid or heat flow, etc. The topics included are vectors and the geometry of space, vector-valued functions, functions of several variables, multiple integration, vector analysis, and second order differential equations. A graphing calculator is required. Course length: One year
Prerequisites: AP Calculus BC | 677.169 | 1 |
Methods for Finance
ISBN :
9781861523679
Publisher :
Cengage Learning EMEA
Author(s) :
Terry J. Watsham
Publication Date :
1 Jan 1998
Overview
This text explains in an intuitive yet rigorous way the mathematical and statistical applications relevant to modern financial instruments and risk management techniques. It progresses at a pace that is comfortable for those with less mathematical expertise yet reaches a level of analysis that will reward even the most experienced. The strong applied emphasis makes this book ideal for anyone who is seriously interested in mastering the quantitative techniques underpinning modern financial decision making. | 677.169 | 1 |
SABC Education is proud to announce its partnership with author Edzai Conilias Zvobwo in making a significant contribution towards mathematics education. SABC Education has sponsored the production and distribution of this debut book entitled "The Mathematical Genius in You".
The Mathematical Genius in You is a ground-breaking, innovative and pragmatic dialogue between the author and the reader focused on how to achieve in mathematics.
It is both motivational and instructional as it takes the reader through the steps on how to bring out the mathematical genius inside the reader. It is an easy read that anyone can go through and understand the principles required for maths success.
The style of writing is simple, and plain English has been used so as to appeal to a wider audience. Learners who can read should be able to fully understand the concepts and principles of becoming mathematically proficient.
Parents can also read on behalf of their children thus ensuring that there is support at home. The parent does not have to be a mathematician but will gain insight into the problem solving methodology and will ask value adding questions which will channel learners to be logical solution finders.
The book seeks to demystify mathematics as a learning area by emphasizing the fact that society as a whole should encourage learners that mathematics can be done. Positive messages should be disseminated as opposed to magnifying the difficulty of mathematics.
The Mathematical Genius in You is a personal and candid dialogue on how to succeed in mathematics. Its objective is to motivate, educate and in the process expose the correct mindset, attitude and attributes necessary to excel in mathematics.
At the formative level this book seeks to produce problem solvers and logical beings that are versatile in the workplace and life in general. The principles exposed in this book can be applied to other learning areas and it is hoped that after reading this book you should able to know where you stand mathematically and then work to improve your status.
If you are a parent you will be shown the attributes and attitudes to emphasize to your children so as to optimize their mathematical potential.
For teachers, this book is handy in the motivation of students to adopt a positive attitude towards maths.
The ultimate aim is for learners to be able to fully acquire math knowledge and skills, combine logical and intuitive reasoning and communicate effectively on interpretations in context to the real world using appropriate mathematical language and notations.
About the Author
Edzai is a high achieving mathematics teacher, motivational speaker, business analyst/IT project manager and Game Mathematician.
He possesses a BSc Honours in Applied Mathematics, Diploma in Business Analysis and Certificate in Data Analysis.
He has a passion for mathematics; this is evidenced by his motivational works across schools in South Africa. He has participated in intervention programmes in the townships through workshops and camps assisting struggling learners.
Edzai has done a lot of work with SABC and their outreach programmes. A lot of MBA students from leading institutions of higher learning have benefited from his tutoring of research methods and statistics. His dream is to demystify mathematics and ensure that learners become good problem solvers.
Edzai is a firm believer of the theory that Africa can be liberated if her people are mathematically literate and apply the problem solving thinking methodology to find solutions to social, political and economic problems that have long hounded the "Mother-Continent". Good problem solvers will eradicate poverty in Africa. To become a good problem solver, one needs to be mathematically trained to achieve this. | 677.169 | 1 |
Simulating Neural Networks with Mathematica cou... Full description
This
Features
Teaches the reader about what neural networks are, and how to manipulate them within the Mathematica environment.
Shows how Mathematica can be used to implement and experiment with neural network architectures.
Addresses a major topic related to neural networks in each chapter, or a specific type of neural network architecture. Paperback. État : New. 1st. 167mm x 19mm x 242mm. Paperback. This book introduces neural networks, their operation and their application, in the context of Mathematica, a mathematical programming language. Readers will learn how to simulate neural n.Shipping may be from multiple locations in the US or from the UK, depending on stock availability. 352 pages. 0.590. N° de réf. du libraire 9780201566291 | 677.169 | 1 |
Exponential Functions Notes
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
0.82 MB | 2 pages
PRODUCT DESCRIPTION
This is a one-sided notes page used to teach the basics of exponential functions.
The notes begin by defining what an exponential function is. Then the equation is given. The student fills in what "a" represents. The information related to "b" is already given. Next, four examples are given where the student is asked to write the exponential equation. On the first two examples the student is simply given the starting value and the rate. The second two examples involve a word problem.
On examples 5 & 6, the student is given an exponential equation and asked to identify the initial value, whether the equation represents exponential growth or decay and the rate of growth or decay.
Finally, the student is given two more word problems that they have to write an equation for and solve. On the first question, the equation is given in the word problem. Key | 677.169 | 1 |
Description
Schaum's Outline of Complex Variables, 2ed
Murray R Spiegel
McGraw-Hill
May 20, 2009
The guide that helps students study faster, learn better, and get top grades More than 40 million students have trusted Schaum's to help them study faster, learn better, and get top grades. Now Schaum's is better than ever-with a new look, a new format with hundreds of practice problems, and completely updated information to conform to the latest developments in every field of study. 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! Schaum's Outlines-Problem Solved.
About Iceberg
The Iceberg Reader revolutionizes the reading experience, making it easier than ever to browse, buy and enjoy books.
Schaum's Outline of Mathematical Handbook of Formulas and Tables, 3edMurray R SpiegelMcGraw-HillAugust 25, 2008Schaum's has Satisfied Students for 50 Years. Now Schaum's Biggest Sellers are in New Editions For half a century, more than 40 million students have trusted Schaum'… more
Schaum's Outline of College Algebra, Third EditionMurray R SpiegelMcGraw-HillAugust 20, 2009Confusing Textbooks? Missed Lectures? Not Enough Time?Fortunately for you, there's Schaum's Outlines. More than 40 million students have trusted Schaum's to help them succeed in the… more
Schaum's Outline of Probability and Statistics, 3/EMurray R SpiegelMcGraw-HillAugust 26 Biology, Third EditionGeorge H. FriedMcGraw-HillJuly 20, 2009Confusing Textbooks? Missed Lectures? Not Enough Time? Fortunately for you, there's Schaum's Outlines. More than 40 million students have trusted Schaum's to help them succeed in the classroom and… more
Schaum's Outline of French Vocabulary, 3edMary Coffman CrockerMcGraw-HillOctober 13 PreCalculus, 2nd Ed.Fred SafierMcGraw-HillAugust 13, 2008Schaum's has Satisfied Students for 50 Years. Now Schaum's Biggest Sellers are in New Editions For half a century, more than 40 million students have trusted Schaum's to help them study faster, … more | 677.169 | 1 |
Prerequisite: Algebra 1 not successfully completed in an earlier grade.
Credits: 2.0 (or 1.0) depending on course
Length: Full Year
Grade Level: 9-12
Algebra 1 is considered the gateway course in mathematics. Students in this course will learn fundamental algebraic skills necessary for advancement into higher level math courses. Topics of study include the real number system, radical expressions, linear equations,linear inequalities, quadratic equations, factoring, and graphing algebraic functions. Real world applications are presented and are an integral component of the algebraic learning experience. High school Algebra 1 is offered as a double-period class that allows for extended time to investigate skills and concepts.
ALGEBRA STANDARDS
Prerequisite:
None - Not proficient or advanced on the Algebra I Keystone Exam
Credits: 0.5
Length: Half Year - One Semester
Grade Level: 9-11
This course was developed with an approach that emphasizes the mathematical structure of Algebra I. Just like the Algebra I Keystone Exam, this course will be broken down into two main modules and sub-units aligned to the standards. Because this course was written with the Pennsylvania and National Common Core Standards in mind, students will receive remediation in Algebra I this semester that will serve as a firm basis for the Algebra I Keystone Exam and future mathematics courses here a at Wilson High School.
CP GEOMETRY
Prerequisite: CP Algebra I, or CP Algebra I-B
Credit: 1.0
Length: Full Year
Grade Level: 9-10
This course emphasizes the mathematical structure of geometry. It allows students to engage in and apply the foundational aspects of lines, planes, and shapes. Topics studied include: methods of reasoning, angle relationships, perpendicular and parallel lines, congruent and similar polygons, properties of polygons, right triangle trigonometry, area and volume of figures, circles, and coordinate geometry.
iSTEM CP GEOMETRY
Prerequisite: Recommendation to program from current teacher
Credits: 1.0
Length: Full Year
Grade Level: 9
Description: iSTEM CP Geometry emphasizes the mathematical structure of geometry. This course allows students to engage in and apply the foundational aspects of lines, planes, and shapes. Students will differentiate between inductive and deductive reasoning, construct transformations, and apply their overall findings from each unit to real world applications Topics studied include: methods of reasoning, angle relationships, perpendicular and parallel lines, congruent and similar polygons, properties of polygons, right triangle trigonometry, area and volume of figures, circles, and coordinate geometry. The integration of technology will be prevalent throughout the course of the academic year.
iSTEM CP ALGEBRA 2
Prerequisite: Recommendation to program from current teacher OR successful completion of Virtual Geometry Academy Summer Course
Credits: 1.0
Length: Full Year
Grade Level: 9
Description: iSTEM CP Algebra II will begin with a review of basic algebra concepts, with an emphasis on solving equations and problem solving. Student exploration of Quadratic Functions will be the main focus of this course, integrating simultaneously with Projectile Motion in Physics. Further topics studied in Algebra II include inequalities, linear equations, polynomials, factoring, polynomial equations, rational expressions, radicals, and complex numbers, quadratic equations, polynomial equations, and logarithms will also be discussed Although students are welcome to purchase their own calculators, the school will provide TI-84 calculators for classroom use to those students who have not purchased one on their own.
CP ALGEBRA 2-A
Prerequisite: CP Geometry
Credits: 1.0
Length: Full Year
Grade Level: 10-11
This course includes all topics from the first-B
Prerequisite: CP Algebra 2-A
Credits: 1.0
Length: Full Year
Grade Level: 11-12
This course includes all topics from the second
Prerequisite: Geometry or Virtual Geometry Academy Summer Course
Credits: 1.0
Length: Full Year
Grade Level: 10
CP Algebra II will begin with a review of basic algebra concepts, with an emphasis on solving equations and problem solving. Topics studied in Algebra II include inequalities, linear equations, polynomials, factoring, polynomial equations, rational expressions, radicals, and complex numbers. Quadratic equations, polynomial equations, and logarithms will also be discussed.Although students are welcome to purchase their own calculators, the school will provide TI-84 calculators for classroom use to those students who have not purchased one on their own.
PRE-CALCULUS
Prerequisite: CP Algebra II, or Algebra II-B
Credit: 1.0
Length: Full year
Grade Level: 11-12
Pre-Calculus begins with a review of algebra topics. It is designed to provide a student with the basics of trigonometry (triangle solutions, inverse relations, trig function graphs, vectors, and trig equation solutions), analytic geometry (line, coordinate geometry, and conic sections), and an introduction to calculus concepts.
HONORS PRE-CALCULUS
Prerequisite: CP Algebra 2
Credit: 1.0
Length: Full Year
Grade Level: 10-11
The goal of Honors Precalculus is to prepare students for AP Calculus. The concepts and topics examined are meant to deepen students' understanding of algebraic concepts and expand their ability to process and think at higher abstract and conceptual levels. The curriculum will provide a comprehensive analysis of Trigonometry and Precalculus, building upon students' prior mathematical knowledge. Topics examined include, but are not limited t Functions, Conic Sections, Matrices, Sequences & Series, the Unit Circle, Analytic Trigonometry, Vectors, Polar Coordinates, Parametric Equations, and Limits. All of the topics covered will be applied to real-world situations through life problems, projects, and writing assignments.
CALCULUS
Prerequisites: Precalculus or Honors Precalculus
Credits: 1.0
Length: 1 year
Grade level: 10-12
This course will begin with a review of pre-calculus concepts. Some of the calculus topics discussed will include limits, differentiation, related rates, curve sketching, graphing, integration, logarithmic functions, exponential functions, optimization and area between curves. Applications of these topics will be included throughout the course. Although students are welcome to purchase their own calculators, the school will provide TI-84 Plus Silver Edition that will be used to reinforce the concepts that are taught.
ADVANCED PLACEMENT CALCULUS
Prerequisite: Honors Pre-Calculus
Credit: 1.0
Length: Full year
Grade Level: 11-12
Advanced Placement Calculus will follow the outline for the Advanced Placement Course AB and BC as prescribed by the College Entrance Examination Board. Students are encouraged to earn college credit and possible advancement by demonstrating their competence on the AP examination in May. Although students are strongly encouraged to purchase their own graphing calculators, the school will provide TI-84 calculators for classroom use to those students who have not, or are unable to purchase one on their own.
STATISTICS
Prerequisite: CP Algebra II or CP Algebra II-B
Credit: 1.0
Length: 1 Year
Grade Level: 10-12
Statistics and Selected Topics is designed to provide a student with the basics of descriptive stats (measures of spread and location, histograms, distribution tables, and regression lines), probability, inferential stats (binomial and normal distributions, and hypotheses testing). Although students are encouraged to purchase their own calculators, the school will provide TI-84 calculators for classroom use to those students who have not purchased one on their own.
ADVANCED PLACEMENT STATISTICS
Prerequisites: CP Algebra 2
Credits: 1.0
Length: 1 year
Grade level: 10-12
This course is divided into four major themes — exploratory analysis, planning a study, probability, and statistical inference. The AP Statistics course will follow the outline for the Advanced Placement Course as prescribed by the College Entrance Examination Board. Students are encouraged to earn college credit and possible advancement by demonstrating their competence on the AP examination in May.
MIDDLE SCHOOL MATH OFFERINGS
MATH 6
Prerequisite: Completion of 5th Grade
Credits: 1.0
Length: Full Year
Grade Level: 6
This is the last course in the Envision Math series which builds the foundation for the study of Algebra. The goals of the course are number and numeration,operations and computations, data and chance, measurement, geometry, and patterns that lead into algebra. It is a balanced instructional program in that it provides for direct instruction as well as exploratory activities and employs multiple methods for basic skill practice. Throughout the course,emphasis is placed on problem-solving, communication, reasoning, and mathematical connections.
CP ALGEBRA 1-A
Prerequisite: Completion of Math 6
Credits: 1.0
Length: Full Year
Grade Level: 7
This is a full year course which provides a conceptual foundation for the study of algebra. It is designed to cover one half of the topics covered in our 1-year CP Algebra I course. Students will begin the year with developing basic number sense of integers, order of operations, exponents, equations, graphing linear equations, linear functions, and inequalities. Students also engage in developing Algebraic skills through the use of real world problems.
CP ALGEBRA 1-B
Prerequisite: Algebra 1-A
Credits: 1.0
Length: Full Year
Grade Level: 8
This is the bridge from the concrete to the abstract study of mathematics. It begins with a review of topics from Algebra I-A. Upon this course's completion, a student will have completed the full content and standards of a full Algebra I course. Topics include graphing linear functions, using slope-intercept form, solving systems of equations and inequalities, factoring polynomials, quadratic equations, rational functions, and radicals. Real world applications are presented within the course content. The review of basic skills, including, but not limited to, operations and conversions of fractions, decimals, percents, and integers, will occur on an on-going basis as the skills relate to the topic at hand.
CP ALGEBRA 1
Prerequisite: Math 6
Credits: 1.0
Length: full year
Grade Level: 7
Algebra 1 is considered the gateway course in mathematics. Students in this course will learn fundamental algebraic skills necessary for advancement into higher level math courses. Topics of study include the real number system, radical expressions, linear equations, linear inequalities, quadratic equations, factoring, and graphing algebraic functions. Real world applications are presented and are an integral component of the algebraic learning experience. | 677.169 | 1 |
Sharpen your skills and prepare for your college algebra and trigonometry exams with a wealth of essential facts in a quick-and-easy Q&A format! Get the question-and-answer practice you need with McGraw-Hill's 500 College Algebra and Trigonometry Questions. Organized for easy reference and intensive practice, the questions cover all essential algebra and trigonometry formulas to the areas of triangles, this book covers the key topics in algebra and trigonometry. Prepare for exam day with: 500 essential algebra and trigonometry questions and answers organized by subject Detailed answers that provide important context for studying Content that follows the current college 101 course curriculum | 677.169 | 1 |
books.google.com - An exploration of conceptual foundations and the practical applications of limits in mathematics, this text offers a concise introduction to the theoretical study of calculus. It analyzes the idea of a generalized limit and explains sequences and functions to those for whom intuition cannot suffice.... Concept of Limits
A Concept of Limits
An exploration of conceptual foundations and the practical applications of limits in mathematics, this text offers a concise introduction to the theoretical study of calculus. It analyzes the idea of a generalized limit and explains sequences and functions to those for whom intuition cannot suffice. Many exercises with solutions. 1966 edition. | 677.169 | 1 |
students truly understand the mathematical concepts, it's magic. Students who use this text are motivated to learn mathematics. They become more ...Show synopsisWhen students truly understand the mathematical concepts, it's magic. Students who use this text are motivated to learn mathematics. They become more confident and are better able to appreciate the beauty and excitement of the mathematical world. That;from the textbook, to the eManipulative activities, to the online problem-solving tools and the resource-rich website;work in harmony to help achieve this goal | 677.169 | 1 |
Don't be overwhelmed by all the maths, you can do a lot in Mahout with some
basic knowledge. The books will help you understand your
data better, and ask better questions both of Mahout's APIs, and also of
the Mahout community. And unlike learning some particular software tool,
these are skills that will remain useful decades later.
As a pre-requisite to probability and statistics, you'll need basic calculus. A maths for scientists text might be useful here such as 'Mathematics for Engineers and Scientists', Alan Jeffrey, Chapman & Hall/CRC. (openlibrary)
Peter V. O'Neil Introduction to Linear Algebra, great book for beginners (with some knowledge in calculus). It is not comprehensive, but, it will be a good place to start and the author starts by explaining the concepts with regards to vector spaces which I found to be a more natural way of explaining. | 677.169 | 1 |
This course was originally designed
as a follow-up course to EMAT 4680/6680 and EM.
The change from EMAT 4680/6680
and 4690/6690 is that each student, or possibly pairs of students,
in the course will select, develop, and carry out a major project issues of school mathematics instruction
is expected.
Alert!For those of you
who have not yet taken EMAT 6680 or 6690, we shall introduce the
basic (and some advanced) functions of the listed software during
the first week of classes. I shall require all participants who
do not already have copies to order student versions of The
Geometer's Sketchpad and Fathom
2.
Communication will be facilitated
via e-mail and the course Home Page and each student should have
an e-mail account.
Requirements for Projects
Develop and carry out one major
project. This must be carefully planned and approved by the instructor.
Objectives
To become familiar with and operational
on modern computer systems.
To use application software to
solve mathematical problems.
To use application software to
create mathematical demonstrations.
To communicate mathematical ideas
that arise from computer investigations using word processing
and web technologies.
To communicate mathematical ideas
via the computer applications.
To use general tools such as word
processing, paint programs, web page construction and spread sheets
to facilitate mathematical investigations and communication about
mathematical investigations | 677.169 | 1 |
books.google.com - Contains a middle school mathematics program designed to prepare students for success in high school algebra and geometry.... mathematics | 677.169 | 1 |
Suitable for modern numerical methods courses for third and fourth years of study. A good basic reference for postgraduates. Skillfully balancing rigor with practicality, this progressive and visually rich text on using MATLAB software explores how this modern and powerful computing technology can - in an elegant and efficient way - solve exacting and interesting problems that have application in science and industry. Encouraging systematic experimentation, it offers an extensive, hands-on study of MATLAB functions and exposes students to a wide range of numerical algorithms - including optimization and regression analysis with applications of symbolic methods - with full explanations of their fundamental principles and illustrations of their application.
"synopsis" may belong to another edition of this title.
From the Back Cover:
Skillfully balancing rigor with practicality, this progressive and visually rich piece on using MATLAB software explores how this modern and powerful computing technology can—in an elegant and efficient way—solve exacting and interesting problems that have application in science and industry. Encouraging systematic experimentation, it offers an extensive, hands-on study of MATLAB functions and exposes the reader to a wide range of numerical algorithms—including optimization and regression analysis with applications of symbolic methods—with full explanations of their fundamental principles and illustrations of their application.
FEATURES/BENEFITS
Developing computing technology used through all algorithms and scripts.
Clear visual interpretation of results using MATLAB graphics.
Use of symbolic methods—Includes a full chapter on the use of the MATLAB symbolic toolbox and its application to numerical methods.
Allows comparison of different approaches to many problems and enhances opportunities for testing and experimentation.
Introduction to the important field of optimization.
Helps the reader understand the broad area of MATLAB application.
Statistical techniques placed in a clear numerical context.
Enables the reader to grasp complex but widely applied concepts.
MATLAB symbolic toolbox—Includes a full chapter on the use of the MATLAB symbolic toolbox and its applications to numerical methods.
The most current MATLAB functions—Includes functions for obtaining specific eigenvalues; for the solution of differential equations (including methods for stiff differential equations); and for regression analysis. | 677.169 | 1 |
Gcse maths no coursework
Computer Studies 7010 is being replaced with computer science 2210 from may/june 2015 onward.So computer studies will no more be available after 2014.The new … There is much confusion over the different terms that you may hear on the news attributed to the thousands of people moving from their home nations to the EU. The ultimate Computer Science Resource Package for schools - KS3/GCSE/A Level - computing resources. Includes 100s of theory powerpoints, video tutorials, … Number Grade Descriptors for the new GCSE Maths specification 1 - 9 Visit weteachmaths.co.uk for the grade descriptors for all of the topics Or fol...
The International General Certificate of Secondary Education (IGCSE) is an English language curriculum offered to students to prepare them for International. DDs school say they do IGCSE in a majority of subjects and they are selling this as an advantage. We have a meeting about course options in a few wee PSA! DoSomething.org Has a TON of Scholarship Opportunities Right Now. SPOILER: college is crazy-expensive. Sorry. Did we spoil it? There are...
Gcse maths no coursework
We value excellent academic writing and strive to provide outstanding paper writing service each and every time you place an order. We write essays, research papers. lots of surd examples, interactive help, free worksheets from GCSE Maths Tutor Computer Studies 7010 is being replaced with computer science 2210 from may/june 2015 onward.So computer studies will no more be available after 2014.The new … A collection of fantastic teaching resource websites that every maths teacher must have in their bookmarks: 1. TES Connect. Thousands of maths teaching resources. The Maths IGCSE course from Oxford Home Schooling guides the student through the basic maths skills in preparation for their exams.
Since the abolition of maths coursework, there is no formal requirement to carry out investigations with your students. However, it is only through investigative work. Like many maths departments around the country, myself and my Head of Department are currently working our way through the new draft Maths GCSE specifications. Science GCSE from 2016. Here are the main points: Totally exam-based. There is no practical exam, although you have to do certain practicals in school.
My undergraduates are unwilling to argue, in person, or in writing. Their essays are often surveys of a scene, descriptions of a landscape. There's nothing. The International General Certificate of Secondary Education (IGCSE) is an English language curriculum offered to students to prepare them for International. Feb 27, 2013 · Hi – am in the same boat. 12 years science and psychology, bit of maths all over the country, just graded "needs improvement" (bilious OFSTED. | 677.169 | 1 |
of Maths Quest 12 Further Mathematics is a comprehensive text designed to meet the requirements of the VCE Units 3 and 4 Further Mathematics course. The textbook also benefits from new supporting calculator companions containing comprehensive step-by-step CAS calculator instructions, fully integrated into worked examples, for the TI-Nspire CAS calculators. The student textbook contains the following new features: * StudyON icons provide links to Concept screens, See mores and Do mores for online study, revision and exam practice. The textbook continues to offer the following award-winning features: * exam practice sections with allocated time and marks. Fully worked solutions to these sections are available on eBookPLUS * technology-free questions * electronic tutorials for key worked examples in each chapter * interactivities * eLessons * full colour with stimulating photographs and graphics * carefully graded exercises with many skill and application problems, including multiple-choice questions * easy to follow worked examples in the Think-Write format * cross references throughout exercises to relevant worked examples * comprehensive chapter reviews with exam-style questions * eBookPLUS references throughout to guide students and teachers to relevant online material. What is eBookPLUS? Maths Quest 12 Further Mathematics eBookPLUS is an electronic version of the textbook and a complementary set of targeted digital resources. These flexible and engaging ICT activities are available to you online at the JacarandaPLUS website ( | 677.169 | 1 |
Solving rate-time-distance and other rate problems
A math homework community created in 1999 by math goodies. Students get help with specific math homework problems. .
A math homework community created in 1999 by math goodies. Students get help with specific math homework problems. . Get an expert homework help on more than 40 subjects delivered by the team of our professional writers & tutors! 247 online help in any kind of academic writing at. . Need math homework help? - you can help your child succeed. Students get help with specific math homework problems. . Math homework help. Hotmath explains math textbook homework problems with step-by-step math answers for algebra, geometry, and calculus. Online tutoring available for. . Studydaddy is the place where you can get easy online math homework help. Our qualified tutors are available online 247 to answer all your homework questions. .
Addition helper step-by-step solutions to your addition problems. Order of operations helper step-by-step solutions to your order of operations problems. . A math homework community created in 1999 by math goodies. Students get help with specific math homework problems. . Free math problem solver answers your algebra homework questions with step-by-step explanations. .
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Mathematics Chapter Notes for IIT JEE, UPSEE and WBJEE
Hariom Singh
Scoring good marks in an entrance examination is an art where more than one subject questions are being asked. In IIT JEE, UPSEE and WBJEE where subjects are common so it becomes very important to choose the subject where you can secure you mark confidently. In general mathematics is a subject which can be the suitable subject to score good marks.
In this article you will find Mathematics Chapter Notes for engineering entrance examinations such as IIT JEE, UPSEE and WBJEE. These Chapter notes will provide complete overview of the chapter for engineering aspirants. During examination time students don't have much time to revise complete book so, they can go through these chapter notes in very short time. These chapter notes include important terms and formulae from Mathematics Section. | 677.169 | 1 |
Browse related Subjects ...
Read More fractions, decimals, ratio sand proportions, percents, roots and exponents, algebraic equations, inequalities, word problems, plotting graphs, and geometry. Fully indexed for locating specific problems rapidly | 677.169 | 1 |
Foundations In Quantitative Business Techniques
Starting with the most elementary mathematics, Foundations in Quantitative Buisness Techniques provides a concise introduction to the core quantitative techniques and their applications in business decision-making. Numerous worked examples and exercises with solutions are integrated throughout each chapter to enable students to assess their progress through the text and gain confidence in the application of the techniques.
"synopsis" may belong to another edition of this title.
From the Publisher:
Integrated and chapter-end exercises with solutions enable students to assess their understanding of techniques Assumes little mathematical knowledge and is easy to follow to enable students to work independently at their own pace Techniques are developed gradually and illustrated throughout with worked examples Covers commercial and financial applications of mathematics Contains helpful guidance on using calculators and spreadsheets
About the Author:
Gordon Bancroft is Principal Lecturer in Mathematics and Statistics at the University of Staffordshire.
George O'Sullivan is Principal Lecturer at the University of Central England Business School. | 677.169 | 1 |
DESCRIPTION: 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 | 677.169 | 1 |
Text:
Our main reference will be the online textbook Diffy Qs by Lebl. You can download the book (for free) or order a copy (for cheap) to be sent to you from this page:
Clickers will be used in some sections of this course. Please confirm with your section instructor about whether you need to get a clicker (available from the bookstore).
Matlab/Octave. You will need one of these tools for some of the homework in this class. See below for instructions about how to get one of these programs for your personal computer. You can also/alternatively use the Math department computer labs.
Grading
Missing a midterm or quiz normally results in a mark of 0. Exceptions may be
granted in two cases: prior consent of the instructor or a medical
emergency. In the latter case, the instructor must be notified within 48
hours of the missed test, and presented with a doctor's note immediately
upon the student's return to UBC. No make-up midterms will be given.
Term marks may be scaled up or down on a
classwide basis, depending on performance on the final exam. This is to ensure
fairness across both sections of the course.
You must submit only your own work. Although you are welcome to study together and discuss the homework with other students, the work you submit (electronically or on paper) must be your own. UBC policies on cheating and plagiarism are extremely strict. If in doubt, enquire before submitting.
Written Homework
Some homework questions will require use of Matlab/Octave to solve problems numerically. See below for instructions on obtaining Matlab/Octave and for TA help hours.
Computer homework information
Getting the needed software
Here are three options for obtaining software needed for the homework.
(1) Matlab is available from UBC IT services for UBC students to install on their own computers.
You do not need any additional toolboxes for this class, but that is up to you.
(2) You can install
Octave. Octave is a free program that is quite similar to Matlab so that most programs are easily portable from one to the other. Installation of Octave on Windows and Linux is fairly simple (in fact if you have a recent Linux distribution, Octave might already be installed). However on Mac computers, installation has been reported to be a bit trickier. We will not help you install software on your personal computer.
(3) You can use the Mathematics undergrad computer labs, located in LSK 121 and LSK 310. The labs are open from 8am-6pm Monday to Friday. LSK 310 is also where the TA help will be available (either on a lab computer, or take your laptop). To use the lab computers you will need to login using a user account. Follow the instructions that will be sent to you by email (and/or ask the TAs during lab hours for assistance).
Lab TA hours:
For help with Matlab, TAs will be available in LSK 310 at the following times (subject to change through the semester, please check back for updated hours): | 677.169 | 1 |
College Algebra - 1.7 MODULAR ARITHMETIC 43 What are the objects Zn actually good for First they are devices for studying the integers Congruence
1.7. MODULAR ARITHMETIC 43 What are the objects Z n actually good for? First, they are devices for studying the integers. Congruence modulo n is a fundamental relation in the integers, and any statement concerning congruence modulo n is equiv-alent to a statement about Z n . Sometimes it is easier, or it provides better insight, to study a statement about congruence in the integers in terms of Z n . Second, the objects Z n are fundamental building blocks in several gen-eral algebraic theories, as we shall see later. For example, all finite fields are constructed using Z p for some prime p . Third, although the algebraic systems Z n were first studied in the nine-teenth century without any view toward practical applications, simply be-cause they had a natural and necessary role to play in the development of algebra, they are now absolutely fundamental in modern digital engi-neering. Algorithms for digital communication, for error detection and | 677.169 | 1 |
Algebra 1
The coupon code you entered is expired or invalid, but the course is still available!
How this Algebra 1 course is set up to make complicated math easy:
This approximately 70-lesson course includes video and text explanations of everything from Algebra 1, and it includes more than 75 quiz questions (with solutions!) to help you test your understanding along the way. Algebra 1 is organized into the following sections:
Operations
Rules of equations
Simple equations
Like terms
Functions
Inequalities
Graphing
Systems of equations
Polynomial basics
Factoring 1 students have told me:
"Thank you Krista for making this so clear and understandable! You are a fantastic tutor. A+++" - Omid N.
"At the peak of my mathematical skill level I reached only into the very early stages of learning Trigonometry. Fast forward to now (my early 30's) and by experience alone, realize the infinite possibilities of being a master in mathematics. I chose Ms. King's Algebra course as a review to move past my previous level and must say that I not only ENJOY the course she offers but am getting more from it learning wise than I could have imagined. I wish to complete all of her classes and am extremely confident that her teaching style can get me there. Other than mere reference of other resources, I plan to stay on track with all of Ms. King's courses exclusively. Thank you for a great learning experience!" - Dean F.
"Great! I love how she teaches!" - Fred B.
"The instructor is very clear, both in ideas and in speech. She moves quickly enough to keep your attention, but not so that you get lost." - Janet M | 677.169 | 1 |
AN
INTRODUCTION
TO
THEORY
THE
OF
NUMBERS
AN INTRODUCTION
TO THE
THEORYOFNUMBERS
BY
G. H. HARDY
AND
E. M. WRIGHT
Principal
and Vice-Chancellor
University
of Aberdeen
FOURTH
AT THE
of the
EDITION
OX.FORD
CLARENDON
PRESS
[email protected] University Press, Ely House, Lon
Nanyang Girls High School
Secondary 1 Mathematics
Enrichment 2003
Algebra: Challenging problems
The following questions are taken from an article written by Tom M. Giambrone, from Indiana University
Pennsylvania.
1.
Omar the Rope Maker decided to make a r
best [(we do our) + (so you can do your)]
Mathematics Resource
Part I of III: Algebra
TABLE OF CONTENTS
I.
II.
III.
IV.
V.
VI.
VII.
VIII.
IX.
X.
XI.
THE NATURE OF NUMBERS
EXPRESSIONS, EQUATIONS, AND POLYNOMIALS (EEP!)
A BIT ON INEQUALITIES
MOMMY, WHERE DO | 677.169 | 1 |
Common Core Checklists – High School Math – Algebra
Compressed Zip File
Be sure that you have an application to open this file type before downloading and/or purchasing. How to unzip files.
0.17 MB | 9 pages
PRODUCT DESCRIPTION
This resource includes 3 documents:
- Checklist for all the Algebra Math Standards (Adobe Acrobat PDF)
- Student roster form with a checklist column for each MATH standard (Adobe Acrobat PDF)
- Editable student roster form with a checklist column for each MATH standard (Microsoft EXCEL 2010 )
These checklists can be used to:
- keep track of which standards you have taught and when you taught them
- record notes about each standard
- keep track of individual student progress for progress reports and differentiation | 677.169 | 1 |
Today's Mathematics in the Middle Grades
ISBN :
9780205433599
Publisher :
Allyn & Bacon
Author(s) :
Art Johnson
Publication Date :
1 Jan 2005
Overview
This text provides current and future middle school teachers with the mathematics content, essential concepts, methodology, activities, and resources to both learn and teach mathematics in grades 6 â 8.
Teaching Todayâ s Mathematics in the Middle Grades focuses exclusively on the middle school learner and the middle school mathematics curriculum. Although each chapter discusses foundational mathematics concepts from earlier grades and previews topics that will follow the middle grades, the emphasis is on the middle school. This selective focus allows for proper development of critical topics in the middle school such as proportionality and algebraic thinking, and the integral role of manipulatives. Assessment practices and problem-solving are emphasized, again, from the viewpoint of effective practices for middle grades students. | 677.169 | 1 |
Editor's Pick
The first in the series of Mathematics professional development titles, Bar Modeling: A Problem-solving Tool, is written for educators who want to know the 'how' and the 'why' of bar modeling – a visual problem-solving heuristic which serves as a foundation to algebraic thinking – as well as the pedagogy of the heuristic. | 677.169 | 1 |
Links
BibTeX entry
@book{Thompson1914,
abstract = {Being a very simplest introduction to those beautiful methods of reckoning which are generally called by the terrifying names of the DIFFERENTIAL CALCULUS and the INTEGRAL CALCULUS},
author = {Thompson, Silvanus P},
title = {Calculus Made Easy},
url = {
year = 1914,
urldate = {2012-04-26}
} | 677.169 | 1 |
9780201642025
0201642000
Marketplace
$4.93
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Summary
The first title in a new series, Fundamental Mathematics through Applications is a shorter version (the first 6 chapters) of Basic Mathematics through Applications. Intended either for a brief course on fundamental concepts or for a quick review before a Prealgebra or Introductory Algebra course, Fundamental Mathematics through Applications truly focuses on content relevant to students. Throughout the text, motivating real-world applications, examples, and exercises demonstrate how integral mathematical understanding is to student mastery in other disciplines, a variety of occupations, and everyday situations. A distinctive side-by-side format pairing an example with a corresponding practice exercise encourages students to get actively involved in the mathematical content from the start. Unique Mindstretchers target different levels and types of student understanding in one comprehensive problem set per section incorporating related investigation, critical thinking, reasoning, and pattern recognition exercises along with corresponding group work and historical connections. Compelling Historical Notes give students further evidence that mathematics grew out of a universal need to find efficient solutions to everyday problems. Plenty of practice exercises provide ample opportunity for students to thoroughly master basic mathematics skills and develop confidence in their understanding. | 677.169 | 1 |
The authors develop students' ability to do abstract mathematics by teaching the form of mathematics in the context of real and elementary mathematics. Students learn the fundamentals of mathematical logic; how to read and understand definitions, theorems, and proofs; and how to assimilate abstract ideas and communicate them in written form. | 677.169 | 1 |
PDF Books Free Download
Day: October 8, 2016
Don't let quadratic equations make you irrational If you are absolutely confused by absolute value equations, or you think parabolas are short moral stories, College Algebra DeMYSTiFied, Second Edition is your solution to mastering the topic's concepts and theories at your own pace. This thoroughly revised and updated guide eases you into the subject, beginning […]
A New York Times Bestseller"Arthur Benjamin . . . joyfully shows you how to make nature's numbers dance."—Bill NyeThe Magic of Math is the math book you wish you had in school. Using a delightful assortment of examples—from ice-cream scoops and poker hands to measuring mountains and making magic squares—this book revels in key mathematical […]
This brand new book includes the most up-to-date information on the Geometry Common Core Regents Exam. Students can use this review guide to prepare for their Geometry Regents Exams. Inside, separate chapters explain and provide practice problems on: the language of geometry, basic geometric relationships (parallel lines, polygons, and triangle relationships), constructions, an introduction to […] | 677.169 | 1 |
This softcover edition of a very popular two-volume work presents a thorough first course in analysis, leading from real numbers to such advanced topics as differential forms on manifolds, asymptotic methods, Fourier, Laplace, and Legendre transforms, elliptic functions and distributions. Especially notable in this course is the clearly expressed orientation toward the natural sciences and its informal exploration of the essence and the roots of the basic concepts and theorems of calculus. Clarity of exposition is matched by a wealth of instructive exercises, problems and fresh applications to areas seldom touched on in real analysis books.
The first volume constitutes a complete course on one-variable calculus along with the multivariable differential calculus elucidated in an up-to-day, clear manner, with a pleasant geometric flavor.
REVIEWS for Mathematical Analysis | 677.169 | 1 |
Product Description:
When Julie Miller began writing her successful developmental math series, one of her primary goals was to bridge the gap between preparatory courses and college algebra. For thousands of students, the Miller/O'Neill/Hyde (or M/O/H) series has provided a solid foundation in developmental mathematics. With the Miller College Algebra series, Julie has carried forward her clear, concise writing style; highly effective pedagogical features; and complete author-created technological package to students in this course area.
The main objectives of the college algebra series are three-fold:
-Provide students with a clear and logical presentation of -the basic concepts that will prepare them for continued study in mathematics.
-Help students develop logical thinking and problem-solving skills that will benefit them in all aspects of life.
-Motivate students by demonstrating the significance of mathematics in their lives through practical applications.
REVIEWS for College Algebra | 677.169 | 1 |
AP Calculus BC?
Question:What should I expect in Calculus all together?
Would it be hard to take AP Calculus BC without taking AB?
Answers:
Calculus BC is similar to Calculus AB, but it's faster paced and it covers a few more topics. You should take BC if you want to put in the extra work, if you want an easier class take AB. But, nobody takes AB, then BC the year after because they cover a lot of the same stuff Since you are taking AP, I assume you are in high school. Calculus is different from every other math. You should expect to learn things such as differentiation, integration, and vectors.
I'm not exactly sure what you're talking about with "BC" and "AB," but I assume you mean that you are thinking about taking the second course without taking the first. It would be extremely hard to do this because Calculus builds on itself. what
This article contents is post by this website user, EduQnA.com doesn't promise its accuracy. | 677.169 | 1 |
Translate To An Algebraic Expression
In this algebraic expression instructional activity, students read short statements and translate them into an algebraic expression. There are approximately 30 problems on this one-page instructional activity. | 677.169 | 1 |
Describing two cornerstones of mathematics, this basic textbook presents a unified approach to algebra and geometry. It covers the ideas of complex numbers, scalar and vector products, determinants, linear algebra, group theory, permutation groups, symmetry groups and aspects of geometry including groups of isometries, rotations, and spherical geometry. The book emphasises the interactions between topics, and each topic is constantly illustrated by using it to describe and discuss the others. Many ideas are developed gradually, with each aspect presented at a time when its importance becomes clearer. To aid in this, the text is divided into short chapters, each with exercises at the end. The related website features an HTML version of the book, extra text at higher and lower levels, and more exercises and examples. It also links to an electronic maths thesaurus, giving definitions, examples and links both to the book and to external sources. less | 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
DEFINITION
1
An ordinary differential equation of order n is an equation that is.. Although such equations should probably be called "derivative equations. For example. In Eq. y = y(x). of the function y(x) with respect to x.Y.CHAPTER 1
Elementary MethodsFirst-Order Differential Equations
1
1
INTRODUCTION AND DEFINITIONS
Differential equations are equations that involve derivatives of some unknown function(s). an introduction to difference equations and an introduction to partial differential equations are presented in Chapters 9 and 11. (1)-(3) are the first and second derivatives... In Eqs.. The terms y' and y" in Eqs. x).. u = u(t. or can be put. respectively. Equation (4) involves partial derivatives and is a partial differential equation.
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. The argument x in y(x) (and its derivatives) is usually suppressed for notational simplicity. (4) the unknown function u is assumed to be a function of the two independent variables t and x. in the form
y"I = F(x. x) with respect to t and x.. respectively. a2u/ate and a2u/axe are the second partial derivatives of the function u(t.y. y("
where y.. y("> are all evaluated at x. In this book we are primarily interested in studying ordinary differential equations. however..y.
(5)
.. that is. (1)-(3) the unknown function is represented by y and is assumed to be a function of the single independent variable x. respectively. that is." the term "differential equations" (aequatio differentialis) initiated by Leibniz in 1676 is universally used.. Equations (1)-(3) involve ordinary derivatives and are ordinary differential equations.
(5) should have the following properties: 1. As an illustration we note that the function y(x) = e' is a solution of the second-order ordinary differential equation y" . the function y(x) = cos x is a solution of y" + y = 0 over the interval ( --. We further observe that y= e.
y"(x) + y(x) = (cos x)" + cos x = . y). In this chapter we present elementary methods for finding the solutions of some first-order ordinary differential equations.
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. . It will be shown in Chapter 2 that y = c. and c2. As another example. In fact._e'=e -e'=0.
3. +x). y should have derivatives at least up to order n in the interval J.. y"-"(x)) should lie in the
domain of definition of the function F. In each of the illustrations the solution is valid on the whole real line
On the other hand. y = V is a solution of the first-order ordinary differential equation y' = 112y valid only in the interval (0. +x) and y = x( is a solution of the ordinary differential equation y' = (1 .is also a solution and moreover
y = c.y = 0. As we have seen. Eq.e + cZe-' for some special values of the constants c. y = e is a solution of the ordinary differential equation
y" . Thus. y"I(x) = F(x. that is.2
1
Elementary Methods-First-Order Differential Equations
The independent variable x belongs to some interval I (I may be finite or
infinite). .
Clearly. e is a solution of y" .
(6)
together with some interesting applications.y =
y' = F(x.cos x + cos x = 0.2x)/2y valid only in the interval (0. F should be defined at this point. that is.. in Chapter 2 we will show that the general solution of the ordinary differential equation y" + y = 0 is given by y(x) = c. (2) and (3) are ordinary differential equations of order 2. . and c2. the function F is given. equations of the form
c. 1). For the most part the functions Fand y will be real valued.y(x).y'(x).e + c2e-' is the "general solution" of the ordinary differential equation y" . and the function y = y(x) is unknown.
. + x). For every x in J the point (x. (5) identically over the interval J. (1) is an ordinary differential equation of order 1 and Eqs. . The differential of a function y = y(x) is by definition given by dy = y'dx. Indeed.y(x).y = 0 valid for all x in the interval
(--. and c2. Also.. any solution y(x) of Eq.y = 0.
2.y = 0. cos x + c2 sin x for arbitrary values of the constants c. y"-'"(x)) for every x in J.e + cZe-' is a solution of this equation for arbitrary values of the constants
0 can be obtained from the function c.
Clearly..y'(x). By the general solution we mean a solution with the property that any solution of y" .
Y"(x)-Y(x)=(e')'.
It is the case with many mathematical models that in order to obtain a differential equation that describes a real-life problem.y)dx or in an algebraically equivalent form. in economics as rates of change of the cost of living.
and linear equations. For example. If this rate of change is known (say. 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.1. equations that can be put into the form
y' = QTY)
or
P(x)dx = Q(y)dy.1.
APPLICATIONS 1. Many natural laws and hypotheses can be translated via mathematical language into equations involving derivatives. the next step is to solve the differential equation and utilize the solution to make predictions concerning the behavior of the real problem. Of all tractable types of first-order ordinary differential equations. We know from calculus that the first derivative y' = dy/dx represents the rate of change of y per unit change in x. in geometry as slopes. equations that can be put into the form
y' + a(x)y = b(x). the differential equation (6) sometimes will be written in the
differential form dy = F(x. two deserve special attention: differential equations with variables separable. that is.1
Differential equations appear frequently in mathematical models that attempt to describe real-life situations. in psychology as rates of learning. by
.
Both appear frequently in applications. First-order ordinary differential equations are very useful in applications. Once the model is constructed in the form of a differential equation. in biology as rates of growth of populations. that is.1
Introduction and Definitions
3
With this in mind. 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. we usually assume that the actual situation is governed by very simple laws-which is to say that we often make idealistic assumptions.
In case these predictions are not in reasonable agreement with reality. the
scientist must reconsider the assumptions that led to the model and attempt to construct a model closer to reality. Let the function y = y(x) represent an unknown quantity that we want to study. in chemistry as reaction rates. and in finance as rates of growth of investments. derivatives appear in physics as velocities and accelerations.
(7)
provides us with an approximation to the actual size of this colony of bacteria. For such a colony. In Eq. However. that is. if k is the constant of proportionality. and so he concluded that the rate of population increase is proportional to the population present.4
1
Elementary Methods-First-Order Differential Equations
experience or by a physical law) to be equal to a function F(x. The solution N(t) can be represented graphically as in Figure I.
. y). it is not continuous and so not differentiable. the function N = N(t) satisfies the first-order ordinary differential equation'
IV=kN. (7) is a mathematical model describing a
colony of bacteria that grows according to a very simple. if the number of bacteria is very large. Y).
(7)
This equation is called the Malthusian law of population growth. N(t) = N(0)e". derivatives with respect to x will he denoted by primes and deriv-
atives with respect to t by dots. Equation (7) is a separable differential equation and its solution N(t) = N(0)e"' is computed in Example 3 of Section 1. The solution.
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.1
It should be emphasized that Eq.
We next give some specific illustrations.
'Since the function N(r) takes on only integral values. Here N(O) is the number of bacteria present initially.
N(r)
N(r) . at time t = 0.) In this instance it is the time that is the
independent variable. Then. T. of Eq. (As is
customary. 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. On the other hand. R.3. we can assume that it can be approximated by a differentiable function N(r).N(0)e"
N(0)
0
Figure 1. Malthus observed in 1798 that the population of Europe seemed to be doubling at regular
time intervals. (7) N stands for dN/dt.I. perhaps oversimplified. since the changes in the size of the population occur over short time intervals.
law: It grows at a rate proportional to the number of bacteria present at any time t. assuming this very simple law of growth leads us to
a very simple differential equation.
That is.1
Introduction and Definitions
5
Clearly. S. Of course.
. For each drug the constant k is known experimentally. and the like. and consequently some simplifications and modifications of real-life laws are often necessary in order to derive a mathematically tractable model.
'This model.2). 1. 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.3)
y(t) = Yoe-r`. As we see from Eq.
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. say every T hours. 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.
The solution of the differential equation (8) is (see Example 3 of Section 1.
0
Figure 1. Rustagi in Int. Technol. give the patient a dose D of the drug. Educ.ky. Math. It goes without saying that a mathematical model that is impossible to handle mathematically is useless. the differential equation will then become more complex. as well as other mathematical models in medicine. the amount of the drug in the patient's body tends
Y
0
Ype -A. However. 2 (1971): 193-203. then the rate of change y(t) of the drug is proportional to the amount present.
limitations of food. y(t) satisfies the separable differential equation
y = . and hence the derivative of y(t) with respect to t is negative.2
to zero as t -> oc.. Sci. 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.1. The negative sign in (8) is due to the fact that y(t) decreases as t increases.
(9)
where yo = y(O) is the initial amount (initial dose) of the drug. (9) (see also Figure 1. is discussed by J.
(8)
Pharmacology
Drug Dosages
where k > 0 is the constant of proportionality.
Chant.' Here G is known as the characteristic learning function and depends on the characteristics of the learner and of the material to be learned. V." implies immediately that the motion of any body is described by an ordinary differential equation.
Hence. 22 (1974): 1156-1174. 2r. then
dt (mv) = kF.
(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. Operations Res. at time T.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. Zionts. it is simple to determine the amount of the dose D. hence. Math.)
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.
(13)
'L. If F is the resultant force acting on the body. In fact.. Recall that the momentum of a body is the product my of its mass and its velocity v. but also.e-k.
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. . 11 (1974): 132-158. 3T. p(t) is the state of the learner at time t.
If we want to maintain the initial amount yo of the drug in the body at the times T. .
. Psycho!.
Mechanics
Newton's second law of motion. 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. the dose D should satisfy the equation
yoe-' + D = yo. Southwick and S. the greater the cost of the
instruction]. G.. 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. J. and before we administer the dose D. the desired dose is given by the equation
D = yo(1 . the amount of the drug present in the body is
y(r) =
yoe-'".
y) = c. respectively. (15).
(14)
Figure 1. or even a function of t and v. Equation (13) is an ordinary differential equation in v whose particular form depends on m and F. that is. F can be constant.1
Introduction and Definitions
7
where k is a constant of proportionality.
Kirchhoff's voltage law states that.
where F and F.3
Consider the one-parameter family of curves given by the equation
F(x.
(16)
gives the slope of each curve of the family (15).1. In view of Eq. we want to compute the orthogonal trajectories of the family (15). "the algebraic sum of all voltage drops Electric Circuits around an electric circuit is zero.4)
LI + RI = V(t).
F. The mass m can be constant or a function of t. are the partial derivatives of F with respect to x and y.
where I = 1(t) is the current in the circuit at time t.3 gives rise to the first-order linear differential equation (see also Section 1. Also.
Computing the differential of Eq.
dy dx
_ _ E. the slope of the orthogonal trajectories of the family (15) is given by [the negative reciprocal of (16)]
dyFr dx F
(15). Thus.
(17)
The general solution of Eq." This law applied to the RL-series circuit in Figure 1. (16).dy=0. 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. we obtain
(15)
Orthogonal Trajectories
Fdx+F. (17) gives the orthogonal trajectories of the family
. a function of t.
Many others are developed in detail in subsequent sections. Additionally.y = 0.4
There are many physical interpretations and uses of orthogonal trajectories:
1. y = sin 2x is a solution of the differential equation y" . 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. 5.
4. In electrostatic fields the lines of force are orthogonal to the lines of constant potential. we present a number of applications in the exercises.
3. answer true or false.
.4).to low-pressure areas.
2. and second.
3.4y = 0. 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. Y = e' + 3e-` is a solution of the differential equation y" .
1.8
1
Elementary Methods-First-Order Differential Equations
Figure 1.
2.
EXERCISES
In Exercises 1 through 8. y = cos 2x is a solution of the differential equation y" + 4y = 0. y = 5 sin x + 2 cos x is a solution of the differential equation
y"+y=0. to expose the reader to the diversity of models incorporating ordinary differential equations. We have given only a few of the many applications of first-order ordinary differential equations. The emphasis there is twofold: first. 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 = (1/2x) e'2 is a solution of the differential equation xy' + y =
xe'2.
is
a
solution of the differential
equation
In Exercises 9 through 14. 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" + y' = 2.e'Inx
15.e'' 12. y"'-5y"+ 6y' =0. 2x. Y = e. and c2. y"'-5y"+ 6y' =0.
9. Y = e' is a solution of the differential equation y' + y = 0.1.. Show that the function y(x) = c.3
10.e2'. are solutions of the ordinary differential equation
y. Show that the function
Y(x) _
10 (x . y = 2 In x + 4
x2y"-xy'+y=2Inx. Y"-2Y'+Y= z(Y-y').
for any real number c. x >. show that both functions are solutions of the differential
equation.1
Introduction and Definitions
9
6.e-' + c.< _ :8t 24mt
1
2
d2s)
s
dt'JJ
14.
1
17.)
. 2x . 16.
8. Show
that the function is a solution of the differential equation. xsecx
11. Show that the functions y.0.(x) = e-' and y2(x) = xe-' are solutions of the ordinary differential equation y" + 2y' + y = 0.xe-' + I is a solution of the ordinary differential equation y" + 2y' + y = 1 for any values of the constants
c. = yin
18.X y=XY. ms
tan x
1
g.
7. If more than one function is listed. y"-(tan x)y'.(x) = 0 and y2(x) = x2/4.y = 0.c)2
forx<-c
forx > c . Show that the functions y.is a solution of the differential equation y' .1
13. a differential equation and a function are listed.
q(k)lp(k).
26.
and this equation passes through the point [k + 1/p(k).
20. Math. 25.1 e-"' k
'°
where xo and to are constants. that is. Find the differential equation of the orthogonal trajectories of the family of straight lines y = cx. Prove that if the family of solutions of the differential equation
y' + p(x)y = q(x). Show that N(t) = cek' for any constant c is a solution of Eq. y = -1 + e. x(to) = xo.2y = e'(1 . (Hint: Show that slopes are negative reciprocals. c) is y .]
'From the William Lowell Putnam Mathematical Competition.cp(k)](x . (11). Y = x= is a solution of the differential equation y"' = 0. See Amer. 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. answer true or false. y = xe is a solution of the differential equation y' .
.is a solution of the differential equation y" = y'(y' + 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.) 30.' [Hint: The equation of the tangent
line to the solution at the point (k.
22.
p(x)q(x) 0 0
is cut by the line x = k. Using your geometric intuition.x). 24.
27. 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. is a solution of Eq. the tangents to each member of the family at the points of intersection are concurrent. y = 28. Show also that this solution passes through the point (to)xo).k). (7).c = [q(k) . guess the solution of the resulting differential equation. Show that the function
x(t) =
1
k
+ k x o . 1954.
In Exercises 22 through 28.y2 = x2. Monthly 61 (1954): 545.10
1
Elementary Methods-First-Order Differential Equations
19.
23. What is the meaning of the constant c? 21.
That is. Before we attempt to discover any ingenious
technique to solve a differential equation. The reader can rest assured that these theorems are accurately presented and that their proofs have been satisfactorily scrutinized. it would be very useful to know
whether the differential equation has any solutions at all. the motion that we are observing.y).
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).y). (1) together with the initial condition (2) is called an initial value problem (IVP). where y is the position (the
state) of the particle at time x. The term "initial" has been adopted from physics.
. and moreover it has only one solution.
THEOREM 1
Consider the IVP
y' = F(x. That is. (1) represents the motion of a
particle whose velocity at time x is given by F(x. it is plausible that if we know its position y6 at time xo. and in fact it should have only one solution (uniqueness). that is. Often the proofs of these theorems require mathematical sophistication which is customarily beyond the scope of an elementary treatment.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. Since we imagine that we are observing the
motion of the particle.
'This theorem and many others that appear in this text are truly important for an over-all appreciation of differential equations. do there exist solutions to the differential equation? For example. The condition (2) is called an initial condition and Eq.y). the differential equation (y')2 + y2 + I = 0 has no real solution since the left-hand side is always positive. Naturally. Starting at time xo the particle will move and will move in a unique way. Such proofs can be found in almost any advanced text on differential equations. 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.1.
(1)
For the sake of motivation. 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. The general form of a first-order ordinary differential equation is
y' = F(x. (2) yo is the initial position of the particle at the initial time
xo. that is. assume that Eq. In Eq. if
Y(xo) = Yo
(2)
we should be able to find its position at a later time x.
y(xo) = yo. the differential equation (1) has a solution that satisfies the condition (2).
yo !5. for any real number
Yo.
Thus. y) = -a(x) are
continuous in
x
Ix-x0I:r Al
By Theorem 1. (5) has a unique solution satisfying the initial condition y(xo) = yo [in other words. yo)].
I x . Show that the IVP
Y' = x' + Y2 Y(0) = 0
(3) (4)
EXAMPLE 1
has a unique solution in some interval of the form .xo s A is contained
in the interval I.xo ' < It. in the notation of Definition I of Section 1.yo). For example.h. that is. one can prove that its solutions
exist throughout any intervai about xo in which the coefficients a(x) and b(x) are continuous. 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.shj.
Here F(x. B> 0
xo
A
about the point (xo. B' A> O. yo) where xo E 1. Then F(x. the IVP
.Y):
x
y . 0). (5).0 <. (5) has a unique solution through any point (xo.12
1
Elementary Methods-First-Order Differential Equations
Assume that the functions F and aFloy are continuous in some rectangle
I(X.h s x :5 h. y) = b(x) . Then there is a positive number h :5 A such that the P/
has one and only one solution in the interval x .(x.1
1=jx:ix-xol<AI and J= Ix: x-xo.
Let us illustrate this theorem by a few examples. Eq.
Solution
Choose a number A such that the interval x .y) = x' + y' and aFlay = 2y are continuous in any rectangle 9t about (0. Show that Eq.
REMARK 1
For linear equations such as Eq.a(x)y and FF.
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. through the point (xo.
Uniqueness of solutions of initial value problems is very important.(x) ° 0 and y2(x) = x2/4 are two different
solutions of the IVP
(6)
Y' = Y12
Y(0) = 0. When we know that an IVP has a unique solution we can apply any method (including guessing) to find its solution. the IVP
y'+ x. For each c > 0. Since FY is not continuous
at the point (0. 0). the IVP (6)-(7) has infinitely many solutions through the point (0. For example. in addition to y.
Y(x) = (x . y) = }y-12.1
has a unique solution in the interval (2.5. y) = y12and FY(x.
EXAMPLE 3 The functions y.
REMARK 2 As we have seen in Example 3 for the IVP (6)-(7). and consequently there is no guarantee that the IVP (6)-(7) has a unique solution.
(7)
Is this in violation of Theorem 1?
Solution
Here F(x. 2). and the IVP
Y'+x-2 1 y=InIxI
y(-1)=5
has a unique solution in the inverval ( --. x).1.2'v=Inx
y(3) = . and Y2.c)2 4
x'5 c
c<x
'
and their graphs are given in Figure 1. we know that the IVP
Y. 0). In fact. 0). we have existence of solutions but not uniqueness. one of the hypotheses of Theorem I is violated.2
Existence and Uniqueness
13
y' +x I2y=1nx
y(l) = 3
has a unique solution in the interval (0. these solutions are given by
10. + xy = 0
Y(0) = 0
(8) (9)
.
5
has a unique solution. Y). we desire to develop
methods for determining this solution either exactly or approximately. Consider again the first-order differential equation
y' = F(x. y(x) = 0 is the only solution of the IVP (8)-(9) and no further work is required in solving this IVP. In particular. if we know that (10) has a solution through a given point (xo. yo).
Theorem 1 gives conditions under which a first-order differential equation
possesses a unique solution.
X
0
C=>CJ
X
Y
0
Figure 1. Then at each point (xo. b) E D with slope F(a. The following method. its slope is F(xo. is useful in finding an approximate solution. yo). called the method of isoclines. yo).
(10)
For a better understanding of the differential equation (10) and its solutions. Knowing that a solution exists. it is useful to interpret y' as the slope of the tangent line drawn to the solution
at the point (x. Just by inspection we see that y = 0 satisfies the IVP. Thus. The totality of all linear elements is called the direction
field of the differential equation (10). Thus for
graphical approximations of the solutions of (10) it is useful to draw a short line
segment at each point (a. the
differential equation (10) assigns a slope y' equal to F(xo. then (without knowing the solution explicitly) we know immediately the equation of the tangent line to this solution at the point (xo. The construction of the direction field can
. yo) will give us. Such line segments are called linear elements. b).14
1
Elementary Methods--First-Order Differential Equations Y
Y
0
C. an
approximation to the true solution of (10) near the point (xo. yo). yo) in the domain D of F. graphically. yo). Hence. A short segment along this tangent about (xo. y).
y) = c have the same slope c. 9. corresponding to the values c = c c2. 4. First we
draw a few isoclines C C2. is the curve F(x.6
be carried out more efficiently by setting F(x. In Figure 1.. see Figure 1. yo).YO)
" C2
Figure 1.2
Existence and Uniqueness
15
Y4
I
IIIIVII'
Figure 1.. Therefore.7]. y) = c and realizing that for any constant c all points on the (level) curve F(x.6). For example. y) = c. c
.. and each one of them has slope c. 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. the isoclines of the differential equation
y2
y' = x2 +
(11)
are the circles x2 + y2 = c. centered at the origin (0. the linear elements of the direction field for all these points are parallel.7
. Through the point (xo.1. yo) we
(xo. IC. 16) and linear elements which give a good idea of the direction field of the differential equation (11). This is the method of isoclines (equal slopes).6 we have drawn a few isoclines (for c = 1... At each point on the circle x2 + y2 = c the differential equation
defines the slope c.
0).
8..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. Y
y' = Y'
Y(1) = I
Y' = Yv3
Y(0) = 0
x + y
Y(l) _ -I
7. answer true or false. G. yo) which is an approximation of the true solution
of the differential equation through that point. (1. 5).Bt)3a
where a.
'A. (-1. 0).y'+xy=3
Y(O) = 0
4. b. 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. 0). the following first-order differential equation was
obtained:
dx dt
ax`
(b . W. Icarus 18 (1973): 407-450. Note that.1.
2.
In Exercises 10 through 13.
until it meets the isocline C2 at a point A2.
9. Proceeding in this fashion we construct a polygonal curve through (x0. and (3. -7). Cameron. we draw the linear element with slope c.xy'+y=3
y(o) = 1
5. Through A. yo) and extend it until it meets the isocline C. x0) through which this differential equation has unique solutions. at a point A.y'= x+y
y(o) _ -1
x y
6. Astronomy In an article' concerning the accumulation processes in the primitive solar nebula. and B are constants. Show that the only solution of the IVP
xy'-y=1
y(2) = 3
is y(x) = 2z . the differential equation (11) has a unique solution through (0.
1.
EXERCISES
In Exercises 1 through 6. 0). from Example 1.
3. Find all points (t(. check whether the hypotheses of Theorem 1 are
satisfied.
. In Figure 1.
(2).
(2)
where c is an arbitrary constant of integration. Use the method of isoclines to find graphical approximations to the solutions of the following IVPs.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. If this is not possible or convenient. The unique solution of the IVP
xy' + y2 = 1
y(-2) = I
is y(x) = 1. Algebraic manipulations enable us to write separable differential equations in the form y' = P(x)/Q(y) or. (1) and then integrate both sides.3
Variables Separable
17
10. The differential equation y' = yJ4 has a unique solution through the point (0. Q(y)dy = P(x)dx. The unique solution of the IVP
y'-xy=1-x2
Y(0) = 0
is y(x) = -x. 0) but not through the point (2. (1).
y' = y .
. thereby obtaining y explicitly in terms of x. 0). then Eq. After performing the integrations
in Eq. more explicitly.1.
12.
(1)
To obtain the general solution of a separable differential equation we first separate the two variables as in Eq.
14. In such equations the equality sign "separates" one variable from the other.
11.
13.
y' = x2 + y2
y(0) = 1
1. it is desirable to solve the resulting expression for the dependent variable y.x Y(O) = 0
15. to obtain
f P(x)dx = f Q(y)dy + c. y' = xy y(1) = 2
16. (2) gives implicitly the general solution of Eq. The only solution of the IVP
y'-xy=I-x2
Y(o) = 0
is Y(x) = X.
Integrating
both sides of Eq. Integrating both sides of Eq. (3). into the forms (3'). respectively.
Solve the differential equation (4). and (5'):
x dx = (5y' + 3)dy
(3')
x22 l dx = -
y I
dy.
forx
1. (4') yields
lnj x2. y # . "it follows." and will be used when
. (3). y # O.11+InI y+11=c'In I (x2-I)(y+1)1=c. (3) is separable because it can be written in the form (3').
As we have seen.(5y' + 3) dy = 0
2x(y2 + y) dx + (x' .l
(4')
e"'dy = e-'sinxdx.
The symbol appropriate. Eq. where the variables x and y are separated. different than zero.
lnIx2.
EXAMPLE 2
Solution The variables of Eq. (4'). and for economy in notation we can still denote it by c with c * 0.
Thus. (4').
stands for "imply" (or "implies"). In what follows we shall use this convention frequently." "then. we have
i (x2 . they can be brought. clearly ±e` is again an arbitrary constant.
Z x2
= y` + 3y 4 c
is an implicit representation of the general solution of Eq.
Taking exponentials of both sides and using the fact that e'" ' = p. (3').'
-In I y + I1+c.18
1
Elementary Methods-First-Order Differential Equations
The following are examples of differential equations that are separable:
x dx .
(3)
(4)
(5)
In fact. we obtain
Solution
fxdx=J(5y4+3)dy+c. (4) can be separated as in Eq.1)(y + 1) = ±e`.I)(y + 1) 1 = e` (x2 .1
Thus.
EXAMPLE 1
(5')
Find the general solution of Eq.1) y dy = 0
e`-'y' = sin x. Since c is an arbitrary constant.
Hence. Since y = dyldt. If we relax the restriction c * 0. (4') is obtained from Eq. it suffices to solve the first IVP. y * -1. Similarly.1. it follows that the general solution of y = ky is y = ce".
Now..8. we have
InIyI=kt+cz> I y I
=ek-=e`e"''y=
± e'ek'
'y=cek'. y = 0. and the proof is complete. Thus. the curves x = ± 1 and y = 0 are
not contained in the same formula. and readers are advised to familiarize themselves with the solutions.
k is a constant
is y(t) = yoek'. However. 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). The solutions of the above IVPs for k > 0 and k < 0 are represented in Figure 1. show that the solution of the IVP
y = . for y # 0.
Y
Integrating both sides..
Y(0) = Yo
is
k is a constant
y(t) =
Yoe-"'
Proof Clearly. (4) by dividing through by the expression (x2 . Using the initial condition.
Since y = 0 is also a solution of y = ky.e".
Y(0) = Y.
The following initial value problems are frequently encountered in applications of exponential growth and decay.
EXAMPLE 3
Prove that the solution of the IVP
y = ky.
. no matter what the value of c is. y(t) = y. the solution to the second IVP follows from the solution to the first IVP by replacing the constant k by .3
Variables Separable
19
(x2-1)(Y+1)=c=> Y=-I+x2c#0.k.1)] for c = 0. We must is not zero. Going
back to the original Eq. and y 0. do not forget that Eq. (4) we see that the four lines x = = 1.1)(y2 + y).1 also satisfy the differential equation (4). is called the general solution. Thus. where c is an arbitrary constant. that is. separating the variables y and t we obtain.
dy=kdt. = c.1 will be contained in the formula
y = -1 + [cl(x2 . and y = . we find that y.
c#0. we require that x finally examine what happens when x = ± 1 and when y = 0 or y = -1. we should ensure that this expression t 1. the curve y = .ky.
Thurstone' obtained a differential equation as a mathematical model describing the state of a learner. (6) y)311 y °(1
-
Here k and m are positive constants that depend on the individual learner and the complexity of the task. If y(t) denotes the state of a learner at
time t. 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.1 that involved separable differential equations. We shall mention a few instances here. respectively.20
Y
1
Elementary Methods--First-Order Differential Equations
Y
Y = Yoe"
Y(0)
Yo
Y(f) -
Yoe-"
r
I
Figure 1.8
APPLICATIONS 1.
Psychology
In 1930.1. L.1
One frequently encounters separable differential equations in applications.
(7)
. or the learning curve of a learner. while learning
a specific task or body of knowledge. If the number of bacteria doubles in 5 hours. Review also the applications in Section 1. For the solution of the differential equation (6).J. 3 (1930): 469-493. Psychol.
Biology
Assume that a colony of bacteria increases at a rate proportional to the
number present. how long will it take for the bacteria to triple?
Let N(t) be the number of bacteria present at time r. Gen. then Thurstone's equation is the following separable differential equation: dy _ 2k dt. Thurstone. dt
'L.3. see Exercise 16.
09
t=kln3=51n2-50. From Example 3 we have N(t) =
N(0) is the initial number of bacteria in this colony.
In 3
Finance
(9)
Equation (9) is clearly a separable differential equation.89hours.1.1.1. Its solution (by Example 3) is
y(t) = y(0)ec"loon
Since y(0) = 5000 (the initial amount invested). What will the amount be after 25 years?
Solution Let y(t) be the amount of money (capital plus interest) at time t.69=7.1 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.
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.
The sum of $5000 is invested at the rate of 8% per year compounded continuously. Then the rate of change of the money at time t is given by
dt
100
Y.945. Since the number of bacteria doubles in 5 hours.
(10)
where F is the resultant force acting on the body and k is a constant of proportionality.3 Variables Separable
21
where k is the constant of proportionality.
. we have
N(5) = 2N(0) N(o)es" = 2N(0)
es' = 2
z' k=5In 2. then Eq. whose units are given in Table 1. In the cgs (centimeter-gram-second) system and in the English system. (10) is a separable
differential equation.28. we find that
y(25) = 5000e(&100) u = 5000e2 = $36.
In Section 1. If it happens that F is a function of the velocity v and does not depend explicitly on time. and if m is a constant.
In addition to its weight. by Newton's second law of motion. Eq (10).
Thus. and so
v(103) = 9 Medicine
8
e
105 cm/sec =
I9 -
/
8
e
km/sec = 6.(air resistance) = (mg .
Here we shall derive a separable differential equation that serves as a mathematical model of the dye-dilution procedure for measuring cardiac output.
dt
(mv) = mg . Find the velocity of the body after i = 10' sec.2v) dynes.1
Table of Units
EXAMPLE 4
From a great height above the earth. in spite of the many simplifications that we introduce.
v(0) = 105.
dv = dt. (Take g = 900 cm/sec'.05 kmisec.(mg .
(12)
The solution of the IVP (11)-(12) is v(t) = 1 [mg . which is numerically (in dynes) equal to twice its speed at any time.)
Solution Let v(t) be the velocity of the body at time t.
From the practical point of view this is a reasonable model.2vo)e-:`]. a body of mass m = 2
kilograms (kg) is thrown downward with initial velocity vo = 105 centimeters per second (cm/sec).
. air resistance is acting upon this body.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. The resultant force acting upon this falling body is
F = (weight) .2v
'
mdt=mg-2v
2v
mgm
Also.
With these data and assumptions.
We also have the initial condition D(0) = Do.
(15)
. the differential equation describing this mixture problem is
ddt) _ . The dye is mixed with the blood passing through the heart. and at each stroke of the heart the diluted mixture flows out and is replaced by blood from the veins. and
rate out =
r. The solution of the IVP (13)-(14) is
D(t) = De-("v)`. Thus. and "rate out" is the rate at which dye runs out of the heart at time t.
where D(t)/V is the concentration of dye in the heart (in other words. if N(t) is the number of atoms of the substance present at any time t. Geology. Since dD(t)/dt
represents the rate of change of the dye in the heart at time t.
It is known.
where "rate in" is the rate at which dye runs into the heart at time t. we have the
equation
dDdlt)
= rate in .V D(t). then
Radioactive Dating
Archaeology. the amount
of dye per liter of diluted mixture). that radioactive substances decay at a rate proportional to the amount of the substance that is present.1. we can now formulate an initial value problem for the amount D(t) of dye in the heart at any time t.3
Variables Separable
23
In the dye-dilution method an amount of dye of mass D. 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. We also assume that the heart is a container of constant volume V liters. This is a special case of mixture problems that we will study in the next section. 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. from experiments.rate out. milligrams of dye at time t = 0.
The quantity (concentration) D(t)/V is useful in measuring the cardiac output. Thus.
(13)
(14)
where Do is the (initial) amount of dye injected at time t = 0.
Paleontology. Arts
N(t) _ -kN(t).
(17). and so on. (17) we obtain e-` = N(t)IN. it is known that each decaying uranium 238 atom gives rise to a single lead 206 atom. 1949).)
"P. If we know the values of k. old paintings. from Eq. 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 value of N(t) is easily computed from the present amount of the
substance. Using Eq.No . [The negative sign is due to the fact that N(t) > 0 and decreasing. N(t). we can utilize some additional information known about some radioactive substances to compute their age. we see that if T is the half-life of a radioactive substance. (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). the time required for half of a given sample of a radioactive
substance to decay.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. In fact. To find the decay constant k. (Vermeer and De Hooghs are famous seventeenth-century Dutch painters.
. we utilize the half-life of the radioactive substance.10 where it is proved beyond any scientific doubt that the paintings sold by Van Meegeren are faked Vermeers and De Hooghs. that is. For example. then
2=e-kT
k=1T
that is. Although the value No is in general unknown.24
1
Elementary Methods-First-Order Differential Equations
where the positive constant k is called the decay constant of the substance. so it has negative "derivative" NV(t). Thus. rocks. fossils.. Van Meegeren's Faked Vermeers and De Hooghs (Amsterdam: Meulenhoff. See also the articles in Science 155 (1967): 1238-1241 and Science 160 (1968): 413-415. Coreman. the decay constant of a radioactive substance is obtained by dividing the natural logarithm of 2 by the half-life of the substance. Half-life is easily measured in the laboratory. the solution of the IVP (15)-(16) is given by (17) N(t) = N°e-". and N°. then from Eq. the number P of lead 206 atoms found in a uranium mineral containing N(t) atoms of uranium 238 is given by
P .] If N(0) = No
(16)
is the (initial) number of atoms of the substance at time t = 0. and tables are available that give the half-life of many radioactive substances.
as well as for smallpox. and
Medicine
1/m the proportion of those who die due to the disease.
(21)
Equation (21) is a differential equation with separable variables. That is. j.
From this we get
y(a) . mumps. This is the case (rather it is very close to being the case. we get
(1/S)dN/dt .1/m
The authors wish to thank Stavros Busenberg for bringing this model to their attention. Y(0) . since there are exceptions to all rules in these matters) for such common diseases as measles. which is now believed to be completely eradicated. then the following
relation can be derived:
dS(t)ldt = -pS(t) + (S(t)/N(t))dN(t)ldt + pS2(t)ImN(t). If we multiply both sides by N/S2 and regroup terms. we can consider one cohort group.p/m.
(20)
Recall that (d/dt)(N/S) = (1/S)dN/dt .p/m
So. so we can write this relation for its solution:
dyldr
oPY . and chicken pox.1/m
P log y(0) .1/m
= a.P/m
Or
°
dt = a. that is.
dy = a. The equation is actually less formidable than it looks.
y(a) .p/m.
. and define N(t) to be the number who have survived to age t. In trying to assess the effect of such a disease.
(19)
This equation was first derived by Daniel Bernoulli (in 1760) in his work on the effects of smallpox.P/m. If p is the probability of a susceptible's getting the disease (0 < p < 1). define y = N/S and rewrite the equation as
dyldt = py . all the individuals born in one specific year.3
Variables Separable
25
Certain diseases" that affect humans can be considered to impart immunity for life.(N/S2)dS/dt = pNIS .
Since N/S occurs as a variable.1/m = ep. and substitute in the lefthand side of this last equation to get
d(N/S)ldt = pN/S .(N/S2)dS/dt. an individual who has had the disease and survived is protected from ever having that disease again.1.tot PY . and S(t) the number who have not had the disease and are still susceptible to it at age t.
For example. The left-hand side. (19) are reasonable on the intuitive level
and lead to the mathematical formulation. pS(t)/m is the rate of death due to the particular disease.26
1
Elementary Methods-First-Order Differential Equations
Solving for my(a). but certainly not fpolproof.
So.1)e°r. Once we have Eq. To see that (19) is reasonable. yielding y(O) = 1.1)e'°. we can use differential equations to get
relation (23). The values of these parameters in the right-hand side of the equation can be determined from census records and statistics on the disease.1)e°^]. Using these constants in Eq. we could expect just about all survivors N(a) to also be susceptible (that is. 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).
REMARK 1
All these assumptions leading to Eq. Eq. The first. of the situation we are examining. recalling that y(a) = N(a)/S(a).1). only one in about eighteen of those twenty-four years old would not have had smallpox. and S(t)/N(t) is the proportion of susceptibles among the cohort group.
£22)
Since. (19). This is given by the second group of terms [dN(t)ldt + pS(t)lm](S(t)/N(t)). [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. Equation (22) now gives my (a) = 1 + (m . and the two basic constants m and p. Thus. (19) is an equality between two different ways of expressing the rate of change of susceptibles.
Formula (23) was derived on the basis of what one believes is a reasonable. every member of the cohort group is susceptible. is the rate at which the number of susceptibles is changing (decreasing). Here we would expect that S(a)IN(a)
. -pS(t). Bernoulli estimated p = 1/8 and m = 8 for the case of smallpox in Paris of the 1760s. we get
S(a) = mN(a)l[1 + (m . or model. where dN(t)/dt is the rate of change of the whole cohort group due to death for all reasons.056. only people who have not had the disease survive). This term includes those who will die from the disease. if the mortality rate of the disease is very high (m . The terms on the right can be collected into two groups. So. 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. dS(t)/dt. we have
my(a) = 1 + (my(0) . by age 24 (a = 24).
EXAMPLE 5 Using the population data that were available to him. and we should make sure that (23) fulfills these expectations. let us examine what its terms mean. Now we need an expression for the rate of removal of susceptibles due to death from all other causes.
(23)
This expression gives the number of susceptibles in terms of the number of survivors to age a of the cohort group. In other words. Finally. at birth. (23). S(0) = N(O). is the rate at which we expect susceptibles at age i to be infected with the disease. relation (19).
Bet. Chemistry Solve the differential equation
dx
dt
= k(a . (b) Compute the decay constant of the substance.x). 4-5 (1956-57): 159. What will the value of x be after a very
long time." and find
the value of I as w . Biology Assume that a population grows according to Verhulst's logistic law of population growth dN
T = AN . (This is true.b>0
which occurs in chemical reactions.2.
22. k.
where N = N(t) is the population at time t.4 . that
is.
23. as t . 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 . In a certain sample 50% of the substance disappears in a period of 1600 years.Y8
1
Elementary Methods-First-Order Dif. for example.»ntial Equations
18. A colony of bacteria increases at a rate proportional to the number present.-. (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.x)(b .
which seems to fit the data collected in a botanical experiment.
.Bt)32
given in the astronomy application in Exercise 8 of Section 1. What will the size of this population be at any time t? What will the population be after a very long time. for radium 226.1). Assume that the rate at which a radioactive substance decays is proportional to the amount present.-?
'Australian J. and the constants A and B are the vital coefficients of the population. that is. Compute the orthogonal trajectories of the one-parameter family of curves
y
cx.a.BN2. If the number of bacteria triples in 4 hours. Botany Solve the differential equation
d!
dw
= 0. as t .) (a) Write the differential equation that describes the decay of the substance.088(2.co?
21.
20.
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.
(beta distribution).
35. In the medicine application dealing with cardiac output assume that V = 500 milliliters (2 liter).
30. Repeat Exercise 34 if it is desired that the money be tripled at the end of
10 years. air resistance is acting upon this body which is numerically (in pounds) equal to three times its speed at any time. A sum of $1000 is invested at 1% annual interest compounded continuously. (A .000?
. = 2 milligrams. 2 seconds.5% annual interest compounded continuously?
34. A = B = D = 0 and C > 0
(normal distribution).
28. A tank contains 50 gallons of brine in which 5 pounds of salt is dissolved. In addition to its weight. 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. B = 0. C = D = 0. A person borrows $5000 at 18% annual interest compounded continuously.
Find the amount of dye in the heart after: 1 second.
Pure water runs into the tank at the rate of 3 gallons per minute. kept uniform by stirring. How much salt is in the tank after 15 minutes?
27. and D. The sum of $15.
33. and D.30 ft/sec (the convention adopted here is that the positive direction is downward).3
Variables Separable
29
25. runs out of the tank at the same rate as the inflow. C. and 10 seconds.1. and A arbitrary 29.
32. Find i if it is desired that the money be doubled at the end of 10 years. The
mixture. Solve the differential equation in the
following cases. r = 120 liters per minute.000 is invested at the rate of 6% per year compounded continuously. B = D = 0. B.1)/C < 1. How long will it take for $2000 to grow to $8000 if this sum is invested at 5.
(exponential distribution). For how long will the body be moving upward?
26. B > 0. From the surface of the earth a body of mass 2 slugs is thrown upward with initial velocity of vo = . 221 seconds. C > 0. and 31.
36. C A/C > -I
(gamma distribution). How much will this person owe the lender at the end of 1 year?
37. and A > -C D.
(x) and set a(x) = ao(x)/a. If we divide both sides of the differential equation by a. then the general solution of the differential equation y' + a(x)y = b(x)
is given by the formula
y(x) = e-M. we obtain the equivalent differential equation y' + a(x)y = b(x).
THEOREM 1
if a(x) and b(x) are continuous functions in some interval 1. ao(x). In fact (recall that (d/dx)[f a(x)d-] = a(x)).[ c + fb(x)efcxidx dx]. and the function f(x) are
continuous functions in some interval I and that the leading coefficient
a.(x).(x).(x) * 0 for all x in 1. The general
solution of Eq.(x)y' + ao(x)y = f(x)
We always assume that the coefficients a. We summarize our results in the form of a theorem.
REMARK 1
. (1) into a separable differential equation. (1)
where a(x) and b(x) are continuous functions of x in the interval 1.
(4)
If you read the method above carefully once more.)d. (1) can be found explicitly by observing that the change of
variables
w = yefa(x)dx
(2)
transforms Eq.
w' = y' efaixwx + ya(x)efa(x)dx
= [y' +
a(x)y]efaawx
(3)
= b(x)efa<xW.
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].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. you will realize that the general solution of a first-order linear differential equation is found in three steps as follows.(x) and b(x) = f(x)/a.
In general.4
First-Order Linear Differential Equations
31
Step I
Multiplying both sides of Eq.
Step 3 Dividing both sides by e5°(')d' yields
y(x) = e-m(x)'[c + fb(x)ef°(')dxdx].2. That is the case in Exercises 1. make sure that the
REMARK 2
coefficient of y' is 1.
Before you apply the method described above. 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. has a unique solution which exists throughout the interval I..
(5)
. where the functions a(x) and b(x) are continuous.
where x. we obtain
[yell"']' = b(x)er°(')dx
[This is a restatement of Eq. you have to divide by the coefficient of y' before you apply the method. 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. we find
yef'(')d' = c + f b(x)ef°(')d dx.] The term el("d' is called an integrating factor. E I. and
Example 2. together with Example 2 of Section 1. (3).
which is the general solution of Eq. This. (1) by e1''.1.2 the solutions of the differential equation (1) exist throughout the interval I. 2 and 5. (1). implies that the
REMARK 4
IVP
y' + a(x)y = b(x)
y(xo) = yo.
Step 2 Integrating both sides.
Dividing both sides by 1/cos x (in other words.(2/x)y = -x.
EXAMPLE 2
Solve the IVP
xy' .
Solution
This is a first-order linear differential equation with
a(x) = tin x and b(x) = sinx.
both continuous in the interval (0.-.
1
y X2
x I
. Although we could use formula (5) to immediately obtain the solution of the IVP. is either (-so.. rr/2).)d.2y = -x2
YO) = 0. 0) or (0.=a-1. x).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.
The differential equation can be written in the form y' . Multiplying both sides by
ea. it is perhaps more instructive to proceed directly by using the steps described in the text. we find
ycos x
1
cosx'
= c . cc).
0<x<2. the interval I.. it follows from Remark 4 that the IVP has a unique solution that exists in the entire interval (0. multiplying both sides by cos x) yields the solution y(x) = (cos x)(c . 00). Multiplying both sides by the integrating factor. ir/2).In cosx).
e-f(2J. Since xo = 1
Solution
-x with x + 0. = e-2 1. Therefore.In cosx.=
we obtain
1
1
cosx '
sinx
ycosx
Integrating both sides. Here a(x) = -2/x and b(x)
lies in the interval (0.d. = e-m
we have
f1/
=
1
X2. where a(x) and b(x) are continuous.
some of which we illustrate below. we obtain
y
1
=c . and a generator that supplies a voltage V(:) when the switch
Circuit Theory
V(! )
Figure 1. which contains a resistance R. an inductance L.
So.10
. Using the initial condition y(l) = 0.4. we find
0=12(c-In1) c=0.
Consider the RL-series circuit of Figure 1.9
APPLICATIONS 1.1.4
First-Order Linear Differential Equations
33
Integrating both sides.10.1
In the literature there are many occurrences of the applications of linear differential equations. the general solution is
y(x) = x2(c .Inx.
and therefore the solution of the IVP is
y(x) = -x2lnx.
yl
x
Figure 1.
The graph of the solution is given in Figure 1.9.In x).
Across the inductor the voltage drop is L(dl/di). The current I = 1(t) in the circuit satisfies the linear first-order differential equation
L dl + RI = V(t). Across the resistor the voltage drop is RI.
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.[c +
F
JV(t)etRcudt
.1. 3.
(6)
This differential equation is obtained from Kirchhoff's voltage law (see the
The reader can verify that the general solution of Eq.L).1) and the following facts: 1. V(t) is the only voltage increase in this circuit.
(7)
Table 1.2
Table of Units
. 2.2 lists the units and conventional symbols of circuit elements that will be used in this book in connection with circuit-theory applications. (6) is given by (see
Exercise 19)
I(t) = e-(R.
electric circuits application in Section 1.
x
Since the curve passes through the point (1.11
. y) is equal to F(x.)e.1. y) are curves y = y(x) whose slope at any point (x. we should have
y(l) = 2. y) is given by 2 . = e"' = x.
Geometry
Find a curve in the xy plane.11). it follows that the solutions of the differential equation
y' = F(x. we obtain
(xy)' = 2x xy = c + xz
'
From y(l) = 2 it follows that c = 1.
Figure 1.
(8)
(9)
Let us solve this IVP (8)-(9). 2). passing through the point (1.4
First-Order Linear Differential Equations
With the geometric interpretation of the concept of derivative in mind.
EXAMPLE 3
Solution The differential equation that describes the curve is
y'=2-y. and soy = x + (l/x) is the curve with the desired properties (Figure 1. The differential equation can be written in the
form
I )' + X y=2
and so is a linear first-order differential equation. whose slope at the point (x. 2).yix. y). Multiplying both sides by eIt .
D = a . the price at which
a-bP=D=S= -c+dP. To compute the "rate out. Clearly. S= -c + dP
(a. EXAMPLE 4 A large tank contains 81 gallons of brine in which 20 pounds of salt is dissolved. Thus. Let us denote by P the "equilibrium" price. Here
rate in = (3 lb/gal)(5 gal/min) = 15 lb/min. (11) that as t -* -. and "rate out" is the rate at which salt runs out of the tank at time t. dP =
dt
a(D .
(10)
where the constant a is called the "adjustment" constant. the amount of salt per gallon of brine at time t.bP. 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'
.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 . that is.
Eq. Let us illustrate this by means of a typical example.
where "rate in" is the rate at which salt runs into the tank at time t.P]
+. d positive constants).
(11)
Observe from Eq.
that is. (10) is a linear first-order differential equation. The mixture. kept uniform by
stirring. In the simple case where D and S are given as linear functions of the price. the price of the commodity P(t)
Mixtures
First-order linear differential equations arise as mathematical models in rate problems involving mixtures.S of the demand D and supply S in the market at any time t. 15 pounds of salt per minute flows into the tank. 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.rate out. (10)
is given by
P(t) = [P(0) . b. that is. Brine containing 3 pounds of dissolved salt per gallon runs into
the tank at the rate of 5 gallons per minute. c." we should first find the concentration of salt at time t. Then y'(t) is the rate of change of the salt in the tank at time t.F.
Solution
y'(t) = rate in .S). that is. runs out of the tank at the rate of 2 gallons per minute.
It is left to the reader (see Exercise 26) to verify that the solutions of Eq.
1 + e) and whose slope at any point (x.
19. V = 8 volts.xy'+6y=3x+1
10. y 9.y = 2x2
6. R = 5 ohms. Find the solution of the IVP
y'-y=b(x)
Y(O) = 1. y _ 1+x2
Y
x
__
X
1 +x2Y
13. where Vo and
w are given constants. What will the current be after a very long time?
22. sin wt.R" le teen
where 1o is the current at time t = 0.
where
IX0.yy'-7y=6x
14. e dx + x' dy + 4x2y dx = 0
1x y-y2=0 sin
11.
18. L = 4 henries. y2dy + y tan x dx = sin' x dx
7.38
1
Elementary Methods-First-Order Differential Equations
5.1 seconds. y'+
3
X-y=(x-1)'
8. Find the solution of the differential equation (6) subject to the initial condition 1(0) = 4.
20. y) is given by
2-y+e X x
16. ydx+x=Y
15. Find the current at the end of 0. In an RL-series circuit assume that the voltage V(t) is a constant Va. Find a curve in the xy plane that passes through the point (1.
21. and
1(0) = 0 amperes.xy'+y=x5
12.
.
b(x) =
x<0
x >_ 0. Show that the current at any time t is given by 1(t) = e° +
(1o
. Verify Eq. Assume that the voltage V(t) in Eq.
17. (7) of this section. Compute the orthogonal trajectories of the family of concentric circles
x2+y2=R2. (6) is given by V. Verify formula (5) and apply it to the IVPs of Examples 2 and 3. xy' . . In an RL-series circuit.
The Bernoulli equation is the differential equation
y' + a(x)y = b(x)y".
24. In an RL-series circuit [see Eq.
Show that the transformation w = y' -" reduces the Bernoulli differential equation to the linear differential equation
w' + (1 .xy = (1 . A tank initially contains 10 gallons of pure water. solve the Bernoulli differential equation (see Exercise 27).1. (1 + x2)y' + 2xy = . Starting at time t = 0 brine containing 3 pounds of salt per gallon flows into the tank at the rate of 2 gal/min. 27. (1-x)y'+xy=x(x-1)2
y(5) = 24
37. V(t) = 3 sin t. (11) of this section. Brine containing 2 pounds of dissolved salt per gallon runs into the tank at the rate of 4 gallons per minute. xy' + y = 2x
y(2) = 2
34. (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. Verify Eq.x2)e°11i'2 y(0) = 0
33. The mixture. and 1(0) = 10 amperes. xy ' -
1
y ' .n)a(x)w = (1 . y' + Xy= -2xy2
31 .n)b(x). A large tank contains 40 gallons of brine in which 10 pounds of salt is
dissolved. kept uniform by stirring. xy'
In x = 0
35.1.
n * 0.4
Flrst.
28.
XI
y= -
2y
1
29. runs out of the tank at the rate of 3 gallons per minute. The mixture is kept uniform by stirring and the well-stirred
mixture flows out of the tank at the same rate as the inflow.Ordar Unaar Difbnntlal Equations
39
23. (x-1)y'-3y=(x-1)'
y(-1) = 16
.2x
y(e) = -1
Y(0) = -1
36. R = 6 ohms.
25.2xy = 4x y'°
Y
In x
=y
2
In Exercises 32 through 37 solve the initial value problem. y' 30 .
32. y' . (6)] it is given that L = 3 henries. How much salt
is in the tank after 5 minutes? How much salt is in the tank after a very
long time?
26.
In Exercises 28 through 31. Compute the value of the current at any time r.
where a.
dT
dt
_ _ k(T. Math. Biophysics In a study related to the biophysical limitations associated with deep diving." Bradner and Mackay.T j. This model gives a good
approximation of the true physical situation provided that the temperature difference is small. Biophys. 25 (1963): 251 72. Show that the general solution of the differential equation
y'-2xy=x'
is of the form
y = Y0(x) + Ae'. Monthly 63 (1956): 414. where the temperature is 40°F. and B are constants. A body of temperature 70°F is placed (at time t = 0) outdoors.
(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. Mackay.
. "Amer. in temperature between the body and its surrounding medium. obtained the following first-order linear ordinary differential equation
y'-Ay=B+be-°`.
where k > 0 is a constant of proportionality. After 3 minutes the temperature of the body has fallen
to 60°F.
41.(x) is a solution of the differential equation that obeys the inequalities
-x
2
1
4x
-Y(x) _ -x xx ?2. b. S. Bradner and R.
where Y. Show that the general solution of this differential equation is given by
B_
where-c is an arbitrary constant.
b
39. Show that the general solution of this differential equation is
T(t) = T + ce-"'. Bull. A.T.'S
40. 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 . YO (x)
[Hint: Find the general solution of the differential equation and observe that the improper integral jo t'e-"dt converges. That
is.40
1
Elementary Methods-First-Order Differential Equations
38. Math. generally less than about 36°F.
03. that is (suppressing the arguments x and y)
df=Mdx+Ndy. 17 (1965): 7-13. "For a detailed study of the differential equation (14). y)dx + N(x.b=0. such as suicide). g.1. y)dy = 0
(1)
is called exact if there is a function f(x. Z. where the functions f.
1.(x) +
w(x)
reduces it to the linear first-order differential equation
w' + [g(x) + 2y. Math. Show
that if y.4
First-Order Linear Differential Equations
41
42. Solve this Riccati differential equation in each of the following cases. Rapoport.
(2)
A. an act that is performed at most once in the lifetime of an individual. see the paper by J.03. Math. y)dy. then the transformation
1
Y(x) = Y. 14 (1952): 159-69. x(t)= -tlI t
-0. Biophys.(x)h(x)]w = -h(x). x (t) =
t
10'b
10
44. [Hint: y = 1 is a particular solution.y)[x(t) + by].5
EXACT DIFFERENTIAL EQUATIONS
A differential equation of the form M(x.
Mass Behavior In a research article on the theory of the propagation of a single act in a large population (in other words. and h are continuous in some interval I. Rapoport" derived the following Riccati differential equation:
y = (1 . Bull.
(14)
Here y = y (t) is the fraction of the population who performed the act at time t. x (t) is the external influence or stimulus and the product by is the imitation component.
. Hearon. Biophys. Bull.(x) is a particular solution of the Riccati differential equation. y)dx + N(x. A Riccati equation is a differential equation of the form
y' = f(x) + g(x)y + h(x)y'."] 43. y) whose total differential is equal to M(x.
= N. then (6) holds. if Eq. 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. yo) in the region where the functions M. Y) = jF M(x. the general solution of Eq. is there a systematic way to solve it?
Answer 1
The differential equation (1) is exact if and only if
M. (1) is exact.
Thus.
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
.y)dx + (-x + 4y)dy = 0
is exact because
d(x2 . y)dy = c
o
J
o
(7)
gives (implicitly) the general solution of the differential equation (1).
Thus. N. Then
AX. Y) = c. provided that the partial derivatives of the function f with respect to x and y
exist. if the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x. are continuous.xy + 2y2)dx +
ay
(4)
(x2 . (4) is given (implicitly) by
x2-xy+2y'=c.
The following two questions are now in our minds. then Eq. For example. then [because of (2) and (1)] it is equivalent to
df=0. y0)dx + r N(x.) If Eq. To justify the answers above. (1) is exact. (1) is exact.xy + 2y') = d (x2 . (1) is
given by
fix. N and their partial derivatives Mr.y)dx + (-x + 4y)dy. the function f(x. the differential equation
(3)
(2x ..xy + 2y2)dy
= (2x . conversely.-
(6)
That is.
Answer 2 Choose any point (x(.
implies that
df = Mdx + N dy.Y)Iiy = M(x. Y) =
JM(t.
o
(T)
and the assumption should be made that the segments from (xo. = 8xy and N. .. . then (6) holds.
where f is the sum of the two integrals in (7) and conversely if Eq. y() in Eq.1. there exists a function f such that df = M dx + N dy f . N and
My. y)dy = M(x.Y0) + M(x.5
Exact Differential Equations
43
M.Y)
0
and similarly fy = N(x. Students sometimes find the presence of the "arbitrary" point (xo. Yo) + Jy N (x.
REMARK 3 As shown. observe that. df = f dx + fdy = M dx + N dy. yo) + Jy My(x. which proves the "if" part of Answer 1. In fact. =f . we have
The point (xo. exist and are continuous. Consequently alternative systematic methods for the determination of f are given in Examples 3 and 4. y)dy
yo yo
= M(x. Therefore. (7) may be chosen judiciously for the purpose of simplifying M(x.y). y = M y and f .. (6). (7) we have
f = M(x. N. = N and since f. y() in the first integral of (7) and the evaluations of both integrals at the lower points of integrations. yo)
and from (x. yo)dt + j y N(x. y) = 2y . = 8xy My = N. y) = 3x2 + 4xy2
''
and N(x. = N.3y2 + 4x2y M. yo) to (x.
Ni. (7) is useful to establish the existence of the function
f and also to verify the test for exactness. from Eq. (1) is exact. yo) and the judicious selection of
values for xo and yo confusing. y) lie in the region where the functions M. Eq.
Solution
Here
M(x. (7) should be
f(x. s)ds = c. To prove the converse. = N
M.
.3y2 + 4x2y)dy = 0.
EXAMPLE 1
Solve the differential equation
(3x2 + 4xy2)dx + (2y . that is Eq.
REMARK 1
REMARK 2 The precise form of Eq.
Eq. since
f = M and f . (1) is exact. yo) to (x.
3y2 + 4x2y).
Integrating with respect to x we obtain
f = x3 + 2x2y2 + h(y). and g(x) would be determined by setting f = M.3y' + 4x2y)dy = 0. = N.
As an alternative construction we could have integrated the expression f. = M = 3x2 + 4xy2. Regrouping terms in accordance with Remark 5.
.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. we have
[3x2 dx + (2y . Taking the partial derivative with respect to y. This omission is permissible since otherwise it would be combined with the constant c when we set f = c. = M and f.
From Example 1 we know that this equation is exact.. = N. In this case the "constant" of integration would be a function of x.
h(y) = y2 . To determine h(y) we utilize the fact that f. Consider
Solution
f. = 4x2y + h'(y).y3) + d(2x2y2) = 0
d(x3+y2-y3+2r2y2)=0 x3+y2-y3+2x2y2=c.
The "constant" of integration is a function of y due to the partial integration with respect to x.) = c.1.3y2
Thus. y) = x3 + 2x2y2 + y2 . and therefore there exists a function f such that f.
Solution From Example 1 we know that this equation is exact.y3
f(x. this last expression yields
f.3y2 + 4x2y)dy = 0. we have
h'(y) = 2y .
and the general solution is given (implicitly) by
x3 + 2x2y2 + y2 .
Setting this equal to N (N = 2y .3y2)dy] + (4xy2 dx + 4x2y dy) = 0
d(x3 + y2 .y3. = N with respect to y.
Note that the solution for h(y) omitted a constant of integration which normally would appear.
EXAMPLE 4
Solve the differential equation (3x2 + 4xy2)dx + (2y . say g(x).
EXAMPLE 3
Solve the differential equation
(3x' + 4xy2)dx + (2y .
4y . we should compute the slope of the curves (10). and for the family (14).y2 . and xy = y with at > 0.
2x + 2y + 2xy' . Thus. We recall from analytic geometry that an equation of the form Axe + Bxy + Cy2 + Dx + Ey + F = 0 represents a hyperbola. In general. zero.46
1
Elementary Methods-First-Order Differential Equations
APPLICATION 1. 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). Differentiating with respect to x. Jr.x + y + 2)dx + (x + y + 2)dy = 0. Calculus and Analytic Geometry.4x . Regrouping the terms in Eq.4x . B2 .aye = a p." For the family (10).4AC is positive.
(14)
The curves given by Eq. p. parabola. (12) in the form
(-xdx + y dy + 2dx + 2dy) + (y dx + xdy) = 0. Thomas.2yy' + 4 .4(1)( -1) = 8.
(12)
we recognize it as an exact differential equation.4y = const.
Thus. the general solution of the differential equation (13) is
x2 .
and so
(10)
Solution First. R > 0.: Addison-Wesley.2xy . Mass. and y arbitrary. both of these families consist of hyperbolas. The
standard forms for hyperbolas are axe . or ellipse depending on whether
B2 . the rotation angle 0 is
"G. = k.
we recognize that Eq.axe = a(3. Alternate Edition (Reading. but after writing it in the
form
( . 508..4AC =
(2)2
. (14) are the orthogonal trajectories of (10).4(1)(-1) = 8. 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). or negative. This equation is not separable or linear.4AC = (. B. B2 .2xy) = 0. 1972).2)2 -.5.1
Geometry
Compute the orthogonal trajectories of the one-parameter family of curves
x2+2xy-y2+4x-4y=c. we obtain
y'=
x+y+2
x-y-2
Thus.y2 .
.4y' = 0.2d(x2 . (3y2 . (13) is equivalent to (the total differential)
(13)
Zd(-x2+y2+4x +4y+Zxy)=0
.
. it is always possible to choose p. (x-y)dx+(-x+y+2)dy=0
y(1) = 1
19. Integrating Factors If the differential equation
M(x. as a function of x only. Since (26) is exact.By . C. y) is an integrating factor of the differential equation (25) if and only if it satisfies the partial differential equation
Nµ. (µM). (x+y)dx+(x-y)dy=0
y(0) = 2
20.
. * 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.
(27)
22. In general. (25). = (µN). (a) Under what conditions will the differential equation (24) be exact?
(b) Using the condition obtained in part (a).
(23)
where c is the parameter and A. 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. Show that the differential equation for the orthogonal trajectories of this family is
(Bx + 2Cy + E)dx + (-2Ax . (27) into a first-order linear differential equation whose solutions can be found explicitly. In this and the following exercise the restrictions imposed on M and N reduce Eq..N)
is a function of x alone. Consider the one-parameter family of curves
Axe + Bxy + Cy2 + Dx '+ Ey = c. (x2 + y2)dx + 2xy dy = 0
18. that is.D)dy = 0. and E are arbitrary (but fixed) coefficients.)µ. and its solutions will also satisfy the differential equation (25). y) dx + N(x.M dx + µN dy = 0 (26) is exact. Show that a function µ = µ(x. If it happens that the expression
N (M.
... D. M.
Y'=
y(1) = -1
y-x+1 -x+y+3
(25)
y(1) = 2
21. y) dy = 0
is not exact.1. that is.N..
(24)
16. . . it is very difficult to solve the partial differential equation (27) without some restrictions on the functions M and N of Eq. B. The function µ is then called an integrating factor of the differential equation (25).5
Exact Difarential Equations
49
15.Mµ. we
can solve it. find the equation of orthogonal trajectories and show that they and the family (23) are hyperbolas.. Solve the following initial value problems:
17. = (M.
d = F (y.e-Y + c and so
y'(x) = w = -y .
Y
which is first order with independent variable y and unknown function w.56
1
Elementary Methods-First-Order Differential Equations
Another integration yields y = b c.x' + c.
EXAMPLE 2
Solve the IVP
y"=Y'(y+Y)
Y(O) = 0
(8)
Y'(0) = -1.. Second-order differential equations of the form
y' = F (Y. In fact. dy
(9)
which is linear. (6) becomes
dw
dy
dw dy _ dw dx . 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'. Eq. its general solution (5) contains the three arbitrary constants c c2. is an arbitrary constant. Eq..
B. (8) becomes
Solution
dw _ w = Y.
(5)
The general solution of an ordinary differential equation of order n contains n arbitrary constants.I + c.
REMARK 1
and c3. For example.
W.eY. Integrating with respect to y we find we-Y = -ye-Y . and as we have seen. 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'.dy w.
. w). from (7) we obtain (using the chain rule)
(7)
y" = dx =
Thus. (3) is
y(x) = c.x + c. As c. (3) is of order 3.x' + czx + c. Multiplying both sides of (9) by a-Y we obtain (d/dy)(we-Y) _ ye-Y.
In fact. the general solution of Eq.
V2II/
(11)
Equation (11) is a differential equation of form A and can be reduced to a firstorder differential equation. Find the concentration in each compartment at any time t.) and Y2 = k(c. The differential equations describing the diffusion are
y.l + e-'.(t) be the amount of the solute in the compartments A.(t) = V.
EXERCISES
Compute the general solution of the following differential equations. we find that c
1. and V2. and 7. Using the initial condition y(O) = 0. .1
Assume that two compartments A. we find that
k
(1
1V2)
k
or
//
11
i
V+i
V.1.
(c2 -. respectively. Thus.7
Equations Reducible to First Order
57
Using the initial conditions.
v'
(10)
e.
y'= -y -l
and y(x) _ -1 + ce '.c. y' + y' = 3
2. from the higher concentration to the lower. and A2. at time t. . we get
c.
Solution Let y.y(4) = 0
.c2)
y'
V2
Differentiating both sides of the first equation and using the second equation. . Then
ca(t) =
ye(t) V. y(5) . = 0. and A2. and the solution of Eq.
APPLICATION 1. and
between
Two
Compartments
c.(t) and y.c2 in concentration of the two compartments.
where k is a constant of proportionality (k > 0). = 0. of volumes V. (8) is
y(x) _ . and A respectively. We leave the computational details for Exercises 5. respectively.(t) = k(c. =
k
V. + k l
\V. Dividing by the volumes of the compartments.)
and
c2 =
k
2
(Cl . 6.
l
+
l l c. Through the barrier a solute can diffuse from one compartment to the other at a rate proportional to the difference c. we find that c.
Chemistry
Diffusion
are separated by a barrier.
1.c.
yzt)
are the concentrations in compartments A.7.c2).
assume that
V(t) = V.
where
26. How long does it take this colony of bacteria to double?
30. Circuits In the RL-series circuit shown in Figure 1. In 1 hour their number increases from 2000 to 5000. how long will it take for the bacteria to triple? 29. A colony of bacteria increases at a rate proportional to the number present. Show that
1(t) __ R2 V
+ww2L2 e' .} (R2
VO
+
sin (wt -
where 4.y =
wL
(R2 + w2L2)" 2 '
cos W = (R2
.14. What will the amount be after 6 years?
. Find the solution of the IVP
Y'+1y=b(x)
y(1) = 0. sin wt and 1(0) = 0. If the number of bacteria doubles in 3 hours.14
28. For this reason the first term is called the transient and the second term the steady state.1
R
+ w2L2)1R
Note that the current 1(t) is the sum of two terms. A colony of bacteria increases at a rate proportional to the number present.60
1
Elementary Methods--First-Order Differential Equations
25.
Figure 1. is an angle defined by
t sin .
27. Compute the orthogonal trajectories of the one-parameter family of curves x2 + b2y2 = 1. The sum of $4000 is invested at the rate of 12% compounded continuously. as t --o m. the first term dies out and the second term dominates.
Desoer and E. Carbon derived from wood taken from an Egyptian tomb proved to have a 10C content of 7. and e = 8 volts. [Note:
1µf = 10-6 f. L = 10 mH. 1969).62 disintegrations per min.
"This is Exercise 5-33 in J. y) is given by y + e`. 0 1971 Addison-Wesley Publishing Company.
(Englewood Cliffs. 169. runs out of the tank at the rate of 2 gallons per minute.]
-This is Exercise 12a in C. 54. Mass. G. Introduction to
System Dynamics (Reading. In Figure 1. H. Reprinted with permission of Addison-Wesley Publishing Company. p. Murphy. Inc.: Addison-Wesley. with the result that they contain 10 disintegrations per minute of "C per g of carbon. p. Inc. kept uniform by stirring. In the circuit20 shown in Figure 1. Kuh. Introduction to Nuclear Physics and Chemistry. calculate and sketch the current i for
t ? 0. find and sketch" the voltage v2 across C2 as a function of
time after the switch is closed.: Prentice-Hall. 1971). Reprinted by permission of Prentice-Hall. S. Initial voltage across C. and C2 is zero. passing through the origin and whose slope at any point (x. [Note: ImA = 10-'A and 1mH = 10-' H]
Figure 1. this '°C decays with a half-life of 5568 years.Review Exercises
61
31.15. T. The mixture. It is known that the half-life of radiocarbon is 5568 years.15
35.. Richardson. Brine containing 2 pounds of dissolved salt per gallon runs into the tank at the rate of 3 gallons per minute. On their death.J. Shearer. 1969). How old is a wooden archaeological specimen which has lost 15% of its original radiocarbon?
32. Harvey. (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. Find a curve in the xy plane.
.16. N. and H. L. Radioactive "C is produced" in the earth's atmosphere as a result of bombardment with cosmic rays from outer space. A. knowing that i(0) = -10 mA and given R = 500 fl. per g. p. 2nd ed. 1 kilohm = 10' fl. The radioactive carbon is taken up by living things. and no fresh '°C is absorbed. What was the age of the tomb? 36.
"This is Exercise 9 in B. A. 143. 33. Reprinted by permission of McGraw-Hill Book Company. A large tank contains 30 gallons of brine in which 10 pounds of salt is dissolved. Basic Circuit Theory (New York: McGrawHill Book Co.
The hawk flies twice as fast as the sparrow. England: Ellis Horwood. Burghes and A. 31. "This is Example 5.1)
(Hint: The first has (x + y)
as an integrating factor. Publishers.
41. 42.17
40. Monthly 40 (1933): 436-37. p. Reprinted by permission of Ellis Horwood Limited.x. Solve the differential equations39
x2-y2+1
Y
x2 . Reprinted by permission of Ellis Horwood Limited.28 The eagle is 50 feet above the sparrow and the hawk is 100 feet below the sparrow. the second is exact. and sparrow are in the air. Classical Mechanics. Modern Introduction to Classical Mechanics and Control (West Sussex. N. If v = u at x = 0. the ball's velocity on hitting the ground is approximately independent of its initial speed. The sparrow flies straight forward in a horizontal line. show that if the building is sufficiently tall. eagle. M.4 in D. "From Amer. 1975). Downs. Math. Math.1 in Burghes and Downs. A hawk. show that the equation dtxldt2 = f can be written as
vd-=f
where v is the particle's speed. How far does each fly and at what rate does the eagle fly?
43. If x is the distance from some fixed point on the straight line. Monthly 50 (1943): 572. integrate this equation
and show that
dv
v2=u2+ 2f .Review Exercises
63
Switch
Figure 1. p. The hawk and eagle reach the sparrow at the same time.]
'"This is Problem 2. A particle26 is moving on a straight line with constant acceleration f.y2
1
'
y
y(2x+y-1)
x(x + 2y . 109.
Assuming a model with constant gravity and air resistance proportional to its speed. "From Amer. Publishers. Both hawk and eagle fly directly towards the sparrow. A ball" is thrown vertically downwards from the top of a tall building.
.
Publishers.
45. A cat30 is running along a straight edge of a garden. sees the cat when it is at its nearest point. Find the current at any time t. 34. p. Reprinted by permission of Ellis Horwood Limited.17 in Burghes and Downs. V(t) = e. L = 2 henries.64
1
Elementary Methods-First-Order Differential Equations
44. and 1(0) = 5 amperes. A dog.volts.
'This is Exercise 2. R = 4 ohms. 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. Find the time that elapses before the cat is caught and show that the cat runs a distance (2/3)b before being caught.
. sitting in the garden at a distance b from the edge. Classical Mechanics.
Our main concern in this chapter.2y = x'. it is a linear differential equation with variable coefficients.
. otherwise.. . (1) is homogeneous..
x
0
(2)
y" + 2y' + 3y = cos x
Y") . When f(x) is not identically zero. 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.
The first differential equation is nonlinear because of the term y2..1.(x).(x)Y' + au(x)Y = f(x)
(1)
. aside from applications. . (1) as a linear differential equation with constant coefficients. ..(x)y") +
a.. If all the coefficients we speak of Eq. Eq... the second
because of the term yy'. and the third because of the term sin y [recall that
sin y = y . a. . y'. ( x ) . .1
INTRODUCTION AND DEFINITIONS
A linear differential equation of order n is a differential equation of the form a. (1) is called
a.. The interval I is called the interval of definition ficient of the differential equation. tion f(x) are continuous functions in some interval I and that the leading coef0 for all x E I. y")..(x) are constants.
(3)
(4)
Equation (2) is a nonhomogeneous linear differential equation of order 1 with variable coefficients.(y'/3!) + (yb5!) . is to develop the elements of the theory of solutions of linear differential equations and to discuss methods for obtaining their general solution.y = 0. Equation (3) is a nonhomogeneous linear differential equation of order 2 with constant coefficients. The following are examples of linear differential equations:
xy' .ao(x) and the funcWe always assume that the coefficients a"(x).. The following are nonlinear differential equations:
y"+y2=sinx
y"+siny=0. nonhomogeneous.. When the function f is identically zero. Equation (4) is a homogeneous linear differential equation of order 4 with constant coefficients.. -. we say that
Eq.CHAPTER 2
Linear Differential Equations
2..
(6) is Eq. (10) with a' = K/m]. 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. 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. = -a sin at ' y. or 0 = 0. For example. called "qualitative theory of differential equations. and 2. For example." 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. and substitution of the results into Eq. matters are not so simple when we try to solve linear differential equations with variable coefficients of order 2 or higher. It will be shown in the applications that the initial value problem (JVP) (6)-(7) describes the motion of a vibrating spring [Eq. This was done in Section 1. However.9 and Chapter 5). dots stand for derivatives with respect to time t. _ -a' cos at. In fact.(t) = cos at when substituted into Eq. In a similar fashion we
can verify that the function y2(t) = sin at is also a solution of the differential
. let us consider the second-order linear differential equation
y+a'y=0. Another way.66
2
Linear Differential Equations
Of course. In fact.
Y(0) = v0. For the sake of motivation. Hence.5.4 of Chapter 1. y. A practical way out of this difficulty is to approximate the solutions of differential equations with variable coefficients.
(6)
(7)
As is customary. is a solution. = cos at y. we shall learn in Sections 2.8). we already know how to find the general solution of any linear differential equation of order I with variable or constant coefficients.. of the Euler type. (2) is easily found to be
y(x) = x' + cx'. Finally. It is easy to verify that the function y. Section 2. (6) reduces the differential equation to an identity. the reader should be told that in more advanced topics on differential equations. is to look for a power-series solution of the differential equation with variable coefficients (see Section 2. no matter what the order of the differential equation is. which will give a better approximation.4. Furthermore (subject to some algebraic difficulties only). 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. the general solution of Eq.
together with the initial conditions
Y(0) = Y.11 how to find the general solution of any linear differential equation with constant coefficients. the variable coefficients of a differential equation can be approximated by constant coefficients. 2. (6) yields -a' cos at
+ a' cos at = 0. in "small" intervals.
(5)
where c is an arbitrary constant.7) or unless we know (or can guess) one of its solutions (see Section 2. y.
Now y(t) satisfies the differential equation (6) and will satisfy the initial conditions provided that c. 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. (6)] How many "different" solutions are there to the differential equation (6)? Is (9) the only solution to the initial value problem. if y. c. involves a second
. + c.1.(i/2a)v. Thus. It would be natural to ask what motivated us to try cos at as a solution. do not. and c2.(0) = 0. =
e'°'. for any constants c. the sum of the two solutions is also a solution. Furthermore. and c2 are chosen properly.2. and c2? [For example. and y2 are solutions and if c. to suspect that if the constants c. y2(0) = 0. This result and its proof are presented as Theorem 1 of Section
2. y2 = e-'°'.2. satisfy the initial conditions. and c2 are arbitrary constants. a solution to the initial value problem is
y(t) = yo cos at + ° sin at.(0) = 0( # vo) and y2(0) = 0 (# yo). the linear combination
Y(t) = c1Yi(t) + c2Y2(t)
(8)
of two solutions is also a solution. (9) is the only solution of the IVP (6)-(7). Or we might ask whether there are simple solutions other than cos at and sin at. since y. + c2y2 is also a solution.
a
(9)
In Section 2. and y2 produce different values for c. and y2(0) = a. this system takes the form
Yo = c. then y(t) as given in (8) will be a solution of the initial value problem (6)-(7). however. = cos at and y2 = sin at. or would some other choice
of y. The preceding comments serve to distinguish linear homogeneous differential equations from other types of differential equations.1
Linear differential equations occur in many mathematical models of real-life situations. y. For linear homogeneous differential equations we have the principle that if y. It seems reasonable..
Hence. Newton's second law of motion. where i2 = . and any constant multiple of any one of these solutions is again a solution.(0) = 1.y.] The answers to these questions will be obtained as we proceed through the remainder of this chapter. for example.
vo = ac2.y2 also solves the initial value problem (6)-(7). [The reader can verify that e`°' and e-'". are solutions of Eq..
APPLICATIONS 2.3 we establish that Eq. then c. then the linear combination c.
Note that the functions y. in general.yo .1. = . although they satisfy the differential equation (6).1
Introduction and Definitions
67
equation (6). y. c2 = 2'y + (i/2a)v.
1(c)]. Assume that at the equilibrium position the length of the spring is L + 1. For convenience. Next we displace the mass vertically (up or down) to a fixed (initial) position. According to Hooke's law.
The differential equation that describes the motion of the mass m is
my = .
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 acceleration of the mass).1
. setting the mass in a vertical motion [Figure 2. There are two forces acting upon the mass: one is the force of gravity.Ky.
F=mg-K(l+y)= -Ky. and at that point we give it a vertical (initial) velocity (directed up or down). known as the spring constant. mg = Kl. Then by Newton's second law of motion. Thus (since mg = Kl). we have
my=F.
Mechanics
The Vibrating Spring
A light coil spring of natural length L is suspended from a point on the ceiling [Figure 2.
where the double dots denote second derivative with respect to time (in other words. and the
positive direction for the vertical (y) axis is downward.1(a)]. we choose the origin 0 at the equilibrium point. It is this motion that we wish to investigate. with the constant of proportionality.1(b)]. and the other is the restoring force of the spring. and consequently linear differential equations of second order have been of primary interest in motion problems. Naturally the stretch 1 is caused by the weight mg of the mass m. That is. the force needed to
produce a stretch of length I is proportional to 1.68
2 Unear Differential Equations
derivative (the acceleration).
or
y+mKy=0. K. which by Hooke's law is K(1 + y). Let y = y(t) be the position of the mass at time t. equal to mg.
(10)
Recall that the mass was set to motion by first displacing it to an initial
equilibrium
(c)
Figure 2.
together with the differential equation (10). (10) and (11)J
y+./=0
y(0) _ .. Thus.2. (11) yo is positive if the initial position is below the equilibrium and negative if it is above.
Solution
Using Hooke's law we compute the spring constant:
K = .1
EXAMPLE 1
Units of Measurement
A spring is stretched 12 cm by a force equal to 360 dyn.1
Introduction and Definitions
69
position. vo is positive if the initial velocity was directed downward and negative if it was directed upward. constitute an initial value problem.
3 cos tt + 5
sin vl0
t
... we have the two initial conditions
Y(0) = vo (11) Y(0) = Yo. together with the initial conditions (11).'n''a.
The IVP describing the motion of the mass is [from Eqs. The differential equation (10). Find the position of the mass at any time t.
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. say y. linear differential equation with constant coefficients. dyn/cm = 30 dyn/cm. (9) the solution of this IVP is
y(t)
5. Also. 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 Table 2. and then giving it an initial velocity v0.1 we list the units used in the two most common systems of
measurement. Equation (10) we recognize to be a second-order. homogeneous.3
y(0) = From Eq. In Eq.
(17)
Figure 2.
(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.4.
G(x + Ax) = -Dy'(x + Ax).
(12)
We want to compute the concentration y(x) of gas in the liquid.[ . say
y(O) = A
and
y(I) = 0. where the vibrating spring is discussed in greater detail. 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).Dy'(x)] . the total amount of gas that disappears by the chemical reaction in the section S is given by
Jf * ky(s)ds.
(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. at any point x.2). That is. 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.Dy'(x + Ax)] =
Jky(s)ds.
G(x) = -Dy'(x). we use Fick's law of diffusion.2
.70
2
Linear Differential Equations
The reader is urged to read Exercise 40 of Section 2.
(13)
To solve this problem we consider a small section S of the pipe between the point x and x + Ax (Figure 2. given that we know the concentration at the points 0 and 1.
Diffusion
Suppose that a gas diffuses into a liquid in a long and narrow pipe.
H(x) = ky(x). Similarly. That is. To compute G(x) and G(x + Ax).
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. 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). Thus. Therefore.
(16)
[ .
xy' + y 0 8. classify the linear ones as homogeneous or nonhomogeneous. (17) becomes
Ax. and state its order. y"+cosy=0
6. (18) as Ax-b 0. A spring is stretched 2 cm by a force equal to 16 dyn. y' + y" = 3x
7. (1 + x2)y" . Derive an IVP that describes the motion of the mass. A mass of 32 g is attached to the end of the spring.+xy=0 3. at time r = 0.
(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.
.2y-3xy'+4y' =0
14. (Note that derivatives are denoted by primes. e'(y')' + 3y = 0
13. we obtain the second-order linear differential equation
y
Dy=O. classify the differential equation as being either linear
or nonlinear. lowercase roman numerals.
EXERCISES
In Exercises I through 19.4y 0
16. (9) to find the solution of this IVP.6y = 0 17. or by arabic numbers in parentheses.
y'(x+Axx)-y'(x)=
fly (x+ 2x). y'+yy"=5
5. 7y' . xy" + 4xy" .2.)
1.yy' =3
18.1
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"'-2=0
15. y"+xy=x
12.e'y' . 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. y" + 5y' . The system is then set to motion by pulling the mass 4 cm above the point of equilibrium and releasing it. x > 0 19. Furthermore. with an initial velocity of 2 cm/sec directed downward. 7y"+3xy' . ys) . with constant coefficients or with variable coefficients. x2y" . 2y"' + 3y" .
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
20.2y(4) + y = 2x' + 3
2. y"+xzy=e' 4. Use Eq. Taking limits on
both sides of Eq.4y' + xy = 0
10.2 = 0
11. y.xy = 1. (sin x)y' + e''y = 1
9.
Use Eq. v(. 7 (1967).rr\/N/K. period 2. If we recall a technique from trigonometry [that an.expres-
sion of the form a cos t + p sin t can be written as A cos (t . at time t = 0.y"dx = [y'(p)]2 .
Jo
Hence. f(x) <_ M.
y(t) = yo cos
1K
t+
vo1/m
sin
fK
m r. A spring is stretched 1. by a 4-lb weight. = 10V cm/sec 24.
22. vo = 0 in. ya = 5 in. no. = cos-'(a/ a2 -+P') = sin ' ((3/ a2 + (32)].5 in. The system is then set to
motion by pushing the mass (note that mg = weight and that g = 32 ft/sec')
6 in. below the point of equilibrium and releasing it.
'Problem 25 and its method of solution are taken from the 1966 William Lowell Putnam Mathematical Competition.
In each of Exercises 22._. obtaining r
y2(T) + 2
0
e-'y'y"dx = y2(0).[y'(0)]2. yo = 3 cm. K = m. Show that all solutions of the differential equation y" + e'y = 0 are bounded as x -> x./sec
23. we must have
21re-y'y'dx = 2
o
y. Amer.
Note that for some p such that 0 s p s T. and phase 4. For this reason we say that the motion is oscillatory or that
vibrations occur. the motion of the mass repeats itself after every time interval of length
2ir vA/K.
The solution of the vibrating spring is given by Eq. the A=
solution can be written in the form
y(r)=Acos( mrwith A = y + mvo/K and 4 = cos-' (y/A ). where a2 + Rz and 4.).
The graph of y(t) is periodic of period tar \/1/K. and 24. K = m. Hence. with
an initial velocity of 4 ft/sec directed upward. (9). consequently. Hint: Multiply the differential equation by e-y' and integrate from 0 to T. Math. vo = 4 cm/sec 25.
. 23.. yo = 0 cm. we say that the motion is oscillatory of amplitude A. K = 5m.. A function f is said to be bounded' as x -> x if there exists a positive
number M such that lim. evaluate the amplitude and the period of the oscillation and sketch the graph of y(t). (9) to find the position of the mass at any time t..4. Monthly 74.72
2
Linear Differential Equations
21.
y2(T) + [y'(p)]2 = y2(o) + [y'(o)P.
Regardless of the values given to the constants c. (1). y'(1) = 2.2. and the solution we seek is
y=2x-1. (3). and c2 such that y = c. and. and y2. + c2y2. = x and y2 = 1 are solutions of Eq. Consequently.
(1)
Solution Because of the simplicity of this equation. c. (1) and Eq.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 as a consequence the general solution of such a differential equa-
. and y2 as constituting a fundamental set of solutions for Eq. (3) shows that they play a special role in that every solution can be expressed
as a linear combination of these two solutions. we speak of y. we consider an example. That is. From Eq.2
Linear Independence and Wronskians
73
2. (1). we can integrate it directly to obtain solutions. there are an infinity of solutions to the differential
equation (1).
EXAMPLE 1
Solve the second-order homogeneous linear differential equation
y' = 0. For example. = 0 and c2 = 1.
if y = y then c. 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. if y is a solution of Eq. = 1 and c2 = 0. we suspect that to solve an nth-order equation requires n integrations.
Intuitively.y. we need two conditions which serve to determine c. if we try to formulate conditions that reduce the infinity of solutions to a single solution.x + C21
(2)
where c. Of course. For this differential equation the general solution contains two arbitrary constants.2 LINEAR INDEPENDENCE AND WRONSKIANS
In Section 2. then c. (1) that satisfies the initial conditions y(l) = 1. suppose that we want a solution of Eq. Thus. and c2. = 2 and c2 = -1. and if y = y2. Integrating Eq.
Equation (3) is referred to as the general solution of Eq. (1). the expression for y given in (3) is a solution of the
differential equation (1). (2) with respect to x. and c2. Integrating Eq. we obtain
y' = c
have
y = c.
2 = y'(1) = c.(1) + c2. (1) with respect to x. (3) we have
1 = y(1) = c. is a constant of integration. c. and c2. differentiating Eq. Since it is possible to express every solution in terms of the solutions y. Hence. we
(3)
where c2 is another constant of integration.. Notice that y. then there exist constants c.
and f2 are linearly dependent on the interval -1 <. . Before we state this test. f . . This contradicts our assumption that f..x + a2x2 = 0
for all x in the interval -1 s x <. and a2.b. the functions are said to be
(a) Show that the functions f. such that
a. In what follows these notions will be made more precise. . am..1.. and f2 are linearly dependent. and f2 are linearly independent
on the interval -1 <.x s b is said to be linearly dependent on a s x s b if there exist
constants a a2. That is. Otherwise.
(b) Otherwise f. This is possible if we choose for example..x 1. . . The Wronskian o f f f2.b... such that
=0
linearly independent on this interval. f are solutions to a linear homogeneous differential equation. f.(x) = 3x + (12/5) and f2(x) _ 5x + 4 are linearly dependent on the interval ( -_).x :5 1.
DEFINITION 1
A collection of m functions f f2.. Hence.
DEFINITION 2
Let f f2. not all of which are zero..
a. f. each defined and continuous on the interval a <.1 derivatives are continuous on the interval a s x <.< x < x. we need the following definition. f be n functions which together with their first n . and a2.
EXAMPLE 2
on the interval -1 s x s 1. + a2 = 0
and
-()(I
The only solution of this system is a. If the functions f f2. = 5 and a2 = .. such that
a. not both zero.3.. 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. there is a simple test that serves to determine whether they are linearly independent or not. But for x = I and x = -1 we find
a. there exist constants a.
.74
2
Linear Differential Equations
tion involves n constants of integration.(x) = x and f2(x) = x2 are linearly independent
Solution (a) To show that the functions f. = a2 = 0.. (3x + 5 I + a2(5x + 4) = 0
for all x in the interval .x <.
(4)
for every x in the interval a <. . . not both zero.
(b) Show that the functions f. and f2 are linearly dependent we must show that there exist constants` at. Furthermore. ..
f..
. The proof of this theorem is neither difficult nor sophisticated. + a.-Ii
ft"-tt
Each of the functions appearing in this determinant is to be evaluated at x. y. the theorem will be stated in its general form and the proof given for a specific case. which distinguishes linear from nonlinear differential equations. we choose not to give a proof under the most general circumstances but rather demonstrate a proof for a specific version of the theorem. This format will be followed in many places in the text.2xcosx...(x)Y' + ao(x)Y = 0.2x
.
(f + f 2 +
+f)'=f+f2
-
+ fk
'For a brief discussion of determinants. f2(x) = cos x. . That is.. f .. . x). 'In Chapter 1 we expressed the philosophy that difficult or sophisticated proofs would be omitted in this text. ...
L
W(f.(x) = x2. is denoted by W(f f..2
Linear Independence and Wronskians
75
evaluated at x......f2.. . the linear combination
ciY1 + c2Y2 + . with the implication that the general proof follows along similar lines of reasoning..sinx.
cosx _ -x'sinx . This principle." are solutions to the same linear homogeneous differential equation.
THEOREM 1
If each of the functions y.
Solution
From Definition 2 and the functions given we compute
W(x2. + cmYm
is also a solution. find W( f f2... y2.. see Appendix A.(x)Y("-') + .
a^(x)Y`"i + a"-. Nevertheless.2.
In the introduction to this chapter we mentioned that linear homogeneous differential equations have the property that linear combinations of solutions are also solutions.
EXAMPLE 3
Given f.cosx. c".. and their application to linear systems of algebraic equations. . x) and is defined to be the determinanl2
fI
f2
.
then for every choice of the constants c c2. is embodied in the following theorem. .x) _
xz
. their properties.
Proof'
The proof depends on two basic properties of differentiation.
fo.x) =
All
fz
f
.
However. H.2 that the Wronskian is either identically zero or never vanishes. Thus. Also. it follows from Section 1.
In Example 2 we demonstrated that the functions x and x2 are linearly independent on the interval -1 <_ x s 1. Monthly 68. Meisters.(x)yi"-" + . "Local Linear
Dependence and the Vanishing of the Wronskian." Amer. the solutions y. .y = 0 are linearly independent because their Wronskian
e`
e
_
-e°-e°= -2
is never zero.x < b are linearly dependent if and only if W(y y2...2. the solutions e` and a-' of the differential equation
y" . which is called Abel's
'For a discussion of the connection between the Wronskian having the value zero and linear de-
pendence of functions.(x)y + au(x)y = 0
defined on an interval a <.. since we have assumed that a2(x) # 0 in L Thus. their Wronskian
REMARK 1
X
1
x21 2x
I
x2
vanishes at the point x = 0 of the interval -1 s x s 1._.
However. Therefore we have the following criterion for testing linear dependence (and linear independence). .(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.. = x and y2 = x2 are solutions of the second-order linear differential equation
x2y" . any interval of definition I for this differential a uation must e d x = . + a.2xy' + 2y = 0.. the reader is referred to the article by G. then their Wronskian is nonzero for some x.. there must not exist a second-order linear differential equation with interval of definition . Rather. Math.
For example. 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. Equation (8). Does this contradict
Theorem 2? No. The reader can verify that y..
.1 < x s 1 that has x and x2 as solutions.2
Linear Independence and Wronskians
77
true: if the functions are linearly independent. y" of the nth-order differential equation
a"(x)y"i +
a. .
THEOREM 2
n solutions y. y x) = 0 for every x in the interval a < x s b. y2. no. 9 (1961): 847-56.
it follows from Eq. + a.4. (5) of Section 1. Also.
Y. .(x)Y1"-u + . (8) that if W is zero at a point x0 on the interval a s x s b.
where each a.(x)Y' + ao(x)Y = 0.Yiy2
= y. ! en et er may.(x) is defined and continuous on a s x s b and a (x) # 0 on s x <.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.
Y.8). 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.
DEFINITION 3
Suppose that y y . y x is never zero on a s x
6
Proof The proof is for the case n = 2... y" are n solutions of the differential equation
a"(x)Y(". ... We then say that these functions form a
fundamental set (or fundamental system) of solutions for the differential equation.)Y2 = -p(Y.
THEOREM 3
I f y y2.(-py2 . y are solutions of the differential equation
a"(x)Y") +
a"_. x).r. then it is never zero. . x vanishes identically on a <_ x<_ b or W(y .
Suppose also that these functions are linearly independent on the interval of definition of this differential equation.
Y2
= Y1Y.
p(x) = a.
For example...(x)Y"-' + . and q(x) = ao(x)/a2(x).gy. For simplicity we show the details for the second-order linear differential equation..qy2) . y.y. + a. then it is identically zero. For brevity we set W = W(y. .Y"Y2
Y.(x)Y' + ao(x)Y
= 0.. and if W is different from zero at a point . the functions e' and
.(-pyi . +
a
.
W' = Y1A' + Y1Y2 . ... the functions cos x and sin x constitute a fundamental set of solutions for the differential equation y" + y = 0.
Using Eq.Y1Y2)
= -pW..(x) # 0.(x)...(Y1Y2 + YiY2) = Y1Y2 .. with a(x) = p(x) and b(x) = 0.(x)/a. but we state the theorem for the more general case. Now
w=
Therefore. Note that both p and q are continuous functions since a.
2x
10. f. x"2. e'.3x.. e"
12.
3. . cf (x). J. Amer. y" form a fundamental set of solutions for a linear homogeneous differential equation of order n. C. integrate the differential equation directly to find the general solution. sin x'. . e". 14x2.
1. . xe'
13. e'. c a constant
In Exercises 16 through 23. 3 (1973).
1. Monthly 80. Y" = sinx
"L. More precisely.2. .
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.
THEOREM 4
If f is a real-valued function defined on the real line R and if there exists a positive
W ( f . x312
8. c are arbitrary constants. f (x).
. compute the Wronskian of the set of functions. Math.
11. . e
7. x2 In x
S. 4 . 6x2.. Inselt. x2. any solution of a homogeneous linear differential equation is a linear combination of the solutions in the fundamental set. In x.x
5. Eggan and Insell' present a different use of the Wronskian concept.3. as we shall prove in Section 2. y"' = 0
18.
Discovering a fundamental set of solutions for a homogeneous linear differential equation is of great importance because. then f is a solution of an nth-order homogeneous linear differential equation with constant coefficients. Eggan and A. sin 3x. x".2
Linear Independence and Wronskians
79
e-' form a fundamental set of solutions for the differential equation y" y = 0. cos x2
4. xe' x2e'
14. no. y"' = x'
19. a"2". e'. cos 3x
1 + x. a".. if y y2.. not all of which are zero. x12. x. ai # X2
16.
16. ft"I) = 0 on R. is the general solution of the
differential equation. then the expression
Y=c1Yi+c2Y2+. x2
2..-+c"Y"
where c c2.. . Y' = ez
17.
EXERCISES
integer n such that f has 2n continuous derivatives on R and
In Exercises 1 through 15. Their results are embodied in the following theorem.. e"
9. 2 .
. .. This existence and uniqueness theorem is. Then there exists a unique solution y satisfying
. there is only one solution (uniqueness) to the differential equation. Coddington.. Define y2 to be the unique
'C.3
Existence and Uniqueness of Solutions
81
43. "-.(x)yt"-. x b and let 30. The main result of this section is Theorem I below. y'(xo) = 0. + a"-.... Biophys.3 EXISTENCE AND UNIQUENESS OF SOLUTIONS
As we pointed out in Chapter 1.
Show that p(v) = 1 and p(v) = In v are linearly independent solutions of this differential equation..1 + . in their study of peristaltic flow in tubes . Monthly 5. yl"-O(xo) = R"-r
Furthermore. in addition. "The Existence-Uniqueness Theorem for an nth-Order Linear Ordinary Differential Equation. be any constants. Raynor. (3) satisfying the initial conditions y(xo) = 1 .:
Prentice-Hall.(x)y' + ao(x)y = f (x)
y(xo) = (3o.(x)y' + ao(x)y
= 0. it is reassuring to know that there is (existence) a solution to the differential equation that one hopes to solve.. y"(xo) = 0. 1961).
. and obtain the general solution. the most important theorem associated with nth-order linear differential equations. 2 (1968).
2. it is possible to prove this theorem using arguments based on a knowledge of calculus only. Flows Barton and Raynor.J.
(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"-."
THEOREM 1
EXISTENCE AND UNIQUENESS
Let a x).$ obtained the following linear homogeneous differential equation with variable
coefficients. no." Amer..
the initial value problem
a"(x)y". Barton and S. 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. ai(x). from one point of view. N.. . Math. + a.. A. 'See E. Bull. the solution y is defined in the entire interval a s x <_ b.
v>0. An Introduction to Ordinary Differential Equations (Englewood Cliffs.. Math... 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 . + a.(x)yl"-" + . to be the unique solution of Eq. The situation is
even better if. 30 (1968): 663-80..
dp+vdp=0..2. or D. . Although we choose not to present the proof.y'(xo) = a.
(3)
We define y. Willett. y( -')(x0) = 0.
in the case of the vibrating spring discussed in Section 2. 1 each of these functions exists and is uniquely determined. these functions constitute a fundamental set of solutions.) = 0..y2 + + c. moves in only one way (uniqueness). the existence and uniqueness of the IVP (10)-(11) of that section simply means that the mass m moves (existence) and.. Define y" to be the unique solution of Eq. y.
y("-')(x(.12. In this case
1
0
1
0
0
0 0
1
.2.
Set Y = c.. satisfying the initial conditions there exist unique constants c c2. Thus. ... can be expressed uniquely as a linear combination of the fundamental set of solutions. c" such that yl"-')(x0) = P . + c.. . y'(x.. (3) satisfying the initial conditions y(x. .4 through 2...Y2 + . there always exists a fundamental set of solutions. It remains to show that these functions are linearly independent. These solutions. By Theorem the initial conditions y(x. For example. .. + c. . y'(x.
.
.) = 0..
y"(xo) = 0. satisfying the initial conditions
. The observations of the preceding paragraph emphasize the fact that for any linear differential equation whose coefficients satisfy the hypotheses of Theorem 1. . then given (3a. y"^-')(xo) = 0. are not always easy to construct in practice. (3) satisfying 0.
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.
P
y = c.1.y. .y. Define y.Y..
REMARK 1
THEOREM 2
If y y2. . this other solution.) = 0. . y" constitute a fundamental set of solutions of Eq. it follows that the function Y is a solution of the differential equation (3).1. + c"Y. there are some forms of linear differential equations for which systematic construction of a fundamental set of solutions can be described. Nevertheless. (3). y.
0
0
0
0
0
0
0
. y"(xo) = 1. in fact. .
. by Theorem I of Section 2.
That is.. . and so on. y( '"(xo) = 1. to be the unique solution of Eq.
. These forms of equations are discussed in Sections 2.
1
Thus.) = 0. (3) . y'(xo) = any other solution. . although guaranteed by the theory.) = 1. We
Proof
.82
2
Linear Differential Equations
solution of Eq.. 0.
2. .. . .8 are devoted to discussing methods associated with the problem of finding a
fundamental set of solutions.
EXAMPLE 1
Consider the IVP
y" .
(4)
c. 1't"-" and then evaluate Y and each derivative at
x = xo.
+ c"Y.
(5)
(6)
(a) Show that the functions y. + c2y2 + .Y'(0) = 8. Now the functions Y and y are both solutions of
the differential equation (3). y" form a fundamental set of solutions for Eq.Y. (c) Find the unique solution of the IVP (5)-(6). . then the ex..(xo) = P. by the unique-
ness part of Theorem 1. For example.. By Cramer's rule (for details.y = 0. where the c. The proof is complete. Sections 2.
+ c"Y"(xo) = ao
c.and y2(x) = ek form a fundamental set of solutions for Eq. .is the general solution of the differential equation y" . Y"..
COROLLARY 1
If y y2. cos x + c2 sin x is the general solution of the differential
equation y" + y = 0. (b) Find the general solution of Eq. with the same initial conditions. their Wronskian is not zero.2y = 0
Y(0) = 1. (3). .
.. (3). and.Y. y = c. the following result is well justified.
pression
y = c. This process leads to the system
c.e' + c2e. (3) boils down to finding a fundamental set of solutions. At this point we realize that the problem of finding all solutions of Eq. Y = y. whose determinant of coefficients is
W(yr
. C. + c..(x) = e .3
Existence and Uniqueness of Solutions
83
calculate Y'.y' .
Since these functions form a fundamental set of solutions. xo). see Appendix A) there is a unique solution for the unknowns c.4 through 2. y".. + c"Yl"-" (xo) = k .yi"-"(xo) + c2y2 -') (xo) + .-
Equations (4) constitute a nonhomogeneous system of linear algebraic equa-
tions in the unknowns c c2.(xo) + c2Y2(xo) + . and y = c. . (5).. are arbitrary constants is the general solution of Eq.Y.
. On the basis of Theorem 2..(xo) + c2Yi(xo) + . (5).y.
. .
a#0. and find its general solution.2e' .(ear)' .e-' + c2ea'. and y2 are linearly independent solutions of Eq.2(e-r) = e.c. (10).
(8)
. we must use the initial conditions (6) in (7) and
y' (x) = . y(O) = 1.84
2
Linear Differential Equations
Solution
(a) By direct substitution of y. (5) is second order.
.+ 3ea=) =
-2e'+12ea'-2e-'-6ea'+4e-r-6ea'=0.2(-2e.(e-')' .
The unique solution of the IVP (5)-(6) is therefore
y(x) = -2e. Second.y'(x) = 2e-' + 6ea' => y'(0) = 2e° + 6e° = 8. Since Eq. + 2C2 =
i
-2.
(9)
If we want to check that (9) is indeed the solution of the IVP (5)-(6).2(-2e-' + 3e') =
(
2e. we must check three things: First. it follows that y.
(10)
Show that y.e -r + 2c2e'
Setting x = 0 in (7) and (8) we find.+ 12e) . we find
(e-')" . and y2 form a fundamental set of solutions for the equation.
y(O)
2e° + 3e° = 1.+ 6e1') .(2e. Third. (b) By Corollary I the general solution of Eq.
I
ear
I
I
-e-'
=2e'+e'= 3e`
2ea'
I
is never zero.2e-' = 0
and
(es')" .(-2e-' + 3e')' .
(7)
(c) To find the unique solution of the IVP (5)-(6).
Therefore y.
y'(0) = 8.2ea' = 0. (5).+ e-' .1 we encountered the differential equation
y+aay=0. c2 = 3. (9) is a solution of Eq (5).+ 3e'. (5) is
y(x) = c.
EXAMPLE 2
In Section 2. and y2 into Eq.c. In fact. Their Wronskian
e_. = cos at and y2 = sin at form a fundamental set of solutions for Eq.2(ea') = 4ear . and y2 are solutions of (5). (5).
(-2e" + 3ea')" . Hence y.
Assume that the motion takes place on the xy plane. that the projectile was fired at the origin 0.7(cos x)y' + y = e15. 2xy" . Mechanics A projectile of mass m leaves the earth with initial velocity vo in a direction that makes an angle 0 with the horizontal.86
2
Linear Differential Equations
For each of Exercises 12 through 17.
Find the solution to the IVP first by physical intuition and then by applying Theorem 1.gt2 + (v0 sin 0)t. y" + 4(tan x)y' .xy = 0
16.
12. (b) Find the general solution of the differential equation. determine an interval in which there exists a unique solution determined by the conditions y(xo) = yo. (x2 + 1)y" + (2x . Initially.1.. (x2 .1.3).
20. Use Newton's second law of motion to show that the position (x(t). the IVP
my+Ky=0
(11)
(12) Y(0) = Yo.
19. (x . Give a physical interpretation of the IVP
y+4y=0
Y(O) = y(0) = 0.
. at time t = 0. sin 0.
(a) Show that
Y1(t) = cos m t.4)y' + 3x3y'
13. y(0) = vo describes the motion of a spring of mass m and spring constant K. the mass is located at the position yo and has initial velocity
VO.1) y' + 3y = 0
18. y'(xo) = y where xp is to be a point of the interval.1)y" + 3y' = 0 14.
Y2(t) = sin
mt
form a fundamental set of solutions of the differential equation (11).
Show (by direct integrations) that the solutions of these initial value problems are given by x(t) = (v0 cos 0)t and y(t) = -. (cos x)y" + y = sin x
+X41y=0
17. As we explained in Section 2. and that the x axis is the horizontal (Figure 2. (c) Find the unique solution of the IVP (11)-(12). 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.
297. .b)y.
As we examine Eq. 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.
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. If
y = e. To find the general solution of Eq. = e' and y2 = e-'°' form a fundamental set of
solutions for the differential equation
y+a2y=O.3
21 Show that the functions y. Magazine 44 (1971): 18. = cosh at and y2 = sinh at form a fundamental set of solutions for the differential equation
y-a2y=0. is a constant and the leading coefficient a * 0. (1) yields
a"We' + a"_.J
2
.4 Homogeneous Differential Equations-The Characteristic Equation
87
Figure 2.b)y' + 2y = 0
[Hint: Compute" the second derivative of the expression (x . y" = Me..
. Find the general solution of the differential equation
(x . This obser-
vation motivates us to try e"'. we need to find n linearly independent
solutions.a)(x .
(2)
e''(a"a" + a"-.
a*0. (1) carefully (especially in the case n = 1).
+ a. Substitution of y = e' in
Eq.X"-'e' +
or
. Show that the functions y.a)(x . then y' = Xe. as a solution of Eq.he' + aoe' = 0
+ alK + ao) = 0.b)y" + 2(2x .a . 56.a"-' +
"See also Math.
22.
a#0. (1).
23. with a a constant. (1).2.. y") = X"e. (1)
where each coefficient a.
we can develop some information about linearly independent solutions to Eq. the problem of finding the roots of the characteristic equation may be far from trivial. (1).
. However. (3). counting multiplicities." For n ? 5.
"For the solutions of the cubic and quartic equations. and the equation f(k) = 0 is called the characteristic equation for Eq. Some elementary methods for finding roots of polynomial equations are given in Appendix C. if n is large. when n 5. 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.k+a0=0.k + ao = 0 and has the root k = -ao/a.k"-' + + a. the characteristic equation cannot be solved (by radicals) in general
unless it is of some special form. The roots of the characteristic equation are called
characteristic roots. Introduction to Complex Variables (Boston: Houghton Mifflin. 1974). y = e' is a solution of the
a"k" + a. the characteristic equation is a cubic equation. and its solution was discovered by Ludovico Ferrari in the sixteenth century. and the formulas for its solutions (called Cardan's formulas) were discovered by Scipione del Ferro in 1515. 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.88
2 Linear Differential Equations
The exponential function never takes the value zero. the interested reader is referred to the text
by E. Grove and G.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. then. exactly n roots. Ladas.k + as is called the charThe polynomial f(X) acteristic polynomial for Eq. Thus..
If n = 1. 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"-. the characteristic equation is a quartic equation. that if k is a root of Eq.
differential equation (1). In spite of the possibility of algebraic difficulties surrounding the solution of
the characteristic equation. If n = 4._.
DEFINITION 1
(3)
We conclude. This fact was proved by Abel and Galois early in the nineteenth century. the characteristic equation is a. (1). A. the characteristic equation is the quadratic equation a2k' + a. indeed. Consequently. therefore.k"-'+ +a.
In Section 2. (1). If n = 2.k + ao = 0 and its two roots are given by the quadratic formula
REMARK 1
z
2az
If n = 3.
EXAMPLE 1
Solve the differential equation y" . and Remark I in Section 2. a. where a and R are real numbers and i2 = . any solution y of this differential equation is of the form
y = c.1 =0=>
0
' A 'A= t1. (5)
The last form follows from the Euler identity. Thus.e' + c2e-'.i sin 0). We then write e' = eca. where [refer to Eq. = e-`.*e.
EXAMPLE 2
Solve the differential equation of he vibrating spring example of
Section 2.1. In this case the complex characteristic
roots occur in conjugate pairs (a .5.1:
+a2y=0.2(# 0). to say that \ is a complex number means that A is of the form a + i(3.
Solution Set y = e'.
where c. Exercises 21 through 29.y = 0.
We have two solutions.4 Homogeneous Differential Equations-The Characteristic Equation
89
We emphasize once more that if. 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.i(3.. e' = cos A + i sin 0 (' e-i° =
cos 0 . = e.2. of the differential equation are real numbers. then e" is a solution of the linear homogeneous differential equation with constant coefficients. For further discussion of this case. two characteristic roots would be a + i(3 and a . = -1) k must satisfy the characteristic equation
)` 2 .iR is called the conjugate of the complex number a + i1). since their Wronskian has the value . it is conventional to write the corresponding solution in an alternative form. In the special case that X is a complex number.
with a2 = Kim. = e' and Y.1. Suppose now that the coefficients a. then (since a2 = 1. First. These two solutions are linearly independent. and c2 are arbitrary constants. The real number a is called the real part of k.a.
b
a=
4bu 262
In addition to the solution (5) there would also be the solution e-(cos Ox i sin (3x).A + bo for which b. < 4b2bo. and the real number p is called the imaginary part of X.ok = e--. a.. Thus.e' = e°'(cos Ox + i sin px).\ is a characteristic root. = 0.
. see Example 2 below. y. (4) and recall that 1/ --K = i %/K-]
«=
262 .
)
31.
In Exercises 31 through 37 write two equivalent versions of the general solution. Naturally. Generally speaking.y'-81y=0 37.
32.y"-4y=0
33.4 Homogeneous Differential Equations-The Characteristic Equation
91
29. y"-49y=0 36.
"See E. air resistance) which act to retard (dampen) the motion and eventually cause the system to come to rest.sinh x. y" + 25y = 0 30.e"' + c2e`x and also in terms of the hyperbolic sine and hyperbolic cosine.y". it is realistic to assume that the spring is subjected to a damping force. Write the general solution of this differential equation in the form y(t) =
c.]
39. Apply the method of Example 2 to write the general solution of
y" + 13y" + 36y = 0
in terms of trigonometric functions. The Vibrating Spring with Damping In practice. cosh x + c2 sinh x. Rev. however. "The Burden of the Debt and the National Income. experiments have shown that the magnitude of the damping force is approximately proportional to the velocity of the mass.y"-121y=0
38." Amer. D.y"-9y=0.y
= 0 can be written in the form y = c. where the income grows at a constant relative rate 0 (0 < a < 1) and D is the total public debt. (b) Show that the general solution of the differential equation y" . provided that the velocity of the mass is small.2.12 Find the characteristic roots associated with this differential equation. [Hint: y2 + 13y + 36 = (y + 4) (y + 9). Econ.
. y"-25y=0
35. Economics Second-order linear differential equations with constant coefficients of the form
b-(3D=0
occur in the Domar burden-of-debt model.
40. (Dec. the damping force is difficult to formulate precisely.16y=0 34. a vibrating spring is most often subjected to frictional forces and other forces (for example. 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 . Domar. Thus. (Hint: See Example 1 and Exercise 30.
1944): 798-827.
and y2 form a fundamental set of solutions.. see C. n.XIe(AI. the damping force is negative when dyldt is positive and positive
when dy/dt is negative. 2.X)e(Ala"2>' # 0. 140. the basic problem is to find a fundamental set of solutions. W. (Boston: Allyn and Bacon. then the nfunctions y. and m which guarantee that the characteristic roots will not be complex numbers. = e ". ... acts in a direction opposite to that of the
mass. Using Newton's second law.
0'z .. 2nd ed. p. i = 1.. .y2+x)
eA"
X1eA."
"For the statement and evaluation of the van der Monde determinant. . In fact. The loose end that still persists is to relate the solutions e' with a fundamental set of solutions of the differential equation.i
ek2'
X2eA2'
X2e(X1.
1
Thus. . 2. Linear Algebra.92
2
Linear Differential Equations
the damping force. However. Consequently.
. In Section 2. The proof for the general case would follow along similar lines of reasoning. 1968). show that the equation of motion in this case
is
my+ay+Ky=0.
w(yl. the Wronskian reduces to a determinant commonly called the Vandermonde determinant.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. say X = X.
THEOREM 1
If all the roots of the characteristic equation are distinct.A2). the evaluation of the corresponding Wronskian would not be trivial. as described.
Proof The proof is for n = 2. where a > 0 is called the damping constant. Describe conditions on a.
(6)
Write the characteristic equation for this differential equation.A2).. y. What conditions will guarantee that the characteristic roots are complex?
2-
5
HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS-THE GENERAL SOLUTION
In Sections 2. we can express the damping force as -a(dy/dt). Thus.3 we demonstrated that for linear differential equations. n constitute a fundamental
set of solutions. K.2 and 2. This we accomplish in the present section via theorems and a number of illustrative examples. i = 1. Curtis.
2.(xe') + c2e'.3). using the differential equation (1).
Y = c1Y1 + c2Y2.2x + 1 = (x .e'.e' + c2e' + c.y'. that is. x = 1. To try to obtain a feeling for the form of the general solution in this case. 2.
Solution
For this differential equation the characteristic equation is
x'-6x2+11x-6=0. The reader can verify that both y.5 Homogeneous Differential Equations-The General Solution
93
EXAMPLE 1
Find the general solution of the differential equation
y'"-6y"+Ily'-6y=0. We introduce
a new unknown function v by the relation v = y' . (1).
This last differential equation is a first-order linear differential equation and can be solved by the methods of Section 1.
Solution
(1)
For the differential equation (1) the characteristic polynomial is x2 .1)(x2 . Thus we obtain (after long division or synthetic division)
V . = xe' and Y2 = e'. Furthermore.
That is.6x2 + I lx .C x+1
1
=e2`(-1)#0.
where y. takes the form
v'=(2y-y)-y'=y'-y=v v'=v 'v=c.
. then we cannot generate a fundamental system of solutions by considering just the exponentials e'"`.x + c2) = c. y2 = e2s.
(x+1)e'
e'
-?.4 to yield y = e'(fe-' c.
which. Whenever this
happens it is always true that x = 1 is a root. The general solution is y = c. = e'.
v=y'-y=c.y.6 = (x . and y2 are
solutions of Eq.
EXAMPLE 2
Solve the differential equation
y"-2y'+y=0.
Therefore.e'.2)(x . Thus x = 1 is a root of multiplicity 2.e'dx + c2) = e'(c. y3 = e''.e'. if the characteristic equation has at least one repeated root. we first consider some special examples.5x + 6)
1)(x .
Hence. 3 and the fundamental set of solutions is y.
If the roots of the characteristic equation are not all distinct. Then v' = y" .1)2.
Note that the coefficients in this polynomial add up to zero.
x" -' e'' cos px. Ordinary Differential Equation (Englewood Cliffs. a # 0. are linearly independent solutions. xe". (3) is described by the
following cases:
CASE 1
Real and Distinct Roots X.e"" + c2e"2+. 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. = 0.40% xec" t'
and
e(a-.y1`) +
+ a.at.2 + a. then the characteristic equation also has the complex conjugate a . and A2 are roots of the characteristic equation a21. and this generalization is stated without proof in the following theorem... X2ex"x"-'eh'. 1961)
. x"-' e°' sin (3x.\ + a. xe°' cos px. Then the form of the general solution y(x) of Eq. .
(3)
Assume that t.J. * X2
y(x) = c. that the real and imaginary parts of the solutions x. N.
COROLLARY 1
Consider the second-order linear differential equation with real coefficients
a2y" + a. It can be shown however.
where c.. we know from the theory of polynomial equations that if the characteristic equation has a complex root. Applying Theorem 2 first to the root a + ip and then to a . . These solutions are
as follows: e". ..
xe'"-. of the differential equation are all real numbers. x`e( -e)-namely
the 2m functions e°' cos px.eca.2. e°' sin (3x. is a root of multiplicity m of the characteristic equation associated with
the differential equation a"y'"' + a"_.a>r
.ip as a root again of multiplicity m.ip we find 2m complex valued solutions. and c2 are arbitrary constants. then there
are m linearly independent solutions associated with X = X.
"For a proof. see Coddington. PrenticeHall.y' + ay = 0. such as A = a + ip. of multiplicity m.."
THEOREM 2
If X = A. ai.5 Homogeneous Differential Equations-The General Solution
95
The results embodied in Examples 2 and 3 are capable of generalization. xe" sin px.
REMARK 1
When the coefficients a.y' + aoy = 0..
it is convenient to set 2a = a/m and b2 = K/m. _
-\/a2
We realize that the form of the general solution of Eq.
APPLICATION 2. + 2
and
.)"]. = cos-'[c.
The characteristic equation associated with Eq. b > 0. sin V b2 . + cse2+ cos 3x + c6e2z sin 3x
cos 2x + c8e . that the "amplitude" of the motion15 decreases as time goes on. the characteristic roots are
X = -a + V a' . and
m > 0 is the mass.e + c. and for this reason we term the motion oscillatory.1
It is suggested that the reader review Exercise 40 of Section 2. Section 2.5 Homogeneous Differential Equations-The General Solution
97
where a. Exercise 40.4]
Mechanics
The Vibrating Spring with Damping
y+aY+-Y=0. (6) depends very heavily upon the nature of the characteristic roots.b2. however. + c. Write down the general solution of this differential equation. K > 0 is the spring constant.5./(c2.a2 t).axe ` sin 2x. / V
: .
X. cos
b2 . We note. To avoid fractions.
The motion of the vibrating spring subjected to damping (retarding) forces is governed by the differential equation [see Eq.4>] d. the equation of motion has the form
y+2ay+b2y=0. We observe that the motion [that is y(t)] oscillates about the t axis. the solution in this case can be written in the form
y(t) = Ae-°' cos [ . Section 2. The three possible cases are discussed separately.4.xe'` cos 2x + c.4 before studying this application. (6). m m
where a > 0 is the damping constant.e
+ c3xe ' +
+ c7e
c4x2e-2. the solution looks as shown in Figure 2.
Solution it follows from Theorems 1 and 2 and Remark 1 that the general solution of the differential equation in question is
y = c.
Graphically. (6) is
y(t) = e-0f (c. (6) is
(6)
k2+2ak+b2=0.1.
CASE 1
b2 >
b > a) The general solution of Eq. Thus.o is the coefficient of y10' in the original differential equation. This situation is referred to as being underdamped
"As in Exercise 21.
Consequently.a2 t + c.2.
A = 1 cr.sin 2x + c.
in reference to Case 1.
.e"" + c2e"2`.
CASE 2 b2 = a2 (z' b = a)
The general solution of Eq. This is Case 2. hence a = b = 2. this case is called critically damped. This case is referred to as being overdamped. and therefore the motion
is classified as critically damped. and hence there is not enough damping to overcome the
oscillations. That is.98
2 Linear Differential Equations
Figure 2.
EXAMPLE 7 It is known that the motion of a certain damped vibrating spring is governed by the differential equation
y+4y+4y=0.
Classify this case as being underdamped. For this reason. the damping force is so strong that it causes the system to "slow down" very quickly. we note that
2a = 4 and b2 = 4. critically damped.4
in the sense that the amount of damping (or the damping constant) is less than the spring stiffness. In this case there are no oscillations and the graph of the solution approaches the t axis very quickly as time goes on. and X2 negative real numbers.
Solution Comparing this differential equation with Eq. (6). (6) is
y(t) = e-°' (c.
with both \.
Since y(t) contains no sine or cosine terms. + c2t). However. we realize that even the slightest decrease in the damping will produce vibrations. or overdamped. (6) is
y(t) = c. we note that there are no vibrations (oscillations).
CASE 3
b2 < a2 (=> b < a) Here the general solution of Eq.
v'"(0) = 4
y(0) = -2._.100
2
Linear Differential Equations
27. y"_ y"_y'+y=0.
28. y(0) = 0. in addition. at time r = 0. Math.
y-x+6y=0
which satisfy i(0) = y(0) = 0 are hypocycloids. The system is then set into motion by pulling the mass 6 cm above the point of equilibrium and releasing it. y+5y+y=0
42.) 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. critically damped. air resistance acts upon the mass.y+9y+4y=0 38.1
Assume that each of the differential equations in Exercises 34 through 43 governs the motion of a certain damped vibrating spring.5 cm by a force of 4 dyn. y+y+y=0
44. Discuss the behavior of the motion as t .y+20y+64y=0 37. (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. y" + y = 0. y" + 3y' + 2y = 0. 4y+ 8y + 4y = 0
differential equations
35.
30. Show16 that the graphs in the xy plane of all solutions of the system of
.
34.6y+4y+y=0
41.
yI q) = 1. y"+4y=0. Find the position of the mass as a function of time.
'Problem 44 appeared on the 1971 William Lowell Putnam Mathematical Competition.2y" + 2y' = 0. Amer. [Hint: See Exercise 40. Section 2.y+8y+ 16y=0
43. 2 (1973).
y
(2) _ 2
32.
y'(0)=5
Y'
y" (0) = 2
(4) = 0
. assume that.
. no. y" . y(-1) = e
31. y+2y+y=0 40. or overdamped. This air resistance force equals m times the velocity of the mass at time t.4.5y+10y+20y=0
39.i +y+6x=0. Classify the resulting motion as being underdamped.
29.
Find the solution of each boundary value problem in Exercises 30 and 31. y'(0) = 0.
yO)=0. Solve this differential equation.y+69+12y=0
36. with an initial velocity of 5V/6 cm/sec directed downward. A mass of 4 g is attached to the end of the spring. Find the position of the mass at any time t.
y(O) = 3. In Exercise 32. Monthly 80. A spring is stretched 1. 33.
"General Solutions of Linear Ordinary Differential Equations.
Solve this problem. Larsson. Leighton.ea B)I=
-
. Birkhoff and G-C. 7 (1958). Kearns. y8(x) = xe".
is
y'(0) = I
YO(x) =
2\/(2r
1
eb-e*. Monthly 65. (a) Show that the solution of the IVP
y"-2(r+(3)y'+r2y=0
Y(0) = 0. no. For further reference. Monthly 65. no.
Define S(x) to be the solution of the IVP
y"+y=0
S(0) = 0. 53.5 Homogeneous Differential Equations-The General Solution
101
45.
"Problem 45 appears in Math. 47.)
46. M.
. 41. Math. One can motivate the form of the general solution in the case of equal roots of the characteristic equation in the following manner.a)I=J
(b) Show that lim. p. and the result will be a differential equation that you can solve. Find" all twice differentiable functions f such that for all x
f'(x) = f(-x). 1962)." Amer. and W. S'(0) = 1
Define C(x) to be the solution of the IVP
y"+y=0
C(0) = 1. Ordinary Differential Equations (Boston: Ginn & Company. "An Analytic Approach to Trigonometric Function. (Hint: Differentiate the expression. Perhaps one of the best examples of this idea19 is the trigonometric differential equation
y"+y=0.
is
Y'(0) = 1
Yo(x) = 1 [et' Of
-e
w2)_)
(b) Show that lim.(x) = xe". no. Baslaw and H.2. Math. 8 (1958). A. C'(0) = 0. "On the Critically Damped Oscillator. Hastings. See also G."
(a) Show that the solution of the IVP
y"-2ry' + r2. 5 (1974). p. Rota. no. Magazine 47. "The outline presented in Exercises 48-53 is extracted from D. D. S.
It is often possible to extract information about solutions to a differential equation without explicitly solving the differential equation. see R. 2 (1975)." Math Magazine 48. y. "See R. An Introduction to the Theory of Differential Equations (New York: McGraw-Hill. 1952).a y=o
Y(o) = 0..-." Amer. use the expression itself.
49. Show that S'(x) = C(x). 51. (Hint: Multiply S' + S = 0 by S'. The following two sections deal with these two special cases.
S(x + a) = S(x)C(a) + S(a)C(x).11 and 2. Show that for arbitrary a. Show that C'(x) = -S(x). and integrate.)
52. then. the initial conditions. 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. and the uniqueness theorem.
2 . it is difficult if not impossible to solve linear differential equations with variable coefficients. is to try to approximate their solutions by assuming that the solutions are power series. and homogeneous).11. then use the fact that S(x) and C(x) form a fundamental set. for which we know already one solution. Unfortunately.
C(x + a) = C(x)C(a) .(x) satisfy certain differentiability requirements. matters are not so simple when we try to solve linear differential equations with variable coefficients of order 2 or higher.10. In Sections 2.)
53. a linear homogeneous differential equation with variable coefficients unless the differential equation is of a very special form.
(Hint: First show that S(x + a) is a solution. (Hint: Use the differential equation. as we will see in Sections 2.6
OVERVIEW
Throughout this book we strive to find solutions of differential equations. at this level. for example.12 we will see that the nonhomogeneous linear differential equation with constant coefficients can be handled in a straightforward manner. Once we solve the homogeneous differential equation the nonhomogeneous equation presents no difficulties (at least in theory). the Euler-type differential equation and differential equations of order 2. 2. In Chapter 1 we learned how to solve a few types of differential equations of order 1 (such as separable. G HOMOGENEOUS EQUATIONS WITH VARIABLE COEFFICIENTS-
. The supposition that the solution is a power series is valid only if the coefficient functions a. Show that C(x) and S(x) are linearly independent. A direct
.S(x)S(x). the supposition that the solution is a power series is nothing more than looking at the Taylor-series representation of the solution. When these requirements are met. and 2. Show that [S(x)]2 + [C(x)]2 = 1. Show that for arbitrary a.12. exact. 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. Apparently.102
2 Unear Differential Equations
48. In general there is no way to solve. linear. explicitly.)
50.
in many cases.a"_ . a ao are constants and a" # 0.xy' + ay = 0.2.7 EULER DIFFERENTIAL EQUATION
An Euler differential equation is a differential equation of the form
a. is reduced to the differential equation
a2Y + (a..r Differential Equation
103
substitution of the power-series solution into the differential equation enables us. 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.M.y = 0.a2)Y + a.. .
(1)
where a".x"y(") + a"-.
(2)
with a0. a and a2 given constants. assume that x > 0. The Euler differential equation is probably the simplest type of linear differential equation with variable coefficients. By the chain rule for differentiation. the differential equation should be solved for either x > 0 or x < 0.xy' + a0Y = 0. ..
Proof
First. 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
.x"-'yc"-n + .
azx2Y" + a. Then the transformation x = e' implies that dx/dt = e.
EXAMPLE 1
Show by means of the change of independent variables above that the Euler differential equation of second order.. to compute as many coefficients of the power series as we please for our approximation. and so dt/dx = e-'.. That is. We have devoted an entire chapter (Chapter 5) to this approach. oo) or the open interval ( . We illustrate this fact for the second-order case. + a.
2.
(3)
where dots denote derivatives with respect to t. We include a brief discussion of power series here to round out the treatment within this chapter.7
Eul. Since the leading coefficient should never be zero. the interval of definition of the differential equation (1) is either the open interval (0. 0).
The method is embodied in the following theorem.
33)iy=0..x<1
40. then the change of dependent variable
y = you
(2)
produces a linear differential equation of order n .x<3
2.
THEOREM 1
If y.2xy' . x > 0
Solve each of the differential equations in Exercises 33 through 40.
5
5
0.
.x<0
35.x>0
34. 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.8 REDUCTION OF ORDER
As the terminology implies. is a known solution of the linear differential equation
a"(x)y(") +
a"-i(x)yc"-n + . (x-4)y"+4y' -X 4y=0. + a.x+2>0
39.(x)Y' + au(x)y
= 0.x>4 38.2y = 0. 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. 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.
5
1Y' + (x 4 1)2Y = 0.x>0
X
37. x'y"' + 3x2y" .108
2
linear Differential Equations
32. 3xy"-4y'+5y= 0.xy"'-x6y=0..I for u'.x<0
36.x2y'"-xy'"=0. (x+2)y"-y' +x+2y=0.
(1)
which has the property that it is never zero in the interval of definition of the differential equation.
Equation (15). will be studied in detail in Section 5. How one obtains the first solution is another question altogether. known as Hermite's equation.0.
Solution We leave it to the reader to verify that y.2. If Laplace's equation is expressed in terms of spherical coordinates and then
. and there are a number of considerations that govern which technique is most efficient for a given situation.
EXAMPLE 5
Let p = 2 in Eq. Applying Abel's formula [Eq.2xy' + 4y = 0. The search for acceptable solutions of Schrodinger's equation for the special case of a harmonic oscillator leads to the differential equation
y" .
(15)
where p is an arbitrary constant. (15).
+
_ay
y
a
+ V(x.2xy' + 2py = 0. (14)] we have
Y2 = (4x2 .X2 + 8y2
.2 is a solution of
Eq.1.
8z
z
Mathematical
Physics
:
az2
. = 4x2 . we
content ourselves with having y2 in integral form.
Since the integral cannot be evaluated in terms of elementary functions.
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.2 is a solution of y" .
Many theoretical studies in physics utilize Laplace's equation.2)
J
(4t2 .4.
There are various techniques associated with the problem of obtaining solutions to this equation. 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.8). A partial answer to this question is presented in Section 2.f'( . Most often at least one of the linearly independent solutions (very often both) of these equations is an infinite series. Given that y.9. A more elaborate and detailed treatment of the question is presented in Chapter 5. 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. use Abel's formula to obtain a second linearly independent solution.e
Reduction of Order
113
in such mathematical studies as approximation theory and probability theory.2)2
di. 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). (15) (with p = 2). z)&
where h is a constant known as Planck's constant.2)2
e2
= (4x2 . = 4x2 . V is the potential energy.2)
exp[. y.2s) ds]dr
fj
(4t2 .
We include the section here simply to round out our treatment of linear differential equations. (7) of this section.4x']w = 0. Solve the IVP
xy"+(1 . (17). Verify Eq.1)y = 0
y(1) = 2e
y'(1) = -3e. Verify Eq.
34.
2.116
2 Lines Differential Equations
28.
29. Verify that v2 = 1 is a solution of the differential equation (8). 33.y' + ay = 0 be a differential equation with constant coefficients. Discussion of these
ramifications is postponed to Chapter 5.
32. Assume that y. (13) of this section.
35.
given that y.
36.
. Set w = v' in Eq.(x) = e' is a solution of the differential equation.
30. There are many ramifications associated with the Taylor-series technique. Show that y.
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. Verify Eq. This method is perhaps the most widely applicable technique for solving (or approximating the solutions of) differential equations with variable coefficients. Show that the reduction-of-order method leads to a (first-order) differential equation with constant coefficients if and only if y/y.2v = 0. where the method is discussed in detail.
Solve this differential equation by separation of variables and then integrate
the result to obtain Eq. is equal to a constant.2x)y' + (x . 31. (8) of this section. (8) to obtain
[(jx' + 1)x]w' + [3(gx' + 1) .
Show that the reduction-of-order method leads to the differential equation
xv"+(3-x)v' . (9). is a nontrivial solution of the differential equation. Perform the integrations required to simplify Eq. = xe' is a solution of the differential equation
y" -4y"+5y'-2y=0. Let a2y" + a.
xo)3 + .3 guarantees us that this IVP possesses a unique solution.xo)". Consider the initial value problem
y" + xy = 0
Y(O) = 1.
We will seek a Taylor expansion for the function y defined by the IVP (1).
. we have
y. . solve the IVP (1).
EXAMPLE 1
Using a Taylor-series approach.)' + fx(x
n(
(xo)
.X(. we obtain
'
(4)
y"'(0)
_ -1 . compactly.(0)(0) _ -1.) +
or."(o) = 0. First.. There are essentially two approaches that one can follow. .
Solution The Taylor (Maclaurin) expansion has the form
y(X)
= y(O) + Y'(O)x +
Y2o)
xZ +
Y2°) x3 + .2. and from (1) we have
Y' = -xY
Y"(0) =
0. and we illustrate each approach via an example.
f 2(x .. None of the methods that we have discussed can be applied to this differential
equation. y(O) = 1.xy"
Differentiating (4).
(2)
Now..
(3)
Differentiating (3).
y(5) _ -3y" _xYy("(o) = 0
Y(6) _
-4y".x(.
(1)
Theorem 1 of Section 2.xy'
--t>
y(b)(O) = 4.
Similarly.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.
y" = -2y' .
y'(0) = 0. recall from calculus that a Taylor-series representation for a function f is of the form
f(x) = f(xo) +.f'(xo)(x . y'(0) = 0.
f(x) = i f
(x .
EXAMPLE 2
Using power series. (5) has an expansion of the form
y = ao + a. . since there are an infinity of terms. we obtain
y' = a..12a.
. (In general.2a2 + a. From (6). .x3+ )
+ (ao + a. we cannot compute the coefficients in the Taylor expansion in the manner of Example 1..
and so on.
Substituting these results in the differential equation (5) yields
(2-x)(2a2+6ax+ 12a4)? +20a. Here we assume that the solution y of Eq. we must have that the coefficient of each power of x must vanish.12a4 + a. + 2a2x + 3a3x2 + 4a.x' +
and
y"=2a2+6a3x+12a. which in light of the result for a2 reduces to a3 = .. . (2)
yields
y=I-3f X+61x6+.r3 + ax' + a.x + a2x2 + a. (8) yields a2 = -7ao and the second yields
a3 = tae . The first of Eqs. therefore. are constants to be determined.2a2 + a. we are led to the infinite system of equations
4a2+ao=0
12a.
(4a2 + ao) + (12a3 . = 0.x' + 5a.)x + (24a.) Substituting our results into Eq. Thus.
Doing the multiplication and gathering together terms with like powers of x.'r1a1.x2+20a.x + ax2 + a9x3 + a4x' + a. the number of terms to be computed would have to be prescribed at the onset.
(5)
Solution In this example we are not given the initial values. . . solve the differential equation
(2-x)y"+y=0.x' +
. .x' +
we have
) = 0.
(7)
In order that Eq. + a3)X' +
= 0.x'+. we cannot find them all in this manner. (7) hold for every x in the interval of definition of the differential
equation.118
2
Linear Differential Equations
We could proceed in this fashion to generate as many terms of the Taylor
expansion as we want.6a3 + a2 = 0
40a.6a3 + a2)x2
+ (40a.oao .fl2a. = 0
(8)
24a. therefore.
(6)
where the coefficients a. .
That is.In (1 + x)y = 0. Furthermore. Thus..iza')XI + (+ (960a0
°18a1)x°
240a. y(0) = 0.960a0
-
14l0a1+
and so on. use the Taylor-series method of Example 1 to solve the initial value problem.
1
i
(9)
a5 .x + (. Finally.5. compute the coefficients out to the fifth power of the x term. y" . y(O) = 2. y' .nao)x2 + (-.
1. y" + y' + ey = 0.ix'
z°ozS + . The next step is to set the coefficient of each power of (x . y"(O) = 0 3.2<x3 + 960' +
+ al(X
ix3
. these
expressions are infinite series and therefore their convergence must be considered. ao and a as it should.
y = a0 + a. y'(0) = 1
4. thus obtaining an infinite system of equations [for
example. y'(0) = 1. We substitute this expression into a solution of the form y =
the differential equation and then rearrange terms so that like powers of
(x .
EXERCISES
In Exercises 1 through 10. since the differential equation (5) is of second order. and we are led to the results
a2 = .2. y'(0) = 0
6.x0)". y"(0) = 0. Example 2 illustrates the Taylor-series method most often used.(sin x)y = 0.
(10)
We note that the solution (10) depends on two arbitrary constants. _ -z°ao . We assume a"(x . y" + xzy = 0.Izal
a4 _ -°0a. Eqs. (10) converge for every x in the interval of definition of the differential equation? These conjectures and observations will be discussed in Chapter 5. In each exercise. y" + (3 + x)y = 0.). do the expressions in parentheses in Eq. (10) are the two linearly independent solutions to the differential equation (5). (8)]. we obtain the coefficients a" from this infinite system.. y(0) = 1. y'(0) = 2
.
2. y(0) = 1. We also conjecture that the parenthetical expressions in Eq.
ao(1 .zaa0 .°ao
a. y"'(0) = 0 5.x0) can be combined. y(O) = 0.9
Solutions of Linear Homogeneous Differential Equations
119
Analogous manipulations apply to each subsequent equation.axz .x0) equal to zero.)x5 +1 . y'(0) = 1.
is called the homogeneous solution for Eq.(x)Y' + a. defined by
Yn = c1Yi + c2Y2 + .
DEFINITION 1
With every nonhomogeneous differential equation (2). + c"Y"..e' + c2e'`. In some texts the homogeneous solution is referred to as the complementary
solution.3y = e'.
x > 0. ]
EXAMPLE I
Find the homogeneous solution of the differential equation
y" .3xy' .5. briefly. + a..(x)Y = f(x)
(1)
or. .
(3)
DEFINITION 2
If the n functions y y2... .
(6)
.(x)yro-n + .
-o
(2)
The function f is called the nonhomogeneous term for the differential equation (1)..
EXAMPLE 2
Find the homogeneous solution of the differential equation
2x2f . (2). the general solution of this equation is
c. It is the general solution of the associated homogeneous differential equation. [Note that the homogeneous solution is not an actual solution of Eq. the homogeneous solution of the differential equation (5) is
Ye = c..10 Nonhomogeneous Differential Equations
121
The general nonhomogeneous linear differential equation has the form
a"(x)Y'"' +
a"_.
(5)
Solution The homogeneous equation associated with the differential equation (5) is
y"-3y'+2y=0. (2).
Using the methods of Section 2. then the function y. a.e + c2e2r. there is an associated homogeneous differential equation defined by
r-o
i a.
(4)
where the c.3y' + 2y = cos x.(x)Yi" = f(x).2. are arbitrary constants.. Thus.. y" constitute a fundamental system of solutions for the homogeneous differential equation (3).(x)Y(') = 0.
Yr.7) is c. This is equivalent to proving that the function Y .° + I ai(x)Y( )
s-0
i-O
i-o
= f(x).)u' = i ai(x)Yt) . (2).
is any particular solution of Eq.. and if y.(x)[c.
i-0 i-0
i-0
.Y. + y0i)
i-o
_ i ai(x)[c. the function y as defined by Eq.. we refer to this function as a particular solution of Eq. Now let Y be any solution of the differential equation (2). We should prove that Y can be written in the form of Eq. .3y = 0. + c2Y2 + .f(x) = 0. (2). + C.x-"2 + c2x'. + C. (7).' + c2yi' + .122
2
Linear Differential Equations
Solution (6) is
The homogeneous equation associated with the differential equation
2x2y" . then the general solution of Eq. satisfies the homogeneous differential equation (3). .3xy' .vD]
= c.
If.
That is. y form a fundamental system of solutions for Eq..i a (x)y(' = f(x) .. = c1Yi + c2Y2 +
Y. somehow or other.. + c"A°
. (2) can be written in the form
y = Yi.y. we have
i-o
I ai(x)Yt1 = I a.y. we find a function that satisfies Eq. and its solution (given in Example 2 of Section 2.
This latter differential equation is an Euler differential equation. Substituting into the differential
equation (2). ai(x)y("
i-o i-o
+ .x. I a. Thus.(x)(Y . (3). (7) satisfies the differential equation (2). (2) and denote it by yo. + Y. I ai(x)Y..(x)Y°' + c2 7..
(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.
i a. (7)...
THEOREM I
If y y2.2 + c2x'. =
c.y.Y. In fact.-o
+ Y.
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. For example. where S is a nonzero constant. + a.. The following example contains the typical features of the method of unde-
. 1)
to denote that any derivative of the function 3x2 on the left is a linear combination of the functions x2. are constants and f(x) is a linear combination of functions of the following types:
1.
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. (cos 2x. We use the symbolism
3x2
Ix'. cos yx. we have 2e` -. a . A (finite) product of two or more functions of types 1-4.124
2 Unear Differential Equations
2. e)
.
Roughly speaking. 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. 3. a. sin 2x1
xe -. the functions 1/x and log x are not of these types.. where y is a nonzero constant.xe' sin x -. e2' sin x. x.. As further illustrations. We also say that the functions x2. where R is a nonzero constant. xe' cos x. 1 on the right.111
5 cos 2x -.
For example.. 5. x. 4. (1) is a linear combination of those functions of types 1-5 that span all the derivatives of f(x). x.y' + apy = f(x). On the other hand.
(1)
where the coefficients a. es'. x. 2. .. any derivative of 3x2 is a linear combination of the functions x2. the function
f(x)=3x2-2+5e' -x(sinx)e'x+5eos2x+xe
is a linear combination of functions of types 1-5. x°.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'
+ . (e'')
-5-. which are all of type
1. e2s cos x). (xe2' sin x. sin &x. a particular solution of Eq. and 1. (xe. where a is a positive integer or zero. 1 span the derivatives of the function 3x2.
respectively). then the particular solution of Eq. On the other hand. On the other hand. 11
(3)
5-{1}
2e'-> {e'}. we compute
(6)
yo=2Ax+B+De'+Dxe' y'=2A+2De'+Dxe'
and substitute these results into the differential equation (2). x. B. C. (For motivation of the method. 1} and {e'}.y = 0. B. and the reader is urged to study this example very carefully. Since e' is.(Ax2 + Bx + C + Dxe') _ -2x2 + 5 + 2e'. With our previous notation.2x2 . x. Now.of the differential equation
y" . and e` (which are of types 1. 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). the derivatives of f are spanned by the functions in the sets
{xi. and so we must multiply the function in this set by the lowest
integral power of x. then all the elements of that set should
be multiplied by the lowest integral power of x. obtaining
(2A + 2De' + Dxe`) .2x2 + 5 + 2e'.y = 0.
where A.
To obtain the undetermined coefficients A. Now if none of the functions in the sets {x2. = Axe + Bx + C + Dxe'.y = 0.11
The Method of Undetermined Coefficients
125
termined coefficients. x.
(2)
Solution The method of undetermined coefficients is applicable to this example
since the differential equation (2) is of the form of Eq. C.. and D. Hence. y" . so that the resulting function is not a solution of y" . x.2. 1} and {xe'}. (2) is of the form
y. we should
multiply e' by x.{x2. but xe' is not. no function in the set {x2. 1. and D are the undetermined coefficients.
y. and so we leave this set as it is.y = 0. (1). x. 1} and {e'} is a solution of the associated homogeneous equation. 11 and {e'} is a solution of the associated homogeneous equation for (2). a solution of y" . and 2. We first compute the functions that span the derivatives of each one of the three terms of the function f. see Exercises 65 and 66. a particular solution of Eq.y = . e' in the set {e'} is a solution of y" . we have
. 1} is a solution of y" .y = 0. (2) is a linear combination of the functions in the sets {x2. obtaining {xe').
(4) (5)
Therefore.=Axe+Bx+C+De'. x. 1. that is.
if any one of the functions in any one of the sets {x2.)
EXAMPLE 1
Compute a particular solution. Thus. and f(x) =
-2x2 + 5 + 2e' is a linear combination of the functions x2.
.
C = -1. x cos x. sin x. xe. A = 2. cos x}
2--. we obtain the following system of equations:
2A . 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. a contradiction will occur in the resulting system of equations when we attempt to compute the undetermined coefficients. xe'}
e2'. Ix.[1l
3x -. it may happen that the coefficients of unnecessary terms will be found to be equal to zero.
.x. e} . To save space. x2e.126
2
Linear Differential Equations
Equating coefficients of similar terms. (2) be abbreviated in accordance with the following format:
-2x'->Ix2. Alternatively.lei'}
x sin x -> {x sin x. Should we assume an inappropriate form for a particular solution. B = 0.=2xz-1+xe.
Since e is a solution of the associated homogeneous equation but xe is not. and a particular solution of Eq. D = 1.
EXAMPLE 2
Find the form of a particular solution of the differential equation
y"+2y' -3y=3x2e'+e'+xsinx+2+3x. 1l.
Thus. we have
Solution
3x2e'-> (x2e.(11
2e'Ie'I. xe instead of e in Example 1).C = 5
2D=2
-B=0
-A = -2. 1}
5 -.Ix"e.
The crucial aspect of the method of undetermined coefficients is that we assumed the proper form for a particular solution (such as.{xe')
(e is a solution of the associated homogeneous equation and xe is not). (2) is
y. (2).
(10).cos x + i x sin x.z x sin x.C2 COS x .sinx + c. When the switch is closed.5 is customarily called an RLC-series Electric Circuits circuit. c.
Now
y(0)=0=> c1+c2=0
Y'(0)=1.2. an inductor of L henries.sin x + 2x .11
The Method of Undetermined Coefficients
129
Finally.c3+2=1 y"(0)= -1=> -c.1sinx . and c3. Differentiating Eq.11. c3 _ -1. we have
y' = .
Thus. We would like to find the current I as a function of time.cosx + 2 . We also want to find the charge Q = Q(t) coulombs in the capacitor at any time t. it suffices to compute Q. a current of I = 1(t) amperes flows in the circuit.1xcosx
2 2
y" = .c. = 0. Here a generator supplying a voltage of V(t) volts is connected in series with an R-ohm resistor. sin x
. and a voltage drop across the capacitor equal to
Figure 2.-1= -1.1
The electric circuit shown in Figure 2. = 0. and a capacitor of C farads. c. It is known from physics that the current I produces a voltage drop across the resistor equal to RI.
APPLICATIONS 2.5
. and the solution of the initial value problem
is
y(x) = . By definition
I = dQ
(11)
Thus. a voltage drop across
the inductor equal to L(dI/dt).c. we use the initial conditions to determine c c2.
we must find the general solution of Eq. Applying this law to the circuit of Figure 2. (13) with Eq.. (16) of Exercise 57.2 Mechanical-Electrical Analogies
. we have
L d1
+ R1 + I Q = V(r). (13). (12). C
1
Reciprocal of capacitance
Displacement External force Velocity
y-Q
F(r) H V(r)
)). Kirchhoff's voltage law states that the voltage supplied is equal to the sum of the voltage drops in the circuit.
Mechanical
Mass Friction (damping)
Electrical
m F. This general solution involves two arbitrary constants.
Electromechanical If we compare Eq. we observe the analogies Analogies between mechanical and electrical systems given in Table 2.I
Charge on capacitor Electromotive force Current
Table 2. Eq. this is one of the basic ideas behind the design of analog computers. the latter condition being obtained from the differential equation (12). For example.5 (with the switch closed). Initial conditions associated with this differential equation are 1(0) = 0 and 1(0) = (11L)[V(0) . 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. To determine these two constants. Q0 is the initial charge on the capacitor and Q(0) = 1(0) = 0. L
aHR
Inductance Resistance
Spring stiffness
K.130
2
Linear Differential Equations
(1/C)Q.2. In order to find the charge Q(t) in the capacitor. because at time t = 0 there is no current flow in the circuit.
(12)
Using the relation (11). These analogies
are very useful in applications. (12) can be written in the form
L dQ+R
z
dQ+CQ=V(t).(11C)Q(0)1. To find the current 1(t). Often the study of a mechanical system is converted to the study of an equivalent electrical system. which is usually easier to study. we impose two initial conditions Q(O) = Q0 and Q(0) = 0.
(13)
This is a second-order linear nonhomogeneous differential equation.
and C are constants.11
The Method of Undetermined Coefficients
133
52. Biophys. Medicine: Periodic Relapsing Catatonia In the development of a mathematical theory of thyroid-pituitary interactions. Danziger and G. Bull. and K are nonzero constants. Prove. find 0.
and
e < C
(15)
a30+a20+a10+0=0. b = 6. Math. Rashevsky.
These equations describe the variation of thyroid hormone with time. Solve the nonhomogeneous differential equation of Exercise 51 if T = . Kneese.
"L. V. Bull. in this case. Herfindahl and A. Math. Elmergreen. Here
0 = 9(t) is the concentration of thyroid hormone at time t and a a a3.
54. "O. Social Behavior Second-order linear differential equations with constant coefficients of the form
y-Ay=B. If K # -1. Rashevsky.
were obtained by Rashevsky27 in his study of riots by oppressed groups..
. the motion is called a free motion. p.4b < 0. 30 (1968): 735-49. Bull. Referring to Exercise 51. Merrill. Here B is an arbitrary constant.
0>C.
56.2.
55. (This phenomenon was observed in the actual experiment with rats..bx= -b. 182. a. for Eq. and (3 = 1. "N.
A>0. 57. L. Economic Theory of Natural Resources (Columbus. 30 (1968): 501-18. 1974). 18 (1956): 1-13. Find the general solution of this differential equation. Section 2. (15). suppose that an external force equal to f(t) acts upon the mass of the spring. Find xr for this differential
equation.) 53. a = 1. Forced Motion of the Vibrating Spring In addition to the damping force assumed in Exercise 40. show that when (1/T2) .
where r. the solutions contain oscillatory terms (linear combinations of sines and cosines). Biophys. K.4. Ohio: Charles E. Math. b. Minerals In finding the optimum path for the exploitation of minerals in the simple situation where cost rises linearly with cumulated production. Discuss the behavior of the solution at t -> oo. Herfindahl and Kneese26 presented the following second-order differential equation: r(3 rK
x-rx. 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. Biophys. Danziger and Elmergreen29 obtained the following third-order linear differential equations:
a36 + a26 + a18 + (1 + K)6 = KC. C. See also N. When the external force is identically equal to zero.
assume that El = I and'solve Eq. 3 (1974).Yo)xex +
e + (1 (a . 11rhese exercises are extracted from A. Farnell.3
y(o) = 5. Describe the limiting motion at t --> x. f(x) = e-x
63. Monthly 54.1)z x
(b) Given that y(O) = yo and y'(0) = y show that c.
. = f(x). an I beam) subjected to a transverse load f(x). f(x) = 1
60. Vibration Problems in Engineering (Princeton. For the forced motion of Exercise 57. Timoshenko. f(x) = sinh x
64.
Solve this IVP. Math. 1928). and cz in the solution of part (a) are
c. and that air resistance acts upon the mass with a force that is 7 times its velocity at time 1. (a) Show that the general solution of the differential equation
y"-2y'+y=e°x.
Elasticity: Static Deflection of a Beam In the consideration of the static deflec-
tion u of a beam of uniform cross section (for example.
is
1
a* 1. there arises the differential equation's
EIdX.134
2
Linear Differential Equations
58. (17) for the
given f. we did not give any indication where the proper form for yo came from." Amer.J.: Van Nostrand. "On the Solutions of Linear Differential Equations. N.
58.
e"
(x)=cex+c zex+ (a .=Yo-(a-1)2
hence
yoex
and
cz=Yi . consider the following exercises. assume that m = 4 g. E being Young's modulus of elasticity and I the moment of inertia of a cross section of the beam about some fixed axis. f(x) = cos x 62."
65. To this end. The mass is also subjected to an external periodic force equal to 5 cos t. 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. . the motion of the mass satisfies the IVP
4y+7y+3y=5 cos t
y(0) = . f(x) = xz
61. The mass is raised to a position 3 cm above the equilibrium position and released with a velocity of 5 cm/sec downward.1)2
a)xe
"S. K = 3 dyn/cm.
In Exercises 59 through 64. no. Certainly. Thus.Yo -A e-
Y(x) =
+ (y.
(17)
where E and I are constants.
is large and gets even larger if w is chosen closer to wo. and c2 in the solution
of part (a) are c.
w2 0
2... (b) If w is "close to" wo.
67. for Eq. sin x + c2 cos x + 1 . however. the solution in part (a) shows that the amplitude of y. sin x + yo cos x + {sin x . Find y. If f(t) . . 57)
y + w2y = Fcos wt. resonance is said to occur when the frequency of the external force matches the natural frequency of the system.(x)yl'i +
a.
(18)
where wo = K/m and F = M/m.
1
66. For this reason wo is called the natural frequency of the free vibration.2. = y. 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) Show that the general solution of the differential equation
y" + y = sin ax
is
sin ax y(x) = c.yo) xe + rr2e.
(x)yc"-n + . requires that the coefficients
.12
Variation of Parameters
135
(c) Show that
lira y(x) = yoe + (y.x cos x). as w .(x)y' + ao(x)y = f(x)
(1)
The method of undetermined coefficients. hence sin ax . like the method of undetermined coefficients. show that -' =
F N2 cos wt.a/(1 . .a sin x
y(x) = y..a2
(b) Given that y(O) = yo and y'(0) = y show that c. sin x + yo cos x +
1-a
2
(c) Show that
lim y(x) = y.a2) and c2 = yo..wo is referred to as resonance. is used to compute a particular solution of the nonhomogeneous differential equation
a. This phenomenon associated with the unbounded growth of the amplitude of y. (16) of Ex.12 VARIATION OF PARAMETERS
The method of variation of parameters. (18) in the case w = w.-. That is.
(a) If w # wo. + a.0 (z>M = 0) the spring is said to be free.
+ U.
(Special case n = 2) By hypothesis. + u2y2 + ..u..
and that u. and to impose appropriate conditions on these functions so that the expression
u. which we present in Theorem 1. and that the function f be one of the types described in Section 2.Y. This theorem is stated for the general nth-order equation.. ..
=0..(x)Y' + ao(x)y
= 0. When we encounter differential equations for which we cannot apply the method of undetermined coefficients.. y. 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.. This method. u" satisfy the system of equations
y.u
y.)u2 + .
(3)
yi"-2)ul + Y( -2)u2 + . (1).y. . as functions (variables) u. Suppose that we know the general solution
of the associated homogeneous differential equation
a"(x)Y(") +
a"-.uR
= 0.
i" qu + Y(2"-. ." evolves from the following consideration.
. and y2. but for simplicity of presentation we prove it for the case n = 2. y1'u. and y2 form a fundamental
system of solutions for the equation a2(x)y" + al(x)y' + ao(x)y = 0.
= 0. which in theory "always works. the system has a unique solution u.+y2uZ+ +y"u.+y2u2 + +y. a . Since this determinant
is different from zero. we use the variation-of-parameters method. + y2uZ +
+ y.. + y("-2)un = 0. (See Appendix
. +
(x) a"(x)
then y = u. + a. and if the
functions u u2. Y.....
THEOREM 1
If y y2..136
2 Unbar Differential Equations
a0.. u"y" is a particular solution of Eq. + u2y2 +
Proof
.. form a fundamental system of solution for Eq. and a of the differential equation be constants.y.. u2. (2).(x)y("-') + .
(2)
Is it possible to treat the constants c.
is a solution of the nonhomogeneous equation (1)? Such an approach has been
shown to lead to meaningful results.11.u. .
' and u2 we determine u. to find. while the method of variation
of parameters involves integration. We first find the fundamental system of
solutions for Eq.. The method of
undetermined coefficients involves differentiation.
utyt + u2y2 + . . And.Yi + u2y2. 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. u2.(x)[u. u u2. (4). Next. finally. (4).) From u.(x)y + ao(x)y. Hence. 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). Therefore. and in some cases differentiation is easier than integration.y. Using these functions. u2.
Y
11
= u"+u '+u "+u =u '+u "+f(x) 1Y12Y2
..[a2()Yi + a. Indeed. u.u. + u.
The last expression is obtained by applying the first of Eqs. (1) (for n = 2) and the proof is complete. + u2Yz + u2y2 = u. . there is no loss in generality if we
set the constant of integration equal to zero. the proof in the general case follows in exactly the same manner.12
Variation of Parameters
137
A. .y.
.]
+ u2[a2(x)y' + a. from (5) we have
Y' = u. u.. Other than the tedious notation. Now
a2(x)Y" + a. and u2 via integration.
iY.2. (3) for u.. (In these integrations.. y satisfies Eq. + u2y2]
= u. + u2Y2
(5)
is a solution of the differential equation..
2Y2
2'Y2
a2(x)
Once again the last expression follows by applying the second of Eqs.. ..
EXAMPLE 1
Solve the differential equation
y" + y = csc X. All that remains to be shown is that the function
Y = uy.' + u2Yi] + ao(x)[uy.. respectively.Y. (2). The reader should observe that Theorem 1 and its proof show us how to construct the particular solution y. we then solve Eqs.(x)y2 + ao(x)Y2] + f(x)
= f(x). . The reason for introducing the method of undetermined coefficients is that it is sometimes quicker and easier to apply.y.(x)Y' + ao(x)Y = a2x)[uiY + u2Yz + a (x)]
+ a. + is the desired particular solution y. we integrate each of the u..
67.
"This is Exercise 1. R is the radius of the earth. [Hint:
v=0forr=x. [Hint: The total acceleration is g .]
32. 1977). Copyright m 1977 by John Wiley & Sons. Thompson. In a vertical takeoff. Reprinted by permission of McGraw-Hill Book Company. 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.. Reprinted by permission of McGraw-Hill Book Company. 3rd ed. 1968). Will the ball clear the fence? Take g = 32.
. where r is the distance from the center of the earth to the particle. Copyright ® 1977 by John Wiley & Sons.. Vector Mechanics for Engineers: Statics and Dynamics.Review Exercises
143
27. Jr.001 v2. ibid. "This is Exercise 11. A batter" hits a pitched ball at a height 4. Reprinted by permission of John Wiley & Sons. "This is Exercise 11. where v is expressed in ft/sec. If the projectile is released from rest and keeps pointing downward. 446. Russell Johnston.0 ft above ground so that its angle of projection is 45° and its horizontal range is 350 ft. Dynamics (New York:
McGraw-Hill.3-2 in Eduard C. 447. p. Part I. Pestel and William T.
29. p.
28. Beer and E. ibid. and g is the acceleration due to gravity at the surface of the earth.001v2..]
31. Derive an expression for escape velocity. 3rd ed. "This is Exercise 11. Reprinted by permission of John Wiley & Sons.0 g for 40 sec. Copyright ® 1977 by John Wiley & Sons. 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). which is travelling in the opposite direction at a constant speed of 45 ft/sec.0. Inc.. p. where g = 32. ibid.2 ft/sec2. determine when and where the vehicles pass each other. A rifle" with a muzzle velocity of 1500 ft/sec shoots a bullet at a small target 150 ft away. 1977).. p. Inc. A short time later it is passed by truck B.16 in Ferdinand P.2 ft/sect. p. ibid.
33. The ball is fair down the left field line where a 24 ft high fence is located 320 ft from home plate. Physics. Reprinted by permission of John Wiley & Sons. The acceleration due to gravity36 of a particle falling toward the earth is a = -gR2/r2.2 ft/sec'.21. What velocity is attained at this time and what is its height?
"This is Exercise 8 in David Halliday and Robert Resnik.' rocket engines exert a thrust that gives it a constant acceleration of 2.2 ft/sec'. Reprinted by permission of McGraw-Hill Book Company. Reprinted by permission of McGraw-Hill Book Company.28. Inc. How high above the target must the gun be aimed so that the bullet will hit the target? Take g = 32. 30. "This is Exercise 6. 454. (New York: McGraw-Hill. p. (New York: John Wiley & Sons. "This is Exercise 14. Automobile A starts" from 0 and accelerates at the constant rate of 4 ft/sec'. 17. 68. 68. determine its velocity after it has fallen 500 ft. Knowing that truck B passes point 0 25 sec after automobile A started from there. p. (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. It has been determined experimentally" that the magnitude in ft/sect of the deceleration due to air resistance of a projectile is 0.
and m is the mass. If the angular velocity of the pole is to be a constant.6. 68. Determine the range.] 39. [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. Show that the steady-state solution (the particular solution) is given by
X =
FO sin wt
m(wa .3-18. 39. at an angle of 40° above the
horizontal. p.
35.
"This is Exercise 1. J. Copyright C 1976 by John Wiley & Sons. where w2 = s/m. find the least time required for the train to go between two stations 82 miles apart. s is the stiffness. If the values of this displacement x and the velocity x are plotted on perpendicular axes.
37. It is estimatedi0 that an athlete.3. ibid.6
34. Inc..144
2 Uneer Differential Equations
h
}
Barrier
Figure 2. 'Ofhis is Exercise 1. 34. p.11 in H. eliminate t to show that the locus of points (x. Reprinted by permission of McGraw-Hill Book Company. Assuming that acceleration and deceleration are uniform. The Physics of Vibrations and Waves. "This is Exercise 1. releases the shot at a height of 7 ft. Find the position of the mass at any time t. then stop
within a distance of 1/2 mile. the maximum height attained by the shot. and the time of flight. 43. A train" can attain its maximum speed of 60 mph in 4 min. [Hint: The differential equation is my + ay + Ky = 0. ibid. Copyright ® 1976 by John Wiley & Sons. ibid. 2nd ed.]
38. Draw a velocity-time diagram. show that the required speed of the truck is i = h9/sin2 0. Reprinted by permission of John Wiley & Sons. p. with a velocity of 34 ft /sec. Reprinted by permission of John Wiley & Sons. i) is an ellipse.
36. The displacementa2 of a simple harmonic oscillator is given by x =
a sin wt.4-42.. ibid. A telephone pole" is raised by backing a truck against it. Show that this ellipse represents a path of constant energy.4-13.w2) '
1117his is Exercise 1.. 1976). in putting a shot. The equation mz + sx = FO sin wt describes" the motion of an undamped simple harmonic oscillator driven by a force of frequency w. Pain. as shown in Figure
2. Inc. p.
"This is Exercise 2.. 19.. Reprinted by permission of McGraw-Hill Book Company. (New York: John Wiley & Sons.
. Reprinted by permission of McGraw-Hill Book Company. An overdamped spring mass system is displaced a distance A from its
equilibrium position and released from rest. p.
ibid.
0.]
(b) Calculate the current through the source for the charge found in part
(a). [Note: the charge oscillates at one-third of the frequency of the source. = (1/54) cost volt.5f
T
Figure 2. Consider" the nonlinear time-invariant subharmonic generating circuit shown in Figure 2.
2
.
. 41.
where A and B are constant. Now show that the general solution for the displacement is given by
X=
Fo sin wt
nt(w2 .(t) = sin 2t volts. Basic Circuit Theory (New York: McGraw-Hill.8
"This is Exercise 4 in Charles A.7. Reprinted by permission of McGraw-Hill Book Company. "This is Exercise 8.7
where wa = s/m. 1969). The inductor is linear and the capacitor has a characteristic
v.
(a) Verify that for an input e.. calculate
the complete response. Knowing that e. = 1 volt.=
1
18q+27q.145
L=1H
Figure 2. 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. p.8 is made of linear time-invariant elements. 40. and that at time t = 0 the state is iL = 2 amp and v. The circuite5 shown in Figure 2. Reprinted by permission of McGraw-Hill Book Company. Kuh. p.
The input is e and the response is v. 333. a response q(t) = cos
(0) coulombs satisfies the differential equation. 332. Desoer and Ernest S.w2) 0
+ A cos wot + B sin Wat.
where A and B satisfy the equations B.
. Inc. 334. ibid.2 in J. Solve the nonhomogeneous equation by assuming a form for the particular solution and then determining the constant coefficients. Inc.Tr
(cos m4i + i sin m4)) is a so-
4)
-m.146
2
Linear Differential Equations
1NH
I P1
T
Figure 2. A'
V
-CC'v . Jr..
45. "This is Exercise 3-3. p.9
42..49 confirm that 4)(4) =
lution of
1 d2(b
__
1
2. N. 102. 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. "This is Exercise 2-2. Reprinted by permission of Prentice-Hall. N. and spring constant K when the driving force (the external force) is a constant FO.: Prentice-Hall.J.: Prentice-Hall. If u and v are linearly independent particular solutions of the differential equation
y"'+Py"+Qy'+Ry=0. 1969). Introduction to Nuclear Physics and Chemistry.
43.
prove' that the general solution is given by y = Au + By. ibid.
[Hint: Differentiate both sides of the equation with respect to x and factor. damping constant b. a JPd.9. 26. "This is Exercise 14 in Bernard G. Solve" the differential equation
(y')' . p. Norwood. Monthly 36 (1929): 112-13. "See Amer.]
"This is Exercise Ila. (Englewood Cliffs.J.6.uv')2
U
and c is an arbitrary constant. Monthly 60 (1953): 264. Harvey. Let' a sinusoidal voltage e. Reprinted by permission of Prentice-Hall.. Math. Math. Inc.(t) = 3 cos (106t) volts be applied at time
t = 0 to the linear time invariant LC-circuit shown in Figure 2. 51. By differentiation. Intermediate Classical Mechanics (Englewood Cliffs. calculate and sketch i(t) fort ? 0. p.3y(y')2 + 4y' = a. p. Reprinted by permission of McGraw-Hill Book Company. Determine01 the general solution for the motion of an underdamped harmonic oscillator having mass m. 2nd cd.SO
47. Reprinted by permission of Prentice-Hall.
2
46. Given i(0) = 1 mA and v(O) = 0. "See Amer. 1979).
one is interested in studying systems of n differential equations in n
unknown functions where n is an integer >_ 2. = a11(t)x1 + au(t)x2 + . + a32(t)x2 + + ay.
REMARK 1
In the treatment of systems of differential equations. Consequently.. are usually (and successfully) studied in more advanced courses. x to denote unknown functions and to use t as the independent variable.1
INTRODUCTION AND BASIC THEORY
The differential equations that we studied in Chapters i and 2 are ordinary
differential equations involving one unknown function. This notation has obvious advantages when we
are dealing with general systems.. we shall restrict our attention to systems of linear differential equations. a22. we had to solve one differential equation which involved one unknown function. that is. including applications and generaliza-
tions. a72. and the functions fl..2(t)x2 + . see Chapter 8. = a11(t)xl + au(()x2 + f1(t) X2 = a21(()x1 + a22(()x2 + f:(t). .. a21.. systems
of nonlinear differential equations.. in special cases alternative notation may be used... As usual. + a. For some results on nonlinear systems. These definitions and theorems
extend easily to systems of n linear differential equations in n unknown functions of the form
X. in each case up to now. Nonlinear systems. it is custom-
ary to use the symbols x x2. linear systems. Since in this book we are mainly
interested in obtaining solutions.(t)x + f2(t)
(2)
X=
a. For many reasons.CHAPTER 3
Linear Systems
3.. In this chapter we mainly consider systems of two linear differential equations in two unknown functions of the form
X. f2 are all given continuous functions of t on some interval I and x x2 are unknown functions of t. the dots over x. . + a1 (r)x + fi(t) X2 = a21(t)x.
(1)
where the coefficients a.(t)x + fi(t). and x2 in (1) indicate differentiation with respect to the independent variable t.. However.. In this section we present a few definitions and state without proof some basic
theorems about the solutions of system (1).
. by approximating each nonlinear equation of the system by a linear one.
DEFINITION 1
A solution of the system (1) is a pair of functions x1(t) and x2(t). makes them identities in t for all t in I.2 is.
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.
. This is an elegant method...
J-1
f(t). when substituted into the two equations in (1). more compactly. but its detailed presentation and justification requires a good knowledge of matrix analysis and linear algebra.0.
In subsequent sections we shall present two elementary methods for finding explicit solutions of system (1) when the coefficients a.
i = 1. perhaps. For the reader's sake we mention here that the method of elimination discussed in Section 3. as the reader can easily verify. can be extended to solving system (2) when its coefficients a.(t) = y1(t) and x2(t) = y2(t). a12. x2(t) = -t . with varying degrees of technical difficulties.
xl(t)
3t . n. each being
differentiable on an interval I and which. a21. Sometimes it is convenient (and suitable for studying more general systems) to denote a solution of (1) by the column vector
For example.3 is recommended for system (2) when n z 3.148
3
Linear Systems
or. . The
matrix method outlined in Section 3. 2.
is a solution of system (3).
z.. .
For example. These methods. are all known constants.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). and an are all known constants.
For any constants c.2(t)x2.
THEOREM I
Any linear combination of solutions of (6) is also a solution of (6). As in the case of homogeneous linear differential equations. Let us consider the homogeneous system
z. and f2 in (1) are equal to zero. To a great extent. On the other hand. This solution is called the trivial solution.(t)l rc. is a solution of (6) for any choice of the coefficients. The reader(caln verify that each of the two vectors
I
e
12e-]
(5)
is a solution of system (4). the system is called
homogeneous. x2(t) = 0.(t) + c2y. a21. each of the two column vectors in (5) is a solution of system (4). and c2..
Ly2(t)J .[c. a.
DEFINITION 3
22e-]
The column vector
[0]
'
that is.e +
is also a solution of (4). Otherwise it is called nonhomogeneous.(t) Lx2(1)] + c..
.(t) fIrx.Lclx2(t) + c2y2(1)
DEFINITION 2
When both functions f.
For example.
(6)
where the coefficients a.
For example. therefore. the linear combination
cl I eJ + c2I 2e2]
. = an(t)x1 + au(t)x2
z2 = a21(t)x. we have the following. the system
X. for any constants c. system (3) is nonhomogeneous.3. = x2
(4)
X2 = . Any other solution of (6) is called a nontrivial solution... and a22 are all continuous functions on some interval 1.2x.1
Introduction and Basic Theory
149
(b) Linear Combination. + a. + 3x2
cal n is homogeneous. x1(t) 3 0.x. and c ry. the theory of linear systems resembles the theory of linear differential equations.
THEOREM 2
The two solutions
and
Lx21(t)J
X"(1)]
x22(1)
of the system
z. The following criterion. we can demonstrate that the two solutions (5) are linearly
independent solutions of the system (4). = a11(t)x. can be used to check the linear dependence or independence of solutions of system
(6).
that is. = 0.e' + c2e2i = 0 and c.
For example. = c2 = 0. reminiscent of the Wronskian determinant. we find that c2e' = 0. Subtracting the first equation from the
second. (t)
x21(t)
x12(1)
x22(t)
is never zero on I. Thus c2 = 0.150
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. The reader can verify that
Ie
J
and
[6e2'. c1 = c2 = 0. the first equation becomes c.
(9)
are linearly dependent solutions of (4).
c2x12(t) = 0 and c. + a12(t)x2
X2 = a2. Otherwise the solutions (8) are called
linearly dependent. which establishes our claim. In fact. Therefore. if
C1 [e"] + C2 L2?2'2'1 = [o
.x21(t) + c2x (t) = 0
for all t in I implies that c.(t)x. + a22(t)x2
are linearly independent on an interval I if and only if the Wronskian determinant
X.e' = 0 and so c.
then c. Consequently.e' + 2c2e' = 0.
.
Let to be a point in I and let x.
is
x1(t) =i e' -2 t . = x2 + t
x2 = .
THEOREM 3
Assume that the coefficients a.3. The reader should recognize the strong resemblance between this theorem and the existence-uniqueness theorem of Chapter 2..) = x10.(t)
(1)
X2 = a2.1
Introduction and Basic Theory
151
For example.t .e
x2(t) = a e .2x.
Using this theorem. and f2 of the
system
z. Then the IVP consisting of system (1) and the initial conditions
x1(t. the reader can verify (by direct substitution) that the unique solution of the IVP
X. = a11(t)xl + a12(t)x2
(6)
x2 = a21(t)x.2.
x2(to) = xm
has a unique solution
(x. the two solutions (5) of system (4) are linearly independent
because
I'E`
e'
2e21
e'I=2e' -
*0
forallt.. + a22(t)x2.
THEOREM 4
There exist two linearly independent solutions of the system
x.u and x. =
a12(t)x2 + f.
.(t)1 lx2(t) . a2 a22 and the functions f. the two solutions (9) of system (4) are linearly dependent
because
I
e2' 3eu 2eu 6e" =6e"-6e"=0.
On the other hand.
Furthermore.(t)x1 + a22(t)x2 + f2(t)
are all continuous on an interval 1. be two given constants. a12.
The following basic existence and uniqueness theorem for linear systems is stated here without proof. + 3x2 +
1
x1(0) = -1
(10)
x2(0) _ -. this unique solution is valid throughout the interval I.
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.
(3)
At this point we use the first equation in system (1) and the value of x. Thus. + 4x2 + 5x3
x3 = -17x. + 3x2. The unique solution of the IVP
z. s. we have succeeded in eliminating the function x2 from system (1). .
The first equation in (1) states that x2 = x. = 0.(t) = c.2x.e' + c2eu.3s.
Solution Differentiating both sides of the first equation with respect to t and using the second equation. which contains one single function and which can be solved. = 2e' + 3e2.5e2'.
x3 = 3e . 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
.
(2)
Thus. given in (3) to find the other unknown function x2 = X.
x2(t) = c.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. + 3x2. + x2 + 7x. = S2 = .
3. = X2
(1)
X2 = . + 3s.
x2 = -9x. + 2x. we obtain
X.
X3(0)
2
x. + X2 + 12X3
X1(0) = 5.
is
x2(0) = 2.
(4)
Equations (3) and (4) give the general solution of system (1).. 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.162
3
Linear Systems
30. = -10x.e' + 2c2e2i.2x.
EXAMPLE 1
Find the general solution of the homogeneous linear system
X.
x2 = e + e. hence
z. = .2x. obtaining in the process the differential equation (2). (2) is
x. The general solution of Eq.
The method of elimination can also be applied efficiently to nonhomogeneous systems.(0) = -1. (7)
A particular solution of Eq.
EXAMPLE 3
(10)
Solve the IVP
z. = -3x.=z2+1=-2x.t. x.+3x2+2.o = ... (6) is x. + 4x2 z2 = -2x. we obtain the general solution of Eq. we have succeeded in eliminating the variable x2 from system (5). into the first equation of the system.t) + 2
z. so
x2(t)=c.= -3t+2.
. The solution of the homogeneous equation associated with Eq. . as the following example illustrates.(t) = c. and substitute the value of x. differentiating both sides of the first equation with respect to t and using the second equation.
. + 3(X.-3z.3. or. (6): x.
(6)
tion and can be solved.=x2+t
C2 = . (6) is of the form x.
(5)
Solution As in the previous example.
(9)
Now from the first equation in (5) we have x2 = z. we obtain
z.e + c2eu. which contains one single func-
'
X.+2x. + 3x2
(11)
(12)
(13)
x.tip =0' -3A + 2At + 2B= -3t+2
and
2A = -3
(8)
-3A + 2B = 2=>A = -. obtaining the nonhomogeneous differential equation (6). solve the second equation of the system for x.h = c.
and
B = -.
x2(0) = 3.2x.1 t-1 .e' + c2e' .
Adding (7) and (8).
Using the first equation in (5) yields
Thus.
Equations (9) and (10) give the general solution of (5).P = At + B
za = A. = -2x. + 3x2 + 1.e'+2c2e'-t-22.Zt .
EXAMPLE 2
Find the general solution of the nonhomogeneous system
z.2 The Method of Elimination
163
of the system for x2 and substitute the value of x2 into the second equation of the system. .
x. and the solution of the IVP (11)-(13) is x1(t) = 7e .c2e-'.+zc2=3. However.3i. and the general solution of the system contains. = 0 * X1(t) = c.
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.
Thus. c2 = -8.8e-'
x2(t) = 7e . + 3x. + 3x2)
_ . We illustrate this by the following example. = z2 = x.(t). The extraneous constants can be eliminated by substituting the solutions into the system and equating coefficients of similar terms.
(14)
x.
REMARK 1
EXAMPLE 4
Find the general solution of the system
x. . we obtain
c. = c.2x. (11) we have
4x2 = i. + 4i2 = . + 12X2 = -3i. two arbitrary constants.3i. if we find x2(t) from the second
.(t) = c.3i.e + 3c2e"' = 4c. ' x.
Thus.e' + c2e-'. Sometimes. =x2+ 1
x2 = x.e' + c2e-'. this follows from Theorem 5 of Section 3.4e"'. = 7.e . we can find x2(t) either from the first
equation of the system or from the second.e' + . we find x. as
expected. the method of elimination introduces extraneous constants.c2e-' + 3c.8x. + 3(i.
x.e' + 2c2e-'. + 4(.e' .
(15)
Using the initial conditions (13) in Eqs. If we use the first equation we find
Differentiating both sides of the first equation with respect to t and
x2(t) = c. . + c2 = -1
c.1.
and so.x.. we obtain
. from Eq.164
3
Linssr Systems
Solution Differentiating both sides of (11) and using (12) and then (11) again. Now that we have x. (14) and (15).1. however.
Solution
using the second equation.ce-' .
x2(t) = c.=0. c. + 3x)
#'
thus. For n = 2.8x. .
Roberts. we find that c3 should be -1. B.
1 k. are associated with the breakdown process.
Flgure 3.
x2(t) = c.
Synthesis
k. is injected
1OThe models and developments presented here have been extracted from E. The albumin in the plasma (and in the extravascular fluids) is changing in the following manner. Reeve and J. referred to as I13'-albumin. E.c2e-' . It is not completely known exactly where this breakdown takes place.3.4. Gen." J.c2e-' + c3. In order to study the albumin process. breakdown. are constants associated with the rate at which the various processes take place.4).
Vascula
x
Y
k2
Extravascular
k.10 Under normal conditions an animal is assumed to have a constant amount of albumin. and synthesis of albumin in animals has led to systems of ordinary differential equations.e' . and some albumin is being broken down.e + c2e-'. and we therefore assume that some breakdown occurs in the plasma and some in the extravascular fluids. In this case.2 The Method of Elimination
165
equation of the system. y. If we substitute these functions into the system and equate coefficients of similar terms. k and k. which introduces a third arbitrary
constant.
. Some albumin is transferring to the extravascular fluids (or. Thus. in its natural state the study of albumin can be viewed in terms of the compartment model given as Figure 3. and k. and k k2. to the plasma). The newly synthesized protein is shown reentering the system via the plasma.
APPLICATIONS 3. Physiol.e' . We refer to the model above as being associated with unlabeled albumin. and k.1
Some investigations into the distribution. Thus the general solution is
x. we have to integrate.e' + c2e-' and x2(t) = c. some investigators proceed as follows.(t) = c. A certain amount of radioactive albumin.(t) = c.represents the number of grams of albumin in the extravascular fluids. k3. 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. and k2 are associated with the rates of interaction. k. x. "Kinetics of I"'-Albumin Distribution and Breakdown. 43 (1959): 415.2. respectively. x represents the number of grams of albumin in the plasma. and this catabolized protein is replaced by newly synthesized
protein.4
In this figure.
y. At any time t thereafter. we must use derivatives
with respect to time. no new protein is being synthesized in this model.166
3
Linear Systems
into the vascular system. thus. determines a unique solution. It"-albumin. The compartment model associated with V`-albumin is
shown in Figure 3. z.x-(k2+k. u = 1. All these actions are continuous in nature. together
substance has been excreted. all the
x = y = z = 0. After "infinite" time. that (16). This solution depends. and u in the figure represent fractions of the total amount originally injected. This system also has initial and terminal conditions associated with it. Since there is only one injection. t --> 0
with the initial conditions. being radioactive. is continuously being broken down as a result of the liberation of radioactive breakdown products. y = z = u = 0.)y
(16)
dz = kax + k.5. there is a certain amount of radioactive substance x in the plasma and a certain amount y in the extravascular fluids. The quantities x. After a few minutes its behavior is assumed to follow that of the unlabeled albumin.5
. all the radioactive substance is in the
plasma.y . Initially. and certain of the breakdown products are being excreted by the animal. We note. thus. and therefore to speak of the rates of change. The process described by this compartment model can be analyzed by studying the following system of differential equations: dx
dt=k. where it mixes with the vascular fluid. The other equations are obtained by means of similar reasoning.
k. however.
Vascular
x
Y
Extravascular
k. t = 0 ' x = 1. The solution of this system is left to the exercises.x and kax g/day.
z
Breakdown Products
u
I Excretion
Figure 3.ksz di du
dt = kz.
The first equation is arrived at by noticing in the model that x is changing due to gaining key g/day and losing k.
and f are positive constants. and 4.e'' and x2 = A2e' satisfy system (1).
3.
.5. For the sake of simplicity. and A are to be determined by demanding that
x. 3. Learning Theory In an article with applications to learning theory.e` + a22A2eu
"S. known as the matrix method. b. Grossberg" obtained the following IVP:
x=a(b-x)-cfy d(x-y)-cfy-ay
x((1) = b. y(O) = a
db
d'
where a. + a22. and k must satisfy the system of equations A. Theoret. =
a12x2
(1)
x2 = a21x. we present the main features of this method in the special case of linear homogeneous systems with constant coefficients of the form
x. d.170
3
Linear Systems
31. we have no difficulty solving systems with constant coefficients of two linear differential equations in two unknown functions.3 THE MATRIX METHOD
As we saw in Section 3.
Illustrative examples will be given for systems of n equations in n unknown functions for n = 2. which was discussed in Sections 2. c. Biol. Mimicking the method of solving linear homogeneous differential equations with constant coefficients. (1969): 325-64.he' = a. Substituting (2) into (1).. The method of elimination is a simple method to use for such systems.eu + a12A2e"
A2ke"` = a21A. let us look for a solution of system (1) in the form
Lx2i (2)
LA2euJ
The constants A A2. J.4 and 2. Grossberg. has the advantage that it can be easily extended to systems with constant coefficients of n linear differential equations in n unknown functions. = A. Show that for every fixed f the unknown function y decays monotonically to a positive minimum. A.2. we find that A A2. The method of this section.
A and A2. (4) is called the characteristic equation of the matrix
an
a2. 0) gives rise to the trivial solution x1(t) = 0. if X. it can be shown that these two
solutions are linearly independent.X
aZl
a12
I
a. the above procedure gives. * k2). Naturally. since the solution (A A2) = (0.
(4)
that is. If the h in (3) is replaced
by a characteristic root.2
a22
(5)
and the roots of the characteristic equation are called the characteristic roots or eigenvalues of the matrix (5).)
Let X. x2(t) = 0 of system (1).
It will be useful to the reader to note at this point that the determinant in Eq. this means that there is a solution (A A2) * (0. 0). We view system (3) as a linear homogeneous system in the unknowns A. and the other from K = K2. = 0. and X2 be the characteristic roots of system (1). Nevertheless. In the case of system (3). we seek a nontrivial solution of this
system. [In matrix analysis. (4') has two distinct roots (in other words. Thus.X)A2 = 0. in
. A2). one can solve this system by taking the following approach. + (a22 .A. and A2 in (2). Eq.x
0.
a22
of the coefficients of system (1) by subtracting X from the main diagonal.
a. (4') does not have two distinct roots (in other words. 0) if and only if
a11 .
(4')
DEFINITION I
Equation (4') is called the characteristic equation of system (1).3. . the corresponding system (3) has a nontrivial solution (A. Substituting these values of A. + a12A2 = 0
(3)
a2. if Eq.
System (3) consists of only two equations. and the roots of the characteristic equation are called the characteristic roots of the system. On the other hand. If Eq. we can obtain two nontrivial solutions of system (1): one from K = K. (4) is obtained from the determinant
a au
a2. we obtain a nontrivial
solution of system (1).(a + a22)X +
a12a2.X)A.
X2
. and A2. (A A2) should be different from (0. if K1 = X2). 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. we obtain the equivalent system
(au . Furthermore. and there are three unknown quantities involved: X.3
The Matrix Method
171
Dividing through by a" and rearranging terms.
= -3.
A2 .=A2.
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. The trick in this case is to look for a second linearly independent solution of (1) in the form
lx.t + b2)e"" '
(6)
where the coefficients a b a2. See Remark 4. + 6x2. any other nontrivial solution of this system (for example.
that is. we still need a second linearly independent solution of system (1). of the solution (8) should satisfy the homogeneous system (3).=0
=>A. = 3 and A2 = 5. =
A2 = 2) would just give us another solution with which (9) is linearly dependent.
z2
__
(a. = 3.172
3
linear Systems
general.)e"" (a.+3A.
The characteristic roots are A. one nontrivial solution (and its multiples) of system (1).8A + 15 = 0.r + b. = A2 = 1.+A2=0 -3A. 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. and A. A. a2.
.+A2=0
A2 = 3A1. As we are interested in finding the general solution of system (1). we have the solution
(9) [e ] [Clearly. When A = A. a13 = 1.
Choosing A. and b.+A.=2x1+x2
X. and a22 = 6)
2-\ -3
is.] When A = A2 = 5. = -3x. system (3) becomes
-3A. the constants A.=0
-3A. that
-A.
6-\
1
0.
we obtain
a. For example. and b2 are to be determined in such a way that (17) is a solution of (15) that is linearly independent of (16). = a2 = 1. = b2.
x. and c2 are arbitrary constants.
where c.=a2
(18)
a2 = _a.1te'el
is not a solution of system (15).)e = (alt + b2)e'
a2e' + (alt + b2)e' = -(a. = (a.
[e e'
te'
+ C2
ll
and
[(t + 1)e'. the general solution of (15)
is
[x2j = that is.
Notice that for a second linearly independent solution of (15). the choice a.)e' + 2(a2t + b2)e. In fact. = c.)e`l
x2j
(alt + b2)e'j '
where the constants a b a2.t + b. we look for a second linearly independent solution of (15) of the form (6).e + (a.3
The Matrix Method
175
Since the roots of the characteristic equation are equal. b.=b2.
a.)e and x2 = (a2t + b2)e into (15). (17) [x . = a2.t + b. + 2a2
or (since the last two equations in (18) are equivalent to the first two]
a. and b2 = 1 yields the solution
re'
(t + 1)e
'
(20)
which is linearly independent of (16).t + b.
a.
Dividing by e' and equating coefficients of like powers of t. + 2b2. + b.e + c2te'
x2 = c.
REMARK 3
t
[e'i
. = 0.e + c2(t + 1)e'. and b2 that satisfies (19) and gives a solution
which is linearly independent to (16) is acceptable.l = [(a. it is not enough to multiply (16) by t as in the case of linear homogeneous differential equations with multiple characteristic roots.
(19)
Now any choice of a b a2. we have
a.
a2 + b2 = . Substituting x.+b.3.t + b. (Why?) Thus. system (15) does not even have a
nontrivial solution of the form
[x2J
[az1e j
.b. Indeed.
a21 = 0. This is also true for more general linear homogeneous systems. = 0.
yl
and
C]
f
. 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. a system of the form (1) with a double characteristic root may have two linearly independent solutions of the form (2). A2 = 1. of the solution (8) satisfy the homogeneous system (3) with a = 1. )\2 .
As we explained in the introductory paragraph of this section. the general solution of system (21) is [ell
x2 J
that is. That is.
.=5x1 +2x2+Zx3
x2 = 2x. unlike the case of linear homogeneous differential equations with multiple characteristic roots. = c. and k = 1. the constants A.4x3
(22)
X3=2x. = 1. respectively. Therefore. and A. as the example below indicates. -4x2+2r3. that is. See Example 5 and Remark 4.
Choosing first A.cl
+ c2 LeJ
x. am = 1. and show how to apply the matrix method to compute their general solution.e'
and
x2 = c2e. a12 = 0. The method of elimination is highly recommended in the case n = 2.178
3 LinearSystems
Also.
EXAMPLE 4
Find the general solution of the system
z. A2 = 0 and then A.
EXAMPLE 5
Find the general solution of the system
8. 0
I
10
The characteristic roots are \. and A2 are arbitrary constants.1\ 0 = 0. but we will not go into the details in this book. = a2 = 1. + 2x2 .
0A1+0A2=0
linearly independent solutions
'A.2X + 1 = 0. = x1
Xz = x2
(21)
Solution The characteristic equation of system (21) is 1 . we obtain the two
poi
Thus. We shall now present two examples with n = 3 and n = 4.
+ u2(t)e5' =
f(t)
til(r)e3r + 3i 2(t)e5' = g(t).]
Lx2p = u.(t)e3.
(38)
where c.
In Example 1 of Section 3. substituting the results into (39). 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. and u2 are functions to be determined.(t)e3. we obtain a particular solution of (37)..
=
3e5'!
[g(t) -f(t)le s`
(41)
From (40) and (41) we find u. That is.
Using Cramer's rule.]
(39)
where u. The general solution of (37) is given by
. (See Section 2. Consider the nonhomogeneous linear system
zl = 2rl + x2 + f(t)
(37)
z2 = -3x.12. and.(t)
+ u2(t)
[3e3.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.182
3
Linear Systems
geneous system and a particular solution of the nonhomogeneous system. The main idea in the method of variation of parameters is to seekf a particularf solution of (37) of the form particular[
x.
xlp = ul(1)e3r + u2(t)es' x2p = u. (t) = g(t)
1e3+
eSil
s'
le'
and
e
= i [3f(t) .) We illustrate the technique by means of an example. + 6x2 + g(t). + 3u2(t)e3'
Substituting into (37) and simplifying we obtain
u.] +
c2
[e]
.3 we found that the general solution of the homogeneous system corresponding to (37) is
j
Ix2h = Cl [. we find
f(t)
ti.
but
1
e"Ib
1
me"dt = Jim
0
b--
e"dt = lim
1".a 3
l
I
o
)
=1lm(3e16-3)=x.
b-. This method resembles in some ways the use of logarithms for solving exponential equations. the
improper integral j edt converges but ledt diverges. -'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.
0
ob
1
D
o 1
¢e-"dt = Jim b--
e'3'dt = lim .1
INTRODUCTION
In this chapter we present another method for solving linear differential equations with constant coefficients and systems of such equations..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.e-"
b-. In fact. For example.
. Then by definition
(t)dt = lim g(t)dt.
3
o
=lbim(-3e '6+3)=3.
4.CHAPTER 4
The Laplace Transform
4. It is called the method of Laplace transform. 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..
Furthermore. the integral (1) converges for s > a. We also write f(t) = `.
DEFINITION 3
A function f is of exponential order a as t tends to infinity if there exist numbers
M.
. then F(s + k) = `.C[e-`'f(t)]. but
first we give the following definitions. and k denote arbitrary constants.
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.
We now quote without proof a theorem that guarantees the convergence of the integral (1).
THEOREM 3
If F(s) = `. Thus. and if f is of exponential
order a as t tends to infinity.P`[F(s)].
The following theorems provide us with useful tools for manipulating Laplace transforms. and T such that
If(t)I s Me"'
when
t ? T.
We note that for a given f the corresponding F is uniquely determined (when it exists). From Examples I and 2 we see that
-w
ll=1
and
Y'Is12J=eu.
THEOREM 2
X[c. a. f(t) + c.o). in each of which f is continuous and has finite left. if F(s) has a continuous inverse f. The proof of the first theorem is a simple consequence of the definitions and will therefore be omitted (see Exercise 60). we will state Theorem 1.
Alternatively. In what follows the symbols c c2.G(s). if f and g are piecewise continuous functions whose Laplace transforms exist and satisfy e[f(t)] = 2[g(t)]. f is of exponential order a if there exists an a such that
Iimlf(t)Ve' = L.and right-hand limits.
THEOREM 1
If f is piecewise continuous on every finite interval in [0.F(s) + c. where L = 0 or L is a finite positive number.Li[f(t)]. Concerning inverse Laplace transforms. then f is unique.g(t)] = c.192
4 The Laplace Transform
bolism T[f(t)] = F(s). then f = g at their points of continuity.
f fu f(u)du dT
P-. one can also use a table of Laplace transforms.ekr] = y[ek"] .1 is presented in Theorem 2 of Section 4.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.2 The Laplace Transform and Its Properties
S
195
s>
k
cosh kt
(xi ) (xii ) (xiii )
(x i v)
s_
k
k'.T[ek" . we have
. the reader realizes that not every function will appear in such a table.1.4.. -
s"F(s) .)
. .T)f(v)dr = fof(( . are continuous and of exponential order.s > k k =
sinh kt
k.k')'
Solution
Using formula (xv) of Table 4. That is. . but
rather we prefer to look up the transform in a table.2[ek9'].ek"] =
k. we do not resort to the definition every time we wish to determine a particular Laplace transform.)(0)
(xxi)
(xxii)
-s"-'f(0) . Naturally.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). In much the same way as one uses a table of integrals or formulas for differentiation.1.k > 0
(\.
ekv ..k ?
(s . Therefore.t)"f(t)
fog(t .3..
EXAMPLE 6 Show that
. Laplace transforms.k)(s .)(s
2ks
. one must develop a capability for expressing
the functions in the forms found in the tables.
.k2
(s .k. Formula (xxii) of Table 4.s > k
k.T[ek"
. if one
is to be proficient at using the table.r)g(-r)d-t
r
G(s)F(s)
F(S)
sl-' F(s)
fo f(t)dt . We take notice of the fact that all the specific functions that appear in Table 4.
Show that
m
4[j l f(u) du d r] = s F(s). so that their Laplace transform exists. The
theorems of Section 4.f(O). Use Theorem 5 to compute T[cos t]. Use Theorem 5 to compute `.k)2 + 68. 70.
d'
64. Use Theorem 5 to compute Z[cos kt].2. The next two theorems provide us with this information.
69.
0
66 Show that C[ek' sin mtl _ (s-k)2+ m2. Use Theorem 5 to compute T[sin kt].
63. In order to apply Laplace transforms to differential equations.
Proof
21RIA = I. Show that T[e" cos mt] =
s-k
m1*
?-s-. If n is a positive integer. or higher) of a function is.
67.
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. Show that e[t cos kt] _
65.
. In both theorems we assume that all functions of t that appear satisfy the hypotheses of Theorem I of Section 4.2 are helpful to us for the purpose of manipulating transforms. If F(s) = _T[f(t)].e-'f'(t)dt.t) f (t)].3 The Laplace Transform Applied to Differential Equations and Systems
199
62.
THEOREM 1
2[f (t)) = sF(s) . show that ds F(s) = `.Z[sin t].P[( .4. 71. we need to know what the Laplace transform of a derivative (or second derivative. show that dsF(s) = `.F[-tf(t)].
43
.
exponential order as t tends to infinity.
f(t)( -se-"dt)
= -f(0) + s 09 e-'f(t)dt J
= sF(s) . Hence. the method of solving differential equations by Laplace transforms
is outlined as follows: Given a differential equation. Formula (xxii) of Table 4.')(0). we apply the Laplace transformation to each term in the differential equation).200
4 The Laplace Transform
Integrate by parts setting u = e-". 2.
Y(f"(t)] = e-"f'(t)io -
f f (t)( -se-"dt)
= -f'(0) + slme-'f'(t)dt
0
= -f'(0) + s. e-"f(t) I .f(0)) = s2F(s) . one must know f(0). Thus.sf(0) . in the final step. we "take the Laplace transform of it" (in other words..
THEOREM 2
"'(t)) = s"F(s) .
Note In the proof of Theorem 2 just given. Hence. That is. in the general case it must be assumed that
lim. since f is of exponential order. f'(0). Similarly. for k = 1.1 indicates that in order to obtain explicit results.. we must have an initial value
. Basically. Then du = -se-" dt and
v = f(t). We then solve this equation for this transform. . setting u = e-" and dv = f"(t)dt.
In the boundary term.. This results in an equation (usually a linear algebraic equation) for the Laplace transform of the unknown function.. and..s"-'f(0)
-s"-Zf'(0)
f("-a(0).f'(0).1. we assume that f and its
derivatives are of. 4v = f'(t)dt. we find the inverse of this transform. n . Therefore. .e-"f(t)=0. For systems of differential equations the procedure is similar and leads to a system of algebraic equations.T[f'(t)] = -f'(0) +s(sF(s) . we know that lim_e-"f(t) = 0.f(0).e-"f")(t) = 0.
ARIA = j
Integrate by parts.
Proof We prove the result for n = 2 and leave the generalization to the reader. du = -se-"dt and
v = f(t). it is assumed that lim. . J (n ._.
2 to compute T[f(t)] where f(t) is the "sawtooth" function of Figure 4.3.4.3y . y(0) = 4 y(O) 14. or the Dirac delta function. happens not to be a function as the word is used in mathematics.1)+9h. the ordinate would be zero up to the time the hammer struck the nail.y . It is conventional that the impulse of the force (that is. the integral of the force with respect to time) is unity.210
4 The Laplace Transform
11.1) + 10h2(t . y+6y+9y=6h. If the graph of this force were drawn. there are advantages to thinking of the unit impulse function as if it were a function.
16.5 THE UNIT IMPULSE FUNCTION
The unit impulse function. one might be interested in describing the force involved when a hammer strikes a nail. it then has some value for a very short period of time and is zero thereafter. With this convention we can approximate the force graphically in Figure 4. y . For example.(t . y(0) = 1
13. Use Theorem 5 of Section 4.(t . say at t = t0 (to > 0).3
4.(t . Since the time interval of application of the force is small.6y = 50 cos t + 30h.(t-3)
Y(0) = 0.
.(t) + 26h.(t.2 to compute T[f(t)] where f(t) is the squarewave function of Example 1. For a first exposition.(t) + 40h.
f(t)
I
t
1
2
3
4
5
6
7
8
Figure 4.4y = 20h. Use Theorem 5 of Section 4. 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.2)
Y(0) = 0.3) y(0) = 1. Y . 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.5) -7. y + by + 26y = 52h. however. y(0) = 0
15.
they transform a given problem into a simpler problem. Using Laplace transforms.11. Naturally.1 we differentiated Eq. (1). the equation
L di + Ri + C
o
i(T)dT = v(t). via Kirchhoffs voltage law.1)+38(t-2)
y(O) = 0. C = 0. in which we have used the relation (11) of that section.
. this requires that v(t) be a differentiable function.1. We do not restrict the treatment to differential equations exclusively. Here we have used lowercase i and v for current and voltage. (1) is Eq. we can solve Eq.11. (1) directly. y+2y+y=S(t)+b(t. to conform to the convention of labelling functions of tin lower case and their Laplace transform in upper case.2y . Note that Eq.
(1)
Equation (1) is referred to as an integrodifferential equation for i. It is worth emphasizing that Laplace transforms have the nice feature that when they are applicable. y .6
APPUCATIONS
In this section we illustrate a few practical applications of Laplace transforms.8y = 85 cost + 68(t .s2 + (R/L)s + 1/LC V(s)
= G(s)V(s).11. R = 40 ohms. y(0) = . respectively. 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 .
Find i(t) for the electric circuit with L = 20 henries. (12) of Section 2. y(0) = 4
14.
Electric Circuits
In Section 2.
EXAMPLE 1
Solution Applying the Laplace transform to Eq. and v(t) = sin t.5) y(O) = -9. (1) to obtain a differential equation for i.05 farad. In Section 2.214
4 The Laplace Transform
13.1 we obtained.3
4.
For the most part. Following this model. it would be convenient if one could obtain an equation relating output to input.G(s)V(s)
Figure 4. Equation (3) is frequently called the input-output relation.+ asin t. that is. and v(t) into Eq. and D = . Thus. 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. chemical. Eq. On the other hand. (2). C. A voltage (input) is applied to the circuit (system) and a current (output) is produced.5.5
. physical) can be viewed in this fashion. Relation (3) is viewed diagrammatically as in Figure 4.t.
1(s)
40 40
(s+1)Z+-i+1
°° `-e
t
i(t)
[(s _+11 )2] +
_ -40te. Then it would be simply a matter of substituting given values for the input and determining the values of the output via this relation. Equation (1) is such a relation. R.4.
Nevertheless. (2) is better suited for these purposes. and in some cases performing integrations. In an abstract way many processes (biological. Eq.6
Applications
215
Substituting the given values for L. it has the advantage that given an input v(t) to obtain the output i(t) requires some algebraic manipulation. C = 0. (2) relates
REMARK 1
the Laplace transform of the output to the Laplace transform of the input. Of course. use of tables of Laplace transforms. but unfortunately the relationship is not simple enough to easily obtain output from input. Engineers and biologists take the attitude that knowledge of the transfer function provides complete information of the system. see Exercises 14 through 17).
System
Input
V(s)
G(s)
Output
t(s) . engineers know the transfer function and wish to analyze the output. B = -4`-0. The phenomenon illustrated by the electric circuit can be viewed in the following way. Biologists observe input and output and try to determine the transfer function from this information (for more discussion in this direction. The function G(s) is called the transfer function or system function.
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. the positive y axis east. see R. Hirzel. Set cos P = a. Rosen.
S
i
System (5) can be solved for X. Planck. The positive x axis is to point south.216
4 The Laplace Transform
It is in this context [Eq. 81.
.g. sin (3 = b. Optimality Principles in Biology (New York: Plenum Press. 'See B. taking into account the rotation of the earth. i(0) = w. Neglecting the mass of the particle.2w(x sin l3 + i cos
(4)
i = 2wy cos
g. and Z since the determinant of coefficients has the value (see Exercise 2)
A = s`(s2 + 4w2). 4th ed.'
Mechanics
A particle is projected from a point on the surface of the earth with a given initial velocity. and the positive z axis points opposite to the direction of the acceleration due to gravity.2wbsY = u
2wbsX + s2Y + 2wasZ = v
(5)
-2wasY + s2Z = w . Thus (see Exercises 3. 1967).
(6)
Our restrictions on s guarantee that .5 # 0. p. 1928).M. We wish to analyze the motion of this particle. y(O) = v. The origin is taken to have latitude P and the angular velocity of the earth is denoted by w.
with initial values x(O) = y(O) = z(O) = 0 and x(0) = u. and 6). (Leipzig: S. and apply the Laplace transform to system (4) to obtain the following system (after rearrangement) (see Exercise 1):
s2X . it can be shown2 that the equations of motion are as follows:
X=2wysin(3
. (3)] that the Laplace transform is utilized in presentday investigations.
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) = . Einfithrung in die aligemeine Mechanik. 4. Y. We choose the origin of a three-dimensional
coordinate system to be at the point from which the particle was projected.
and (12). > 0 also survive according to this rule.(t) of these specimens still in the population.
(13)
Equations (13) would represent the motion of a particle relative to a still earth.4. we refer to this age classification as zero age.sin 2wt) + v a sine wt
+ Zw (2b2wt + a2 sin 2wt) (b2w2t2 + a2 sine wt). 1967).. if m(t. Then n. (11).sin 2(ot) + 2w2 ab(w2t2 y(t)
sin 2 wt)
(10)
w b sine wt + v sin 2wt .ab(2wt ..).) f(t . there results
x=ut.
.
Furthermore.Transforms (London: Van Nostrand.'
In studying the population growth of some species. say n(0). all of a certain age. we consider that individuals of age zero placed in this population at some time t.(t) and n(0) are connected.
'For a more complete discussion of this illustration and other applications of the Laplace transform.) of these individuals are placed into the population at time t. At a future time t there are n.
x(t) = u (2a2wt + b2 sin 2wt) + v b sine wt
2w
w
. where
m(t) = nt(t. For convenience. there are a certain number of them. That is. (10). Guide to the Applications of the Laplace and Z.sin 2wt) z(t)
2tn ab(2wt . we assume that at time t = 0.
y=vt. 8. with the use of a survival function f (t) by the relation
Population Growth
n.t.
z=wt-1gt2. see Gustav Doetsch.
2w2
Note that if one considers the limit as w tends to zero in Eqs. and 9). then at a future time t > t.(t) = n(0) f(t).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. there will be m(t) of these individuals left in the population.w a sine wt
w
+ 4w2 a(2wt .
We wish to determine the total
population n(t) at some later time t.
n(t) = n(0) + ct. zero-age individuals are
T. In fact.
It is easy to see that if ro = 34.000e-' + ro(1 .218
4 The Laplaee Transform
Consider now that age-zero individuals are placed in the population at a rate r(t).000e-' + roe-` e d r
f
/
= 34.000 wild rabbits in Rhode Island.
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. In the time interval from T to T + AT. 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. commonly called the replacement rate. The consideration is as follows. At time t = 0 there are n(O) zero-age individuals of a certain species given. that the survival function for these rabbits is a-'.)OT.e-). Certainly one could approximate the answer for n(t) by splitting the time interval [0. T. that
EXAMPLE 3
is. we can say that the approximation tends to n(t) as AT tends to zero. individuals are placed in the population. According to the survival law.
placed into the population at the rate r(t).000. r(T. must be such that T s
T + AT. t] into intervals of length AT and adding up the number of survivors for each interval. As time goes on.T. r(TI)AT f (t . then n(t) = 34.000e-' +
-(-Jroe'')di
= 34.
.000e-' + roe-' I e' \
ICJ
o
= 34.
Suppose that in 1976 there were 34.) of these individuals will still be present in the population at time t. Thus.
(14)
Equation (14) is a model for studying the growth of populations governed by the considerations above.
n(t) = n(0)f(t) + j
r() f(t -
)dT. Naturally.000 for all t. and that wild rabbits are introduced into the population at a constant rate ro. the approximation is made more accurate by making the time interval AT smaller.
(s)
G.(s)
Y2(s)
= G.(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. A controlled system is one in which
a specific activity is executed so as to maintain a certain fixed activity of the system. w constants. 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.(s)G2(s)
Y. a number of other transfer functions can be derived which
characterize the behavior of the entire feedback system.(s) .6
14.(s)G2(s)Y.(s)
Feedback Loop
Figure 4. we can write
Y0(s) = G. (16) in each
of the following cases:
(c) wo < a2 (b) wo > a2 (a) wo = a2 13. Suppose that v(t) = vo (a constant). From the transfer functions
G. Determine i(t) from Eq.(s)Y.(s) = G. Such a feedback system can be represented diagrammatically as in Figure 4. Determine i(t) from Eq.220
4 The Laplace Transform
12. Using the definition of transfer function. w # wo). Show that
YO(s)
=
G. Suppose that v(t) = vp sin wt (vo.(s)
1 + G.(s) and G2(s). where Y.
Controlled
Error Signal
Controller
GI (s)
Control Signal
System
Output
G2(s)
r
Ye(s)
Y.Y0(s) and 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. This other system is the controlling system.(s).(s)G2(s)
Y.(s)G2(s)'
r
Hint:
a_a bl
c
bc
.6.(s)G2(s) is called the open-loop transfer function.
(t) cos t
'Many illustrations of this sort can be found in Rosen.4.(s)G=(s)*
Regulators For a "pure regulator" the input is a constant (thus it is generally chosen to be zero).(s)G2(s)
is the closed-loop transfer function.6
Applications
221
15. One such system is the eye.
Many biological systems are modeled as feedback systems or as regulators. The transform of this signal is denoted by Yt (s).7. 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.'
REMARK 2
REVIEW EXERCISES
In Exercises 1 through 4. Show that
Y0(s)
_
G2(s)
Y1(s)
I + G.
. 4t5 + 3t sin t
2. Efficient techniques for doing this are as yet not known. Show that
Y(s)
Y. and there is an additional signal which represents the effects
of fluctuations in the environment of the system.(s)
G2(s)
Y.
[Hint: Follow an "initial signal" from input to output. Such feedback systems are represented diagrammatically as in Figure 4.. Optimality Principles.(s)
Figure 4.7
16.]
17.
1.(s)
-
1
1 + G. find the Laplace transform of the given function. h.
Ye(s)
G. Show that
Y"(s) = G (s)
Yf (s)
s
is the open-loop transfer function. (s)
Y.
xy = 0 Bessel's equation: x2y" + xy' + (x2 . 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.
Airy's equation: y" .p2)y = 0 Chebyshev's equation: (1 . 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. especially in the process of solving some of the classical partial differential equations in mathematical physics.2xy' + 2py = 0 xy' + (1 .x2)y" . Before we explain how to obtain series solutions for the above as well as other differential equations with variable coefficients.
We shall refer to this as the method of series solutions. Second-order linear differential equations appear frequently in applied mathematics. It is common to refer to these equations by the name that appears to the left of the differential equation. All these equations can be solved by the method of series solutions.x)y" + [c .CHAPTER 5
Series Solutions of Second-Order Linear Equations
5.xy' + ply = 0 Gauss's hypergeometric equation: x(1 . Following are some of the most important second-order linear differential equations with variable coefficients which occur in applications.x2)y" .x)y' + py = 0 Laguerre's equation: Legendre's equation: (1 .(a + b + 1)x]y' Hermite's equation: y" . 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. we review some of the properties of power series that will be used in this chapter.try' + n(n + 1)y = 0
aby = 0
With the exception of Airy's equation.
.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.
_oa"(x . such as Er_ox2n `/2".. For such series the radius of convergence is determined by using the ratio test of calculus. If R =
it
converges for all x. It is within this interval that all operations that we will perform on series. the series is said to diverge at the point x. it is important to find all points x for which the
series converges.x0)2 +
+ a"(x . and 24. (2) and (3) are not applicable in general.
(3)
provided that the limit in (2) or (3) exists.. Finally. Eqs.. series (1) converges only at its center x = x0.x0) + a2(x . This term is denoted by R and is given by the formula
R=
or
1
lm
(2 )
R = lim
4" a". the series may converge or diverge. and the point x0 is called the center of the power series.2
Review of Power Series
225
5.. . In this case the value of the limit is called the sum of the series at the point x.x0)" converges at a specific point x.x0)" +
is called a power series in powers of (x .x0)". then repeat the procedure for x = R + x0.R + xo < x < R + x0. .. At the endpoints of the interval
(4).
.x0 I > R. To do this we compute the radius of convergence of the power series. for x = -R + x0 and x = R + x0.5. if 0 < R < -. we set x = . See Exercises 3.(x .x0 I < R. Given a power series (1). or the entire real line if
R = -.. We also say that (1) is a power series about the point x0. if
N
lim I a"(x. (4) and diverges for I x .)
If R = 0. that is. is called the interval of convergence of series (1).
"-o (1)
The numbers a0. for . . (For series in a form other than (1). that is. If this limit does not exist.R + x0 in
(1) and check the resulting series by one of the known tests. in our attempt to find series solutions of differential equations. The interval (4). To determine the behavior at these points.2 REVIEW OF POWER SERIES
A series of the form
ao + a.. . are called the coefficients of the power
series. 9.x0) and is denoted by
i a"(x . are legal. We say that a power series E. a". a a2. the series converges in the interval I x .x0)"
exists.
xo)
f(x) = j n(n . Finally. . (c) Here a" = 1/n!.1)" and from formula (2).xo I < R.
1
lim V-1 I
).xo)"
"-o
for.1 1 = 1. x .. series (b) converges for all points x in the interval .
and both of these series diverge. the sum of the series exists and defines a function
f(x) = I a"(x .x0)". the derivatives f (x). f "(x). then for every x in the interval of convergence . x ..
(5)
The function f(x) defined by power series (5) is continuous and has derivatives of all orders.xo )n
n-2
2. for x = 0 or x = 2.
and so on. one can see directly that the series becomes
j (-1)"(± 1)" = j ( -0 -0
1/n!
I
1)". (b) Here a" _ (. series (c) converges for all x.1 j < 1. that is. -1 < x . of the function f(x) can be found by differentiating series (5) term by term.. .
. That is.1 ! > 1.1)a"(x .1)"
(c)
Solution
(a) Here a" = n" and from formula (2). Furthermore.1 < 1 or 0 < x < 2. have the same radius of
convergence R as the initial series (5). x . Then
R=lim
1/(n + 1)!1
=limn+1)=x. I
Illm 1 .
If R is the radius of convergence of a power series In_oa"(x . these series for f'(x). for x < 0 or x > 2. that is.
f (x)
na"(x .1'
Thus. The series diverges for j x .. It is more convenient to use formula (3) in this case. For I x .
1
1
E
R
1
limn
Hence series (a) converges only for x = 0 and diverges for any other x.x0 I < R. that is.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 .
Thus. f"(x).
However.
(6)
which holds for every integer k. na"(x .x0)k . The operations in (a).(x .)"
j 3(n + -0
-2
j 3(n + 2)a"-ix".2
Review of Power Series
227
In the process of finding power-series solutions of differential equations.x0)" . subtract. (6) says that we can decrease by k the n in the general term a"(x .o
(b) i a"(x .x0)" =
b"(x . we have to add. all the coefficients of the series must be zero. (b).x0)"-' _
and
(n + 1)a". For example.i b"(x .
In particular.o ". The additional restriction for power series is to perform the operation
within the interval of convergence of all series involved. we have to combine series whose general terms are not of the same
power.xo)" provided that we increase by k the n under the summation symbol.(x . The simplest way to prove (6) is to write out the two series term by term. Thus. x . then
a" = b"
for n = 0. 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. The basic idea behind the change of index is incorporated in the following identity:
"-0
i a"(x . 1.-..
(c) a(x .
. In words.
and equate two or more power series.xa)"+
"-o
(d) If i a"(x . a"(x .b")(x . in addition to taking derivatives of power series.x0)"
-0
-0
(a" . in the first illustration the index n in the summation on the left is replaced by the
.
The reader should note that these changes in appearance of the summations are very much like making a simple substitution in a definite integral. multiply.x0)" = i (a" + b")(x .x0)"
e-o
as"(x . in practice.x0)" _
n-k
a.xo)"
for all x in some interval .xo)
". and vice versa.. 2. if a series is identically equal to zero.xo I < R.5.x0)" + i b"(x .x(. For example. and (d) were performed in one step because the
general terms of the series involved were of the same power. .
(a) "-o
a"(x .xo)"-k.x0)". These operations are performed in a fashion that resembles very much the corresponding operations with polynomials.
-05 .
4'
_0 n!
S. Evaluating f(x). the quotient of two polynomials. For example. where j + 1 = n. as we can see from their Taylor-series expansions.
Let us give some examples of analytic functions. Thus.. a "rational function. .
EXERCISES
In Exercises 1 through 6.. f (xo) = a f'(xo) = 2a2.
(
(2n)!
I)" xm
x"
X. A function f is called analytic at a point x0 if it can be written as a power
series
f(x) = j a"(x .
sin x. the function 3x2 . determine the radius of convergence of the power
series. the Taylor-series expansion of any polynomial function has only a finite number of nonzero terms. Also. Also.228
5 Series Solutions of Soeond-Order Uneer Equations
new index j. . while the function
x2-5x+7
x(x2 . the functions e'. . . and -3. 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. and so
it converges everywhere.. 7
(2n + 3)
." which will be widely used in this chapter.
(x + 1)" _0 n + 1
S'
.xo)"
"-0
(7)
with a positive radius of convergence. f")(xo) = n!a"
Hence.
1. since the derivatives of order higher than n of a polynomial of degree n are equal to zero. . f (x). 3. . Every polynomial is an
analytic function about any point x0. we define the concept of "analytic function. Finally." that is.
"-o
3"x"
3. j 3"(x . 1.7x + 6 is analytic
everywhere. and cos x are analytic everywhere. Within its interval of convergence the power series (7) can be differentiated
term by term. and in general for n = 0. Indeed.. f'(x).9)
is analytic about every point except x = 0. The resulting summation is that on the right
with the index being called n again instead of j. at the point xo we obtain f(xo) = ao. is an analytic function at every point where the denominator
is different from zero.1)"
"-o
2. 2. a" = f"1(xo)ln! and the power series in (7) is the Taylor-series expansion
f(x) =
"-o
n x (x .xo)"
(8)
of the function f at the point x0.
the form of the solutions will depend very much on the kind of point that x0 is with respect to the differential equation.x0)2 Q0(s)
(3)
are analytic at the point x0.(x).1 are regular singular points. where x0 is a real number.
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 .(x)
(2)
are analytic at the point x0.(x)/a2(x) and a0(x)/a2(x) are analytic at every point except where the denominator vanishes. After cancelling common factors. the rational functions a.
EXAMPLE I Locate the ordinary points. the coefficients a2(x).
In Exercises 9 through 15 the student is asked to verify that all singular points of the differential equations of Table 5. and irregular singular points of the differential equation
(x` .x0) a2(x)
and
(x . according to the following definition.1 see Table 5.x2)y" + (2x + 1)y' + x2(x + 1)y = 0. 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. regular singular points. which gives their ordinary and singular points in the finite real line. A point x0 can be either an ordinary point or a singular point.
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.1. As we will see. then x0 is called a singular point of the differential equation (1). 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. a.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 .x0). If at least one of the functions in (2) is not analytic at the point x0. then x0 is called an irregular singular point of the differential equation (1).
(4)
. The points at which the denominator vanishes are singular points of the differential equation. and all other real numbers are ordinary points. If at least one of the functions in (3) is not analytic at the point x0.
(5)
'
a.
Since both of these expressions are analytic at x = 1. and -1 is a regular singular point and which is an irregular singular point for the differential equation (4). For xo = 0. we need to examine the two functions in (3).x2. and -1 is an ordinary point of the differential equation (4).1.5. the two functions in (3) become (x + 1)
2x+1 _ 2x+1
X. The study of solutions near irregular singular points is difficult and beyond our scope.x2
x-1
and since both of them are analytic at x = -1 (their denominators do not vanish
at x = -1).. the two functions in (3) become
(x-1)
+X =x2+1
and
(x-1)Zxzx
X)=x-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. Our aim in the remaining sections of this chapter is to obtain series solutions about ordinary points and near regular singular points.x2
x2(x . Finally.(x)2x+1
a2(x)
2x+1
ao(x)x2(x+1)=
x' .1)
and
(x+1)Zx2(x+1)-(x+1)2
z .
ao(x) = x2(x + 1).x2
x2(x . 1.
and so
al(x) = 2z + 1. for xo = .1
Solution
Here
a2(x) = x° . we conclude that the point x. 1 None 0
±1
Table 5. we conclude that the point xo = 1 is a regular singular point for the differential equation (4).
. hence we conclude that the point x0 = 0 is an irregular singular point for the differential equation (4).1)(x + 1) ' a2(x)
x-1
It follows from (5) that every real number except 0.x2
1
x' . = -1 is a regular singular point for
the differential equation (4). 1 All points All points except xp = 0 All points except x0 = ± 1
Singular points
None
0 ±1
0. For xo = 1. To see which of the singular points 0. 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.
a.
(6)
with positive radius of convergence.x0 I < R. and R2.(x)la2(x) and ao(x)la2(x) have power-series expansions of the form
a..2(x . Our task here is to compute (or approximate) this unique solution. if (6) is the solution of the IVP (1)-(3).. of the series (6) can be obtained in terms of ao and a.
(3)
Let us recall that if the coefficients a2(x). is an ordinary point of the differential equation (1). and R2. The point xo is usually dictated by
the specific problem at hand. The coefficients a" for n = 2. 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. The following theorem describes the form of any solution of the differential equation (1). and R2 are the radii of convergence of the series (4) and (5).3. in particular the form of the unique solution of the IVP (1}-(3).
EXAMPLE 1
Find the general solution of the differential equation
y".
( 4)
and
ao(x) a2(x)
B"(x
for I x
2
(5)
with positive radii of convergence R..x0) "
f or I x
. then the general solution of the differential equation has a power-series expansion about x0. The functions (4) and (5) are in particular continuous in the interval I x .(x) _
a2(x)
A "(x
. by the existence theorem. Finally. More precisely.. if R.. 3. the IVP (1)-(3) has a unique solution throughout the interval I x . x0 is an ordinary point of the differential equation (1) when the functions a.(x). is an ordinary point of the differential equation (1) when a2(x0) # 0.1)y' + 2y = 0
about the ordinary point x0 = 1. In general. by direct substitution of (6) into the differential equation (1) and by equating coefficients of the same power. = y.x0) " . (Solutions about an Ordinary Point) If x.x0 I < R. then the radius of convergence of (6) is at least equal to the minimum of R. and therefore. where R is the smallest of R.
(7)
. then as = y.xo I < R . and aa(x) are polynomials in x.5. a"(x .
THEOREM 1
Y(x) = i. and a.xo)"..4
Power-Swiss Solutions About an Ordinary Point
233
in some interval about any ordinary point x0. Theorem 1 of Section 2.xo I < R
. then a point x(. and R2.
.1)a"(x "_0
na"(x-1)"
1)"-2 = j n(n . and ao(x) = 2. we need the radii of convergence R. (7). It is easier to add the
. Hence.2(x .(x)
Here a. y' should be multiplied by . Writing the expressions in columns is very handy in preparing the series in the summation process.1)"-`
2na"(x . we obtain
y' = >na"(x-1)"-'=
and
y" = j n(n .
(8)
with a positive radius of convergence. will remain unspecified. In fact. and 2y of the differential equation in a column as follows:
Y' = 2 n(n . it should contain two arbitrary constants. solution (8) will converge for all x. Thus.1)a"(x n-2
We are now ready to substitute y. a.1)
and
ao(x) = 2 a2(x)
and so R. = R2 _ -.1)". n-2
1)n-2
2(x . will be expressed in terms of ao and a. The coefficients of series (8) will be found by direct substitution of the series into the differential equation.1). y'. while every other coefficient a a .2(x . Differentiating series (8) term by term. the sum of the three series on the righthand side must be set equal to zero. For the sake of maintaining easy bookkeeping. To find a lower bound for the radius of convergence of the series (8). That is.(x)
and
so(x)
a2(x)
a. and y" into the differential equation (7). As we can see from Eq. (7) has a power-series
expansion about xa = 1. and R2 of the power-series expansions of the functions
a.1)a"(.1)"
2y=2F.
Y(x) =
"-o
a"(x .2(x -1)y'.
Since (8) is the general solution of the second-order equation (7). .5 Series Solutions of Second-Order Linear Equations
Solution By Theorem 1 the general solution of Eq.1) and y by 2. the radius of convergence of the series (8) is also
equal to -.
-0 -0
The sum of the terms on the left-hand sides is zero. Hence.a"(x-1)"= i2a"(x-1)".(x) = 1. because y is a solution of the differential equation (7).1)y' _ -2(x . the coefficients ao and a..1)
na"(x . we write the terms y".(x) _ -2(x ...
a'-(X)
a2(x)
. .
....
(11)
and from (10) we obtain
_
a"'2
2(n ....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.1)
a"
for
(n + 2)(n + 1)
(12)
From Eq.2(x .2 to be calculated once a" is known.3
6!
=
0.
_
as
2-5 8...2(x .......3 ao = 8!
ao
. we can express the coefficients a2...a0.1)y'
(n+2)(n+1)a"...5..2na" + 2a"](x .6..5.. we obtain
0 = (2a2 + 2ao) + j [(n + 2)(n + 1)a". (12) we find
a3=0... That is..4. -0 n-I
Adding the left-hand sides and the right-hand sides of these three equations... Hence.
.....On
2na"(x .....7
23.1)"..... ..3
ae = .2na" + 2a" = 0
for
n = 1.2. In fact..-5 a` = -
ao
a = 0.. With this in mind.. a3. all its coefficients must be zero.
The right-hand side of this equation is a power series that is identically equal to zero..3 ao
2
23....3
22.7....1)'
"-o
=2a2+
.
a3
2 22 a4 =43a2= -43ao= -4iao
2.
2a2+2ao=0
and
(9)
(n + 2)(n + 1)an.......
ab
6.1)"
2y= 2a"(x-1)"=2ao+ j2a"(x-1)"...... of the power series in terms of the coefficients ao and a....... Using Eq.. from (9) we have
a2 = .
(10)
The condition (10) is called a recurrence formula because it allows a".4 ......2(x. (9) and the recurrence formula (10).8... we rewrite the three series above in the
following suitable and equivalent form:
y" = j (n + 2)(n + 1)a".3
6 5 ....3 ...5.2 .
1)2 are two linearly independent solutions of Eq.1) + a2(x .
Hence.
(13) (14) (15)
Solution
Since the initial conditions are given at the point 0.= -
(2n)!
ao.(x)
Thus....
n = 2. = 0.1 and 1 .
n = 1. and ao.(x ..2.(x-1)+ao[1-(x-1)2-2
(x-1)`-263(x-1)6-.
n-o
1-x
x
IxI <1
a . we must compute the radii of convergence of the powerseries expansions of the functions a.1)6 +
=a.. the general solution of the differential equation (7) is
y(x) = ao + a.. Therefore..
REMARK 1
As we expected. the general solution involves the two arbitrary
constants a. the reader can verify that
2a2-a.].
aL.236
5
Series Solutions of Second-Ordor Linear Equations
Thus.=0
.x)y" .(x ..(x)la2(x) and ao(x)/a2(x). the IVP (13)-(15) has a unique solution of the form:
y(x) = i a"x".y' + xy = 0
Y(0) = 1
y'(0) = 1.. series (16) converges at least for j x I < 1. . Hence. Observe that
a'(x)=
a2 (x)
and
-
1
= .Ix". (7).1)' EXAMPLE 2
Solve the IVP
(1 .
and
a. The only singular point of the differential equation (13) is x = 1.1)4 + a6(x . we are interested
in a solution of the IVP (13)-(15) about the point xo = 0. By direct substitution of (16) into (13) and equating coefficients.. 3. the functions x . (22/4!)(x . . and so the point xo = 0 is an ordinary point.1)2 + a.
"=o
(16)
If we want at this time to find a lower estimate of the radius of convergence of the power series (16).(x .
and from (15) we find that
a1= 1..4
Power-Series Solutions About an Ordinary Point
237
and
n a.1)a"x"-' _
(n + 2)(n + 1)a".convergence of the solution (17) is equal to -.2(n .1)a"_1x"
(22)
. In
REMARK 2
general..5.2x"
(21)
. we always have a recurrence formula that we can use to compute as many coefficients of the power-series solution as we please. Of course.a. this is a luxury that is not always possible.
(n + 1)n
From the initial condition (14) we obtain ao = 1. we compute enough coefficients a" of the power-series solution to obtain
a "good approximation" to the solution.1)a"x"-' = 2 n(n
"-2
y"
n(n . In Examples 1 and 2 we were able to compute all the coefficients a" of the power-series solution.1 =
n = 2.
(17)
By Theorem I the radius of convergence of the power-series solution (17) is at least equal to 1. Compute the first five coefficients of the power-series solution y(x) = E..'. However.
1
a2
1
3
1
21
31
n1
Thus... the radius of.3.
Y'(x) = i na"x"-'
na"x"-'
y"(x)=In
"-o
n .oa"x" of the IVP
EXAMPLE 3
y"-2x'y'+8y=0
y(o) = 0
y'(0) = 1. . In fact..I . the solution of the IVP (13)-(15) is
y(x)= ia"x"=
++1
G--x"=e.2x'y'
2na"x` .2
z
a". However. it can be larger.
(24)
From the initial conditions we obtain ao = 0 and a. there is no known closed-form expression for the general coefficients a" for all n.) and (6a.)x are the contributions of the series
(21) and (23) for n = 0 and n = 1..X3 + a4X4 +
=x-4x7+6x4+
In this example. we can compute as many
coefficients as time permits.X + a2X2 + a.l)a"_. = 6.
APPLICATIONS 5. + 8ajx". In fact. Finally.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. Legendre's equation appears in the study of the potential equation in spherical coordinates. Thus.. Then a2 = 0 and a3 = -. from the recurrence formula (24) for n = 2.a and
(n + 2)(n + 1)an+2 .2(n . = 1.
Legendre's Equation
The differential equation
(1 .
"-z
where the terms (2a2 + 8a(. = 0 and so a.4. Hence. a3 = -. + 8a. For example..3. However..x2)y" .
(25)
where p is a constant.2xy' + p(p + I)y = 0. + 8a2.2(n 8a" = 0
for
n = 2. (25) are very important in many branches of applied mathematics.. the potential equation
z z
y2
z
axe + a
+ az .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 . is Legendre's equation. using the recurrence formula.0'
. a2 = -4ao. we find 12x4 2a. The solutions of Eq.
y(x) = ao + a.
and
aa(x) = p(p + 1).
x <1. the point xs = 0 is an ordinary point for the differential equation (25). When suitably normalized (as we shall explain below).(x) = -2x. They appear.F -so 0 a4)2
If we are interested in a solution that is independent of. these polynomial solutions are called Legendre's polynomials.
z = r cos 0. Now we proceed to obtain two linearly independent solutions of Legendre's
equation about the point x. Since a2(0) = 1 * 0.x2
"_o
= -2z(1 + x2 + x° +
)
=-2x2nx <1
and
a (x) a2(x)
.
becomes
a2V
y = r sin 0 sin 4). where 0 is a function of 0 only. for example. in quantum mechanics in the study of the hydrogen atom.4
Power-Series Solutions About an Ordinary Point
transformed to spherical polar coordinates
x = r sin 0 cos 4). Legendre's polynomials are widely used in applications.
.
Using the change of variables x = cos 0 and replacing 0 by y. a. we need to compute the radii of convergence of the Taylor-series expansions about zero of the functions a.
(26)
To find a lower bound for the radius of convergence of the solution (26). (25) about the ordinary point x0 = 0 is a polynomial. we obtain
Legendre's equation (25)..5.x2.$ of the form V = rPO. When p is a nonnegative integer. we find
de2 +cot0d +p(p+1)0=0. = 0.p(p + 1) =
p(p + 1)(1 + x2 + x° +
1 .x2
_
"-0
p(p + 1)xx2'.
1 a2v 2aV 1 a2V cot 08V are + r ar + r2 802 + r2 a0 + . one of the solutions of Eq. Here a2(x) = 1 . We have
a`(x)
a2(x)
2x
. The form of any solution of the differential equation (25) about the point xs = 0 is
Y(x) _
n-0
a"x".(x)la2(x) and ao(x)/a2(x).
.2n + 2)
(2n)!
ao
(42)
a2n+
'
(2n + 1)!
The general solution of the differential equation (40) is
y(X) = F.1)2] x2"+. (p .-0
a2"+. 2.... T2(x). that is.
The differential equation
Hermits's
y" . the point x.
(40)
Equation
where p is a constant.. one of the solutions is a polynomial of degree n. (p2 . two linearly independent solutions of Chebyshev's equation are p2 (p' . When suitably normalized. As we shall see. we obtain a polynomial solution called a Chebyshev polynomial and denoted by T .12)(P2 ...3x are
the Chebyshev polynomials Ta(x). the reader can verify that for n = 1 .2xy' + 2PY = 0. .. For example.2)2] x2
Y. .3) .2n + 2) x2. 2x2 .]
+ a' [.
+ j (.5.(2n .(x). T.1)(P . is Hermite's equation. .2n + 1)
(2n + 1)!
+. expansions in terms of Hermite's polynomials.(2n . = 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.2) .4 Power-Series Solutions About an Ordinary Point
243
Hence.22) ... (40) has a polynomial solution about the point
xo = 0. and T3(x). these polynomial solutions are called
Hermite's polynomials. (p . x.
]
. If we multiply this polynomial by 2"-'.2) (2n)!p .I)" 2"(P . The Hermite polynomials are very important in quantum mechanics in the investigation of acceptable solutions of the SchrOdinger equation for a harmonic oscillator.1. [p2 .
and
2 P(P . respectively. (X) = 1 +
(2n)!
(38)
and
Y2(x) = x +
"-I
(P2 .(x). (36) and (37) that if p is a nonnegative integer.xzn+
y(x) = aol 1 +
(-1)" 2 P(P . Eq.32) .. a"x" = F.. if the constant p is a nonnegative integer. and 4x3 . (2n + 1)!
(39)
It is clear from Eqs. Clearly. Hermite's polynomials are also useful in probability and statistics in obtaining the Gram-Charlier series expansions.
a. By direct substitution of (41) into (40) and equating coefficients..
-0
or
"-0
F. the polynomials 1.
34. y" .x2)y" . from which we delete its center x0 (see Figure 5.
In Exercises 31 through 33. Verify the recurrence formulas (36) and (37). and V(t) = 0. Verify the recurrence formulas (33). assume that L = 20 henries.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. (1 -x2)y"-2xy'+2y=0. (1 .xy' + 9y = 0
32. (1 . In an RLC-series circuit (see the electric circuit application in Section 2.(1-x2)y'-xy'+4y=0
In Exercises 34 through 36.
27.246
5
Series Solutions of SsoondOrder Linear Equations
23. compute the Legendre polynomial corresponding to Legendre's equation.05
farad. compute the Chebyshev polynomial corresponding to Chebyshev's equation.2xy' + 6y = 0 37. R = (60 + 20t) ohms.x2)y" .2xy' + 12y = 0. (1 -x')y"-2xy'+6y=0.
26. Verify the recurrence formulas (42) and (43).xo I < R for some positive number R.y"-2xy'+4y=0
36. 30.
5.
In Exercises 25 through 27.x0 I < R. y"-2xy'+2y=0
35.
28. This set consists of the interval I x . 24.(1-x2)y"-xy' +y=0
33. A deleted interval about xo is a set of the form 0 < I x .1).
25.11. C = 0. Compute a recurrence formula that can be used to
evaluate approximately the current 1(t) in the circuit. The power-series solution of Exercise 15 converges for all x in the interval
-1<x<1.
31.(x)y' + ao(x)y = 0
(1)
in a small deleted interval about a regular singular point x0. compute the Hermite polynomial corresponding to Hermite's equation. 29.
.1). The radius of convergence of the power-series solution of Exercise 18 is at least equal to 1.
in general. (1) is of the form
Y'(x) = I x .x0)"
. and R2.
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. However. and assume that the expansions (2) and (3) hold.
(2)
(x . Then the quadratic equation
x2+(A0-1)X+B0=0
is called the indicial equation of (1) at x0.x0 I < R.x0 I'" i a"(x . Our problem in this section is to compute (or approximate) these two solutions near every regular singular point.
DEFINITION 1
Assume that x0 is a regular singular point of the differential equation (1). ? X2 in case that both roots are real numbers. one of the solutions of Eq. and R2.
3)
with positive radii of convergence R.1
Let us recall that when the point x0 is a regular singular point of the differential
equation (1).x0 < R 2.
"-0
(5)
. we need the following definition.x0)". then the functions
(x .x0) a'(x)
a2(x)
and
(x .x0 )2 ao(x) =
a2(x)
"0
B" (x
for I x .x 0 )"
for I x
. Let X.5
Series Solutions About a Regular Singular Point
247
x0-R
xo
x0+R
Figure 5. where R is the smallest of R.5. 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. the differential equation (1) has two linearly independent solutions in the deleted interval 0 < I x . and X2 be the two roots of the indicial equation
X2+(A0-1)X+B0=0.x0) a 'x) _ a2(x)
A "( x
. Then. Since the point x0 is a singular point of the differential equation (1).
(4)
indexed in such a way that k. its solutions. are not defined at x0.x0)2 ao(x)
a2(x)
have power-series expansions of the form
(x
and
( .x0 I < R.
The following three examples correspond to Cases 1.
-0
(6)
with bo = 1. then
y2(x) = CY.(x)=xx-x2=l-lx
a2(x)
and
2x2
2
2
(x -
xo)2 a2(x) = x2 2x2 = -
1
.
"-o
(8)
with bo = 1.xo I + I x .248
5
Series Solutions of Socond.x2)Y' . (7). then
Y2(x) = I x . where R = min (R R2).(x) In I x .xo I < R.xo 112 i b"(x .xo j"2 F.
"=o
(7)
with bo = 1. (5) are called Frobenius series.
CASE 3
If A. Since a2(0) _ 0. (1) in the deleted interval 0 < I x . The constant C is sometimes equal to zero. and 3 of Theorem 1. Since
(x-x)a.xo I < R is found as follows.xo)".
CASE I
If X. 2.y = 0
(9)
near the point xo = 0.1. Series of the form of Eq. the point xo = 0 is a singular point of the differential equation (9).
CASE 2 If X.xo)". and is valid in the deleted interval 0 < I x . . then
Y2(x) = y.8. (See Exercise 29. First we compute the solution (5). and the method of finding such solutions of differential equations is customarily called the method of Frobenius.(x) In I x .X.x2.(x) = x .X2 # integer. a. as the case may be.)
As in the case of ordinary points.Order Linear Equations
with ao = 1. the coefficients of the series solutions above can be obtained by direct substitution of the solution into the differential equation and equating coefficients. = A2 + (positive integer). and ao(x) _ . I + I x . or (8).xo J
2
b"(x .
Solution Here a2(x) = 2x2. = X2. A second linearly independent solution y2(x) of Eq. b"(x . A second solution can be computed from (6).
EXAMPLE 1
Compute the general solution of the differential equation
2x2y" + (x .xo)". A second solution can also be computed by using the method of reduction of order described in Section 2.
2 = 0.. ?'2.. for
x < 0 or x > 0. and so a second linearly independent solution y2(x) is of the form (Case 1)
Y2(x) = I x I
b"x". We first compute the coefficients a of the solution (10).
The roots of the indicial equation are . Since R.(x) = x j a"x".2 and 1. However. the solution (11) is not defined at x = 0. 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' _
. that is. the two roots of the indicial equation do not differ by an integer.na.5. and we must index them so
that X.
"-a
(10)
with ao = 1.5 Series Solutions About a Regular Singular Point
249
are analytic functions (with radius of convergence equal to -).
X. it follows that the power series in (10)
converges for all x.(n + 1)a"x' 2 =
-na"-ix"-
-y =
"-o
-a"x"-'
(2n(n + 1)a. It is defined (by Theorem 1) in the deleted interval 0 < I x I < oo. = R2 = w.1 =0.. Since X. that is.ao)x +
. and therefore the indicial equation of'the differential equation (9) at the regular singular point 0 is X2 + (2 .1)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. Here AO = 2 and Bo = .
2X2-X. = 1
and
a2 = .a Jx""
0 = (ao ..
"-o
(11)
with bo = 1.X2 = 2. . the point
xo = 0 is a regular singular point of the differential equation (9). + (n + 1)a .2. that is. one solution of the differential equation (9) is of the form
y.
By Theorem 1.2.
+
(x/2)" n!
x"
n!+
= x
1/2
or
Y2(x) = x-v2e'
(13)
Now we must compute the solution (11) when x < 0.=1
and
(91)
Since x < 0 in Eq. we look for a solution in the form
Y2(t) = I t
I -"I i b"t". (9') is the same as for Eq. (9) becomes
2t2y-(-t-t2)y-y=0.252
5 Series Solutions of Second Order Unser Equations
Thus. we have
Y2(t) = t-ia 7. The indicial equation of Eq. Its roots are again
X2--i
We only have to find the solution corresponding to X2 = -2 because for
X. we use the transformation x = -t in the differential equation (9).
x Y2(x)x a(1+2
= x -la
x"
2"n!
x2
22
21+. As before. To do this.. 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.a
x <0. (9). and so
Y2(x) =
(-x)-"ae. Eq. b"t". = 1 we have already found a solution.
"=o
Because t > 0 here.
t > 0. we see that for x > 0 or x < 0.
Butt = -x. we obtain (the calculations are omitted)
Y2(t) = t-'ae-`a.
"-o
Computing the coefficients by direct substitution of y2(t) into the differential equation (9'). we have Y2(x) = I x I-"evz
(14)
. we have t > 0 in Eq.
(13')
Combining (13) arid (13'). (9).
X. (9').
"-o
EXAMPLE 2 Compute the general solution of the differential equation
(x 1)2y. the general solution ofEq. and c2 are arbitrary constants. (13) and (13'). Eq. This was observed in the previous example in Eqs.1. Here AO = -I and Bo = 1.2 l2
-1-t
a2(t) = t2. (9) is
Ax)=c.=]\2= 1. Therefore. Since
t = x .
. the indicial equation is
X2-2X+1 =0
and
1`. . we obtain x = t + 1. It will be very convenient to use the following remark in the sequel. (17) becomes
t2y-(t2+t)9+y=0.
y(x) = I x I" j a"x". we obtain for x > 0 or x < 0. + y = 0
(17)
near the point xo = 1. Thus.xI1+5 7.
Solution
Since computations about the point zero are simpler. Combining (15) and (16)..
REMARK 1
It can be shown that if
y(x) = x"
. = 1'
it follows that to = 0 is a regular singular point.5. we set
t = x . and y" = y. (18). (2n+3)]+c2 X1 1aen
where c.x)y.I and find the general solution of the resulting equation near 0.(x2 . and because
(18)
ta. (1) for x > 0.
The point to = 0 is a singular point of Eq. y' = y.5
Series Solutions About a Regular Singular Point
253
Thus. a"x" "-o
(15)
is a solution of Eq. then
y(x) = (-x)"
a"x"
(16)
is also a solution for x < 0.(t)=t-(12+t)_
a2(t)
and
.
The solutions of Bessel's equation. then it is proved in physics that the temperature u = u(r. For a discussion of this method see Section 11.
1
(28)
Here we assume that the temperature is independent of the height of the cylinder. 0. The important method of separation of variables in solving partial differential equations is also mentioned in the process.p2)Y = 0.=k u. in general. Before we study the solutions of Eq. (28) is a constant that depends on the thermal conductivity and. 6) at any time t satisfies the following partial differential equation (in polar coordinates):
1
1
u. we assume that Eq. called Bessel functions.
u(r. the function R will satisfy Bessel's equation (27). and Tare to be determined. t) = R(r)0(0)T(t). According to this method. 0. Substituting (29) in (28) and suppressing the arguments r.
Bessel's
Equation
The differential equation
x2Y
+
xy'
+
(x: .5.4.
(30)
.
I
In x
I+I. (28) has a solution u(r.
(29)
where the functions R. are very important in applied mathematics and especially in mathematical physics. a function of 0. is Bessel's equation of order p. we obtain
R"OT+rR'OT+ RO"T=kROT'. 0. t) at the point (r. t) which is a product of a function of r. 0.. That is. 0.
(27)
where p is a constant. -.
Temperature Distribution in a Cylinder
If we know the temperature distribution in a cylinder at time t = 0.1
The method of Frobenius is a powerful technique for obtaining solutions of
certain differential equations which occur in applications. +
Y:(x)=12(x2+4+40.
/
I
7!
`
X. + tX x 2 4
/
+
APPLICATIONS 5. and a function of t. the materials used in making the cylinder. We now employ the method of separation of variables to solve the partial differential equation (28). and t. we obtain
Therefore.258
5 Series Solutions of Second Order Linear Equations
Taking bo = 1 and equating coefficients to zero. As we will see below. (27) we show briefly how Bessel's equation appears in a specific application. The quantity k in Eq.
In fact. But
4. we must also solve Eq. say . we obtain
R' = dR dx _ X Y = .
a.
(36)
.. Thus. Hence.IT'
kT
33)
Now the left side of Eq. T' + X2kT = 0
and
(34)
r2R" + rR' + (A2r2 .(x) = x. while the right side is a function of t.p2.(x) = x2 . we find Bessel's equation (27). The only way that this can happen is if both sides of Eq. (33) is a function of r alone. (27) near the point xo = 0.
Since a2(0) = 0. We now return to find solutions of Eq.A2. (35).
a.my = x2 y
r
Substituting R' and R" in Eq. the point xo = 0 is a singular point. we assume for simplicity that p ? 0.
(30) as follows:
rZR" R'r2T'_ R+rR kT
0
(31)
The left side of Eq. while the right side is a function of 8 alone.
(35)
The functions 0(0) and T(t) of the solution (29) are easily found by solving the simple differential equations (32) and (34). (31) is a function independent of 0. (30) in such a way that the variables separate. To find the function R(r) of the
solution (29). both sides of Eq. If we make the transformation x = lAr and set y(x) = R(r). Although p could be a complex number. (33) are equal to a constant. Then we obtain the equations
0" + p20 = 0
and
(32)
R" R'_r2T'_ p: r2R+rR kT
or
R" R +
1R'_E! rR r2
. (35).(x) = x2.y'
dx
dr
r
and
R" =
d (Xy) dx dx dr
.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.p2 )R = 0.X)
z
=1
and
x2 a°(X) :
p + x2.5. say p2. we can write Eq. We have
a. (31) are equal to a constant.
(50)
Choosing ao = 1.4x + x2. (50) we see that if p is a nonnegative integer k.x. As we will see below. 1.(x). when 2p is not an integer. we obtain
(2)2. if the constant p is a nonnegative integer. the coefficients a" vanish for n >.(x)... and L. respectively. a second linearly independent solution of Bessel's equation is of the form (39) or (40). Such solutions of Eq..k + 1.x3 are
the Laguerre polynomials Lo(x). is Laguerre's equation.
(47
)
Thus. L.. L2(x). the functions J°(x) and J_P(x) are linearly independent solutions of Bessel's equation. This polynomial multiplied by k! is denoted by L. near the point x0= 0. respectively..(P-n+1)
(n!)2
X. is a polynomial. the reader can verify that
a"P(P-1).
(51)
and converges for all x.2 = 0.(p-n+1)
(n. (49) where p is a constant.x)y' + py = 0. For example.
J
°(). The Laguerre polynomials are useful in the quantum mechanics of the hydrogen atom. The point x0 = 0 is a regular singular point of the differential equation (49) with indicial equation k2 = 0..)2
Thus.18x + 9x2 . one solution of Laguerre's equation is
Y. and can be obtained by direct substitution of the appropriate series into the differential equation.
Laguerre's Equation
The differential equation
xy" + (1 . Clearly.2. (46) and observing that f(n + 1) = n!. 2 . Setting p = 0 in Eq. From Eq..)l -
(48)
The function J0(x) is a solution of Bessel's equation (27) for p = 0. Furthermore. Since X. 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. and 6 . In this case the solution (51) is a polynomial of degree 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_.(x)=1+
P(P-1). one of the solutions of the differential
equation (49)..
. = 1.
Jo(x) = _ ((n.0n!1'(n-p+1) 2
x_
(-1)"
(X)2--p. the differential equation has a solution of the form
y. JP(x) is always a solution of Bessel's equation (27).(x).0(x) and is called a Laguerre
polynomial.. When p = 0 or when 2p is a positive integer.(x) _
"-o
a"x"
for
-<x <
forn=1. 1 . When suitably normalized.
Hypsrpeometric (52) Equation
where a. and c. b. 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. and c are constants.
F(a. Since c is not an integer.
The two roots of the indicial equation are 0 and 1 ..k.c..x)y" + [c .
(54)
The series solution (54) is called the hypergeomeiric series and is denoted by F(a. b. b. (c+n-1)
n!
(55)
and converges in the interval I x I < 1.aby = 0.
c(c+1). . It is important to note that many functions
can be obtained from the hypergeometric functions for various values of the constants a.. 2.. we obtain the solution
Y1(x) = 1 + ! b_x +
a(a +21c(b(b
1)1)x2 + . 2. b"x".5 Swiss Solutions About a Regular Singular Point
The differential equation
Gauss's
x(1 . -x) = In (1 + x). x).
The reader can verify that
a"
n
= 1.011)=1
limFla. -x) = (1 + x)°
F(k + 1. For example. by Theorem 1. b.b. Eq. the two roots do not differ by an integer.
.c. choosing a° = 1. Thus.b.al =e`
F(-a.(a + b + 1)x]y' ..
x I < 1. b. (52) has two linearly independent solutions of the form
Y#) =
a"x"
and
y2(x) = I x 11-` F. The indicial equation of (52) near the regular singular point x° = 0 is
X2+(c-1)A=0.
(53)
Hence..5. Let us assume that c is not equal to an integer. 1. 1. . x) is a polynomial of degree k fork a nonnegative integer
xF(1. Thus. 3.b. is Gauss's hypergeometric equation or simply the hypergeometric equation. c.
9)Y = 0
has two linearly independent solutions near x = I of the form
27. 24.
In this case.is)Y = 0
near the point x0 = 0 involves a logarithm.y' +4x'y=0.5.1 I +
b"(x . The indicial equation of the differential equation
x2y" .1)".
25.
29.
Y2(x) = Y1(x) In I x I + I b"x".
Y2(x) = Y1(x) In I x . Find
the two linearly independent solutions and in the process verify that C = 0 in formula (8). 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". The indicial equation of the differential equation
x2y" .()3 + x2 + x)y' + (4x + 1)y = 0
at xo = 0 is
K2-2K+ 1 = 0.1)y"+(2-x)y'+y0
has two linearly independent solutions near x = 1 of the form
Y1(x) = i a"(x .
26.(X2 + x)y' + y = 0
at xo = 0 is
.\2-2X+ 1 = 0.
.5
Series Solutions About a Regular Singular Point
265
23. The differential equation
x2y" + xy' + (x2 .1)". One of the solutions of the differential equation x2y" + xy' + (x2 .
28. Show that the point x0 = 0 is a regular singular point for the differential equation
xy" . the roots of the indicial equation differ by an integer. The differential equation
(x.
5.'
'See also Amer.7 are as described by Theorem 1.268
6 Series Solutions of SecwW-Order Unear Equations
as found in Section 2. The above can be used as a motivation for series solutions about a regular singular point. Monthly 67(1960): 278-79. Math.
. Section 5.
xo being the initial time and yo and y.(x)Y' + ao(x)Y = f(x).(x). when the independent variable x represents a space variable (as in the diffusion applications in Section 2. First.3). as we shall see in the
examples and exercises below. constitutes an initial value problem (IVP). constitutes a boundary value problem (BVP). a BVP may have one. The conditions (3) which are given at the end
points (or boundary points) of the interval a s x <_ b are called boundary
conditions. 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 2.1). 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. However. However. the initial conditions. The procedure for solving a BVP is similar to the procedure used in an IVP. In most IVPs the independent variable x of the differential equation usually represents time. or infinitely many solutions. where the coefficients a2(x). (2) The differential equation (1). 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. 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.
y(a) = A
and
y(b) = B. The form of the boundary
conditions at the end points a and b may vary widely. a. For example. ao(x) and the function f(x) are continuous in
some interval a s x s b with a2(x) * 0 in this interval.
(3)
where A and B are two constants. The following examples illustrate this point. together with the initial conditions (2). Consider the second-order linear differential equation
(1) a2(x)Y" + a.CHAPTER 6
Boundary Value Problems
6. There is a large body of theoretical material that has been developed for boundary value problems. Our presentation will bring out some essential features of the basic theory by means of simple yet instructive examples. 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.
. The differential equation (1). none. together with the boundary conditions (3).
1
Introduction and Solution of Boundary Value Problems
271
EXAMPLE 3
Solve the BVP
y"+y=x
y(0) = 2.
for
0:5 xsir
(10)
y(ir) = a .6. or infinitely many solutions. + 7r = IT . This problem. c. = 2.
Y(IT) t.
Solution The general solution of the differential equation (13) is y(x) = c. the BVP (10)-(11) has
infinitely many solutions:
y(x) = 2 cos x + c2 sin x + x.2y'(0) = -2
and
c.2' ..c.
y(O) = 2-:> c. + 6c2 = 3. = 0 and c2 = .2y'(0) = -2. .
(12)
The boundary conditions at the end points a and b need not always be of the type considered in the previous examples. it is very important to know under what conditions a BVP has a unique solution. no solution.2. = 2 while c2 remains arbitrary.2' c.
Of course. cos 2x + c2 sin 2x.
EXAMPLE 4
Solve the BVP
y"+4y=0
for
0sxsrr
(13)
(14) (15)
y(0) . We give a simple example. We have y'(x) _ -2c. c. cos x + c2 sin x + x. sin 2x + 2c2 cos 2x.
Thus. Solving this system we obtain c. Therefore.
(11)
Solution The general solution of the differential equation (10) is
y(x) = c. and using the boundary conditions.3y'(7r) = 3.4c2 = -2
Y(7r) + 3Y'(n) = 3 #. in
. we find that
Y(0) . They can contain combinations of y and its derivatives at the points a and b.
Now. = 2
and
y(7r) = IT . the BVP (13)-(15) has the unique solution
y(x) = 12 sin 2x. In this case.
We only develop here a simple but interesting result about the BVP consisting of the differential equation (1) and the simple boundary conditions (3). 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. if the determinant in (19) is equal to zero. Then the general solution of the differential equation (1) is given by
Y(x) = c1Y1(x) + c2Y2(x) + Y.y. we have the following. and c2 of the general solution should satisfy the following linear system of algebraic equations:
c1Y1(a) + c2y2(a) + Yp(a) = A
c.(b) + c2y2(b) + y.2 y = b2 (18)
has exactly one solution if the determinant of the coefficients satisfies all a12
a21
a22
# 0.(b) = B1
or
c1Y1(a) + czy2(a) = A . Let y.(x)
(16)
In view of the boundary conditions (3).(x) and y2(x) be two linearly independent solutions of the homogeneous differential equation corresponding to the differential equation (1) and let y.(a)
(17)
c1Y1(b) + c2Y2(b) = B . and let y.
(19)
On the other hand.y.272
6 Boundary Value Problems
general.y.(x) and y. Summarizing the facts above. the following statements are valid:
(a) If
Y1(a)
Y2(a)
Y1(b)
y2(b)
.
THEOREM 1
Let y. We recall (Appendix A) that a linear system of the form
a21x + a. the constants c.(x) be two linearly independent solutions of the homogeneous differential equation corresponding to the differential equation (1).(x) be a particular solution of the differential equation (1).(x) be a particular solution of the differential equation (1). is very difficult and beyond our scope.(b)
Now it is clear that the BVP (1) and (3) has as many solutions as the system (17). Then.
2 .
Finally.1).(b) y2(b)
1
0
=0
-1
0
cos a sin rr
and
y.a
1
2
= 0. y2(x) = sin x.(a)
y.1
.
(22)
Let us apply Theorem 1 to the BVPs in Examples 1.(x) = x. and y.(a) A . because
y.(a) y.rr 1 . narrow pipe (Figure 6.1
Suppose that a gas diffuses into a liquid in a long. respectively.(a)
y2(a)
0
y.1r
1
#0.y.(a) y. where y.6.
0
2
The BVP (8)-(9) has no solution.(x) = cos x.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 <.rr
-2
APPLICATIONS 6. The BVP (4)-(5) has a unique
solution because y.(a) A .(b)
2 -0
rr . and 3.1.(b) y2(b)
cos 0
sin 0
sin -w 2
1
1
0
1
cos
IT
=1#0.(b) B .y. the BVP (10)-(11) has infinitely many solutions.y.(b)
y2(b)
cosy sin a
cos 0
-1 0
=0
and
y.(a) y2(b)I
y. 2.
-1
cos.b. 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.y.y. the BVP (1) and (3) has no
solution or infinitely many solutions in the interval a s x s b.r
cos 0 sin 0
1
2
-1 1 . because Icos0 sin0 ly. depending on
whether the determinant
yl(a) A .(a) y2(a)
y.(b)
B .(b) B .yo(b)
is * 0 or equal to zero.(b)
cos 0 2-0 cos. (b) If the determinant in (21) is equal to zero.
c2 = f0L/2T.1.
(25)
Since the ends of the string are kept fixed at the points x = 0 and x = L of the x axis. A taut string.x). y(x) satisfies the BVP
yY(0) = A.
(26)
EXAMPLE 5
Solve the BVP (25)-(26) when f(x) = fo is constant.
Hence.Tu" = f(x). lying on the xy plane. u(x) must also satisfy the boundary conditions
u(0) = 0. answer true or false.
But
and
u(0)=0'c.27x2 = 2Tx(L . + c2x .274
6 Boundary Value Problems
initial point 0 (and is independent of time).
Air) = -1.ZTx2.
EXERCISES
In Exercises 1 through 6.
Deflection of a Taut String
y=0
y(1) = 0. under a tension T. as we proved in Section 2.
1. y(x) = cos 3x .+c2L-2T L2=0. A transverse (perpendicular to the x axis) load f(x) is applied to the string and produces a deflection u(x). 2. and the BVP (25)-(26) has the unique solution
u(x) = 2T x . is stretched.1. c.
(23) (24)
in Exercise 16 the reader is asked to solve the BVP (23)-(24).
Solution The general solution of the differential equation (25) is
u(x) = c.
.
u(L) = 0. Then. between two fixed points x = 0 and x = L on the x axis. Y('n) _ I. = 0.=0
u(L)=0=> c. y(x) = cos 3x is a solution of the BVP
y"+9y=0
Y(0) = 1. Then it can be shown that the deflection u(x) satisfies the differential equation
.sin 3x is a solution of the BVP
y"+9y=0
Y(0) = 1.
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.
16.
(2)
The BVP (1)-(2) is always satisfied by the trivial solution y(x) = 0. the important issue in this type of BVP is to determine values of x for which
the BVP (1)-(2) has a nontrivial solution. is governed by the BVP
y"(x) = 0
Y(0) = c Y(I) = c2 (provided that the diffusion coefficient D is constant). However. especially problems in mathematical
physics.
15x52
y(1) = A. to c2 through the medium. Show that the BVP
x2y"-3xy'+3y=1nx. and the two boundary conditions at the end points a and b are of the form
o1Y(a) + a2Y'(a) = 0
P. Solve the BVP consisting of the differential equation (23) and the boundary conditions (24). respectively.
. Solve this BVP and show that the concentration y(x) changes linearly from c. a is a real parameter.
T = 2. These values of x are called the
eigenvalues of the BVP (1)-(2).
6.Y'(1) = 2. Solve the BVP
1 :5x!52 xzy" . 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. and c2. y(2) .276
6 Boundary Value Problems
14.Y(b) + R2Y'(b) = 0.
17.
18. Solve the BVP (25)-(26) when
f (x) = x. The corresponding nontrivial solutions are called eigenfunctions.
(1)
where p(x) (> 0) and q(x) are continuous in some interval a 5 x 5 b.
15.3xy' + 3y = 0.
2 EIGENVALUES AND EIGENFUNCTIONS
From the point of view of applications. y(2) = B has a unique solution for all values of A and B. The BVP (1)-(2) is also called an eigenvalue problem (EVP).
L = 1.2y'(2) = 4y(1) .
y(rr) = 0
eigenvalue.6. We have
y(0) = 0 ' c. and c2.
EXAMPLE 1
Compute the eigenvalues and eigenfunctions of the EVP
y" + Ky = 0
Y(0) = 0.
(7)
Solving the system of simultaneous equations for c.
c. Since k > 0.
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. and its general solution is
CASE 1
y(x) = c.e-*' + c2e*' = 0.
for
0 s x <. (7). we see that
(8)
y(0) = 0
and
c.e-"x + c2e". = 0
' c2rr = 0
c2 = 0. + c2 = 0
and
(6)
y(rr) = 0' c. and c2. Then
c2 = 0 and the solution (5) is identically zero. = c2 =
0. Remember that we want to find solutions that are not identically equal to zero. + c2x.
CASE 2 \= 0 In this case the differential equation (3) becomes y" = 0.
Using the boundary conditions (4). the solution (8) is identically zero. and so c. we find that c. This implies that the EVP (3)-(4) does not have negative eigenvalues. zero. The characteristic roots of the differential equation (3) are
\.e*') = 0. and
its general solution is
y(x) = c. a-"° # e".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.k2. Hence. = 0.
K < 0 Set k = . + c2rr = 0
Thus.key = 0.
(5)
Next. or positive. we find c. from Eq.(e-" . (6) we obtain c2 = -c and inserting this value into Eq. where k > 0. which implies that a = 0 is not an
. we use the boundary conditions (4) to determine the constants c. Then the differential equation (3) becomes y" .rr
(3)
(4)
y(rr) = 0. So in solving the eigenvalue problem (3)-(4) we have to examine three cases. In fact. the form of the solutions of the differential equation (3)
depends on whether X is negative.
.. In this case..... c.. Thus. . .
We now employ the boundary conditions
(9)
y(o)=0=> c. must be zero. = 0. 2. c. or positive. that is.. cos kx + c2 sin kx. . where k > 0. and k = n.=0
and
y(7r) = 0 c.
.. Then y = c2 sin nx. To find the eigenfunctions corresponding to these eigenvalues. c2 * 0. = 0 and c2 sin krr = 0. . Thus.k is negative. Hence..
(11)
Hence.
(10)
Equation (10) gives all the eigenvalues of the EVP (3) -(4).
for n= 1. zero. cos kir + c2 sin kzr = 0 ' c2 sin kIr = 0.. 2.
for 0sx<1
y'(1) = 0.278
6 Boundary Value Problems
X > 0 Set X = k2... the form of the solutions of the differential equation (14) depends on whether the quantity 4 ..
(12)
y" = sin nx
eigenvalues and eigenfunctions. Recall once more that we try to compute
nontrivial solutions of the differential equation (3). and therefore the eigenvalues are
given by
X = n2
for
n = 1. 2. . hence. up to a nonzero constant multiple (that is. 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. Now k = k2.. 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. 2. c2 sin k7r = 0 if sin kzr = 0 or ka = nrr for n = 1. k = n for n = 1.. 2. . we use (9) with c.. Indeed. the eigenvalues and eigenfunctions of the EVP (3)-(4) are given by
n = 1.
(14)
(15)
Solution
The characteristic roots of the differential equation (14) are . 2. Then the differential equation (3) becomes y" + k2y = 0. a nontrivial solution is possible only if c2 can be chosen different from
zero. and its general solution is
CASE 3
y(x) = c.
X. (16) and (17) for n = 0 give us o = 4 and yo = 1.
(13)
.
.
In Example 1 it is clear that property (20) is satisfied...
THEOREM 1
The EVP (1)-(2) possesses an infinite sequence of eigenvalues
a a2. 1. any function f(x) that is twice continuously differentiable and satisfies the boundary conditions (2) has the following eigenfunction expansion:
f(x) _
(f. The inner product of the functions f and g is denoted by (f. ..
(22)
where series (22) converges uniformly in the interval a <_ x s b..
(18)
y = cos nax
n =0.
n = m. 2. we state a theorem that summarizes some important properties of the eigenvalues and eigenfunctions of the EVP (1)-(2).. g) = 0.. -
+00
(20)
and a corresponding sequence of eigenfunctions
Y Y2..
.
. In the EVP (3)-(4) we found that [see Eq.
DEFINITION 1
Let f and g be two continuous functions defined in the interval a s x <_ b... 2.R
x s b...280
8 Boundary Value Problems
If we observe that Eqs.1.Yoy...
.n2
and
for for
n = 0. . we conclude that the eigenvalues and eigenfunctions of the EVP (14)-(15) are given by the equations a = 4 + n2.
(19)
Finally. To this end we first define the concepts of inner product and orthogonal functions.. g) = J f(x)g(x)dx. (13)]
y = sin nx
for n= 1.
. the eigenvalue and eigenfunction found in Case 2.2. with the property that in the interval a
10. we say that the functions f and g are orthogonal..
if nxm
if
(Y >Ym) =
1. Let us verify (21). g) and is defined by
(f.
(21)
Furthermore.
that is.
When (f.
(24)
(25)
U .. y.. property (22) indicates that if f(x) is a function that is twice continuously differentiable (in other words.c=u = 0
and
u. For example.
(26)
Let us demonstrate this in the case of Eq. which is acceptable since the eigenfunctions are defined up to a nonzero constant multiple. (21) be satisfied. Let u(x. then
f( x) _
2
(f(x)..") we obtain that
r
0. t) denote the temperature in
'See R.
(23)
and this series converges uniformly in the interval 0 <_ x s rr. 2nd ed. f(O) _
f(irr) = 0. see Section 11. we have to multiply each eigenfunction in (13) by (2/7r)".2.+ uYY = 0
(Laplace's equation).
(13.. (24).2 Eigenvalues and Eigenfunctions
281
Computing the inner product (y. initial-boundary value problems which involve one of the following partial differential equations:
u. (New York: McGrawHill. 2
In order that Eq. sin nx) sin nx =
2
(ff(x) sin nx dx sin nx.6. = «:u_
(heat equation)..7).
APPLICATION 6. Thus. f" exists and
is continuous) and satisfies the boundary conditions (4). Assume that the bar is oriented so that the x axis coincides with the axis of the bar.)
With respect to these eigenfunctions.
The Heat Equation
. 1963).
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..
sin nx sin mx dx =
if if
nm
n = m. Fourier Series and Boundary Value Problems. The series (23)
is known as the Fourier sine series' of the function f(x) in the interval 0 s
xsa. V. by the method of separation of variables.1
Eigenvalue problems occur frequently in problems of mathematical physics. Churchill.2. consider an insulated uniform bar of length 1. 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. In particular.
(y.
(wave equation). that is. they appear in the process of solving. ym) =
1o
'r .
.x s 1.
X"
X -X
and
1 T'
a2T
or
X"-AX=O
and
(31)
' . that is. That is.
and Eq.
(33)
X(0) = X(l) = 0. (27). t) = 0. and (28) of the form
u(x. t) in the bar.
u(0. t) = u(1. we look for a solution u(x. t) = X(1)T(t). (30) is a function of x alone.t) = X(0)T(t)
which imply that
and
0 = u(1. and its right-hand side is a function oft alone.
Using the boundary conditions (28). that is. we find
(32)
0 = u(0.4). The
constant a2 is called the thermal diffusivity and depends only on the material of which the bar is made.
(27)
and that the ends of the bar are kept at zero temperature. 0) = f(x).1 or Section 11. t) of the initial-boundary value problem (24).
u(x. (24) becomes
XT' = a2X"T
or
X" X
1 T' a2 T
(30)
The left-hand side of Eq.5. For example. (24) for 0 < x < 1. t) = X(x)T(t).a2AT = 0. To determine the flow of heat u(x. = XT' and u = X"T. Then u satisfies Eq. The only way that this can happen is if both sides of Eq. assume that the initial temperature f(x) in the bar is a known function of x.
0 <.
We have
(29)
u.
(28)
Using the method of separation of variables (see Section 5. (30) are equal to some constant X. we need to know some initial and boundary conditions.282
6 Boundary Value Problems
the bar at the point x at time t.
(27). c.
n = 1. ... (35). t) = exp(-a2n'Tr2l-2t)sinnx
(37)
is a solution of the heat equation (24) and satisfies the boundary conditions (28). n7rx . Y('rr) = Y'('T)
. Since A has one of the values given by Eq. is given by
T (t) = exp(-a2n'ar2l-2t).
n = 1..
(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.(x. . t) to the initial boundary value problem (24). All that we have to do now is to choose the constants c.. the function X(x) of the solution (29) satisfies the EVP consisting of the differential equation (31) and the boundary conditions (33). . . 2. (34). A -1 0<x<-1. we now indicate what remains to be done to find the solution u(x. y(2) = 0
2..2
Eigenvalues and Eigenfunctions
283
Thus. 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).7. y"+\y=0for 0sxs1
Y'(0) = 0.
1
. Furthermore. . Y" + Xy = 0for -2sx<_2
Y(-2) = 0.
1. up to a constant multiple.
EXERCISES
In Exercises I through 8. 2. We leave it to the student to show that the eigenvalues and eigenfunctions of the EVP (31).6.rtz
a=
and
/'
n=1.y=0for0:s xs2
y(0) = 0. .. n = 1 . . 2. The procedure just outlined is carried out in detail in Section 11. Y(2) = 0
4..
(35)
For the sake of completeness.. compute the eigenvalues and eigenfunctions of the
EVP. That is. and (36)..
(34)
XX(x) = sinnIx. y" + Jay = 0for0 <_x s ar
Y(0) = Y'(0). Y(I) = 0
3. y"+1. (33) are given by
nz.
(36)
In view of the equations (29). 2.2. and (28). (32)... each of the functions
u.
n = 1. the solution of Eq.
such computational difficulties can be easily overcome if the accuracy desired is within the capability of the device used. the differential equation y" . is not what is meant by numerical solutions of differential equations.CHAPTER 7
Numerical Solutions of Differential Equations
7. we simply substitute the given value of x into the appropriate formula and compute the corresponding value of y. the differential
equation y" + y = 0 has y = c. matters are not always that simple.4 7. For example. or cos x. The reader perhaps recalls from calculus that many functions do not have antiderivatives. 1+
L
1.e' + c2e-x as a formula for its solutions. our methods have provided us with formulas for the solution.
Our description of determining the numerical value of the solution of a differential equation. This is the
case in the description of the solution of y" . It is natural to adopt the attitude that these
formulas supply all the information about the solution that we need. if we want the value of y corresponding to a specific value of x in the cases above. and hence it is impossible to represent the indefinite integral of these functions by a formula in terms of elementary functions. Even in those cases where the solution is represented by such relatively simple functions as e`. the formula for y may itself impose computational difficulties. sin x + c2 cos x as a formula for its solutions.1
INTRODUCTION
Up to now our treatment of differential equations has centered on the determination of solutions of the differential equation.xy = 0 above. and the differential equation y" . sin x.. the accuracy desired in the numerical value of y may cause some computational problems. Rather. 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. More specifically. In this same context there are many differential equations for which it is impossible
. Of course...1
1
as a formula for its solutions. Unfortunately.5.(3n-2)x3"
(3n)!
+c2x+
2.8 . although it does point out one aspect of the problem. with the present-day proliferation of pocket calculators and the widespread availability of digital computers.xy = 0 has
y=c.. For instance.(3n-1)
(3n+1)!
x3".y = 0 has y = c. Furthermore.
Hence. 1966).
y
(x0.'
In our subsequent discussion.1 exists only in theory (from the existence-uniqueness theorem).1
Introduction
287
to obtain a formula for the solution. The Analysis of Numerical Methods (New York: John Wiley & Sons. Suppose that we
want to know the coordinates of the point (x y). Since our treatment is admittedly introductory and brief. let us imagine that the curve in Figure 7.1
If y is not known in terms of x and the solution curve of Figure 7. we assume that the differential equation in question is of the form y' = f(x. Furthermore. Keller.1 represents the solution to the IVP (1)-(2). In such cases we "solve" the differential equation by applying certain numerical methods. questions arise concerning the accuracy of the results. Any numerical method involves approximations of one sort or another.y)
Y(xo) = Yo. from one point of view there is no loss of generality in making this assumption.Yo)
17
Figure 7. B.7.
(1) (2)
To obtain a feeling for the goal of numerical methods. Isaatson and H. In other words. y). In the case that y is given
explicitly by a formula involving x. how much error is involved in the approximation? Although we present a few
definitions and results concerning the error involved. There are at present a large number of numerical methods for solving differential equations. we touch upon only a few of the elementary methods.
'For some answers to questions about the accuracy of the methods. we concentrate on initial value problems. We note that all differential equations can be written in this form or as a system of differential equations of this form. to avoid cumbersome treatment of integration constants.
. any reasonable attempt
at answering such questions here would take us far beyond our goal of presenting a brief introduction to numerical methods.? The answer to this question is the basic goal of any numerical method. the interested reader is referred to E. knowing x. it is simply a matter of knowing x. Consequently. is it still possible to determine y. Thus we investigate the IVP
y' = f(x.
y)dx .
to approximate y2 for some given x2 (> x.. +
y
1. 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.f(x. and hence
Ax to be known exactly. If it is desired to obtain n points in addition to the initial point.). The Euler method is to approximate Ay by the differ-
ential dy.. Equivalently.2). + h. 2..xo)/n.
x
Rgure 7..
7. where Ox = x.yo.):
X1 = x.
These changes are denoted by Ax and Ay. Thus. then Ox should be quite small (see Figure 7. Naturally.
value x.yo)Ax = yo +
f (xo.y)dx. with x = b and y. . = yo + Ay. 3.yo). .2
.2 EULER METHOD
Referring to Figure 7. 2. We can then approximate y. if this
approximation is to be reasonably accurate. directly from the information given. The sequence of
close to the solution curve. Of course. y. This process can then be case.. then h is given by the formula when plotted. Now dy = (dy/dx)dx = f(x.
i =
(1)
(2)
s. one considers x. this process is carried out in a specified interval with a prescribed spacing between the x coordinates.)Ox = y.288
7
Numerical Solutions of Differential Equations
We will say that we have solved a differential equation numerically on the interval [xo. ..x0). . = xo + Ox and y. an approximation of the solution when x has the (x.1 and a specific value of x. we can write x. y2 = y. . In this One can then use x. we can consider that in transferring from the initial
point (xo. Usually.y._.x0 and Ay = y.). . 3. by approximating Ay.b] when we have obtained a set of points {(xo. yo)(x. is given by the Euler method to be yo + f(xo.
yj = y. n i = 1. + repeated to obtain additional points. y0) to the new point (x y) we have induced changes in the coordinates. n. hopefully lie very (b . Given the IVP (1)-(2) of Section 7. . one can calculate y. The Euler method can be summarized by the following formulas of the coordinates of the points (x. + f(x.2. say
x then y...
although it could be displayed geometrically by plotting each of the points. Thus.0.0.) Thus.0).2.040013). and y.6. we present the intermediate calculations to assist the reader in following the details. the more calculations we require.8. is that the smaller h is.3. (0.x s 1. We choose n = 5. (1.0. (0.1.008). of course. The more points we ask for. (0. it is to be considered that if one were to plot the points of our numerical solution and connect these points with straight lines.2
Euler Method
Yo
Yt
Yx
Ys
h
Rgure 7.
Solution We were not told how many points to obtain.0). Where necessary we rounded numbers
off to six decimal places.3
This solution The "solution" to the IVP is then the sequence of points is usually displayed in tabular form. the collection of points ((0. The results are displayed in Table 7.242857)} 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).)
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.
. but again we skirt the issue since it would take us beyond our
treatment. Even though we are only interested in x.. (There are problems of accuracy associated with round-
off also. h = (1 . Here and in subsequent applications.7.2.0)/5 = 0. the choice is at our disposal. therefore. (See Figure 7. the smaller the value of h will be and so the more accurate our results will be. The price to be paid.0. the resulting polygonal curve would be an approximation to the graph of the solution on the interval 0 <.4.
R ax
'The authors wish to thank Thomas M.4. Green for the development of the flowcharts and the
computer and calculator exercises.7. y). While this is no guarantee of the accuracy of the results. we decrease the value of h = Ox and generally improve the accuracy of the method. Of course.y). we state the following theorem without proof. and n = 5. ay
(. until the computed results [that is. y = 0.1 can be carried out with the aid of a computer or programmable calculator. of and of are defined and continuous on
the rectangle R = {(x. A program for Example 1 would use the initial values x = 0.y).R ay
maxIaf+fafIsL. The error resulting from the use of a numerical method. p-). 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. This is due to the limitation of all computers that can carry only a specified number of digits. such as the Euler method.
With regard to the error in the Euler approximation. Generally. many programmers find this practical rule very useful for a first guess. each time doubling the size of n. another type of error. xo s x s xo+a... called round-off error.
Suppose that the functions f(x. (x. Let M =
max I f(x. to approximate a true solution is called truncation or discretization error.
(=. b = 1.R
such that
maxi -I<.2 A flowchart of the steps involved is shown in Figure 7.
THEOREM 1. This implies that the final digit in any number represented in the computer may be in error due to round-off. tends to become more significant as h gets smaller. one way to obtain a better approximation of the solution is to use
another. As n increases.2
Euler Method
291
The calculations presented in Table 7. if we increase the value of n.y). 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. more accurate method with fewer steps. Let K and L represent two positive numbers
(2. the accumulated round-off error can more than offset the gains made in improving the accuracy of the method. In particular. yo-b s y s y0+b}. We shall do this in the remaining sections..K. y) I and y = min (a. 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). increasing n (decreasing the step size) reduces the truncation error. y).
. However.
12. y'=y2-2x2+1
y(0) = 0
y'=2xy'
y(0) = 1
7. 9.
y' = siny + x
y(0) = 0
5. Write a computer or calculator program to repeat Exercise 1 using (a) h = 0. Compare your result with the actual value. Given the IVP y' = 3xy2.2.2. 5 with the approximate set obtained in Exercise 6.. 3.2 to solve each of the IVPs on 0 s x s 1. 10. 0. use the Euler method with h = 0.3
Taylor-Series Method
293
If E. y(jo) = 0. . y'=a-y-y+1
Y(0) = 1
3.y(0) = 0. The
.9 and in Chapter 5 we demonstrated how the Taylor series can be used to generate series representations of solutions of a differential equation. Repeat Exercise 12 with h = 0. Write a computer or calculator program to repeat Exercise 12 with
h = 0. 2. Compare with the exact results.
7.5. on the interval 0 < x s 1.01 and compare with the results of Exercise 3.7. use the Euler method with h = 0.1.
where a = L(e'K .125..0625. 0. In Exercises 1 through 6.1 and (b) h = 0..
14. y'=x-2y
y(0) = 3
4.3 TAYLOR-SERIES METHOD In Section 2.85).). . and in other cases it is not possible (or perhaps not feasible) to obtain the complete series. Write a computer or calculator program to repeat Exercise 3 using (a)
h = 0. 0._ y.
13. y.1 and (b) h = 0.25. represents the error made in the ith step by approximating the actual solution
y(x. 4. allowing h to decrease until there is no change in the first three decimal places of the approximations for y(1).01 and compare. maintain four-decimal-place accuracy. Solve the IVP y' = x + y.
8.
1._. Use the
Euler method with h = 0. Solve Exercise 6 exactly and compare the actual set of points (x.).
11.
2. with the results of Exercise 5.... Write a computer or calculator program to repeat Exercise 5 using (a) h = 0. y'=3x2-2y
y(0) = 0
6.sah.) by the Euler approximation y.01 and compare with the results of Exercise 1. is printed and not the other points Adjust the program so that only (x._.2. + hf(x.1)12K. = y.
n
In all problems. then
E. on the interval for i < n.25 to approximate y(0. i = 1. In some cases it is possible to generate the complete series.1 and (b) h = 0.
EXERCISES
i = 1.
0) h2 +
= Yo + f(xo..
. For example. In this language the Euler method is a Taylor approximation of order 1.
x . .Y))
x-YO
r-YO
of + afdy
ax
ay dx
= f (xo.
Y(xo) = Yo. Yo) + fr(xo.b].y)}^-o that lie close to the solution curve on the interval [xo. We recall that the Taylor-series representation of the solution y about xo is
Y(x) = i
Setting x = x. + h. Given the IVP
y' = f(x...f (xo. .
Y'(xo) = y'(x)
= f (xo.xo)". .
then the differential equation (1) and the initial condition (2) can be used to determine each y()(xo).. +
h'. and consequently y.
If we assume that all partial derivatives of f(x. = x(..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. Yo) . . : 5 x : 5
(1) (2)
we again wish to obtain a sequence of points {(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.. In general.y) of any order exist at (xo..y). k = 2.
Y("nxo)
= Y(xo) + Y'(xo)h +
L o) h2 + Yi o) h3
2!
+ . we have
Y(x1) = Y(xo + h) =
o
h
(x . 3.. if the Taylor series is approximated by
Y(xo) + Y'(xo)h + .Yo). Yo)
Y (xo)
dx
(Y (x)) ==.
we say that the approximation is a Taylor approximation of order 1.m
Y-YO
= dx(f(x.
+-xo
r-rO
. Yo).
8. Thus. incorporating the intermediate calculations.-. The larger 1 is. (0.)h2
y..-.Y.0).331469)).0. Thus we anticipate that the Taylor approximation of order
.Y.0).) = Y(x.-.
In general. we have
.
Solution We choose h = 0.016).2.
For the Taylor approximation of order I it can be shown that the error is proportional to h'.-.
f(x.)h +
=y._. 0.Y.2. (0.) = so on. (0.161911). and Y"(x. the more accurate we anticipate the results to be.
where Y(x. (1.Y(x.4. In any application of the Taylor method.7. (0. We illustrate the method by repeating Example 1 of Section 7.-.'-.).
+
{Mx.) = y Y'(x. Thus. is the set
1(0.. From the differential equation (3).) + Y'(x.2. = y(x. we obtain
Y' = x2 + y2
y" = 2x + 2yy' = 2x + 2y(x2 + y2)
=2x+2x2y+2y3. 0.2. our numerical solution of the IVP (3)-(4).064154).)
2
Y.-.
We present the results in Table 7. by the Taylor approximation
of order 2 with h = 0.3
Taylor-Series Method
Subsequent derivatives can be obtained in like fashion.)h
h2
+ (2x. 0.6. + 2x?_. + h)
h
and x2. Once again the computations involved increase with increasing 1. on the interval 0 s x s 1.2. one must specify the order I of the approximation.
Y2 = Y(x2) = Y(x. + h) . +
2. Knowing we can then calculate y2 in a similar manner. +
(x2
y. Additional points are obtained in the same way.) =
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.
Y) f(x. y) +fy(x. y)f(x.2 to solve the IVPs on
0sxs1.3
Taylor-Sari" Method
297
2 is more accurate than the Taylor approximation of order 1.
EXERCISES
In all problems. we change steps 4 and 5 as in Figure 7.5. We have only to make minor modifications in our flowchart for the Euler method to produce a flowchart for the Taylor-series method.Y)h [fx(x.Y)f(x. Namely.1
. maintain four-decimal-place accuracy. Of course a higherorder Taylor approximation would be even more accurate.
f(x. The reader will be asked to compare the various methods of this chapter in Review Exercises.
y'=x -y
y(0) = 1
4.Y)1 h2/2
Figure 7.2 to Produce Flowchart for Taylor-Series
Approximation of Order 2.
CALCULATE
4. Y)+fy(x.
y' = siny + x
y(0) = 0
5.Y)h + Lf (x. and therefore the numerical results of each method can be compared to the actual results as well as comparisons with the other methods. Some exercises involve IVPs that can be solved exactly.
y'=x2+y
Y(0) = 0
2.
1.y +f(x.5. The forms presented are those
commonly called the Runge-Kutta methods. Naturally. Changes in Flowchart Presented in Section 7. In Exercises 1 through 8. the higher the order.
y'=x2+er
y(0) = 0
3. the more laborious the calculations.
y' = 3x + 2y y(0) = .7. y)l h2/2
ASSIGN
xt-x+h
y. Special forms of "higher-order
methods" are presented in the next section. use the Taylor approximation of order 2 with h = 0.
y'=xy
y(0) = 1
S.
1 and (b) h = 0. yo) + [f(xo.2) of Exercise 9 and with the exact value of
y(0. [Hint: Calculate y" and y"' directly from the differential equation.4 RUNGE-KUTTA METHODS
In the description of the Taylor approximation of order 2. Repeat Exercise 6 with (a) h = 0.
14. Repeat Exercise 1 with (a) h = 0. yo) f(x(.1 and (b) h = 0. yo)} 6
Based on our knowledge of the Taylor series (or convergent series in general) from calculus.
7. y(0.] 11.
13.
In Exercises 13 through 15 write a computer or calculator program for the Taylor approximation of order 2.2) of this exercise
is to be compared with y(0.1 and compare with the corresponding
results of Exercise 10 and with the exact results (see Exercise 11).' yo)f(xo. Repeat Exercise 9 with h = 0.1
9. yo) f(xo. Also compare with the Eule method (Exercise 12 in Section 7.01. yo) + 2f:r(xo. y(.(x(. yo)l2f(xo.y(xo) + f(xo. yo)) 2
The Taylor approximation of order 3 would have the additional term
{fa(xo. Compare with the results of Exercise 9 and with the exact results. Use the Taylor approximation of order 2 with h = 0.1 and (b) h = 0.01..298
7
Numerical Solutions of Differential Equations
7.2). yo) + fy(xo. Use the Taylor approximation of order 3 with h = 0..
12. Repeat Exercise 3 with (a) h = 0. we know that the accuracy of our approximation improves with
.) + f (xo.
y' = e'' sin y y(0) = 0.2 to solve the IVP
y'=x+y
y(o) = 0
on 0 ss x :s 1. Repeat Exercise 10 with h = 0.2 to solve Exercise
9. For instance. y0)12
+ Mx. yo)(flx0. Compare with the exact results.1 and compare with the corresponding results of Exercise 9 and with the exact values. we wrote
y(xo + h) . 15.
y' = x2y y(O) = 2
8.
10.2).01. yo)h + V .
1957).072509). We present here the Runge-Kutta formulas of order 3 and
order 4. + 4k2 + k. + 1. 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. S. (0. + 12 h.2 to
solve the IVP
y. 0.021371). For a development of these formulas and an analysis of their accuracy. the description of the higherorder terms gets more and more complicated and the associated calculations more profuse. (0. = y.k(2)
k3=hf(x.)
y . in the hope that they help the reader to follow the development. With this value of y.+2k2-k.002667).174273).2. and Kutta generalized and refined the approximations of orders 3 and 4.
EXAMPLE I
Use the Runge-Kutta approximation of order 3 with h = 0. Numerical Analysis (New York: McGraw-Hill.
RRunge first developed the approximation of order 3. y. = hf(xny.
.+h. 0. Kunz.2 of the IVP (5)-(6) on 0 s x s 1 is given by the set
Solution
{(0. we obtain y2 from formulas (1)-(4) by setting i = 1.). 0). 0.4.8. + 6(k.
As before.=x2+y2
y(0) = 0
(5) (6)
on o:5xs 1. 0. (0.
The Runge-Kutta approximation of order 3 is defined by the following collection of formulas:
k.4
Runpe-Kutta Methods
the number of terms retained. 0. Once again we show the intermediate calculations. (1.)
(1)
k2 = hf(x. from formulas (1)-(4) by setting i = 0.7.6. This process is repeated until we have the desired number of points.350721)}. The alternative proposed by these methods involves evaluating the function f at certain judicious points rather than evaluating the specific partial derivatives.3). (0. the reader is referred to K. Our numerical solution using the Runge-Kutta approximation of order 3 with h = 0.y. On the other hand. it is convenient to display our solution (see Table 7.
(3)
(4)
The scheme then is to obtain y. and with x.
y.
(11)
We employ formulas (7)-(11) in exactly the same manner as the Runge-Kutta formulas of order 3. use the Runge-Kutta approximation of order 3 with h = 0. 0).6 in order to produce a flowchart for the Runge-Kutta method of order 4.021360).002667). 0. 0.+14).
Solution
As in the previous 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<_1.072451). We illustrate with an example. (0.6. maintain four-decimal-place accuracy. 0. 0. Use the Runge-Kutta approximation of order 4 with h = 0. (1. In Exercises 1 through
8. y. The Runge-Kutta method is highly accurate (small truncation error) even
when n is relatively small (small round-off error).)
y.4. we demonstrate our results in a table (see Table . + ih. y.+212+21.2 would be changed to those
in Figure 7. = hf(x.
{(0.
EXERCISES
In all problems. 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.
It can be shown that the error involved in the Runge-Kutta approximation of order k(k = 3.7.-. + 212)
14=hf(x.2.350257)}.8. (0. = hf(xi.
Steps 4 and 5 of our flowchart given in Section 7. + Ih.y. 0.. + i 1.) (7)
(8) (9)
(10)
1.+}(1.+h.=y.+1.4).2.)
12 = hf(x. Our solution of the IVP (12)-(13) on 0 :s x <_ 1 using the Runge-Kutta approximation of order 4 is the set
(0.
.
EXAMPLE 2
Solve the IVP
y'=x2+y2
Y(0) = 0
(12)
(13)
and 0 <_ x <_ 1. 4) is proportional to h*. (0. For these reasons it is a widely used method.174090). 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.
and k is constant for
fixed body posture. vo is the initial velocity of the wave.3c.
. t = x. where U is a unit vector in the
direction of V.
(1)
In this equation g represents the acceleration due to gravity (g = 32. This improved the existing world mark by an astounding 2 feet 78 incl. In his article Brearley attempts to disclaim these critics. Brearley.
A novel and entertaining application of differential equations is presented An Outstanding Long Jump in an article by Brearley. The equation of motion in vector form is
M
dt. however. x is a space variable whose axis is
parallel to the direction of flow of the river. The results are rather cumbersome. Magazine 45 (1972). r = (v0 + co)t .(4) . J. where p is air density. the center of mass G of the long jumper describes a path that one can study as a projectile moving through a resisting medium.(V . The direction of D is opposite to that of V. Water Waves (New York: Interscience. so numerical methods might be more appropriate as a means of solution (see Exercises 1 through 4). 'M. "The Long Jump Miracle of Mexico City.es.ca)/ir. Beamon of the United States set a world rewrd of 29 feet 212 inches in the long jump.16 ft/sec2). and D is the air drag on the long jumper. + cgS
vo
3co
gBgB2c2I = 0. Many c. by co = (gyo)"2 (note that co has units of velocity). R.x. V = I V 1. We present here Brearley's model only and recommend that the reader consult the article for Brearley's complete argument. co is related to the initial depth y. V is the velocity vector of G at any instant. and the magnitude of D is given experimentally as kpV2. is an unknown function that is related to the
depth y of the river by the approximation c. 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). c.=Mg+D. Furthermore. Stoker..
'J. Stoker. N. 1957). Hence.308
7
Numerical Solutions of Differential Equations
7.itics attributed the length of the jump to the thinness of the air at Mexico City. After takeoff." Math. D = -pkV2U. g is the acceleration vector due to gravity. and t is time.' derives the
following differential equation:
(vo + co)'-dc (-l
d
.6 APPLICATIONS
Flood Waves
In his study of the behavior of flood waves in rivers. Equation (1) is a Bernoulli equation and hence could be solved by the method associated with that type of differential equation (see Exercise 19).°
During the 1968 O'jmpic Games in Mexico City. S represents the slope of the river bed.
(2)
where M is the mass of the long jumper.
19
. At time t after takeoff. yo = 15 ft. vo = 2. we shall modify the problem by making an assumption that Brearley did not make.pkv(u' + v')"`.
(4)
We also have the initial conditions u(0) = u0.6
Applications
YJ
mg
Figure 7.pku(u' + v')"
M
(3)
dt
= -Mg .984 kg/m'
(at Mexico City)
(7)
(8)
M = 80 kg
(approximate mass of R.
(9)
Equations (3) and (4) are difficult to solve in their present form and hence can be studied numerically. treated as a system.45 m/sec
vo = 4.7
The path of G is considered relative to a two-dimensional coordinate system
(x. Equation (2) can be written as the following system du M = .7. Brearley gives the
values
uo = 9.182 m'
p = 0. However.15 m/sec
(5) (6)
k = 0.
take the following values: B = 500 ft. can more conveniently be solved numerically as indicated in Section 7.4 ft/mile. y) with the origin located at the position of G at t = 0 (takeoff).)
EXERCISES
For the differential equation (1) applied to the Pawcatuck River in Rhode Island. (See Exercises 5 through 8. respectively (see Figure 7. Equations (3) and (4).7). S = 0. Beamon). v(0) = vo.5. u and v represent the velocity components of G parallel to the
x and y axes.
(7).T. Solve Eq. and co = 15.
14.
8.(0) = 22.2 by:
9.15). the Runge-Kutta method of order 3. Using the values given in (5). Observe that the
expression involved in the solution is not convenient for obtaining numerical values for T. the Euler method. 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-
. The differential equation (11) can be solved by separation of variables and
partial fractions. In Eq. solve Eq.(u .
6. 12.
(10)
which is a differential equation involving only the unknown function u. (3) we assume that v = . the Taylor approximation of order 2. so we obtain
M du
= .)
(11)
Assuming that M = 100. and (9). the Runge-Kutta method of order 4. A = 3 x 10-4. the Runge-Kutta method of order 3. (11) numerically on 0 :s t s 1 with h = 0.3. T. 4. the Taylor approximation of order 2.1. = T(0) = 30.pku[u2 + .6 mph. represents the temperature of the medium surrounding the mass.
10. (11) by these techniques. the Euler method.(u . the Runge-Kutta method of order 3.
11. The rate of loss of heat from the mass is given by the differential equation
M
J= -A(T' . 3.
7.2 by:
5. solve the differential equation (10)
numerically on the interval 0 s t s 1 with h = 0.15)2]" 2. and T. the Taylor approximation of order 2.
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. (8).2 by: 1. the Euler method. 2. = 10. With this information and c. (1) numerically on the interval 0 s k s 1 using h = 0. the Runge-Kutta method of order 4.310
7
Numerical Solutions of Differential Equations
mph. T.1.
13. the Runge-Kutta method of order 4. solve Eq.
'C.2 cos a)"
/
(14)
where 0 is the angle that the pendulum makes with the vertical. solve the IVP (12)-(13) numerically on 0 :5 t s 1 with h = 0. G is a universal constant.6 Applications
311
time.A. the following differential equation occurs:
de = .2 by the Runge-Kutta method of order
3. C.2 using the Taylor approximation of order 2. (14) numerically on 0 s t s 1 with h = 0. I is the length of the pendulum.
16. If we set v = du/de.J 1(2 cos e . the following differential equation occurs':
d'u
GM
(1
(15)
In this differential equation r. the equation of the orbit is assumed in the form r = r(e). Segel. Solve the IVP (12)-(13) exactly by separation of variables. 0 are polar coordinates. Solve Eq. Using the values of A and k given in Exercise 14. Astronomy If one takes into account the general theory of relativity when studying planetary orbits. A (a constant) ions per unit volume are produced. and e is a small parameter.u
It
v
16
For simplicity we take GM/h' = 1. and 0(0) = eo = a. (16) numerically on 1 s u s 2 with h = 0. Mathematics Applied to Deterministic Problems in the Natural Sciences (New York: Macmillan.7.01 and v(1) = 1. h is twice the area swept out by the radius vector in unit
time. Lin and L. where the constant k is called the constant of recombination. Physics In the study of the simple pendulum (see Chapter 8). 1974). This leads to the
1VP
dodt=A
. We assume that the gas is initially (t = 0) un-ionized.000 and k = 5 x 10-6.2 by Euler's method.kn'
(12)
n(0) = 0.
15. E = 0. we can write
d'u/de' = v(dv/du) and the differential equation (15) takes the form
dv _ (GM/h')(1 + Fu') .
17.
(13)
If we take A = 100. M is the mass of the sun. If we assume that g/1 = 2 and at = 7r/6.
. Positive ions and electrons recombine to form neutral molecules at a rate equal to kn2. solve Eq. g is the acceleration due to gravity. u = r-'. compare the numerical results of Exercise 14 with the actual results. and for t > 0.
show that
y
43x
11.6
Applications
313
6.2 if it is given that v = 700 cm/sec when x = 0. KIR = 100. 1968). Use the results of either Exercise 6. or 9 to determine whether or not the relationship in Exercise 10 is valid for all t 1.2. and R = 2 10-5. an approximate equation for free oscillations of such a resonator is of the form
.
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. 7. solve this equation on the interval 0 <_ x s 1 by the Runge-Kutta approximation of order 3 with At = 0.
Exercises 13 through 15 relate to the following initial value problem (see Example 2. 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.. Uno Ingard. Setting v =
(E)
rewritten in the form
d and vdx = drx in Eq.
.-(D). 883. Theoretical Acoustics (New York:
McGraw-Hill. 8.2. 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. Morse and K.5):
dy
(4 -y)[(x cos x) In (2y . Section 1.8) + 1]
x sin x '
dx
(F)
Al) = 2
(G)
'This is Exercise 3 in Philip M. Reprinted by permission of McGraw-Hill Book Company.d'tx 4-P-li +pldtl2J+Kx=O
12. (E). Solve the initial value problem (C)-(D) on the interval 1st :s 2 by the Euler method with h = 0. If (x(t). p.2.y(t)) is a solution of the initial value problem (C). 9. show that this equation can be
dv__1 /
dx
R Ix + v(1 + 13v2)
rRly
Taking R = 900/(K/R)..2. 7.7.
If we account' for the nonlinear dissipative effects in the neck of a Helmholtz cavity resonator.
10.
15.2 and compare the
approximate results with the exact results. 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 = 0.314
7
Numerical Solutions of Dlferential Equations
13.2 and compare the approximate results with the exact results.
14. Solve the initial value problem (F)-(G) on the interval 1 s x s 2 by the
Euler method 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.
.2 and compare the approximate results with the exact results.h = 0.
Also. Fortunately.
(2)
. to approximate them by linear differential eq rations. Our aim in this chapter is to present some elementary qualitative results for nonlinear differential equations and systems. x2)
X2 = f2(t.
8. separable.CHAPTER 8
Nonlinear Differential Equations and Systems
8. The questions raised above belong to that branch of differential equations known as qualitative theory.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. x2. For example. We are often primarily interested in certain properties of the solutions and not the solutions themselves. x2). the situation is not as hopeless as it seems. The same is true for nonlinear systems. exact) can be solved explicitly. 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. the coefficients and initial conditions of dif-
ferential equations are often determined approximately.
(1)
subject to the initial conditions x1(to) = 4
xx(to) = xi. 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. only a few types of nonlinear differential equations (for example. x1. homogeneous. 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. However.
(t. y1.
and the initial conditions (4) are equivalent to
x2(to) = y' (7) x. x2) satisfies a Lipschitz condition (with respect to x. -
(3)
(4)
(5)
In fact. and L2 such that
If(t.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. x1.
(Existence and Uniqueness) Let each of the functions f.
the differential equation (3) is equivalent to the system
Xl = x2
(6)
x2 = f(t. x) be continuous and satisfy a Lipschitz condition (with respect to x. The constants L.
DEFINITION 1
Let f(t. Y Y2)I < L. y). Y.
subject to the initial conditions
Y(0°) = yo. x1.(to) = y°.316
8 Nonlinear DHfrential Equations and Systems
The same theorem applies to second-order differential equations of the form
y = f(t. Ix. and x2) in the region 9t defined by Eq.t°I s A
(t.Y2I for all points (t. 9(to) = Y. x1. x2). I + L2 Ix2 . x x2) and f2(t. and x2) in 91 if there exist constants L. x2) be a function defined in a region 9R of the three-dimensional Euclidean space R3. Then the PVP (3)-(4) has a unique solution defined in some interval a < t < b about the point to. x2).y. When the partial derivatives of/ax. and df/8x2 exist and are continuous. it can be shown that f satisfies a Lipschitz condition.f(t. y2) in 9t. x1. Before we state the existence and uniqueness theorem we explain what is meant by a Lipschitz condition. and L2 are called
Lipschitz constants. x x2):
I x1 . . (8) with x° = y° and x112 = y'. x1. x x2) . and x2) in the region
THEOREM 1
= I
t .
. x x2) be continuous and satisfy a Lipschitz condition (with respect to x. if we set
y = x1
and
y = x2. We say that f(t.
COROLLARY 1
Let the function f(t.(t.
EXERCISES
1.8.n + x2 does not satisfy a Lipschitz
condition (with respect to x. (1) where the functions f and g are not time-dependent is called an autonomous system.
How about in the region
0sx. and x2) in the region
-1 sts1. x x) = x.
respectively. y°.
has a unique solution?
3. Y). 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. This result is very important in physical applications.
What is the largest error we make in evaluating the solution y(r)?
8. Second-order differential equations of the form
z = F(x.
2.
0:5 x2s 1?
-1 <t<1.
3sx.51.3 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. With the hypotheses of Theorem I it can be shown that the solutions of system (1) depend continuously on the initial conditions (2). because we can never measure the initial conditions exactly. 54. Show that the function f(t. "small" changes in the initial conditions (2) will result in small changes in the solutions of system (1).
Y=y+yia
Y(10) = y°
0<_x2<1. z)
(2)
. For what points (t°. Y = g(x. Y).3 SOLUTIONS AND TRAJECTORIES OF AUTONOMOUS SYSTEMS A system of differential equations of the form
x = f(x. y') does Corollary 1 imply that the IVP
Y(ro) = Y. Therefore.
y. sin t) trace out the unit circle x2 + y' = 1 an infinite number of times. y)
The solutions of Eq. As this
in the interval -.oo < t < oo is a solution of the system
= -y. In fact. Y) x = Y. while a trajectory is a curve in the phase plane that is described parametrically by a solution of system (1). y)
(6)
f(x. we can get a lot of information about the trajectories of a system even when we cannot find the solutions explicitly. if (xo. y. This was
actually done in the example above.
(5)
This solution is a helix in the three-dimensional space t. the solution (3) defines a curve in the three-dimensional space t. Another solution of system (5) is x = sin t. To see this.
y. (6). while an explicit solution of the system is impossible. With the assumptions on f and g made above. y = sin t. yo) E R. the parameter t from the two equations in (3). And. . in some cases we can find the trajectories explicitly. (6) give the trajectories of system (1) through the points of D. If we can solve Eq. f(x. x = cos t. y(t)) trace out a curve in the xy plane called the trajectory or orbit of the solution (3). x. Even
. then we obtain explicitly the trajectories. y) 0. the trajectory of this solution is the circle x' + y' = 1 in the xy plane. we can find the corresponding trajectory by elimi-
nating.
Y(to) = Yo
(4)
Clearly. Let us emphasize again that a solution of system (1) is a curve in the three-dimensional space t.318
8
Nonlinear DUtsrential Equations and Systems
can be also written as an autonomous system by setting
y = F(x. For the sake of existence and uniqueness of solutions. For example. the xy plane is usually called the phase plane. y) is called a phase of the system and. However.2. If we regard t as a parameter. Then. Thus. 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. assume that in some region D. therefore. the pair (x. y = . we obtain the first-order differential equation
dy = dx
dy dt dt dx
g(x. and satisfying the initial conditions
x(to) = xo.< t < oo. If we know a solution (3) of system (1). yo) is any point in 9t and if to is any real number. In the study of physical systems. if possible. using the two equations in (1) and the chain rule. defined in some interval a < r < b which contains the point t0. it can be shown that system (1) has exactly one trajectory through any point (xo.
the points (x(t). different solutio is of a system may have the same trajectory. Then by Theorem 1 of Section 8. as t varies
x' + y' = 1.cos t. x. x.
y=X. there exists a unique solution
x = x(t). the points (cos t. then as t varies in the interval a < t < b. The trajectory of this solution is again the unit circle
shows. Y = y(t) (3) of system (1).
EXAMPLE 1 Find the critical points and the trajectories of the following system of differential equations:
x= -y. yo) is a critical point of system (1). (6) nor Eq. y) to comput. (0. are of special interest. Eq. Thus. (12). draw the trajectory corresponding to the solution of (10) with x(0) = 1. (6). if x represents the position and y the velocity of a particle moving according to the differential equation (2). Such points are called critical points (or equilibrium points) of system
(1). 0) is the only critical point of (10). yo) in the xy plane (the phase plane). we can utilize the first-order differential equation dx f(x. The trajectory corresponding to this solution is the single point (x0. y(t) =
yo is a solution for all t. it follows that x = xo is a position of equilibrium of the motion. and hence
(9) x(t) = x0. (0. The points (x0.
Y = F(x. if g(x. Thus. For such points neither Eq. As we mentioned above. Similarly.
(7) is valid.8.3
Solutions and Trajectories of Autonomous Systems
319
if we are unable to find explicit solutions.the trajectories through the points of S. x = 0. y(t) = 0 is a solution of (8) for all t. y0) is a critical point of (8). Eq. y) (7)
dy g(x.
initial conditions
y=x. (6) to compute the slope of any trajectory through a point of D. Thus. we can use Eq. Then
x = y.
y(O) = \.
(10)
In particular. we see that the trajectories of (10) for y * 0 are the solution curves of the
differential equation
-= -x' dx y
y#0. yo) in the phase plane. With such an interpretation.
(11)
Solution The critical points of (10) are determined by the two equations
-y = 0. 0) is also a trajectory that is contained in Eq. (12) for c = 0. we see that critical points are identified with positions of equilibrium.
The general solution of this separable differential equation is
xz+y'=c'. y) * 0 in some region S. then x(t) = x0. which is a one-parameter
family of circles centered at the origin. where both functions f and g are zero. (2) can be easily put in the form of system (1)
by setting z = y.
(12)
For y = 0 we see from (10) that x = 0. if (x0. Hence. y)
(8)
If (x0. gives all the trajectories of system (10)
. Using Eq. it follows that yo = 0. Clearly.
Section 8. From Eq. For example. x = x.z=y.
x(0) = 1.z= -x.1). 7.1
(see Figure 8.
x(0) = 1. when y > 0. x = x. y = 2y. the first equation in (10) implies that x < 0.
2y
3. y = -2y. find the critical points and the trajectories of the
system. 1 = -x.y= -x
x(0) = 1.y=Y
2. x(0) = 1.
4.
y(0) = -1
Y(0) = 1
6. compute the trajectory of the solution of the IVP.
EXERCISES
In Exercises 1 through 4.
y(0) = 0
Y(0) = 0
8.320
8 Nonlinear Differential Equations and Systems
Rgure 8. Indicate the direction of increasing t. (12) we see that the trajectory of the solution of the IVP (10)-(11) is the circle x2 + y2 = 4.Y.it =x. This means that x(t) decreases (and so it moves in the counterclockwise direction). y = -x . x = x.
1.5 contains many worked-out examples in which we find the shape of all trajectories of linear systems. The arrows on the trajectories indicate the direction of increasing t and can be found from system (10).
. centered at the origin with radius equal to 2.y=x
5. y = x + y.
In Exercises 5 through 10.z=y. Indicate the direction of increasing t.
y = g(x.
(1)
We recall that a point (xo.Yo)2 < e
(4)
for allt>-0. 19.y= -4sinx
15. is called stable. the solution (2). 0) is the only critical point of the system
X=ax+by. Since the derivative of a constant is zero. y(0)=1
x(o) = 1. 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. y(t)) of (1) which at t = 0 satisfies
1x(0) .x=y.x= -x+y.4
Stability of Critical Points of Autonomous Systems
321
9.y= . i = -y. transform the differential equation into an equivalent
system (by setting x = y) and compute the equation of the trajectories.y= -x . or the critical point (xo.
if and only if ad . it follows that if the point (xo. A. z + sin x = 0
14. we give the following definitions.be * 0. z + x = 0 16. yo) is a critical point (or an equilibrium point) of system (1) if f(xo. More precisely. x(0)= -1. y(o) = -1 In Exercises 11 through 14.x + x' = 0
In Exercises 15 through 18.8.
8. x = -x2 + y2.xo)2 + [y(t) . A.
11. every solution (x(t). (a) Show that the point (0.x0. find the critical points and the equations of the trajectories of the solutions of the system.
y=cx+dy
(b) Show that the system has a line of critical points if ad .x=x-xy. yo) = 0. x .
DEFINITION 1
The critical point (xo. r = x. If this is true. yo).x0)2 + [Y(0) .be = 0. 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. then the pair of constant functions x(t) .4 STABILITY OF CRITICAL POINTS OF AUTONOMOUS SYSTEMS
Consider again the autonomous system
z = f(x.Yo)' < 8
exists and satisfies
(3)
NO .y.x = 0 18. yo) is a critical point of (1). y = try
12. y = -sinx
13. Y(t) ° Yo (2) is a solution of (1) for all t. yo) = 0 and g(xo.y+xy
17.
. x . x = y.
10.
(x0. The concepts of stable.322
8
Nonlinear Differential Equations and Systems
DEFINITION 2
A critical point (x0. while instability means that a small change in the initial conditions has a large effect on the solution. (b). asymptotic stability means that the effect of a small change tends to die out. 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. every
solution (x(t).2(a). Is it asymptotically stable?
Y=x
(7)
Solution We shall apply Definition 1.Yo)
x
(a)
(b)
(c)
Figure 8.
.
(5)
(6)
DEFINITION 3
A critical point that is not stable is called unstable.c2 cos t.x0]2 + [Y(0) . sin t . and (c).
Y.
is stable. Choose 8 = E. Roughly speaking. 0) of the system
x= -y.
lim Y(t) = y0. y(t)) of (1) which at t = 0 satisfies [x(0) . and unstable critical point are illustrated in Figure 8.2
EXAMPLE I
Show that the critical point (0. cost + c2 sin t y(t) = c. Every solution of (7) is of the form
x(t) = c. respectively. Let e > 0 be given. asymptotically stable. stability means that a small change in the initial conditions causes only a small effect on the solution.Yo]2 < au exists for all t ' 0 and satisfies
lim x(t) = x0.
it follows that the critical point (0. 0) of system (7) is stable. 0) of system (9) is asymptotically stable. (c) The critical point (0. b = -1. it is not surprising that the stability character of the critical point (0. the solutions of system (11) are of the form
x=Ae". An effective technique in studying the auton-
.324
8 Nonlinear Differential Equations and Systems
which cannot be true for all t ? 0. both roots of Eq.be = 0. Since a transformation of the form X = x : xo. (12) is real and positive or if at least one of the roots has a positive real part. 0) as a critical point. we have the following theorem. (12) are real and negative or have negative real parts. the character-
istic equation is (here a = 0. Hence. 0) of system (11) depends almost entirely on the roots of Eq.
When the autonomous system (1) is linear with constant coefficients. Its roots are 1.
y=cx+dy.
(11)
where a. Its roots are X.] As we explained in Section 3. using Theorem 1(c) it follows that the critical point (0. (12) are real and negative or have nonpositive real parts. In Example 2. b. b = 0. = X2 = -1 (a double root). Since they are real and negative.yo transforms the autonomous system (1) into an equivalent system with (0. 0) of system (11) is stable if. We shall assume that ad . we can obtain the solution explicitly. 0) is the only critical point of (11). Since one of the
roots is real and positive. Then the
point (0. = 1 and a2 = -1. using Theorem 1(b) we see that the critical point
(0. Y = y .
y=Be". that is. the characteristic equation is (here a = -1. 0) is an unstable critical point of the system (10). b = 4. (0.
The three examples considered were all linear.3. In Example 1. 0) of system (11) is asymptotically stable if.(a + d)X + ad . 0)
of system (11) is easy to study.
(12)
where X is a root of the characteristic equation
X 2 . we assume. 0) is a critical point of system (1). and d are constants.
In Example 3. c = -2. c = 0. both roots of Eq. Therefore.. [See Exercise 19(a) in Section 8.1 = 0. 0) of system (10) is unstable. 0) of system (11) is unstable if one (or both) of the roots of Eq. yo) is a critical point of the system. when
X=ax+by. In fact. Its roots
are ±i. without loss of generality. and only if.be f 0. and d = 0) X2 + 1 = 0. c = 1. using Theorem 1(a). that (0.
The stability character of the critical point (0. Consider again the autonomous system (1) and assume that (x0. c. and only if. Since they have zero real parts.
THEOREM 1
(a) The critical point (0. (b) The critical point (0. and
d = -1) a2 + 2X + 1 = 0. the characteristic equation is (here a = -3.3. and d = 3) X2 . (12).
be # 0 and F(0. furthermore. y)
x2 + y2
. with ad . 0) is to approximate it by a linear system of the form (11). G(x.
EXAMPLE 4 Show that the critical point (0. y)
(. 0) = G(0. The result that we are about to state is of this nature. 0) of the nonlinear system
z= -x+y+ (x2+y2)
(15)
Y= -2y-(xz+yz)aa
is asymptotically stable. = -1 and k= = -2. the functions F and G are continuous.8. and ad .
Clearly.be =2 # 0. Hence. the point (0. Its roots are
k.Y2). 0) of the "linearized" system (11) is asymptotically stable. The linearized system is
x= -x+y. 0) of the nonlinear system (13) is asymptotically stable
if the critical point (0.
F(x. the condition (14) means that the linear system (11) is a good approximation of system (13).
(16)
The characteristic equation of system (16) is Xz + 3X + 2 = 0. 0) = G(0.0
G(x. and that
lim F(x.
THEOREM 2 (a) The critical point (0. 0) of the nonlinear system (13) is unstable if the critical point (0. y) = . (b) The critical point (0.
y= -2y. the point (0. Assume that system (1) is of the form
i = ax + by + F(x. have continuous first partial derivatives. y = cx+dY +Gx y). 0) = 0. 0) is an
asymptotically stable critical point of (16). d = -2. the linear system (11) has solutions which themselves are "good" approximations to the solutions of system (1).
. the conditions (14) are satisfied. 0).
(14)
[Roughly speaking.(x2 + y2)'a. 0) is also an asymptotically stable critical point of the nonlinear system (15). In many cases one can prove that if the approximation is "good". that near the origin (0. 0) is just a
stable point of system (11).] Then the following result holds. Since they are both negative.0. Y) ryl
(13)
. 0) is a critical point of (13). c = 0.4
Stability of Critical Points of Autonomous Systems
325
omous system (1) near the critical point (0.
This theorem offers no conclusion about system (13) when (0. 0) = 0. 0) of system (11) is unstable. [Hence (0. by Theorem 2(a). y) = (x2 + .lim
i-.] Assume. b = 1. and F(0.
Solution
Here a = -I.
0) of the
system is stable.c= -2. 0) of this system is asymptotically stable and therefore both populations are headed for extinction. Assume that the system
5.z=5x-6y. 0) = 0. the point (0. Show that the critical point (0.y= .b =4. F(x.y= -2y. the point (0. asymptotically stable. 0) of each of
the systems in Exercises 1 through 3 is stable.y= -x 2. We express
y --).sin20)
x2 + y2
and
x and y in polar coordinates: x = r cos 0.x=x-y.
y= -2x+3y. G(x.y
3. y = r sin 0.y=5x-y
8.y=4x-y
11. with F(0.x= -y. The linearized system is
z= -3x+4y.x . y) = -xy.0
G(x. By Theorem 2(b). 0) of the nonlinear system
x=-3x+4y+x2_y2
-2x+3y-xy
is unstable.y=6x-7y 10. Since one of them is positive. or unstable.x= -3x+5y.x=3x-2y.y2. where x is the desirable population
and y is a parasite. y)
x+y2
r
=
Hence.x= -x+y. x= -x+y.
6.
1. 0) is also an unstable critical point of the nonlinear system (17).x=y.1 = 0. y) = x2 . 0) is an unstable critical point of (18). 0 is equivalent to r -+ 0) F(x. determine whether the critical point (0.
EXERCISES
By using the definition.0 and r
r2 cos 0 sin 0
= r cos 20
+0
as r .
(18)
The characteristic equation of system (18) is k2 .y.y=x+3y 7. determine whether the critical point (0. Its roots are as = 1 and \2 = -1. Then (x .
Solution
(17)
Here a = -3.y= -x+y
x=4x-6y.y= -x
4.
. 0) = G(0. and d = 3 with ad .
In Exercises 4 through 10. or unstable. z= -3x+4y.be = -1 4
0. y) _ r2(cos2 0 .
y=8x-lOy
represents two competing populations.326
8 Nonlinear Differential Equations and Systems
EXAMPLE 5 Show that the critical point (0. asymptotically stable.y= -2x+3y 9. the conditions (14) are satisfied. x=x .
x=
-x-x2+y2. p2 > 4q. y = 0.y= -y+xy
17.
where X is an eigenvalue of the matrix
(c
y = Be'.y2)5
15. by means of examples. if possible.]
In Exercises 13 through 17. and investigate the stability character of the resulting system at the critical point (0.6y + 1.q>0.
Y = y .y=6x-7y-xy
14.p2=4gandp<0 24. 0) is the only critical point of system (1). x= y+x2-xy. all possible phase portraits of the linear autonomous system
x=ax+by.andp>0 21.y= -y+2xy
Transform the differential equation t + pi + qx = 0 into an equivalent system by setting x = y.p2<4gandp<0
19. q > 0. and this is equivalent to assuming that
ad . p2<4gandp>0
25. x = 3x . In this section we first present.
.
18.x=5x-6y+xy.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.
13. the solutions of system (1) are of the form
x = Ae".x=x-xy.7y + 1. 0) by means of X = x . Show that the critical point (1.1. whether it is asymptotically stable or unstable. p2>4q.
which appears in Turing's theory of morphogenesis is asymptotically stable.
cx+dy
(1)
near the critical point x = 0. andp < 0
20.y + (x2 .3.p=0andq>0
8. We shall assume that (0.
d).
[Hint: Transform the critical point to (0. 0) in each of the following cases. 0) is a critical point for the given system and determine.5
Phase Portraits of Autonomous Systems
327
12.y= -2x+3y+y2
16.be * 0.8.1. 1) of the system
x = 5x . p2>4gandq<0
22.2y + (X2 + y2)2.
y = 6x . (See Exercise 19. show that the point (0.3. y = 4x . Section 8.p2=4gandp>0 23. As we explained in Section 3.
In this case we obtain two distinct portraits. that is.. that is. (2). ] There are five different cases that must be studied separately.=k+il. [The assumption ad .\.
X2 = -il
with
I * 0. k is a root of the characteristic equation
A2-(a+d)>.2=k-il
with
k<0
or
k>0.
>.
It is sufficient to illustrate these cases by means of specific examples.
y=x-2y.
A2<A.>X.
(3)
.
CASE 1
Real and distinct roots of the same sign.
CASE 3
.)
Complex conjugate roots but not pure imaginary.<0
CASE 2
or
A.
EXAMPLE 1
(A2 < \. < 0) Draw the phase portrait of the linear system
x= -2x+y. depending on whether or not the
ac
is equal to zero.
A.
1`1 = il. = A2 = A
matrix
with
A<0
or
A > 0.
Real roots with opposite sign.
(2)
The phase portrait of system (1) depends almost entirely on the roots \.
CASE 5
i. (2).+ad-bc=0.<0<A2.
Equal roots. that is.
Pure imaginary roots. and A2 of Eq. 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.
CASE 4
d-b
1.328
8
Nonlinear Differential Equations and Systems
that is.>0. that is.be * 0 implies that A = 0 is not a root of Eq. that is.
First.. it is clear that every trajectory of system (3) approaches the origin (0. + 0 and c2 = 0. for c. for c. y = x > 0. and c2.3. y = c. From (6) with c2 > 0 and c2 < 0.e-'. Furthermore.5
Phase Portraits of Autonomous Systems
329
Solution Here the characteristic roots are X.8.3 shows a few
Figure 8.
and when c. = c2 = 0. all the solutions (5) have the common trajectory. We want to find the trajectories of all the solutions given by (4) for all different values of the constants c. + c2e-3. we have the solutions x= -3% y = -c2e-3'
(6)
For c. To obtain the other trajectories explicitly. except the pair y = -x > 0 and y = -x < 0. These four trajectories are half-lines shown in Figure 8. all the trajectories of system (3). approach the origin tangent to the line y = x.e-'.3. The general solution of system (3) is x = cle-. < 0. When this approach is complicated (as in this example). 0) as t approaches infinity.e _ c2e-3. respectively. Figure 8. The arrows on the half-lines indicate the direction of motion on the trajectories as t increases. y = 0. we must eliminate t from the two equations in (4) and investigate the resulting curves for all nonzero values of the constants c.e-' . we can still obtain a good picture of the phase portrait of the system as follows. we find the solutions (5) Y = c. all these trajectories approach the origin tangent to the line y = X. (4)
where c.e ' + c2e
'
c. and c2. y = x < 0. we have the solution x = 0. When c. > 0. from (4). x = c. we also find the trajectories y = -x < 0 and y = -x > 0. whose trajectory is the origin (0. _ -1 and K2 = . Similarly.3. 0) in Figure 8. = 0 and c2 # 0. In fact. + c2e ''
-*1
asr ->m.c2e-3' c . all the solutions (5) have the same trajectory.c2e-u
x
c. we have y c. and c2 are arbitrary constants.3
.
Thus. When c. # 0 and c2 # 0.
the trajectory of the solutions (10) is the half-line y = x > 0. except that the direction of the arrows on the trajectories is reversed and the trajectories approach the origin as t . (10) For c. as in Exercise 1.
x = c. The arrows on these trajectories point away from the origin.
y=2x-2y. approach infinity as t -. The arrows on these trajectories point toward the origin because both x and y approach zero as t . * 0 and c2 = 0.
For c2 > 0.
. This observation follows from (9).. we obtain the solutions
x = c. Figure 8. all trajectories are asymptotic to the line y = j x as t tends to 00.
(9)
and when c. < 0 < X2) Draw the phase portrait of the linear system
s=3x-2y.4). we have
y x
2c.
which implies that all trajectories are asymptotic to the line y = 2x as t tends to -°°. < 0.330
8
Nonlinear DiNerential Equations and Systems
trajectories of the phase portrait of system (3).
When X.e
+
(7)
Solution Here the characteristic roots are X.e
+ c2e2'
2c.e-'.
2c.
(8)
where c. The general
Zc2e". This type of critical point is called an unstable node.e. = 0 and c2 * 0. This type of critical point is called a stable
node.
while for c2 < 0.z
asr-> -°°.e-" + k2
c.+ C2eL. solution of (7) is
I and x2 = 2. °°.e-' + k2e2i
c. When c.°D. On the other hand.4).
Hence.
-_ X
2c. the trajectory is the half-line y = j x < 0 (see Figure 8.
while for c. + c2e
31 .e ' + c2
1 ->2
ast->°°. > 0. The arrows on the trajectories indicate the direction of increasing t. # 0 and c2 # 0.e-'. all the solutions.
EXAMPLE 2
(X.4 shows a few trajectories of the phase portrait of system (7).
y = 2c. + 12c2?
=
c. > X2 > 0. except the solution x = 0. y = !c2e'. y = 0. In the case of an unstable node.e -t + c2eh
!2ex. as can be seen from (10).
For c. since in this case both x and y tend to ±°° as t-> °°. y = 2c.°°.e-' + c. we have the solutions
x = c2e. the trajectory of the solutions (9) is the half-line y = 2x > 0. and c2 are arbitrary constants. the trajectory is the half-line y = 2x < 0 (see Figure 8. the phase portrait is exactly the same.
e 2`.5. = X2 = -2 < 0. As we see from Figure 8. = X2 = k < 0 and the matrix (a
C
X
d b
AI
is zero)
Draw the phase portrait of the linear system
x = -2x. (12) where c.5 except that the direction of the arrows on the trajectories is reversed. only two trajectories approach the origin. it follows that a saddle point is an unstable point of the system. and c2 are arbitrary constants.4
This type of critical point is called a saddle point. all slopes are possible. Since one of the roots of the characteristic equation is positive. The phase portrait of an unstable node is as in Figure 8. and the general solution of the system is
x = c.
y = -2y.8. y = c2e 2'..) The phase portrait of system (11) has the form described in Figure 8.
. the rest tend away from it.4. (Thus.y = c2x with slope c2/c.
EXAMPLE 3A
(X. The trajectories of the solutions (12) are the half-lines c.5
Phase Portraits of Autonomous Systems
331
Y .Ix<
Figure 8.
(11)
Solution Here the characteristic roots are X. This type of critical point is called a stable node when X < 0 and an unstable node when x > 0.
c. = K. From (14) we see that all trajectories approach the origin as t --> w. obtaining
y=fi=x+xln c.te" + c.Y.5
EXAMPLE 38
a < 0 and the matrix (t2_ '\ d b \ i is not zero)
Draw the phase portrait of the system
x = -x. the positive and negative y axes are the trajectories of the solutions (15) (see Figure 8.6
. = 0. _ -1 < 0.6). When c.
y = c. are arbitrary constants.
(16)
Figure 8. For c.e
(15)
Clearly.e". # 0. we can eliminate t between the two equations in (14).
(14)
where c.
(13)
Solution Here the characteristic roots are a. and the general solution of the system is
x = c.
y = -c. and c.e-'. we find the solutions
x = 0.
y = -x . * 0.332
8
Nonlinear Dl ferential Equations and Systems
Figure 8.
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
.. (-2ar. ..13). . Since x = 0 is the angular
displacement of the rod from the equilibrium position OA (see Figure 8. 0). 3ar.. ±2ar. x increases
Figure 8.
Since sin x = 0 for x = 0. we study the critical point (0. the critical point (0.
+ 3!5!
x3
x5
we can approximate (13) by the linear system
(15) y = . 0).12). 0).13
. 0) only. 21r.342
8
Nonlinear Differential Equations and Systems
Let us denote for simplicity the constant g1L by w2. the critical points are
(14) (0... . vertically upward.w2 sin x = 0. ( . Since
sinx=x z = y. . 3 r . .ar.. 0) and (ar. the differential equation (12) is equivalent to the system
X = y. The critical points of system (13) are the solutions of the simultaneous equations
y=0
and
.tar. 0).. 0) is stable and a center for system (15) (see Figure 8.
(13)
In what follows we present a phase-plane analysis of system (13). all of them are located on the x axis. If we now set 0 = x and 0 = y. First. the pendulum is pointing vertically 7the pendulum is pointing downward. (n. 0). and when x = ar. :t 7r.w2x.. as we see from the first equation in (15). Hence. 0). . we conclude that when x = 0. Because of the periodicity of the function sin x it suffices to study the nature of the critical points (0.. (2'n. -ar. The characteristic equation of (15) is X2 + w2 = 0 with characteristic roots
X = ± wi (pure imaginary). The arrows on the trajectories point in the
clockwise direction because. Clearly.
y = -w2 sin x..
±1.. The trajectories in this
case are the heavy loops shown in Figure 8.r.
Y=w2(x-Ir)
(16)
and has a critical point at (w... By Theorem 1(d) of Section 8. . 0) to (0. We can transform the critical point (7r. 0).w2 = 0 with characteristic roots
± w.
Next we study the critical point (a. Now.14
. The closed trajectories
in Figure 8. . 0) by setting v = x . ±2. and hence there is no gain or loss of energy as in the case of a focus). The Taylor-series expansion of sin x about x = a is given by
sin x=
-(x-.. 0) for n = ± 1. 0) is either a center or a focus point for the nonlinear system (13).n in (16).13). .14..14 correspond to the oscillatory motions of the pendulum about the
stable points (21rn. 0) is a saddle point and therefore an unstable equilibrium point for system (13). n = 0. it follows that the critical point (0. however. System (13) also has a saddle point at each of the
critical points ((2n + 1)a.
(17)
The characteristic equation of (17) is Xz . it follows that (0.
Rgure 8. Then (16) becomes
v = Y.
Y = w2V. 0) is also a saddle point and hence an unstable equilibrium point of the linearized system (16) (see Figure 8.)5
3! 5!
Hence. Since the roots are real with opposite sign. ±2.. Thus. from Theorem 1(b) of Section 8. (. ±2.5 (whose hypotheses are satisfied here). 0).8. it can be shown that it is a center (the simple pendulum is a conservative system.. 0). 0) of (17) is a saddle point and hence an unstable equilibrium point. System (13) also has a center at each of the critical points (27rn. the linearized system is
X=Y. we conclude that (a. 0) for n = ±1..r)+(x-a)'(x-.. In this case.5.8
Applications
343
when y is positive.
= 2 aresin Zw < tr. For the sake of describing the phase portrait. 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.. and they are given by
x Y = ±2w cos 2.
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...14. The trajectories in this case are the wavy curves on the top and bottom of Figure
CASE 2
8. we obtain the family of trajectories
2 'y2-w2cosx=c.
CASE 1
x = 0) is attained when y = 0 and
In this case the pendulum oscillates between the extreme angles ±xm.
(19)
The three types of curves shown in Figure 8.cosx) = yo
y2 + 4w2 sing 2 = YO.
. I yo I < 2w Then the maximum value of the angle x (remember that
xm. Assume that y = yo when x = 0. Then from (18) we find that
2 c=2Yo-w 2
1Y2 .
yo I > 2w In this case the pendulum makes complete revolutions. The trajectories in this case are the closed curves shown in Figure 8.14 correspond to the following three cases.w2
y2 + 2w2(1 .
(18)
where c is the integration constant.14. it is convenient to express c in terms of initial conditions. In fact.344
8 Nonlinear Differential Equations and Systems
The equations of the trajectories of the solutions of system (13) can be found explicitly. The equations of these trajectories can be found from (19) if we substitute yo by 4w2.w2cosx = 1Y0 .
cy + dxy .
y = .Exercises
EXERCISES
1. Study the stability of the system at the point (0. Biology In the predator-prey interaction that we discussed in the text. In each case.
3. 0) for n = 0. 0) and determine whether (0.
In each of the following cases. 0) when µ < 2 and when µ > 2. c. [Hint: Set x = X + c/d. Show that (na. y = Y + a/b. [Fore.
saddle. a/b) and study the nature and stability of the critical point (0.
where a. The differential equation
6+k6+w2sin0=0. determine whether (0. study the stability of each critical point and determine whether it is a node. 0) is a node. are the only critical points of the system.
describes the motion of a simple pendulum under the influence of a frictional
force proportional to the angular velocity 6. we assumed that the prey population has an abundant supply of food. or focus. In the special case
z=3x-xy-2x2. or focus.1)z + x = 0.bxy . Assume now that the prey has a limited supply of food available.
k>0. ± 1..] Find the critical points of the system (negative populations are not permissible). In this case it is reasonable to assume that the predator-prey interaction is described by a system of the form
x = ax . saddle.. = e2 = 0. 0) is a node. and e2 are nonnegative constants.. (a) k < 2w (b) k = 2w (c) k > 2w
4. Transform the differential equation into an equivalent system by setting i = y. b. 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. study the stability of the system at (0.
where x and y are measured in hundreds of organisms.x2. 0) is the only critical point of
the system. ± 2. Show that (0.e. Linearize system (2) in the biology application in the neighborhood of the critical point (cld. or focus.]
2.e2y2.
y= -y+2xy-y2. and d are positive constants and e. 0) of the linearized system. Electric Circuits The differential equation
k + µ(x2 .
. saddle. we obtain system (2) in the biology application.
x=4x-9y.
6. x = 4x . Show that the IVP
i=2x2-y+t
y=x+2y2-e
X(O) = 1. 2. determine whether the critical point (0.y= -x
9. Show that (0.l0y
y+xy
10.4x-y+xy
12.8y.y=4x-y
15.
x=x. Determine the nature of the critical point (0. Find the critical points of the system
y=x+2y. 0) is the only critical point of the system
x=2x2-xy.
y=5x-Joy
-(x2+y2)v2. and graph a few trajectories. saddle point. y = x .
13. y = 4x .2y
14. y = 5x .
y=2x-y.8y.x= -3x+2y. y = 5x . x = 2x .3x + 2y.(x2 + y2)
In Exercises 13 through 16. asymptotically stable. x = .9y + (x2 + y2)3. focus.
.
-2x+y.Joy .9y.y
7. X = 4x .9= -x -y
11. or center.y= -x . x=3x+8y.10y
17.
x=2x2-xy. 0) is a node. 0) of the system
x=4x-9y+ (x2+y2)2.
3.y
16. y = x . Find the critical points and the equations of the trajectories of the solutions of the system
x=x-4y.
y(0) = 0
In particular.2.y=5x. x=3x+8y-xy. x= 3x+8y.
and indicate the direction of increasing t.
4.
In Exercises 6 through 12. Find the critical points and the equations of the trajectories of the solutions of the system
y= -x2+2y. determine whether the critical point (0. 0) of the
system is stable.x= -3x+2y+x2-y2.8 Nonlinear Differential Equations and Systems
REVIEW EXERCISES
1. draw the trajectory corresponding to the solution with initial conditions
x(0) = 1. or unstable. x = 2x .
5.2y
8.
y(0) = 0
has a unique solution defined in some interval about the point to = 0.
Q and x2 = L)
Figure 8.
.2-3. 1974). Reprinted by permission of Prentice-Hall.Exercises
347
18. ® 1974. Determine' the stability of the linear time-invariant network shown in Figure 8. Assume that the initial voltage on the capacitor is va. Reprinted by permission of Prentice-Hall.)
Figure 8. Determine the stability of this
network. = 1 and x2 = Q. p.15. 455.J. 0 1974.16.15
19.2-2 in Behrouz Peikari.. Let the initial current through the inductor be io. = . (The state variables are x. Consider' the linear time-invariant RLC network shown in Figure 8. Inc.: Prentice-Hall. Fundamentals of Network Analysis and Synthesis (Englewood Cliffs. N. ibid. Inc.16
This is Example 11. (The state variables are x. 'This is Example 11.
Difference equations are the discrete analogs of differential equations. . Yk.... 1.. is an equation of the form
F (k. For example.3Yk. .lyk = V. As we will see in this chapter..l. .
DEFINITION 1
A difference equation over the set of k-values 0.. and k = 0. 2.1 . they appear as mathematical models in situations where the variable takes or is assumed to take only a discrete set of values.2 + 5Yk.4yk = -2k + 5
yk-1 .
Yk. Yk. 2. which
involve instantaneous rates of changes.l .1.
.n) = 0.
. . 1.2.2 + 6Yk. The following are examples of difference equations over the set of k-values
0. Yk. the theory and solutions of difference equations in many ways parallel the theory and solutions of differential equations..1
INTRODUCTION AND DEFINITIONS
Difference equations are equations that involve discrete changes or differences of the unknown function. Yk.
3Yk.. the sequence
Yk = 2*.8Yk = 0
Yk. . 1.
(1)
where F is a given function.) = 0
. n is some positive integer. of the unknown function.. 1.1 + 6yk = 0
Yk+3 . Yk.
. or derivatives.. .2yk.
DEFINITION 2
A solution of the difference equation
F(k. 2.. k =0. This is in contrast to differential equations.
(1)
is a sequence Yk which satisfies (1) for k = 0.CHAPTER 9
Difference Equations
9. .l. .. 2....
when the function f is identically zero. are given functions of k defined for k = 0. When f is not identically zero.(k)Yk.4 + yk = k
hand.
.3 ..2 = 1 k"
Such equations are called nonlinear.
DEFINITION 3
A difference equation over the set of k-values 0..
In fact. (2) as
a linear difference equation with constant coefficients. Eq.k)Yk = 1
kyk. . where n is some positive integer and the coefficients an. . . For example..1 + yk =
k2
Yk.n-1 + . (4). (4). none of the following difference equations can be written in the form of Eq. is said to be linear if
it can be written in the form
(2) an(k)Yk.. Eqs. 2.2
1
2+y2 + Yk. a0.
If all the coefficients an.. but Eqs. 1.2yk = 0..1 . an .. . and (7) are linear difference equations with constant coefficients.
5yk.t . we speak of Eq... otherwise. 2.. and (7) are nonhomogeneous.(1 . ao are constants.1+ao(k)Y0 = f(k).Y2k = 0
Yk Yk. As in the case of ordinary differential equations. + a. .4Yk.n + an-1(k)Yk. Eqs.2 + 2Yk.an.2 2k = 0
for
k = 0. (3) and (6) are homogeneous. (5) and (6) have variable coefficients. . (3). . 2... (2) is called nonhomogeneous.. 2.2 2k = 2 21 . 1. Eq.3yk = 0
(3) (4) (5)
(6) (7)
Yk. The following are examples of linear difference equations over the set of k-
values 0.l ..
On the other
Yk-2 ... 1.
Yk.yk. it is a linear difference equation with variable coefficients.2Yk = 2k -1 .3 . (2) is called homogeneous.1
Introduction and Definitions
349
is a solution of the difference equation
Yk. together with the function f. but Eqs. . For example. (2).9.2 + Yk = 0 Yk. 1..1 = 3k .. (5)..
For example. That is. .
(9)
. We should remark here that over a set of k-values different from 0. Also.. from one generation to the next.. (5) is not of order 1 since the coefficient ao(k) vanishes when k = 1. in order to define the order of the
linear difference equation (2). it is more realistic to assume that the size N of the population is a step function. Eq. . Eq. A simple model for such a population will be to assume that the increase in size. Also over the set of k-values 1 . For some populations. . where k is the constant of proportionality. As we have already seen in Chapter 7.(1+a)Nk = 0. of a power series solution y(x) = E%o a x" in terms of the coefficients ao and a. For example. ... . The basic idea there is to approximate the differential equation by a difference equation. (5). and Fibonacci sequences will be given in subsequent sections.1) assumes that the rate of change of a population is proportional to the population present..1. and Eq. 1 . this is the case in some insect populations. Then
Nk+. (7) is of order 4. electric circuits. k = 0 is a singular point. we must make the assumption that the coefficients
a. Let N denote the size of the population of the kth generation.. difference equations are useful in connection with the numerical solution of differential equations. its order over this set is 1 .2 and Remark 1 in Section 9. Eqs. then N(t) = kN(t). We say that k = 1 is a singular point for Eq. 3. 3.1. which enabled us to compute the coefficients a a3. however... where one generation dies out before the next generation hatches.(k) and ao(k) are nonzero for all k = 0. Further applications on these topics and applications to business. Eq. Eq.
In this section we see how difference equations appear as mathematical models
describing realistic situations in biology. (2) is said to be of order n. Here N(t) is assumed to be a differentiable function of time. In this case. 2.3). Eq. is proportional to the size of the former generation.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. 1. (6) has no singular point and its order is 3.. For Eq. probability theory. For example... 2.(k) vanishes when k = 0.
APPLICATIONS 9. . .
form
(8)
where a is the constant of proportionality. (4) is of order 2.. if N(t) denotes the size of the population at time t. . (5) has no singular points over the set of k-values 2.. 2. (5) and (6) may not have any singular points. Equation (8) can be written in the
Nk+. Eq. and optics. (3) is of order 1. (6). . But Eq. (6) is not of order 3 because the leading coefficient a.
Biology
The Malthusian law of population growth (see the biology application of Section 1.Nk = aNk.350
9
Difference Equations
For reasons that will become clear in subsequent sections (see Remark 1 in Section 9. Difference equations were also encountered in Chapter 5 as recurrence formulas. 4.
where No is the initial size of the population. together with the boundary conditions Vo equal to a given constant and V..1
R
= Vk. V.. Each resistance in the horizontal branch is equal to R and in the vertical branches equal to 6R. Equation (9) assumes that the size of the population depends on the population in the previous generation..2.2-13Vk. . the sum of the currents flowing into a junction point is equal to the sum of the currents flowing away from the junction point. 2.2 +
Using Ohm's law. In fact.1+6Vk=0. draw a line parallel
Optics: Rays Guided by Lenses
. 2.9. n satisfies a linear homogeneous difference equation of order 2 which.
Consider the electric circuit shown in Figure 9. .Vk.1
Consider the converging thin lens L shown in Figure 9.. In some cases it may be more realistic to assume that
Nk. = 0. = 0.1 .. .1
Introduction and Definitions
351
which is a linear homogeneous difference equation of order 1....2 + R
V11 . Applying this law at the junction point corresponding to the voltage Vk. n-1. + qNk = 0. I = V/R. we have
Ik.. Its solution is
(see Section 9. and the shaded region indicates the ground where the voltage is zero. Assume that Va is a given Electric Circuits voltage. 1 . according to Kirchhoff's current law. We want to find the voltage V k for k = 1 . . A ray PQ passing through L will be refracted by the lens in the direction QR.Vk. We will show that the voltage V k for k = 0.l = 4.
Figure 9. will give the value of Vk for k = 1 .I .
(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.2 + pNk. 2. . which can be found as follows: From the center 0 of the lens.0
6R
(11)
Equation (11) can be simplified and reduced to
6Vk.4)
Nk = NO(1 +a)k. n -1. .1. the above equation can be replaced by
VI .
satisfies the linear difference equation
fYk. =
mk f Yk =
. as shown in Figure 9.1 Yk + mk.2 . as we are about to prove. That is. of the line segment QkQk.mkf). is that the displacement Yk of the ray. of the ray PQo through the system of lenses.4. Thus the coordinates of the point A are (f. The two lines meet at the point R.. Let us denote by mk the slope of the segment Q.3
.2
to the ray PQ.3. Now consider a system of identical converging thin lenses. The result.(2f-d)Yk. (See Figure 9. and its equation is y = mx.) Then the line OA also has slope mk.-.
(12)
where f is the focal length of each lens and d the distance between lenses. with respect to the optical axis x. each of focal length f and spaced d units apart on the x-axis. Since Qk has coordinates (O.yk).. it follows that the slope mk+..l + fYk = 0. we want to find the y-coordinate yk of the point Qk.Q2 .352
9
Difference Equations
I
x
F
L
Figure 9. We want to study the path QQQ. draw a line parallel to the
lens...Q. through which the refracted ray passes. is given by
mk. from the focal point F of the lens.
(13)
QO
P
Qk
Qk + I
I Qn
ex
4+d+
f
Lo
Lk
Lk + l
Ln
Figure 9. as it traverses the lenses.
9.l + gkyk = rk.
... . . B are given constants. On the other hand. . (1) by a2(k) and setting
a1(k) a2(k)
.
Hence the value of y2 is uniquely determined in terms of known quantities.2 . We must prove that yn+2 is also uniquely determined.. A solution of Eq.
where pk. we find
y_2 = -P Y.. 2. qk. setting k = 1 in Eq. Using the initial conditions (4) we find
Y2=ro-p0B-q0 A. and rk are given sequences of numbers with qk + 0 for all k = 0. .
f(k) k a2(k) ..
In fact. qk. .
. has exactly one solution. 2. 2. 1..
.2
Existence and Uniqueness of Solutions
355
Consider the second-order linear difference equation
(1) a2(k)Yk....
(2)
we get
Yk-2 + PkYk..
Proof
For k = 0. (3) and using the values of y1 and y2. . 2.. 2.
THEOREM 1 EXISTENCE AND UNIQUENESS
The initial value problem
Yk.. (1) is assumed to be of order 2. .. we find that y3 is uniquely determined.
.2 + al(k)yk. where the coefficients a2.. from Eq.1. a ao and the function f are given functions of k defined
for k = 0. and rk are given sequences of numbers. Next. 1. 1.. (3) with k = n..2 + PkYk. and qk # 0 for all k = 0. (3) becomes
Y2 + poy1 + g0yo = ro... the proof of the corresponding theorem for difference equations is almost trivial. y 1 .1 and the result follows.
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. of the solution
yk are uniquely determined. we should continue by mathematical induction as follows: Assume that the terms yo. . Since Eq. 1. .. and so forth.Pk. .
yo = A.1 + aa(k)Yk = f(k). y. yl = B.r
for k=0.
ao(k)
a2(k) = qk. the coefficients a2(k) and ao(k) are nonzero for all k = 0. (2) is a sequence yk which satisfies Eq. (2) for
all k = 0.l + gkYk = rk. Eq. For a rigorous proof that every term yk of the solution is uniquely determined. 1. and A. Dividing both sides of Eq.
where Pk..
. Every new pair of rabbits does exactly the same. k = 1. if yk denotes the kth term of (11).. 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. They were discovered in the famous Fibonacci rabbit problem... . 2.
(11)
called the Fibonacci sequence. 2. k = 0..2.
.1.2. each term after the second is the sum of the two preceding terms. 3.2 = Yk.. are said to be linearly dependent if
there exist constants A and B. 2. the sequences
ak = 3k +
12
5
and
bk = 5k + 4. Therefore..
(2)
The rabbit problem first appeared in Leonardo "Fibonacci" da Pisa.5.. called the Fibonacci numbers. + yk. not both zero.13.'
16.
(1)
Otherwise the sequences are said to be linearly independent. then the Fibonacci sequence is the unique solution of the IVP
Yk. 1.
(12) (13)
y. Show that the unique solution of the IVP (12)-(13) is given by (14)..=y2=1. have many fascinating properties and appear mysteriously in many diverse situations.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.. 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. k = 0...)
DEFINITION 1
Two sequences ak and bk.
9 .8.. (See Definition 1 of Section 2.2. How many pairs will there be after k months if we begin with a pair of newborn rabbits?
17.
W+ (1+
L\
2
J
In Exercise 17. .358
9 Difference Equations
Fibonacci Sequence
In the sequence
1.. Liber abaci (1202). such that
Aak+Bbk=0
for
k=0. 1. For example.1.3..
. 2.3
Linear Independence and the General Solution
359
are linearly dependent because (1) is satisfied with A = 5 and B = -3.. . such that (1) holds. 2. 3k) = 2k. .
.
(4)
Setting k = 0 and k = 1 in (4). 2.
DEFINITION 2
The Casoratian of two sequences ak and bk.. 2. On the other hand.. k = 0. 2. Let us compute the Casoratians of the sequences in (2) and (3). 1. We have
C{3k+ 5 5k+41
.. we find
A+B=0
and
(5)
But the only solution of system (5) is A = B = 0. This contradicts the hypothesis and establishes our claim that the two sequences in (3) are linearly independent. 1.
(3)
are linearly independent. the Casoratian plays for difference equations the role that the Wronskian plays for differential equations.9.1
3k. Otherwise.. 1
= 2k3`' .. the sequences
ak = 2k
and
bk = 3k.(5k + 4)
\\\\\\13k+ 5?/
=+3k+ S )(5k
=0
and
j2k
5
for all k = 0. . k = 0..
A2k + B31 = 0
for
k = 0. Before we present a simple test for linear dependence and independence of solutions of linear homogeneous difference equations.bk) and is defined to be the determinant
C(ak.' bk bk
(6)
As we will see. 1.. they are linearly dependent. we need the following definition.
2A + 3B = 0..
3'
C(2k.bk) _
ak ak.
//
\\
3k+
5
12
5k +4
12
3(k+1)+
5(k +1)+41
9). not both zero.. 1.2`3k = 3(2k3k) . 1... That is.2(2 3k)
= 2 k 3 k = 6' # 0
for all k = 0. which means there exist constants A and B. . is denoted by
C(ak.
Yk2)) # 0
for
k = 0. 3. 15. 0. 2.
Yo
Y(V)
Yo
y(2)
= 0.
Otherwise the Casoratian is identically zero.. are linearly independent if and only if their
Casoratian
(2) CUM +Yk) _
-
A
Ykl. 30.
yk 1
:
1.. of the linear homogeneous difference equation
Yk. .2) = 0
(12)
(13)
Since. in particular. 1. We
should prove that
C(y(k`). 1. 0..2.. 0. .9. In fact. 0.
(10)
where qk # 0 for k = 0.
are solutions.
Proof Assume that the solutions y." and yk2) are linearly independent. . and. 0." + By.
and
Yk2' : 0.. 2. Their Casoratian is
11
C(yk ). .
On the other hand. 1.
. it follows.
Consider the homogeneous system
Ayo" + Byo) = 0
Ay.. 1.... .1
Yk2'
Yk2. from Appendix A.3. B). 0. . yk')
0
if k=0 if k#0
k = 1. . although
Yk2'=2yk`)
for
THEOREM 2 Two solutions.1
is different from zero.. that the system has a nontrivial solution (A. 6.2 + Pk Yk«1 + qk Yk = 0. . 2. 2. on the basis of Definition 1.. 1. yk" and y(k2). by (11).. the solutions ykl) and yk2) are linearly independent. the determinant of the coefficients of the system is zero..3
Linear Independence and the General Solution
361
considered over the set of k-values 0.
. 2.
means there are numbers A and B. by Definition 1. from
(16).y. we find
go(AYo ) + Byo2>) + Po(AYi'> + BY(2)) = 0
or
A(Po Yi'> + qo Yo >) + B(P0 y.
Conversely. we have the following theorem for the general solution of a linear homogeneous and a linear nonhomogeneous difference equation of order 2.
This contradicts our assumption and the proof is complete..2> = 0. 1.. yk2)) # 0. 21 . As in the case of differential equations. it follows that the solutions y(I) and yk2 are linearly dependent. using the fact that yk'> and yk2> are solutions of (10).
(14)
Now.2> + qo yo >) = 0.
and (14) becomes
Ay(I) + By.
(16)
Assume A * 0.
Since A and B are not both zero. we have
y( I) = .
We must prove that yp"and yk2> are linearly independent. 1.Yk `2)) = yo> cn . The proofs are straightforward.. such that
Ayk'> + Byk2> = 0
for
k = 0. This contradiction proves the claim that
C(Yk'>.
(15)
One can show by induction that
Aykl) +Byk2) = 0
for
k = 0.
.. not both zero..2 .362
9
Difference Equations
Multiplying both sides of (12) by q0 and both sides of (13) by po and adding the results. The case B * 0 is treated in a similar fashion. Then.. assume that
C(Yk". Otherwise they are linearly dependent which.
yk'>
B A Yk2>>
and so
C(yu> .. Yk2)) # 0. See Exercises 9 and 10.2>ycu
k
k Yk-l
0.(Po Yi'> + qo )Q0o'>)
and
22> = -(po yti2) + qo y( o')).
15\k = 0.2 .2 = 0. Substituting (2) into Eq. we looked for a solution of the form y(x) = e'... We present the main features of the method by means of examples. Here the idea is to look for a solution of the form
Yk =
.
EXAMPLE 2
Find the general solution of the difference equation
3Yk.
Solution
(1)
With differential equations..2Yk = 0
Solution Characteristic equation: 3X2 . Hence.
where c. 1. .366
9
Difference Equations
9.
Ik2-8X+ 15=0. are Jk. = 3 and A2 = 5.5).
(2)
where X # 0.k. (1).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.5Yk.3.
and dividing by lkk. . They are linearly independent because their Casoratian
3'
3k
5k
5k"
is never zero for k = 0.2-8Yk. called the characteristic roots.1 + 15yk=0. Hence
Ykl) = 3k and
yk2) = 5k
are two solutions of Eq. and c2 are arbitrary constants.
.
(3)
Equation (3) is called the characteristic equation of the difference equation (1). the
general solution of Eq. we find
Xk+2
-
8Xk. . The roots of the characteristic equation (3).1
.
EXAMPLE I
Find the general solution of the linear homogeneous equation
Yk.. 2. by Theorem 3 of Section 9. (1). (1) is
Yk = c13k + c25k.
2k + c21 -
EXAMPLE 3
Find the general solution of
Yk.(.
(9)
.1 + i)k + C2(. as in Example 1. That is.
Characteristic roots: K _ -1 ± i.
(7)
Solution Characteristic equation: K2 + 2K + 2 = 0..
But here the characteristic roots are equal. (4) has the solution y°' = 3k (5) (and all its multiples). However.4 Homogeneous Equations with Constant Coefficients
367
Characteristic roots:
5_
25+245±7
6
6
2
={-1/3'
General solution: yk = c.
and yk'). + 2yk = 0. = 1\2 = 3. in the case of equal roots. but it does not have two linearly independent solutions of the form Kk.i)k.
that K must be a root of the characteristic equation
(4)
Solution Looking for a solution of the form yk = K". we find
x2-6k+9=0. (4) has the solution
Yk2' = k3k.)
Yk = c. and so Eq. (4) is
EXAMPLE 4
Find the general solution of the equation
Yk_2 + 2yk. In general. the general solution of Eq. (Show it. the product kyk'> gives always another solution which.
General solution:
Yk = C. Eq. consider the difference equation
a2 Yk-2
+ alYk l + a0 Yk = 0. together with yk".1 .
Hence. (7) as a linear combination of two real and linearly independent solutions. are linearly independent.l+9yk=0. in addition to (5).9.2-6yk.
REMARK 1
(8)
We will show here how to write the general solution (8) of Eq. yk2)
(6)
are linearly independent. a.3* + c2k3k.
where C. the general solution of Eq.ice are arbitrary constants which. -ice) sin k9
= C. we can still denote by c.(a+ib)k + C2(a-ib)k = c. (7). the general solution of the difference equation (9) is
Yk = c. In summary. Then they must be complex conjugate.
(13)
From (11).
-1 + i = r(cos 0 + i sin 0). Now the crucial step is to write the complex number a + ib in complex polar form. = c. and c2. (9). 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. In polar form. In Exercise 38.
Taking complex conjugates.
where r and 9 are determined by Eqs. we find
a + ib = reB.k + ao = 0
(10)
are complex numbers. + c2 and C. for
economy in notation.i sin k9)
= rk(c.rk cos k8 + c2rk sin k9.388
9
Difhrence Equations
with real coefficients. with
r
-r<9sn.rk cos k9 + C2rk sin k9. and assume that the roots of the characteristic equation
a2X2 + a. using Euler's identity.
r
sin 9 = b.ib=re °
Thus.rk(cos kO + i sin k9) + c2rk(cos k9 . That is. where a and b are real numbers. = ic.
Let us apply the above results to Eq. (9) is Yk = c. The characteristic roots are -1 ± i. (10) are of the form a ± ib.(reiB)k + C2(re-ie)k
= c.+c2) cos ke + rk(ic.
. we also have
a .
where the modulus r of a + ib is given by
(11)
r=
a2 -+b'. . That is.
(12)
and the argument 9 satisfies the equations
cos 8 = a. (12) and (13). in the form
a + ib = r(cos 9 + i sin 9).rkeik° +
c2rke-k°
= c. the roots of Eq.
2.Uk =
Yk-1
Xk-.
43. A certain colony of bacteria undergoes a 5 percent relative increase in size per hour..
Use Eq.1).2. .. Investigate the displacement yk of the ray as it traverses the lenses.
-Y=
Xk
C(Xk.. Section 9. [Hint:
Uk-.]
47.
41.. Consider the second-order linear equation
yk-2 + P.1) are all equal to R. Find the voltage Vk for k = 1 .000 students... If the college currently has 20. = 2. = 1. .Yk)
XkXk-. 2. = 1. assume that d = 4f.
find the general solution. + gkyk = 0. q * 0
to approach zero as k --' x. (9) of Section 9.3 with C. as with differential equations. 1..1. what will the enrollment be after 10 years?
44.
.1. where qk * 0 for k = 0.3).4(k + 1)yk-.1. Assume that Xk is a solution and xk * 0 for k = 0. However. .372
9 Difference Equations
are necessary and sufficient for all solutions of the difference equation
Yk-2 + PYk-1 + qyk = 0.1]. n . what will its size be after 12 hours?
45. In the electric circuit application (see Section 9. if we know a nontrivial solution of the homogeneous equation. (12). But there exists no general method for solving linear equations with variable coefficients of order higher than one. . If the initial size of the colony is 1000. 2.. Assume that the resistances in the electric circuit application (see Section
9.n . and y.. we can use a transformation to obtain an equation of lower order.. ..
46.
42. + (2k + 3)yk = 0.. 1. Given that yk = I is a solution of the equation
(2k + 1)Yk-2 .. find the voltage Vk for
k= 1.1. In the optics application [see Eq. Section 9.. It has been estimated that the student enrollment in a college undergoes a 5 percent relative decrease in size per year. Reduction of Order First-order linear difference equations with variable coefficients can always be solved explicitly (see Exercise 20.J'k-. y.1. Show that the transformation
Yk = Xkuk
produces a first-order linear difference equation for uk.
where 3 is a nonzero constant. With this correspondence in mind. sin 8x listed under types 1-4 in Section 2.l + a0yk = 0. 2. where y is a nonzero constant. 3.
9. where S is a nonzero constant. (1) be constants and that the nonhomogeneous term rk be of a special form.5 NONHOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS
Consider the linear nonhomogeneous difference equation with constant coefficients
a2Yk. That is. we know that the general
solution of Eq. (2) and a particular solution of Eq. one can find the form of a particular solution of a difference equation
.. In Section 9.5 Nonhomogeneous Equations with Constant Coefficients
373
9. cos yk. k°. however. the method of undetermined coefficients usually requires less labor.4 we explained how to obtain yk. sin 8k. the method of variation of parameters imposes no restriction on rk. es`. that when
we have a choice.
(3)
where yk is the general solution of Eq. From Theorem 4 of Section 9.) On the other hand. a and ap of Eq. + aOYk = rk
(1)
and the corresponding homogeneous equation
a2yk. it applies to equations with constant coefficients always. (1): the method of undetermined coefficients and the method of variation of parameters.5. cos yk. will be completely known if we show how to obtain a particular solution of Eq. yk is the general solution of the homogeneous equation (2).5. (1). 4.2 + a1yk. It should be remarked. (1).2 + alyk. The method of undetermined coefficients requires that the coefficients a2. Hence Y. As in the case of differential equations.9. where a is a positive integer or zero. pk. (1).
The sequences k°. sin 8k correspond to the functions x.3. pk.11.
cos yx. and to equations
with variable coefficients only if we know the general solution of the corresponding homogeneous equation.
(2)
where a2ao # 0. (See Section 9. there are two methods which can be used to find a particular solution of Eq.1
UNDETERMINED COEFFICIENTS
This method applies successfully when the nonhomogeneous term rk in Eq. A (finite) product of two or more sequences of types 1-4.1. 5.
yk = Yk + YPk . (1) is the sum of the general solution of Eq. and yk is a particular solution of the nonhomogeneous equation (1). (1) is a linear combination of sequences of the following types:
1.
.6B3k = 3 25 . (4) and equating coefficients of similar terms.11 and treating terms of the form ¢k much in the way we treated ea` in Section 2.
A=1
. Using notation similar to that of Section 2. where n is the smallest positive integer such that k"31 is not a solution of Eq.
Yk = A21 + Bk3k
(9)
is a particular solution of Eq. first we find the general solution of the corresponding homogeneous equation Yk-2 . (5) can influence the form of a particular solution of Eq. Therefore. 752 = 5.2 .35
812A2k+3B(k+1)1 + 15(A21 +Bk3k) = 3 25 . We now have
[A22+B(k+2)35-2]_ 81A2 k'`+B(k+1)35"1+ 15(A2k+Bk3k) = 3.2k-5.
Characteristic equation: K2 .374
9
Difference Equations
in a manner similar to that used in Section 2.
Characteristic roots: K.11 for a particular solution of a differential equation. The rule is to multiply by k".8yk.8Yk.11. (5) since the solution of Eq. Homogeneous solution: yk.I + 15yk = 0.
EXAMPLE I
Find a particular solution of the nonhomogeneous difference
Yk. we have
3
2k*{2k}
3k -> {3k} -a {k
35}. (5). = 3. (4).
-5
(7) (8)
where in (8) we multiplied 3k by k because 3k is a solution of Eq.5 3k
11
3A2k .8K + 15 = 0. = c. + 15yk = 3 2k . The undetermined coefficients A and B will be determined by substituting (9) into Eq.5
3k.5 3'
3A = 3
and and
-6B = -5
B=5/6.
equation
(4)
Solution
As with differential equations. (4).3' + c25k. (5).
(6)
Here rk is a linear combination of two sequences of type 2.
Suppose we invest $500 principal at 8 percent simple interest.5.P b l ems m compound interest often lead to simple difference equations.
Now.380
9
Difference Equations
Eqs. the general solution of Eq. (29) and (30). and c2 are arbitrary constants.
Thus. and their Casoratian
is
I
1
k'
(k+l)z1=(k+1)2-k2=2k+1#0
for
k=0.1. Here we have no choice of method. Thus... the general solution of Eq. (29) because they are solutions. (30) is
Yk=Y2+Yk=c. We must use variation of
parameters because the coefficients of the equation are variables. The two sequences in (28) are linearly independent solutions of Eq.
yk =
=o
(2n + 3)
1
1
2n+3)
(n + 1)2
-o(
k2 . (30).
where c.2. the unique solution of the IVP (30)-(31) is
Yk=2+
k2
4k3 -3k2-k
6
4k3 -k
6
Business:
9.3 APPLICATIONS
Mathematics of . = 0 and c2 = 1/2. we find 1 (n + 1)2
1
k2
k-. using formula (27). At the end of the first year the bank will add (0. we find that
rk=2k+3.(n + 1)2
(n +2)'-(n + 1)2
_
-a
(n + 2)2 [k2-(n+1)2]=k. at the end of the second year
. we find c. (30) by the leading coefficient (2k + 1)... to Fmance
First let us recall the meaning of simple and compound interest.+c2k2+4k'-3k2-k
6
Using the initial conditions (30)._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. as we just proved.
1 + c2k2. Dividing both sides of Eq. Next we will find a particular solution
of Eq.08) 500 = $40 to the account. (29) is
Yk = c.
we find
A8 = 500(1 + 0. we earn interest only from the principal investment of $500. from (32).
EXAMPLE 7 I invested $500 at 8 percent annual interest compounded quarterly. and so forth. What will the amount be after 2 years?
Solution P = 500.
Proof
(32)
Since A. or
annuities.02. together with the interest.02)8 = 500(1.20 = $520. bringing the total to $580. Show that the value A4 of the account at the end of the kth period will be
Ak = P(1+i)k.02) 500 = $10 to
the account. at the interest rate of i per period. 1. with interest compounded quar-
terly. That is. 2. is kept there for k time periods.82. (See Example
7). and the interest is 4 = 2 percent
per quarter.
Also
A = P.02)8 = $585. after 2 years the account will be
worth $585. At the end of the first quarter (each quarter is 3 months). so 8 percent interest compounded quarterly really means 2 percent per quarter. . Continuing this process. Hence.9. suppose we invest $500 at 8 percent compound interest. i = 0.(1+i)Ak = 0. k = 0. bringing the total to 510 + 10. at the end of the second quarter the bank will add (0.20.
Theorem 2 is a basic model for compound interest. This interest is now compounded with the principal and from now
on will also earn interest. it follows that
Ak_.. Interest rates are usually quoted as annual rates. Hence. that is.
Theorem 3 gives the formula for periodic payments of a fixed amount. . = Ak + iAk.
..
THEOREM 2
Assume that the amount P is deposited in a savings account and.02) (510) = $10. and iAk is the interest during the next period.20 to the account. compared to $580 in the case of simple interest. the bank will add (0. denotes the value of the account at the end of the kth period.82. On the other hand.
The solution of this IVP is given by (32). k = 4 2 = 8 periods.
or
Ak_.5 Nonhomogeneous Equations with Constant Coefficients
381
it will add another $40.
if yk = E is a solution. Equilibrium and Stability A constant E is called an equilibrium value for the difference equation Yk+2 + PYk+I + qyk = r. after we eliminate y. 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 -q>0. xk+1 = -3xk + 4yk
Yk+1 = -2Xk + 3Yk
59.k'
Yk _ -
k+2)'-(k+1)'.
Linear
56. 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.
58. The equilibrium value E is stable if every solution of the equation approaches E as k-k -.Yk .2k
7Yk + 1
Yk+l = 5Xk .2yk '.
57.
In Exercises 57 through 60. = 3xk . find the general solution of the system by the method of elimination (see Exercise 56).(k
(k+2)3-k'
+ 1)' k'Yk+1
+(k+2)'-(k+1)3
(k + 1)' .q+p+1>0. q #-0. 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. Show that. xk+.386
9
Difference Equations
Then use the method of variation of parameters to find a particular solution of the equation
Yk+2 .6Xk
-
60.
. xk+I = 4xk.1
61.al2a2l)xk al2tk .a22rk + rk+1
After we find xk we compute yk from the first or second equation. we obtain the second-order linear equation with constant coefficients
Xk+2 .2).Yk
Yk+1 = 2xk + Yk
xk+l =5Xk-6yk+ 1
Yk+l .(all + a22)xk+l + (a11a22 .
On each play of the game.
In Exercises 14 and 15. tennis. If the initial size of the colony is 1000. That is.Y2=2
11.1).7Yk -1
Xo = 1. Yo= -2
16.9
Difference Equations
10. A colony of bacteria undergoes a 10 percent relative decrease in size per day.4 ' Y.
Xk+1 = -3Xk . Principle of Superposition Show that if yk11 is a solution of
Yk+2 + PkYk+1 + gkyk . the probability that A wins is p.1. In the electric circuit application of Section 9. Yk+i + Yk = 3
yo=5. Probability: The Gambler's Ruin Two people. solve the IVP. A and B. YO = -1
XO=z.2k+1 Yo = 1. Xk+1
= 5Xk + 6yk + I
3 3
Yk. and the outcomes of the successive plays are
. what will its size be after 5
days?
17. and the probability that B wins
is q.8Yk+1 + 15yk = 3 ' 2k .
13. Use the principle of superposition (see Exercise 11) to find a particular solution of the equation
Yk+2 . play a game (cards.4Yk
Yk+l = 2rk + 3Yk
15.
14.I = -6xk . If V0 = 10 volts and
V10 = 0 volts.y1 =1. find the voltage V5. where p + q = 1. the winner wins one dollar.rk1
and yk21 is a solution of
Yk+2 + PkYk+l + gkYk =
rk21.
18.1 (see also Figure 9. There are no ties. chess) in which their skills or chances of winning are rated as p to q. 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 . assume that each resistance in the horizontal branch is equal to 1 fl and each resistance in the vertical branch is equal to 211. Y1 = 0.
then aykl" + by12j is a solution of
J
Yk+2 + PkYk+1 + gkYk = ark) + br'kl
12.
. 83. + Yk = Yk+2 . Electromagnetics: Classical and Modern Theory and Applications (New York: Marcel Dekker.. Reprinted by permission of Prentice-Hall. Inc. Poularikas. p.
26.
20.Y4 + 24. New Jersey. She borrowed $25. If the ray enters the first lens at height 2 cm and at an angle of 352. What is the monthly payment?
In Exercises 20 through 23.1 = Y2k
22 Y2 +Y4+
+Y2k-Y2k+1 .700 at 9 percent compounded monthly.
. Show that the current i in the nth loop satisfies the difference equation
and solve for i.A. N.Y2 + Y3 .A.1
1`k+l
Yk = . what is the probability that A
will ruin B by winning all of B's money?
19.. yl + Y3 +
+ y2k.9. N.Y. Mary has a 25-year mortgage on her house.I
21. In the network shown in the figure. Cadzow. 1979).
'This is Exercise 2. verify the summation property of Fibonacci numbers. 533. Discrete-Time Systems: An Introduction with Interdisciplinary Applications (Englewood Cliffs.` assume that all resistances except R( on the end are of the same value R. Inc.
(_ 1)k+l
25. p.: Prentice-Hall. The borrower is to pay the loan
back in one lump payment at the completion of s4 years.5 Nonhomoganaous Equations with Constant Coafflciants
389
independent.
'This is Exercise 11-9. yI . An optical system2 of identical converging thin lenses is constructed with d = 5 cm and f = 16 cm. Aseltine. 1958). + (Yk+1 + 1 23.
'See Section 16-1 in J.J. Find the necessary value of r. Englewood Cliffs.D. Used with the permission of McGraw-Hill Book Company. and if they continue to play until one of them is ruined. Transform Method in Linear System Analysis (New York:
McGraw-Hill Book Co. Reprinted by permission of Marcel Dekker. plot the ray for the distance of five lenses.2 in S. 246. If A starts with a dollars and B with b dollars.4-2 in J... p. 1973). Y1 + Y2 + . Suppose an investor' wishes to double her money in six years by loaning it out at a rate of r compounded yearly. Seely and A.
1
INTRODUCTION
In connection with the solution of the heat equation in Section 6. Fourier.1. 2. in such a way that a given function f can be expressed as a trigonometric series of the form
f(x) _
(1)
This. 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. thus leading him to the false conclusion that any arbitrary function can be expressed as a series of the form (2).2. the theory of Fourier series has been developed on a rigorous basis and has become an indispensable tool in many areas of scientific work. they were considered to be nonsense by many of his contemporaries.CHAPTER 10
Fourier Series
10. Since then. l > 0.
(1)
which are the building blocks of Fourier series.. 2. This is not true.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. as we will see in Section 10. for n = 1. 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. . Fourier's work lacked mathematical rigor.4. . Consequently. 3. we still have to show how to choose constants b. 3. However. when Fourier published his results. B. 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
10.
this means that the graph off repeats itself in successive intervals
of length T. the functions nax nTrx
cos . For example. if it exists.. and in general (see Theorem 1) the fundamental period of each function in (1) is 21/n. 61T. Periodic functions have many periods. the fundamental period of sin x and cos x is 2a..
n = 1..10.
THEOREM 1
The functions cos px and sin px. is a period. are
periods of sin x and cos x.. and the motion of a pendulum are examples of periodic functions.1. . are also periods of f.. .1(a) has period 2 and that in Figure 10.1(b) has period 1. The functions sin x and cos x are simple examples of periodic
functions with period 27r.. More generally.and
sin .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. 47r. is called the fundamental period of f For example. 27r. 3.
. This is because any positive number. 3T.. . no matter how small. the vibrations of a spring. 2. are periodic with fundamental period T = 21/n. The function in Figure 10. Alternating currents. On the other hand.. 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 Smallest positive number T for which Eq. a constant function has no fundamental period.
where 1 is a positive number.. In particular. then
f(x) = f(x+ 7) = f(x+2T) = f(x+3T) =
.. sound waves. it follows from (2) that if f has period T. p > 0.
which means that 2T.
Y
i4TiT--o
£--c
0
l1
1 23
-1 0
Figure 10. (2) holds. A constant function is also periodic with period any positive number. are periodic with fundamental period 27r/p.
Then the statement f(x+ T) = f(x) for all x is equivalent to cos (px+pT) = cos px or. More precisely. 2. The proof for sin px is similar. for all x. Thus T = 2n7r/p. 3. with n = 1.
THEOREM 2
The functions
cosnIX
and
sinnx
for
n = 1. n = 1 ..sin px sin pT = cos px.
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. 2. (For a review of odd and even functions. .. Next we turn to the question of orthogonality of the functions in (1). (recall that p and T are positive). k.x s 1. pT = 2mr.x s 1.
n
(4)
(5) (6)
J
1
#k.3. 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.
n. cos px cos pT . Recall the definition of orthogonal functions given in Section 6. Assume that T is a period of f(x) = cos px.. Clearly.
Since the functions in (1) have (fundamental) period 21/n.
J
cos n j x cos k l x dx
= 111
sinnrxsinklxdx={0
if if
n# k. it follows that each one has also period 21.
(3)
But (3) is true for all x if and only if cos pT = 1 and sin pT = 0.392
10
Fourier Series
Proof We give the proof for the function cos px.2. see Section 10. we have
Theorem 2. ..2.5... after expanding the cosine
of the sum px + pT. This means that any two distinct
functions from (1) are orthogonal in -1 <.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. That is.)
. ... each of the functions listed in (1) is periodic with
fundamental period 27r/(n7r/1) = 21/n. and so the fundamental period (the least positive) of cos px is 27r/p.
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)
The hypotheses of the above theorem' are known by the name Dirichlet
conditions.
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. Such discontinuities (where the left. However. and to the
average (f(x -) + f(x + )1/2 of the left. there is a huge class of functions for which (4) fails at the points of discontinuities of the functions.and right-hand limits at each point x where f is discontinuous. although not the most general sufficient conditions known today. An example of a piecewise continuous function is shown graphically in Figure 10.sinn"xl=f(x-)+f(x+) gg+ i (acos-7. That is. W.. a piecewise continuous function on an interval I has a finite number of discontinuities on I. 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. necessary and sufficient conditions for (4) to hold have not been discovered. are. then
f (x) if x is point of continuity of f
i x is point of if discontinuity
nax+b. Advanced Calculus (Reading. These conditions.2. Kaplan. if f satisfies the Dirichlet conditions. for example. As we will see in Theorem 1. Examples are also known of functions whose Fourier series diverge at "almost" every point.c from the left. Hence. Mass. nevertheless.) = lim f(x + h). 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. in each of which f is continuous and has finite left. The notation f(c -) denotes the limit of f(x) as x .c from the right.402
10
Fourier Series
where the coefficients a and b are given by (2) and (3).:
.and right-hand limits exist but are unequal) are called jump discontinuities. 1973). If f is continuous at c.
f(c .
Similarly we write f(c+) to denote the limit of f(x) as x .and right-hand limits.
'For a proof of Theorem 1 see. Then the Fourier series off converges to the value f(x) at each point x where f is continuous. Sufficient conditions for (4) to be true were given by Dirichlet in 1829._
!
/
(6)
2
Addison-Wesley Publishing Co. Fourier was wrong in asserting that (4) is true without any restrictions on the function f. generally satisfied in practice. Clearly.
It follows from (6) that (4) is. 3. from (6'). Using (5) we can write (6) in the form
fix ) + fix +)
2
as
2
+ i (a. (6') agrees with (6). then.to indicate that the right-hand side is the Fourier series of the function to the left. we will continue using the symbol .2
where a and b are given by (2) and (3). if x is a point of discontinuity of f. then (4) is true for all x. 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. On the other hand. In fact. if x is a point of continuity. in
general. the sum of the Fourier series of fat I is
F(I -) + F(1+)
2
. The periodic extension off can also be utilized in finding the sum of the Fourier series of fat the endpoints ± 1. and the left-hand side of (6') reduces to f(x). Furthermore. the sum of the Fourier series of f at each of the endpoints ±l is Z(f(l-) + f(-1+)]. and 4 of the last section are relevant to this section as
well.4 Convergence of Fourier Series
403
Figure 10. if f is continuous everywhere and satisfies the Dirichlet conditions. which agrees with f on I. we have f(x -) = f(x +) = f(x). provided that f and f ' are piecewise continuous on I. cos mrx + b sin narx
(6')
which is true for all x. The Remarks 2. satisfies the Dirichlet conditions. the Fourier series off and its periodic extension are identical.
. false at the point where f is discontinuous. In fact. if F denotes the periodic extension of f.10. Then the periodic extension of f. The conclusion of Theorem 1 is also true for functions f which are only defined
on an interval I with endpoints -1 and +1.
.
a_r
0
f (x) cos nx dx = . which in this case is 0...
Solution The function f. ±Tr. the only points in the interval -Tr s x s IT where for
f' is not continuous are x = 0. . It is identical to f everywhere except at the discontinuities of f.and right-hand limit. At each of the points 0.
Sketch for a few periods the graph of the function to which the series converges..3. whose graph is known as a square wave of period 21r and amplitude 1. Here I = Tr and
f (x) with
1)
+ i (a cos nx + b. ±2Tr. Therefore.
Y
1
I
-2rr
[-rr
0
rrl
2rrl
x
Figure 10. ±2ar. [Note: f'(x) = 0 in -Tr < x < 0 and
0 < x < Tr.
-it<x<ar
1.2. . See Figure 10.] The graph off is sketched in Figure 10.10
Fourler Series
EXAMPLE I
Find the Fourier series of the function
AX) __
1. satisfies the Dirichlet conditions (the hypotheses of Theorem 1) with 1 = Tr. .. the Fourier series of
f converges to f(x) at each point except 0...and right-hand limits at
these points exist and are finite. where the value of the Fourier series is zero.
0<x<Tr '
f(x + 27r) = f(x). ±n.
±2rr.1. ..4..
C
n=0.. the graph of the function to which the Fourier series off converges is now completely known.3
.-
* cos nx dx + . In fact..
Next we compute the Fourier series of f. From Theorem 1..* cos nx dx = 0.
The function f is continuous everywhere except at the points 0. where f is not even defined. sin nx). and x = ±a. the Fourier series converges to the average of the
left. the left. ±Tr.
2Tr.8
. . 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.10
Fourier Series
Y
Figure 10. where f is discontinuous. Therefore. ±4rr.6 shows the graph of the function to which the Fourier series of
f converges. the
Fourier series converges to the value IT. Theorem 1 applies and Figure 10.rr) = g(x)..
n = 1 .6. . 2. we compute the Fourier series of f.3 with c = 0 and i = -rr. Next. The Fourier series off is
f(x) .) At all other points the graphs off and the function to which its Fourier series converges are identical. 3.
g(x+2.
-a s x < rr.. it is advisable to use formulas (4) and (5') of Section 10. Since the interval 0_ x < 27r is not symmetric with respect to the origin.5
The function in this example is different from the function
g(x) = x.2 +
with
(a cos nx + b sin nx). At the points 0." x cos nx dx =
1r
n=0
10. (See Figure 10. ±2a.
a =1
Tr Jo
f(x) cos nx dx = 1 f.
Figure 10. ..
In the interval 0 s x < 2Tr.
coslx n
(2)
.x < Tr. -Tr <. 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. -1 :5 x<1.f(x+2) = f(x)
converges to f(x) everywhere. The function
f(x) _.
26. -Tr s x s 7r.f(x+21r) = f(x)
does not involve any sine terms.f(x+21r) = f(x)
is equal to f(x) everywhere.1 s x s I.
25. f(x+2Tr) = f(x)
converges to f (x) everywhere.5
Fourier Sine and Fourier Cosine Series
24.x s 1. .
27.2.f(x+27r) = f(x)
does not involve any sine term. The Fourier series of the function
f(x) = x2. 0 s x < 27r. The Fourier series of the function
f(x) = x2.
30.1 in connection with the solution of the heat equation.x s 7r.5
FOURIER SINE AND FOURIER COSINE SERIES
As we saw in Section 6.10.1 <. The Fourier series of the function
f(x) = x..
10.
29. The Fourier series of the function
f(x) = I x 1. The Fourier series of the function
f(x) = x2. .f(x + 2) = f(x)
is continuous everywhere.f(x + 2) = f(x)
satisfies the hypotheses of Theorem 1.
28. The function
f(x) = V. -Tr <.
5x . 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. for example. x2. the labor of computing the
Fourier coefficients is reduced. Sedss
A Fourier series of the form (1) is called a Fourier sine series. Before we establish the above claim.
(ii) odd + odd = odd. Then
F(-x) = f(-x)g(-x) = f(x) (-g(x)) = -f(x)g(x) = -F(x).
DEFINITION 1
A function f.7 are even. For example. (iv) odd x odd = even. Geometrically speaking.7
. 1. then in the interval 0 < x < I we have the choice to represent f as a Fourier sine series or a Fourier cosine series.x2 + x4 cos 2x are even and sin ax. The functions whose graphs are sketched in Figure 10.410
10
Fourl.
Let us prove.
which proves that F is odd. (iii) even x even = even. Set F = fg.. for such functions. and if f and f' are piecewise continuous there. a Fourier series of the form (2) is called a Fourier cosine series. even and odd functions have the following properties:
(i) even + even = even. 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. whose domain is symmetric with respect to the origin. we will review the concepts of odd and
even functions and see how. With respect to the operations of addition and multiplication. (v). I x 1.) Assume f is even and g is odd. x.x2 sin 4x are odd. and those in Figure 10. 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. In this section we will show that if a function f is defined in the interval 0 < x < 1. With respect to integration. 3 .8 are odd. (The others are proved in a similar fashion. the functions cos ax. (v) even x odd = odd.
the integrand in (6) is odd and in (7) is even. . from (3) and (4).
a cos n j x.4 are satisfied for the functions g and h. Furthermore. the Fourier series of an even function is reduced to
f (x) -.2 +
where the coefficients a. if f and f are piecewise continuous on 0 < x < 1.
0 < x < 1
g(x + 21) = g(x)
(12)
and its odd periodic extension by
h(x) _ {
l-f(-x)..
and
n = 0.. 2. we find
a. 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. 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.
b_
u
f(x) sin
n = 1. 2. = 0. . the Fourier series of an odd function is reduced to
f(x) -.. g.
.
n jx
dx. the hypotheses of
Theorem 1 of Section 10.
(9)
Odd Functions When f is odd. for even or odd functions. 3.412
10 FounerSeries
Hence.
f(-x). the formulas for the Fourier coefficients use the values of the function in the interval 0 < x < I only. since h is odd. we define its even
periodic extension by
g(x) = If(x). then g and g' and also h and
h' are piecewise continuous on -1 < x < 1.
(13)
Note that the functions f.
.
It should be remarked that. Then. are given by (8). Similarly. and h agree in the interval 0 < x < 1.
mrx
(10)
Hence.i b.
If a function f is defined only in the interval 0 < x < 1. Since g is even. -1<x<0
0 < x < l
h(x + 21) = h(x). 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. and to the average
h(x-) + h(x+)
f(x-) + f(x+)
. Therefore. 1. sin
where the coefficients b are given by (10). -1<x<0'
f(x)..
is shown graphically in Figure 10.9.1) + n sin
2
J cos
I=
. .q + j [n4 (cos Z .1 < x < 2
Solution The even periodic extension off is sketched in Figure 10.
-6
-5
-4 -3
O
-2 -1
o
2
O
Z7--G
4
5
o
6
iX
I
3
Figure 10. ±4.
f(x) -. the Fourier cosine series off is n2x f(x) 2 + a cos
with
a. referred to here as g. = fa
1
f(x)cosn2xdx
= t xcosn2xdx
2. ±2.
Solution
I = 2. .
EXAMPLE 3
Compute the Fourier cosine and the Fourier sine series of the f(x)
function
Ix. /n=0
n ir' I cos 2 Hencc.10. ±6. From this graph we see that Theorem 1 of Section 10.4 is applicable. 0<x<1 0. From (14). 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.
) AX)
0<x<1
0. 2..
The only discontinuities of g are at the points 0.
The even periodic extension of f.
1 + n-Tr sin 2
n = 1.9
.
Y
.9.414
10
Fourier Series
EXAMPLE 2
Sketch the even and the odd periodic extension of the function
x.. and ±1.. . 3. and the odd periodic extension is sketched in Figure 10.
..11 (which is easily constructed from Figure 10. Next we compute the Fourier sine series of f. ± 2. The graph of the function to which the series converges is shown in Figure 10.5
Fourier Sine and Fourier Cosine Series
415
Figure 10.. ±5. t 2. t 3. From (15). From Figure 10...
AX)
nn 4 sin ni-Z 2-
n rr cos 2) sin 2 . ± 3.11
. The graph of the function to which the series converges is sketched in Figure 10..9 and from Theorem I of Section 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
.10..10
t3.9). ± 1.. t 4. . we
conclude that the Fourier series converges to 0 at the points t 0.
y
Figure 10. .10..4 is applicable. The only discontinuities of h are at the points 0. t 2. and converges to z at the points ± 1.4. is shown graphically in Figure 10.
2
nlr
n rrx
The odd periodic extension of f. referred to here ass h. From this graph we see that Theorem 1 of Section 10. we have
AX) .
JU
Hence.12. At every other point x the
series converges to g(x). . with I = 2. ..
Show that..
Compute the Fourier sine series of the function
fx(x) = 1 +x. and a repetition period of T seconds. Each pulse has a height H.
Compute the Fourier cosine series of the function
f(x)=1-x.
.4 in R. Englewood Cliffs. 1977).. Inc. B..4 in ibid. Electrical Network Science (Englewood Cliffs.
0<x<-rr. p. Reprinted by permission of Prentice-Hall.
n-'
(-ln
2
cosnIrx=2Ix14
14. Kerr.
0<x<a. New Jersey. 335. Englewood Cliffs.
16. Use the result in Exercise 13 to prove that
1 1 1
n=
15.J. Reprinted by permission of Prentice-Hall. 'This is Example 6. N.: PrenticeHall.
1(r)
17. 195.
18. Inc.
H
F-1
-W
I
F-1
Fl'
t
III--TH
'This is Exercise c6. Find the Fourier series' of the half-wave-rectified cosine shown in the figure. a duration of W seconds. for -1 s x < 1. p. Find the Fourier series expansions of the pulse train shown in the figure. New Jersey.420
10
Fourier Soria
13.
the potential (Laplace) equation. and 11.2). the initial value fixes the value of the arbitrary constant in the general solution. If one considers a first-order linear partial differential equation (to be defined in Section 11.
At first glance. The general theory of partial differential equations is beyond the scope of this book.10.
. In the case of ordinary differential equations. In this chapter we present an elementary treatment of partial differential equations. each coming
about by assigning a different value to the arbitrary constant. hence this general solution can be interpreted geometrically as a set of two-dimensional curves. the results are very different. That is to say. a partial differential equation seems to differ from an ordinary differential equation only in that there are more independent variables. 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.1
INTRODUCTION
Partial differential equations are equations that involve partial derivatives of an unknown function. However. The general solution can be interpreted geometrically as a collection of three-dimensional surfaces. in general it is quite different. although the study of partial differential equations frequently utilizes known facts about ordinary differential equations. one of the simplest types of ordinary differential equation is the
first-order linear equation (see Section 1. and we make no attempt to develop it here. and the wave equation.4). For instance. 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. 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. Uniqueness of
solution (in other words. The general solution of this ordinary differential equation contains one arbitrary constant. 11.
Unfortunately. the unknown function depends on a single independent variable.3. 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.CHAPTER 11
An Introduction To Partial Differential Equations
11. Example I illustrates this fact. In contrast.4.6 through 11. In this latter category we concentrate on the classical equations of mathematical physics-
the heat equation.
Y)](-1).
au
ax ay
(a) u = f(x . u = 0 when x = y = 0).y)"=> au = n(x .
Thus.y)^.y)"-'(1).y) => ax ` [f'(x . In Section 11.Y)](1).
au
au ax+a =n(x-y)^'-n(x-y)^'=0. where n is any positive integer. demonstrates that we cannot expect (in general) to obtain uniqueness of solution by specifying a particular curve that the general solution must contain. there are some similarities and analogies between ordinary differential equations and partial differential equations. but rather we will discuss particular situations.+ -au= 0. on the other hand.
Solution
(a) u = f(x . where f is any function having a continuous derivative.
.
au
ay
au
= [f'(x .y). In particular.y)"-'(-1). where c is an arbitrary constant. We will not attempt to investigate all of the ramifications of this issue. there is an infinity of solution surfaces (corresponding
to the values of c) that contain this point. then there is an infinity of solution surfaces (corresponding to the values of n) which contain this curve. If we specify that the solution surface contain the origin (in other words. u = 0 when y = x).2 we will discuss a few of them.
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. for partial differential equations.
au
ay
= n(x . More specifically. One point cannot uniquely determine the arbitrary function f. Appropriate conditions that produce uniqueness of solutions usually depend on the form of the partial differential equation. The solution in part (b).Y) . In spite of the above differences.
and
(b) u = (x .f'(x . suppose that f(x .
au
ax
+ ay = f'(x .y). if it is specified that the solution surface contain the curve y = x in the xy-plane (in other words.
Thus.
au
(b)
u = (x .Y) = 0. Example 1 illustrates that.422
11
An Introduction to Partial Differential Equations
EXAMPLE 1
Show that the following functions are solutions of the differential
equation
.y) ° c(x . uniqueness of solutions is not (in general) accomplished by simply specifying a point or a curve that the general solution must contain.
where n is a positive integer are solutions
of the partial differential equation
au ax
ay Y
= 0.. where f is any function possessing a continuous derivative. denoted by u. Furthermore. we define the partial differential equation (1) to be linear if F is linear as a function of the variables u. for example..
. is a solution of the partial differential equation du + 2
= 0. (1) we have used the subscript notation for partial differentiation.2
DEFINITIONS AND GENERAL COMMENTS
To simplify the notation. u.... and if g possesses a continuous
derivative. 2. y. ur = u..
is.
_ au
u
az
u . u. u u... z. = u. u. ) = 0... u-.. Show that u = f(2x . and no more than three independent variables. show that u = g(3x + 2y) is a solution of the partial differential equation 2 ax xy-plane.
. If g is any function such that g(0) = 0. denoted by x.... where f is any function possessing a continuous
derivative.. 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. u.. and (b) u = (x + y)^.
and soon.. u. With these constraints we can define a partial differential equation to be a relation of the form
F(x. that
_
a2u
axeY
a"'
_
a3u
ax3 .. u. .. 0).. 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. and so on.
Just as in the case of an ordinary differential equation. y.
(1)
In Eq. Show that (a) u = f(x + y).2
DaflnItions and General Comments
EXERCISES
1. Equation (1) is said to be quasilinear if F is linear as a function of the highest-order derivatives. 0. . u. Give an argument to support the statement that
there is an infinity of solutions which contain the curve y = .. u".. u.y)....y
.
dq 3. . that is.
au
3
au
aY = 0..
u. F is a linear combination of the unknown function and its derivatives.11.x in the
11.. we restrict our attention to the case of one unknown function. Give an argument to support
the statement that there is an infinity of solutions which contain the point (0. = u. u u..
Y.
+ a. (8) it is understood that the function f and the coefficients a. Eq. and (5) are nonhomogeneous. and u is unknown. and Eqs. (8) is called a partial differential equation with variable coefficients. (8) we mean a continuous function u. we will limit our discussion to linear partial differential equations of order one or two.+u. z) = 0.
(9)
. (3).
EXAMPLE 1
Find a solution u = u(x.. + a. + a. Y.
Equations (2). + a9(x. With very few exceptions. + 8u. + 2u = 0..1. By a solution of Eq. z)u = f(x. If every one of the coefficients a.424
11
An Introduction to Partial Differential Equations
The following are examples of partial differential equations:
u. y. y. which. y. and Eq.y) was a solution of u + u.
7u. .. z)u. (3). y..(x. y) of the partial differential equation
u. Eq.(x.-2x2-3z
u = x2
(2)
(3)
5xyu. Equation (6) is quasilinear. the partial differential equation (8) is called homogeneous. + a.
(8)
In Eq.. = 0.. (5).. (8) is called a partial
differential equation with constant coefficients. (4) has
variable coefficients. (4)
and (7) are homogeneous. Thus.(x..o(x. is a constant. Note that Eqs. Y. reduces Eq. are known. z)u. in Example 1 of Section 11. Equation (6) is nonlinear because it is not of the form of Eq. y.. y.(x. + 3u = 2xey. with continuous first.3zu. If at least one of the a. Y. is not a constant. z)u + a2(x.. = x + Y. (8) to an identity. y.. of the independent variables x.(x. z)u.(x. otherwise it is called nonhomogeneous (or inhomogeneous). + a. z)u. and (7) have constant coefficients.
If f(x. z)u
+ a. z)u. z). when substituted in (8). we demonstrated by direct substitution that
u = f(x . (8). (2).=3u. z)u. + a6(x. z)u. z. y.and second-order partial derivatives. and the rest are linear.
(4)
(5)
(6)
(7)
Equations (2) and (5) are first order and the rest are second order. Thus our most general partial differential equation can be written in the form
a. Y.
(x. since
afa(y)
= 0. If we set
f(x.11. notice that even if c were not a constant but a function of the variable y. Since we seek the most general form possible for the solution.
. would still be a solution of the partial differential
equation (9).=z+x. z)ds.
(10)
We note that the "constant of integration" is denoted by c. When verifying this.). z) = JYf.
u = xyz + 2x2y + f(x.
(12)
Solution First we integrate partially with respect to y (treating x and z as constants) to obtain
ux = yz + xy + f. We conclude that Eq. z). Next we integrate partially with respect to x (treating y and z as constants) to obtain
u = xyz + gx2y + Jfi(s. is an arbitrary function of the variables x and z. (9) "partially with respect to x" (in other
words.
(11)
where f is an arbitrary function of y. 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. where f.
then our solution takes the form
g(y. f and g are to have continuous first and second partial derivatives with respect to their arguments. z). we need only substitute this expression for u in Eq.
where f.
(13)
where f is an arbitrary function of x and z. treating the variable y as if it were a constant) to obtain
u=zx2+xy+c. z) =f2(y. is a solution of Eq. y. (10) is not the most general
result possible unless we emphasize that c is to be replaced by an arbitrary
function of y. and g is an arbitrary function of y and z. (10).
EXAMPLE 2
Find a solution u = u(x. as given in Eq. z) + g(y. In order to verify that u.z)ds + f. z). (9). (9). then u as given by Eq. we integrate with respect to x.2
Definitions and General Comments
425
Solution We begin by integrating Eq. such as f(y).(s. with c replaced by f(y). (10). z).(y. is an arbitrary function of the variables y and z. we write
u = 2x2 + xy + f(y). z) of the partial differential equation
u.
n = 1..-u.= cosy + ex
u.. Each specific assignment of the arbitrary function(s) in the general solution gives rise to a particular solution of the corresponding partial differential equation. and determine whether it has constant or variable coefficients. s = 2.=0
13.. In Example 2. u. u.
3.
Even though it is difficult (if not impossible) to make all-inclusive general statements about partial differential equations.u..11
An Introduction to Partial Differential Equations
As in the case of ordinary differential equations. s = 3. = 0 17.-3zu+u.. in Example 1. n = 2. = 3u
7.. u... u. = 0
S. (Not necessarily the same
set of (s .=3x2+4y
15.. and the general solution contains the arbitrary function f.+5xu..
In Exercises 11 through 20.. we call the solutions (11) and (13) the general solution of Eqs.-u. = 0
.=sinx-siny
16. determine whether it is homogeneous or nonhomogeneous.5u.
1.
11. n = 1..=0 8. Thus.. u.) Thus. f depending on the variables x and z..y.y2
4. and g depending on the variables y and z. which depends on the single variable x . s = 2..
2. z) = ye]. u. assume that u is a function of the two independent variables x and y.. each of which depends on (s .=0
14. + u ..u.1) variables applies to each arbitrary function.+2zu = 3u. 3u .3u... u.=27y+z2 S.+17u=0
5. u.u..=cosz 10. . u.y + 4u. and
u = xy. The general solution to a linear partial differential equation of order n.u. for an unknown function depending on s independent variables. and the general solution contains the two arbitrary functions f and g. .
EXERCISES
For each of the linear partial differential equations in Exercises 1 through 10. which depends on the single variable y. and the general solution [given as solution (a)] contains the arbitrary function f. u = 0
18. (12) [for the particular choices f(x. (9) [for the particular choice f(y) = e'].1. In Example 1 of Section 11.+u.. z) = z cos x
and g(y.... u. u. the following is a reasonable claim for the partial differential equations to be treated in this text. Integrate each equation to obtain the general solution.
u=2x=+xy+e'
is a particular solution of Eq.1) variables..
12. involves n arbitrary
functions.= 3x' . + 2 x2y + z cos x + ye..
is a particular solution of Eq. state its order.. (9) and (12) respectively.
is a solution of the partial differential equation 0... u. =y 29.. = cos y + e'
20..3 The Principle of Superposition
427
19. + a. y.x
23. 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. +
Classify this equation utilizing all definitions of this section that are appropriate.]. + 24.(x... 32. Thus.11.. z)u. y. for the operator of Eq.u. A[u.u. 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. + c2u2] = c. z)u._.. u. In some instances the methods will be further illustrated
in subsequent sections.
21. = x2 + z
27.]. (8) we write. u.. (8).
22. 4. Our partial differential equation has the appearance of Eq...
(1)
We speak of the symbol A as an operator. 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... u.. The shearing stress and the normal stress in an elastic body are obtainable from Airy's stress function. + a9(x. Integrate each equation to obtain the general solution.
A[u] = a. A[u.(x... we write Eq. = sect y
In Exercises 21 through 30.
34.. assume that u is a function of the three independent variables x. u. y..=0
24. Section 11. and u2. = 0
25. u =y+ 3x
28.=2
30. u.
sec' x
31. u.3
THE PRINCIPLE OF SUPERPOSITION
In this section and the next two sections we outline some general ideas and methods of solution.
11.2.
. = 0
26. 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. y. u. For simplicity..A[u.. 33. u.
DEFINITION 1
An operator A is called a linear operator if A[c. = sec'. (8) in the abbreviated form
A[u] = f. c2 and for every permissable (in other words.. where 4.. z)u + .] + c2A[u2] for every choice of the constants c. u. 4. and A[c. + c2u2] make sense) choice of functions u. (8). Elasticity. and z..
. (ii) If u.. c. be any constants. that is...
Suppose that u is a function depending on the variables. + c2u2 +
+ cmum]
= c. + u. = xyz is a particular solution of the equation u..u. and that u satisfies the partial differential equation
EXAMPLE 1
uxy = z + X. then E . where f and g are arbitrary func-
tions. z) + g(y.u. then u = u..... then u = u. A very important property of linear partial differential equations is contained in the following principle.. is a solution of the equation = f. = 0 is given by u. . {A'' [u1] + c2A[u2] + . any linear combination of solutions is a solution. Two important consequences of the principle of superposition are as follows:
(i) If u u2.... respectively.. but the methods of solution are not as concrete or systematic for partial differential equations as for ordinary differential equations. = f where
... .. then 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. The same approach is sometimes useful in trying to solve a
nonhomogeneous linear partial differential equation. the equation
A[u] = 0 is called the associated homogeneous equation.. .. are solutions of A[u] = 0 and c c2.u.. + c2u2 +
A we have
A[u] = A[c. .A[u. . A[u.
and the proof is complete. . are any
constants... and up is a
particular solution of A[u] = f. x. z). if u....
Principle of Superposition
are..u.]
c1f +c2l2 T
+cJ.] = f A[u2] = f2. is the general solution of the homogeneous equation A[u] = 0... (ii) u. + c.. u.428
11
An Introduction to Partial Differential Equations
DEFINITION 2
Given the nonhomogeneous partial differential equation (1).. is a solution of A[u] = f. c.. . ... y. ..] + c. solutions of the equations A[u. the sum of a solution of the homogeneous equation and a particular solution is also a solution..
(2)
Show that (i) the general solution of the associated homogeneous equation u.
Let f f2. u.. = f(x. f be any functions and let c c2... + up is the general solution of A[u] = f... that is. is also a solution of A[u] = 0. is a solution of A[u] = 0 and if up is a particular solution of A[u] = f. ... As in the case of ordinary differential equations.. If A is a linear operator and if u u2. . and z.
= X(x)Y'(y). u. and Z are to be determined.y = X(x)Y"(y)
Substitution of these results in Eq. Y. U. In this method we try to write the solution as a product of functions. two-.4 SEPARATION OF VARIABLES
A frequently used method for finding solutions to linear homogeneous partial differential equations is known as separation of variables.
then
u. where the functions X. The basic ideas and manipulations involved in the method are illustrated in the following examples.6.11. Y are to be determined. For example. = aO [u]
(the heat equation). depending on whether the Laplacian operator is one-. u. each of which depends on exactly
one of the independent variables.r = X"(x)Y(Y). = c2O [u]
(the wave equation). where the functions X.7. two-. the one-dimensional heat equation is discussed in Section 11.
11. or three-dimensional.
. and the twqdimensional Laplace equation is discussed in Section 11. or three-dimensional. (1) leads to
X°(x)Y(Y) .
EXAMPLE 1
For the partial differential equation
(1)
find a solution in the form u = X(x)Y(y).
(the potential or Laplace equation). The one-dimensional wave equation is discussed in Section 11. For the partial differential equation
we would try to write the solution in the form u = X(x)Y(y).
Solution If
u = X(x)Y(y). we would try to write the
solution of the partial differential equation
u = X(x)Y(y)Z(z).
0 = A [u]
In each case the equation is called one-.
u.X(x)Y"(Y) = 0.4
Separation of Variables
433
Remark The classical equations of mathematical physics are
u.. = X'xx)Y(y).8..
we can write
X(x)
X(x)
and
Y(Y) =
Y(Y)
a. In fact. it must
happen that changes in the variable y do not affect the expression
Y
either.ru + c2e -. the expressions L (X) and Y "(Y) must X(x) Y(Y) be constants. they must be the same constant.rte
X> O
C1+C2X.e .11
An Introduction to Partial Differential Equations
Divide this latter equation by u = X(x)Y(y) (assuming that u # 0) to obtain
X.
. (4) Equations (3) and (4) are ordinary differential equations with constant coefficients and can be solved by the methods of Section 2.5 to yield.
X"(x) Similarly. if (2) is to be an equality."(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.
el cos VC
A<0.
.
X"(x) . changes in x should not affect the expression The net conclusion X(x) is that in order for (2) to be an equality.
X(x)
A=0
+ c2 sin N/ -\x.
c.
C3 cos
x=0
y + C 4 sinVy.
Thus.X Y(Y) = 0.XX(x) = 0
and
3)
Y"(Y) . If the constant is denoted by A.
C3C'Y + c4e "
A>0
Y(Y)=
C3+c4Y. we note that changes in y will not X(x)
have any effect on the expression X( ).
A<0 . Thus.
= X(x)Y(y)Z'(z).
A>0
0
X(x)Y(y)
(c
c.5X(x)Y(y)Z'(z) = 0. Thus 3X'(x) (6) X(x)
and
-
2Y'(y)
Y(y) +
5Z'(z)
Z(z) -
(7)
.0
3X'(x) 2Y'(y) + SZ'(z) (5) Z(z) ' Y(y) X(x) Using the same type of argument as in Example 1. we conclude that the only way that Eq.
e
A + c4 e Y
" 1.
uy = X(x)Y'(y)Z(z). hence we cannot specify the form of the solution. say N.
Solution
If
u = X(x)Y(y)Z(z)
then
u. (5) can be an equality is that both sides of the equation equal a constant.-2u. these conditions usually dictate the value of A and the form of the solution (see Sections 11.10).
Substitution into the partial differential equation yields
3X'(x)Y(y)Z(z) . we have
3X'(x) X(x)
or
2Y'(y)
Y(y)
5Z'(z)
Z(z) . -5u.
I (c3 cos Vy + c4 sin V y).6-11.4
Separation of VarlabMs
435
Thus.
EXAMPLE 2
For the partial differential equation
3u.
U
(ce
J
x + c2) (c3 e -.=0.\.
Dividing by u = X(x)Y(y)Z(z) (assuming u # 0).
find a solution in the form u = X(x)Y(y)Z(z).
and
u. In many practical problems there are other conditions that the solution must satisfy.2X(x)Y'(y)Z(z) .x
A < 0.ru
crr) (c3 CJ').
Without further information we have no way of knowing the value of.11. cos N/--lx + cI sin
->. = X'(x)Y(y)Z(z).
(10) is Z(z) = c. u.
Y(Y) -
Once more we argue that in order for (8) to be an equality.")'si'."
EXERCISES
In Exercises 1 through 22. Equation (7) can be rewritten as 5Z'(z) 2Y'(y) (8) Z(z) .-3u. (9) is Y(y) = ce("`2)Y. assume a solution in the form u = X(x)Y(y). Show that the equation "separates." and find the differential equations that X and
Y must satisfy.51t) =
ke)b3)x-(W2)!-[(" -")15]t
In Examples 1 and 2.
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). the variables "do not separate. It is not the case. we must have 2Y'(y)
Y(y)
= µ>
(9)
and
XSZ(z)=µ. Then ut = X'(x)Y(Y). and the solution to Eq. however.=0
2.436
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.+3u.=0
. the partial differential equations have constant
coefficients. U.. Therefore a solution to the partial differential
is
u = (c. therefore we conclude that the method does not work for this partial differential equation. The solution to Eq. but the method can also be applied to equations with variable
coefficients. = X'(x)Y'(y). that is. Example 3 is a case in point.").el('.
Z(z)
(10)
where µ is a constant.e(")') (C2e(w2)Y) (C3el`".4u. There are many equations for which the method does not apply. 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.
1.el"3j`. It is not possible to algebraically manipulate this latter equation to a form P(x) = Q(y).
z.
where a2 is a constant. Acoustics The nonlinear partial differential equation
(u. one considers the velocity potential u(x. Given
u. F.(u. Separation of variables for partial differential equations normally relates to the method described in this section.
64. Y.) It can be shown that u satisfies the three-dimensional wave equation (see the remark following Exercise 50. (The term potential is usually used in physics to describe a quantity whose gradient furnishes a field of force.)"
lu. p. = azu
occurs in the study of the propagation of sound in a medium.
66. and T. (a) Set u = X(x)Y(y) and determine the differential equations for X and
Y. Supersonic Fluid Flow In the study of the supersonic flow of an ideal compressible fluid past an obstacle..
. In this case the gradient of u yields the velocity of the flow. 459.
65. the velocity potential satisfies the equation
(MIwhere M(>1) is a constant known as the Mach number of the flow. Section 11. + u) = 0 x+y
occurs in the one-dimensional isentropic flow of a compressible fluid. 2 (New York: lnterscience
Publishers. Set u = X(x)T(t) and determine the differential equations for X and T. Hilbert. Solve these differential equations.. Set u = X(x)Y(y)Z(z)T(t) and determine the differential equations for X. t). let us assume a solution of the equation
'R.. 1962). Courant and D. Isentropic Fluid Flow The second-order linear partial differential equation
a u. where gradient F is the vector [F. + a2U = U.
(b) Solve the differential equations of part (a). 67.3)
U + U. Methods of Mathematical Physics..' a is a constant which depends on the fluid.y. To illustrate.u. y.+ U =
.. Set u = X(x)T(t) and determine the differential
equations for X and T. . There are other techniques that may also be called separation of variables.4 Separation of Variables
439
62.. Acoustics In the study of the transmission of sound through a moving fluid.
63. a is a constant and represents the velocity of sound in the medium.
c
I
where the constant c represents the velocity of sound in the medium. Fy. vol.].11. Set u = X(x)Y(y) and determine the differential equations for X and Y. Z...
or both.
Thus. or one or more of the initial conditions may be missing.. f. x = 0 and x = 1. thus setting it in motion. f3(x). The distance along the string will be denoted by s and as usual
ds =
(dx)2 + (dy)2.8 The Homogeneous One-Dimensional Wave Equation: Separation of Variables
441
where a. y) are assumed to be known functions and A..(x).6
THE HOMOGENEOUS ONE-DIMENSIONAL WAVE EQUATION: SEPARATION OF VARIABLES
Consider that we have a string that is perfectly flexible (that is to say. The main prerequisite for utilizing the method we discuss is that the method of separation of variables be applicable to the partial differential equation. If no other forces are acting
on the string. For a given problem.
as
ax
+ dye
ax
. and Eqs. one or more of the boundary conditions. y). In Sections 11. we have the situation illustrated in Figure 11. f..
may be infinite. where T represents the (constant) force due to tension. Consequently. While we do not solve the general case of Problem 1. we do provide the necessary ingredients for the solution of many problems that fall into the category of Problem 1. Equations (2) and (3) are called the boundary conditions.9 and 11.
11 .
In Problem 1. y) * 0. 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. . a6(x. y. and f. the tension in the string is approximately constant.
If we assume that the displacement. y) = 0. then approximately
as
= 1.11. y).10 we consider special cases of Problem 1 with F(x. is small enough so that (ax)2 is a very
small quantity (in comparison to 1).
. I or m. f. In Sections 11.(x.6-11. f. for specific cases. that is. how to solve Problem 1 with F(x. and F(x. . we obtain the associated homogeneous problem..(y). (4) and (5) are called the initial conditions.x).. .1.8 we demonstrate. f2(y). f2. The string is to be stretched
and attached to two fixed points on the x-axis. .. 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). each equal to zero. A6 are known constants. if we set F. the
length of the string is approximately unchanged (since s . The string is
then given an initial displacement and/or an initial velocity parallel to the y-axis.
then the mass of the portion of string in Figure 11.x and taking the limit as Ax tends to zero. Since the displacement is considered to be small.tan 9z = y'(x + Ax). ax ax ax as
At the point (x + . y) is given by
. = T ax2
.
. y + Ay). sin °2 . R.-O Ox
lim 1 R=0. has the property
. Expanding y'(x + Ax) in a Taylor's series. however.x. the vertical component of the force T is
T sin 02.1 is given by pds.T sin 0. Thus. namely.
az where the remainder.11 An Introduction to Partial Differential Equations
The vertical component of the force T at the point (x. the vertical component
is
z
T
+T
ax.
Ax + R. we obtain the equation of motion of the vertical displacement of the string. which is approximately pOx.
azy
P
:
at.Tay= -Tay as = -TaY.
If p represents the mass per unit length. for motion in the vertical direction.
Dividing by . T as . we have by Newton's Second Law )of Motion
pdxaa
=
Tas+ Tay1x+R+l-Tax)
= T ds Ox + R.
Thus.. are discussed in Sections 11. (1) is frequently referred to as the equation of the vibrating string. t>0
u(x. A standard approach to solving this initial-boundary value problem is to use the separation of variables method of Section 11. T are unknown functions to be determined. Note that the wave equation is an example of a hyperbolic partial differential equation. Section 10.4).
(5) (6)
It can be shown that if f and g satisfy the Dirichlet conditions (see Theorem 1.7 respectively. 0<x<1.t`
(8)
. 0) = f(x). t 0. (1) the one-dimensional wave equation. -czu.
(1)
where c2 = Tip is a constant according to our assumptions.
0 < x < 1. 0 < x < l
u.c=u = 0. (2) to be of the form
u(x. the initial-boundary value problem (2)-(6) has a unique solution. 0) = g(x). The wave equation is one of three partial differential equations known as the classical equations of mathematical physics. For obvious reasons.11.(x.=0.6 The Homogeneous One-Dimensional Wave Equation: Separation of Variables
Conforming now to our convention of denotirig the unknown function by u.
with X a constant.
t > 0. The other two. (2) yields
XT"-c2X"T=0. Substitution of (7) into Eq.
u. t) = 0. t z 0 u(l.
(7)
where X. the potential (Laplace) equation and the heat equation. To this end we assume a solution of Eq. A typical initial-boundary value problem for the one-dimensional wave equation is the following. t) = X(x)T(t). It is also customary to call Eq.
T"
c2X"
T
X = X. Eq.4 and Fourier series (Chapter 10). we rewrite the equation of motion in the form
u . t) = 0. Consequently we have two separate problems:
T" _
T . 0 < x < 1
(2)
(3) (4)
u(0. Condition (3) represents the initial position of the string and condition (4) is the initial velocity. 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.8 and 11.
therefore. then. t) _ X. X(0) = 0.
This problem can be solved by the methods of Section 6. we consider an alternate approach. is to be a solution of the eigenvalue problem
X".
. the function X. condition (4) demands
narx
!
Hare
sin
(1 b = g(x)
(14)
This... If f is of this form.3.2. where B is a constant...
(10)
and the corresponding eigenfunctions
1
.
T=0
(11)
With K given by Eq... to yield the eigenvalues
z:
n jz
. . restricts g to be of the form B sin nlx .444
and
11 An Introduction to Partial Differential Equations
c2X
X
= a.. 3 . 3. too. The
function X.2.. cos
nc
t + b sin
n cc t.(x)T (t)....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
..(x)T (t) is a solution of Eq..
!Z
H ence
T(t) = a. (2) that satisfies conditions (5) and (6). 2. What about conditions (3) and (4)? Let us investigate condition (3) when u(x.
(12)
where a and b.
n
=1 . c=
X(0) = X(1) = 0. the boundary condition (6) indicates that X(1) = 0. thus. Eq. 2 .
(9)
The boundary condition (5) demands that X(0)T(t) = 0 for all t ? 0. 3. 2.).
Conditions (13) and (14) place too great a restriction on the permissible forms for f and g. where A is a constant. Similarly.
n = 1. (8) takes the form n2arZ
T" +
. We conclude that for each specific value of it (n = 1. are the integration constants in the general solution.
u(x. with it not specified but otherwise considered fixed.
n=1...X X=0. (10).
as is easily verified. u(x. 0) = x(ir .5)
-1
a" =
1
Of(x)sinnjxdx.4u_ = 0. (16) and (17) respectively. Consequently. t) = 0. X (x)T (t) is a solution of Eq._. to be a solution of Eq. We will not investigate this question here but rather emphasize the method of solution.. Naturally there is the question of whether or not this infinite series converges.
that is. 0 < x < IT. .(x.
0 < x < n. (2) is a linear partial differential equation. that the Fourier sine series for f(x) in the interval 0 s x s I be nrx
a" sin
. 2.
u(a.
(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.
u(0.6 The Homogeneous One-Dimensional Wave Equation
445
Since X"(x)T (t) is a solution of Eq.
(18)
EXAMPLE 1
Solve the following initial-boundary value problem.. given by (11) and T given by (12).
U11 . b" is given by
b"
nac o 8(x) sin n x dx. t) satisfies conditions (5) and (6).11. t) = 0. it seems reasonable to expect that I. We consider
u(x.).x). t) = i X (x)T(t). 3. (2). (2) for each value of n (n = 1. a" is given by (see Section 10.
. t) will satisfy condition (3) provided that
a" sin
mrx
= f(x).
(16)
Likewise.
(>O'
u(x. 0) = 0. condition (4) will be satisfied provided that
b"Isinnx = g(x)
In other words. and since Eq. u(x.
t ? 0.
(15)
with X.
u.
0 < x < IT.
t ? 0. (2).
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
(5) gives the initial distribution of temperature.
t > 0. we find that X is to be a solution of the eigenvalue problem
X"-aX=O. t) = 0.4).
X(0) = 0. ... 3. t>0.
(9)
(10)
X(x)=sinnjx..
u(0.
EXAMPLE 1
Solve10 the initial-boundary value problem
0<x<1. We seek a solution in the form u(x. we obtain the one-dimensional heat equation
(1)
where a = klcp is known as the diffusivity. t) = 0.
The eigenvalues and eigenfunctions of problem (6)-(7) are. u(l.. 2. Note that the heat equation is a
parabolic partial differential equation. t > 0.
n27r2a
(8)
l2
.
(2)
(3) (4)
(5)
u(x.
and T is to satisfy the differential equation
(6) (7)
T'-XT=O. 0 < x < 1. (8) is T(t) = c. Using the method of separation of variables.. problem (2)-(5) has a unique solution.e Thus. x(l) = 0. Section 10.
.
u(x.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.
For k as given in (9). t) _
X (x)T"(t)
1OIt can be shown that if f satisfies the Dirichict conditions (Theorem 1. respectively. 0) = f(x). .2. the general solution of Eq..
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).
n = 1.3. n=1.
Some of these higher-dimensional problems can be handled
in precisely the same manner as in this section. For this reason we have restricted our attention to the one-dimensional heat
equation. show that the twodimensional heat equation can be written in the form
U. u) = cpq(x. See Exercise 24. u) = F(x. t.
0<x<1.
0 < x < 1. L.11. Berg and 1. if r is negative. 1 = a = 1. This case can be thought of as heating of the rod caused by a chemical reaction that is proportional to the local temperature. t > 0. Section 11. (a) q(x.
(19)
where h(x. t). p.
u. 0) = f(x). Automatic Heat Control The initial-boundary value problem
u. If r is positive.and threedimensional problems that lead to relatively simple solutions (see Exercises 26. u) is the rate of production of heat energy per unit volume per unit time. and 32 through 36). u). however. t) = u(1.
(19) with q = -au. the initial-boundary
value problem for v is solvable by the methods of this section.
"P.7 The One-Dimensional Heat Equation
Consequently one can consider heat diffusion problems in two and three
dimensions also. however.3. the analysis is
complicated by the fact that the corresponding eigenvalue problems involve Bessel and Legendre functions.1
27. . a heat source exists).9 for a specific example.
.
26. a(> 0) a constant. [Hint: Set u = e-' v(x. the rod is giving off heat (thus.
u(x. t. There are. W. Section 11. t>0
t > 0.]
28.
Set up and solve the initial-boundary value problem consisting of Eq. t) = 0. then the equation" of heat flow is
u. McGregor. The following special cases are of interest. t). t). 32. 1966). =
a
(ru.
u(l. t) = Au(0. :)u. Elementary Partial Differential Equations (San Francisco: HoldenDay.=au_. the rod is receiving heat (thus.
(18)
[Hint: See Exercise 50.(0.au_ = q(x. u) = r(x. 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. a heat loss
exists). This situation corresponds to heating caused by an electric current through the rod and gives rise to a nonhomogeneous problem. If the temperature is a function of the radius only. 0) = x(1 . (b) q(x. some special types of two. These functions are considerably more difficult to deal with than the sine and cosine functions heretofore encountered. t.). u(0. t. u(x.x).1) = 0. t.
t) denotes the
. Just prior to contact.
. with the hot rod in the middle.). Two rods of the same type of material. solve this initial-boundary value problem. describes the temperature in a rod whose left end (x = 0) is thermally insulated (that is. are placed face to face in perfect contact.11
An Introduction to Partial Differential Equations
where A is a constant (A # 1).).2 for the material. a = 1. 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. long.
30.
33. one of the rods has constant temperature 100°C.I < A < 1. (b) Show that there are no real eigenvalues in the case A < . 31. The outer faces are maintained at 0°C.
(b) Show that
(r 2u.. from this common face twenty minutes after contact is made.02.
1
where A is a constant. long. The rods each have constant temperature. two at 0°C and one at 100°C. Three rods of the same material.
2<x<1 .
32. 0<x< 2
U. no heat is lost through this end. The rods are placed end to end. (a) Find the temperature distribution as a function of x and t after contact is made.1. 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). the other has constant temperature 0°C throughout. (a) If . (a) Referring to Exercise 50(c) of Section 11. each 10 cm.
29.3 and the Remark which
follows that exercise. =
r (ru) . find to the nearest degree the temperature at a point on the common face and at points 5 cm. hence ux = 0). [Hint: See Exercise 29. 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. The initial distribution of temperature in the rod is f(x).] (b) If a = 0. 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. If u(r. are each 10 cm. =
a
r
(r2U. Solve the initial-boundary value problem (2)-(5) when
/Tx) = 1
IA.
t) = 0. Initially the temperature is 100°C throughout the sphere. and consider the flow of heat in the differential portion of the sheet depicted in Figure 11. 34.5j.. is made of material for which a = 0.r < R. 35. if we set
v(r.
0 <. Find the temperature
(to the nearest degree) at the center of the sphere 20 minutes after the
cooling begins. we would have v(0. 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. 36. Repeat Exercise 33 for g(r) = RZ . then u(r.]
11.2. 0) = g(r). Repeat Exercise 33 for g(r) = A. u(r. therefore.2. (b) Find u.1 1. Using the same arguments as in Section 11.r2. t) = ru(r. and the sphere
is cooled by keeping the surface of the sphere at 0°C. [Hint: See Exercise 33.
t > 0.8 THE POTENTIAL (LAPLACE) EQUATION
We begin this section with a consideration of heat flow in two dimensions. t) = 0. t).8 The Potential (Laplace) Equation
461
temperature in the sphere as a function of the radius r and the time t only. A solid sphere of radius 10 cm. a constant. t) satisfies the problem (see Exercise 32)
u. = a(ru). :>0
u(R.
Y+Ay ayY
kDt
or
1
Thus.7.. 0<r<R. Refer
to Section 11.
The conducting material is in the shape of a rectangular sheet of constant
thickness D. 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
. (a) Set up and solve the
corresponding initial-boundary value problem for v.
It is reasonable to assume that u is bounded at r = 0.7 for the appropriate definitions and the experimental laws associated with heat flow.
Y)
(x+Ax. for R.2
By Law 2 of Section 11.y)
Figure 11.462
11
An Introduction to Partial Differential Equatlona
(x.
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 . Note that the potential equation is an elliptic partial differential equation. In addition to the steady state temperature for two-dimensional heat flow. this rate of gain of heat is also given by
cpDOxAy at . Hence. = a(uu + ayy). of the xy plane. a typical problem associated with the potential
equation is a boundary value problem. u is independent of time and u. the temperature u is a function of x and y only
(that is. Consequently.x and Ay tend to zero.
where a = k/pc. say R. 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.
If at a future point in time.7.. = 0). one says that steady state conditions
have been achieved. + uyy = 0. there are many other physical applications in which the potential equation occurs. we obtain the two-dimensional heat equation
u. and that the values of u are known on the
boundaries of this region. Thus. the partial differential equation for the
steady state temperature for two-dimensional heat flow is
u. a common situation is that u is to satisfy Eq.
.
(1)
Equation (1) is called the two-dimensional potential (Laplace) equation. (1) in a fixed bounded
region. commonly referred to as Dirichlet's
problem. In this case and for that of two-dimensional heat flow (as well as others). One application is that u represents the potential of a two-dimensional electrostatic field.
10)..-1
where the functions T.
Solution If we set F(x. t) = i
.
Consequently. t) = 0 is solvable and that the
"boundary terms" in the integration by parts process [see Eq. we can write
T JO = (u.
t > 0.
0 < x < 1.
EXAMPLE 2
Solve the initial-boundary value problem
0<x<1. By Theorem 1 of Section 6. t) = 0 in Eq.(x)dx. (10) and (11). As in Example 1. the constant c in Eq.
(14) (15) (16)
u(0.(1) are to be determined. t) = 0. (5) and T . The latter requirement will be met if the boundary conditions for the original problem are u(0. xx
. 0) = fjx). (12). we note that
TJO) = fu(x. we then have the initial-boundary value problem of Section 11. (8)) either vanish
or are known as a function of t.2. The method of solution outlined in Example 1 is useful for many problems of the type encountered in Sections 11.9 Nonhomog.nsous Partial Diftarsntial Equations: Mathod I
471
From Eqs.(t) is given by Eqs.7.11.
(17)
. t)
T
(x) is given by Eq.
If this is not the case. and 11. t) = 0. 40 = J u(x./2-/l sin
n.(X) = -. 1>0
u(x. we set +.6.
and we seek a solution of problem (12)-(16) in the form
u(x. the solution of the initial-boundary value problem (1)-(4) is given by
u(x. it is sometimes possible to introduce a change of variables that makes the boundary conditions have this form (see Section 11.
t > 0.
u(l.
0
(11)
Finally. (5) and (6).
(12)
(13)
u.6. t) = 0 and u(1. 11..8.
0 < x < 1. t) = 0. 0) = g(x). (10) is given by
c = T JO) = J f(x)4.(x. It is important that the
associated problem obtained by setting F(x.
since the only criterion is that Problem 1' be a solvable problem.6.(0.A. Find the displace-
ment of the string as a function of the time i and the distance x.5 in the form A[u] = F(x. y)
A3v(x. 11.
t > 0.7.(1. y)
A3v(1. Take c = 1. t).(y) . y) .A6w. or 11. Y) .10 Nonhomogeneous Partial Differential Equations: Method II
475
displacement of the string as a function of the time t and the distance x. 26. Substitution of (1) into Problem 1 yields
PROBLEM 1'
A[v] = F(x.
u(x.v(0. y).A3µ'(1. .
10
IOl GENEOUS PARTIAL DIFFERENTIAL EQUATIONS:
METHOD
We write the partial differential equation (1) in Problem 1 of Section 11. 0) = f3(x). 0) A6v. Once w is chosen and v is calculated. in other words..(x.(0. 0) and (1. y). y) = v(x. y) + w(x.
. where A is a linear operator. 0).
A. Set
u(x. y) = f2(y) . t) .w(x.(x.au_ = F.(1.
(1)
where v(x.
t > 0.
u(0.(x) .A. For a given problem we may have considerable flexibility in the choice of w.A[w].11. (1).(x. A string is stretched between the points (0. t) = f. y) = f. y) + A2v.w(0.
0 < x < I.9 or by one of the methods in Sections 11. u = w + v. 0). y) is an unknown function to be determined. t> 0. 0 <x<1. The string is
initially at rest and is subjected to the external force 7r2x. 0) = f3(x) .A2w. Take
c = 1. t) so that Problem 1' reduces to a problem that is solvable either by the method of Section 11.
u(l.
11. and w(x. The idea is to choose a specific w(x. t) = 0. Y) + A4v.8. y) is to be a new unknown function. (1).
EXAMPLE 1
Find the solution of the initial-boundary value problem
u.A4w. 0) = f. the desired solution u is obtained from Eq.
where a.
t > 0.
0<x<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. and midnight corresponds to t = 24).(x. 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 angular displacement. = f(x.a/2 s y s a/2. p are constants. If a beam of uniform cross section has its axis coincident with the z-axis. The following initial-boundary value problem corresponds to a special case of the telegraph equation (Exercise 60.12) (the time t is measured in hours.11
An Introduction to Partial Differential Equations
14. t). u.
u + a2u = uu.
17. 1 AM corresponds to
t = 1. 1)
=v(x. t>0
0 <X < 1. 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).6).t)+11-xP
\\\
1
3!
. Section 11.x <_ 7/2. u(x. 0) = 0. y) _ .a/2 <. Set
u(x. Y) (4)
is known as the (two-dimensional) Poisson equation. Section 11. Solve the new boundary value problem as a first step. Consider that the surface of the earth is warmed by the sun so that the temperature. (4) with Ax.6) has the end x = 0 fixed and is initially at rest in its equilibrium position.)
16.
. for the beam satisfies Eq. t) = u(1. Take c = 1. t = y + a/2. 0) = 0. y). For convenience of the eigenvalue problem.
u. and note that Poisson's equation becomes
u + u = -2.2 and u = 0 on the boundary of the area of intersection
of the beam with the xy-plane. set s = x + a/2.
u(0.
)2
puN. Torsion of a Beam The two-dimensional nonhomogeneous Laplace equation uu + uy.
u(x. Find the stress function of a square
beam of side a units. then linear elasticity theory shows that the stress function. Suppose that a circular shaft (refer to Exercise 51.]
15.
t > 0. (Assume that the effect of time units in hours rather than seconds has already been accounted for and that no further modifications are necessary. t) = t3/3!. at the surface is given by 28 cos i2 (t . [Hint: The original rectangle is . set up and solve the
initial-boundary value problem for the temperature in the earth's crust when o = 1 = 1.
r. assume that u is a function of four variables x. u-. =
54x.. Show that u = x' + cos (x + 3y) is a particular solution of 9u .
u(r. indicate whether it is homogeneous or nonhomogeneous and whether it has constant
coefficients or variable coefficients.. + uY = 5x + 3y. 8..
r..r)
produces a nonhomogeneous equation (for v) with homogeneous initialboundary conditions (the solution of which involves Bessel functions). and f(r) = 100(101 . 2xy [Hint: set v = u. and f(r) is the initial distribution of temperature in the pipe. = 3xy 3. < r < r2. B is the temperature of the air (or medium) surrounding the pipe. < r < r2.. u. therefore. t>0.
.
r. + u. state whether it is linear or quasilinear.. t).. uu + 5u. if it is linear. 18. B = 0°C. u.=z'+2xy-t
6. 0) = f(r).5y2u. = a
A.
t>0.
r2 = 101 cm..
where A is the temperature of the fluid. and we wish to in-
vestigate the loss of heat through the sides of the pipe. take a = 1 and show that the
substitution
u(r... state the order of the partial differential equation. Integrate the equation to obtain the general solution. = 0
In Exercises 5 and 6. . t) = B. t > 0
u(r2.11.. For the case A = 100°C. y. u .u. + 3u = 0 2... Loss of Heat through the Sides of a Pipe Suppose we have a cylindrical pipe filled with a hot fluid (of constant temperature). The pipe can be considered a hollow cylinder of inner radius r outer radius r2.
5.]
7.
REVIEW EXERCISES
In each of Exercises 1 through 4.
1. u.2(u.. the temperature in the pipe satisfies the initial value problem u. t) + 100(101 .)2 + 2u2u. and. z and t..r).10 Nonhomogeneous Partial Differential Equations: Method 11
481
and determine an equation for the eigenvalues of the initial-boundary value problem for v(x. = 100 cm. Show that u = 3xy + x' is a particular solution of u.. t) = v(r. | 677.169 | 1 |
Product Description
Each lesson plan lists the primary concepts taught, learning objectives, materials needed (with page numbers referenced for the student materials), teaching tips, and assignments for the student to complete. The solutions-keys for both the student book and the (sold-separately) "Tests & Resources" book are included in this guide. General notes on preparing lessons and administering tests, a scope & sequence, lesson-to-worksheet correlation chart, and "appearance of concepts" list are also included. 399 pages, softcover.
I loved the elementary levels and was very happy to see the Pre-Algebra and Algebra 1 just when I needed them. I found them (teacher's guides) for both levels very difficult to use. The pages with answers are 1/3 the size found in the students books making it very hard to see. They didn't show how they came up with answers so I had to figure out making it very time consuming. I used both for my two oldest children and felt no different giving it two years-hoping I would grow accustomed to them. Hopefully they will revise and correct these issues. The curriculum is advanced but without a good teachers guide, it makes it more difficult | 677.169 | 1 |
Publication Date: January 25, 2012 ISBN-10: 1133112285 ISBN-13 :978-1133112280 Edition: 2
This book is for teachers that most calculus textbooks are too long. In writing this book, James Stewart asked himself: What is essential for scientists and engineers for the three-semester calculus course? Basic calculus: early transcendence, Second Edition, provides a concise main concepts of calculus teaching focus, methods, and support the precise definition of these concepts, patiently explained carefully graded. This book is only 900 pages – two-thirds the size of Stewart's other calculus textbook, but it contains almost all of the same theme. The relatively simple the main condensation Expo, and some of the features on the book's website, Although the volume is smaller, has a modern book, covering technology and combination of materials to promote understanding of the concept, although no prominent in Stewart's other books. The basic Calculus: Early beyond the eye with the same attention to detail, innovation, meticulous and accurate, the the Stewart textbook text in the world's most selling calculus. | 677.169 | 1 |
Precise Calculator has arbitrary precision and can calculate with complex numbers, fractions, vectors and matrices. Has more than 150 mathematical functions and statistical functions and is programmable (if, goto, print, return, for). | 677.169 | 1 |
This book is an introduction to combinatorial mathematics, also known as combinatorics. The book focuses especially but not exclusively on the part of combinatorics that mathematicians refer to as 'counting'.
The book consists almost entirely of problems. Some of the problems are designed to lead you to think about a concept, others are designed to help you figure out a concept and state a theorem about it, while still others ask you to prove the theorem.
Other problems give you a chance to use a theorem you have proved. From time to time there is a discussion that pulls together some of the things you have learned or introduces a new idea for you to work with. Many of the problems are designed to build up your intuition for how combinatorial mathematics works. There are problems that some people will solve quickly, and there are problems that will take days of thought for everyone. Probably the best way to use this book is to work on a problem until you feel you are not making progress and then go on to the next one.
About the Authors
Before his death in 2005, Ken Bogart was in the final stages of completing an NSF-sponsored project about the teaching of combinatorics through Guided Discovery. | 677.169 | 1 |
Product Description:
This book explores the standard problem-solving techniques of multivariable mathematics — integrating vector algebra ideas with multivariable calculus and differential equations. Provides many routine, computational exercises illuminating both theory and practice. Offers flexibility in coverage — topics can be covered in a variety of orders, and subsections (which are presented in order of decreasing importance) can be omitted if desired. Provides proofs and includes the definitions and statements of theorems to show how the subject matter can be organized around a few central ideas. Includes new sections on: flow lines and flows; centroids and moments; arc-length and curvature; improper integrals; quadratic surfaces; infinite series—with application to differential equations; and numerical methods. Presents refined method for solving linear systems using exponential matrices.
REVIEWS for Multivariable Mathematics | 677.169 | 1 |
This lively introductory text exposes the student to the rewards of a rigorous study of functions of a real variable. In each chapter, informal discussions of questions that give analysis its inherent fascination are followed by precise, but not overly formal, developments of the techniques needed to make sense of them. By focusing on the unifying themes of approximation and the resolution of paradoxes that arise in the transition from the finite to the infinite, the text turns what could be a daunting cascade of definitions and theorems into a coherent and engaging progression of ideas. Acutely aware of the need for rigor, the student is much better prepared to understand what constitutes a proper mathematical proof and how to write one. Fifteen years of classroom experience with the first edition of Understanding Analysis have solidified and refined the central narrative of the second edition. Roughly 150 new exercises join a selection of the best exercises from the first edition, and three more project-style sections have been added. Investigations of Euler's computation of ζ(2), the Weierstrass Approximation Theorem, and the gamma function are now among the book's cohort of seminal results serving as motivation and payoff for the beginning student to master the methods of analysis.
Designed for students having no previous experience with rigorous proofs, this text can be used immediately after standard calculus courses. It is highly recommended for anyone planning to study advanced analysis, as well as for future secondary school teachers. A limited number of concepts involving the real line and functions on the real line are studied, while many abstract ideas, such as metric spaces and ordered systems, are avoided completely. A thorough treatment of sequences of numbers is used as a basis for studying standard calculus topics, and optional sections invite students to study such topics as metric spaces and Riemann-Stieltjes integrals.In this text which gradually develops the tools for formulating and manipulating the field equations of Continuum Mechanics, the mathematics of tensor analysis is introduced in four, well-separated stages, and the physical interpretation and application of vectors and tensors are stressed throughout. This new edition contains more exercises. In addition, the author has appended a section on Differential Geometry.
This text for a second course in linear algebra, aimed at math majors and graduates, adopts a novel approach by banishing determinants to the end of the book and focusing on understanding the structure of linear operators on vector spaces. The author has taken unusual care to motivate concepts and to simplify proofs. For example, the book presents - without having defined determinants - a clean proof that every linear operator on a finite-dimensional complex vector space has an eigenvalue. The book starts by discussing vector spaces, linear independence, span, basics, and dimension. Students are introduced to inner-product spaces in the first half of the book and shortly thereafter to the finite- dimensional spectral theorem. A variety of interesting exercises in each chapter helps students understand and manipulate the objects of linear algebra. This second edition features new chapters on diagonal matrices, on linear functionals and adjoints, and on the spectral theorem; some sections, such as those on self-adjoint and normal operators, have been entirely rewritten; and hundreds of minor improvements have been made throughout the text.
Contains the exercises and their solutions for Lang's second edition of "Undergraduate Analysis." The variety of exercises, which range from computational to more conceptual and which are of varying difficulty, cover several subjects. This volume also serves as an independent source for those interested in learning analysis or linear algebra.
These notes were first used in an introductory course team taught by the authors at Appalachian State University to advanced undergraduates and beginning graduates. The text was written with four pedagogical goals in mind: offer a variety of topics in one course, get to the main themes and tools as efficiently as possible, show the relationships between the different topics, and include recent results to convince students that mathematics is a living discipline.
An introduction to complex analysis for students with some knowledge of complex numbers from high school. It contains sixteen chapters, the first eleven of which are aimed at an upper division undergraduate audience. The remaining five chapters are designed to complete the coverage of all background necessary for passing PhD qualifying exams in complex analysis. Topics studied include Julia sets and the Mandelbrot set, Dirichlet series and the prime number theorem, and the uniformization theorem for Riemann surfaces, with emphasis placed on the three geometries: spherical, euclidean, and hyperbolic. Throughout, exercises range from the very simple to the challenging. The book is based on lectures given by the author at several universities, including UCLA, Brown University, La Plata, Buenos Aires, and the Universidad Autonomo de Valencia, Spain.
Was plane geometry your favourite math course in high school? Did you like proving theorems? Are you sick of memorising integrals? If so, real analysis could be your cup of tea. In contrast to calculus and elementary algebra, it involves neither formula manipulation nor applications to other fields of science. None. It is Pure Mathematics, and it is sure to appeal to the budding pure mathematician. In this new introduction to undergraduate real analysis the author takes a different approach from past studies of the subject, by stressing the importance of pictures in mathematics and hard problems. The exposition is informal and relaxed, with many helpful asides, examples and occasional comments from mathematicians like Dieudonne, Littlewood and Osserman. The author has taught the subject many times over the last 35 years at Berkeley and this book is based on the honours version of this course. The book contains an excellent selection of more than 500 exercises.
With a fresh geometric approach that incorporates more than 250 illustrations, this textbook sets itself apart from all others in advanced calculus. Besides the classical capstones--the change of variables formula, implicit and inverse function theorems, the integral theorems of Gauss and Stokes--the text treats other important topics in differential analysis, such as Morse's lemma and the Poincaré lemma. The ideas behind most topics can be understood with just two or three variables. The book incorporates modern computational tools to give visualization real power. Using 2D and 3D graphics, the book offers new insights into fundamental elements of the calculus of differentiable maps. The geometric theme continues with an analysis of the physical meaning of the divergence and the curl at a level of detail not found in other advanced calculus books. This is a textbook for undergraduates and graduate students in mathematics, the physical sciences, and economics. Prerequisites are an introduction to linear algebra and multivariable calculus. There is enough material for a year-long course on advanced calculus and for a variety of semester courses--including topics in geometry. The measured pace of the book, with its extensive examples and illustrations, make it especially suitable for independent study. | 677.169 | 1 |
EMT1103 Mathematics for Architects
Course Description
This course introduces the student to the practical imperatives of the architectural profession. The student covers elementary mathematics applied in problem solving of architectural design challenges.
Objectives/aim
To provide an introductory treatment of mathematical concepts fundamental to architectural problems.
To consolidate and advance the material covered in Pre-University Mathematics.
To develop the analytical and critical thinking abilities fundamental to problem solving in architecture. | 677.169 | 1 |
Chap 1 the number system
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Unformatted text preview: FIRST YEAR CALCULUS
W W L CHEN
c
W W L Chen, 1982, 2005. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It 1
THE NUMBER SYSTEM 1.1. The Real Numbers The purpose of the first four sections of this chapter is to discuss a number of the properties of the real numbers. Most readers will be familiar with these properties, or have at least used most of them, perhaps sometimes unaware of their generality. We do not propose to discuss here these properties in great detail, and shall only give a brief introduction. Throughout, we denote the set of all real numbers by R, and write a ∈ R to indicate that a is a real n...
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This note was uploaded on 02/01/2009 for the course MATH 2343124 taught by Professor Staff during the Fall '08 term at UCSD. | 677.169 | 1 |
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Solving Exponential Equations using Logarithms.
This innovative activity is designed for Algebra, PreCalculus, or College Algebra
Included:
✓ Task Cards: There are two sets of 20 Task Cards one with and one without QR codes. Students do not need to be on the internet to use the QR's but must have a device with a QR reader installed. Cards #1 – 10 do not need logarithms or calculators to be solved, but, of course can be done that way. Cards #11 – 20 do need logarithms or calculators to be solved.
✓ Master List of Questions which can also be used as an assessment or class worksheet
✓ Student response sheet
✓ Answer keys
✓ Additional Quiz or Worksheet with ten questions and room for students to show work | 677.169 | 1 |
Description: This document is a resource that integrates a text, a large number of videos (more than 270 by last count), and hands-on activities. It connects hand calculations, mathematical abstractions, and computer programming. It encourages you to develop the mathematical theory of linear algebra by posing questions rather than outright stating theorems and their proofs. It introduces you to the frontier of linear algebra software development. | 677.169 | 1 |
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Bridging The Gap
This Bridging The Gap page is mainly aimed at year 13 students who have been accepted by Oxford to read for one of the Mathematics degrees, a joint school or Computer Science. The page contains various material - reading and exercises - covering some of the most important topics relevant to a student about to begin study at university. Some of the material is on a typical further mathematics A-level syllabus, but other topics are also included. This page is still being implemented, with many other articles being planned or in the pipeline.
Of course, anyone thinking about mathematics at university, or anyone studying further mathematics, or any school teacher is welcome to make use of these articles. The codings in the notes of M, M+ and M++ determine respectively whether the subject matter is highly relevant, of some importance, or tangential to the first year of the Oxford mathematics courses. Also the questions are graded from (A), the easiest, to (C), the hardest.
If you have any comments about this page, whether they be about errors spotted, or suggestions for future articles, then do please email the Schools Liaison Officerwith your thoughts. (I am sorry, but currently there are no solutions available for the exercises).
Documentation from previous Bridging Courses in Mathematics
Copies of these bridging notes can be bought from the Institute for £10 per copy - the price includes postage to UK addresses. Send a cheque for £10 payable to "The Mathematical Institute, Oxford", together with your address, to "Dr Richard Earl, Mathematical Institute, Andrew Wiles Building, Radcliffe Observatory Quarter, Oxford, OX2 6GG".
Study Guide - How do Undergraduates do Mathematics?
This is intended for incoming first-year undergraduates, but may be of interest | 677.169 | 1 |
Students find the optimal price for an insurance company premium in this game by interpreting data and applying their understanding of linear and quadratic models. [Access requires setting up a (free)... More: lessons, discussions, ratings, reviews,...
Students shoot a ball and earn higher scores the closer they are to hitting a target. They can improve their scores by interpreting data from the linear model that relates the distance of the target t... More: lessons, discussions, ratings, reviews,...
A story from the middle school classroom: Ihor describes how he scheduled a "contest" for students to show what they know about slopes and y-intercepts using the Green Globs software. He also provides... More: lessons, discussions, ratings, reviews,...
Students explore the relationship between equations and their graphs in this hands-on learning environment where they investigate, manipulate, and understand linear, quadratic and other graphs. They ... More: lessons, discussions, ratings, reviews,...
The SimCalc Project aims to democratize access to the Mathematics of Change for mainstream students by combining advanced simulation technology with innovative curriculum that begins in the early g... More: lessons, discussions, ratings, reviews,...
DC Proof is a PC-based educational software package. It is designed to facilitate students' transition to proof-based mathematics at the undergraduate or advanced high school levels. An interactive tu... More: lessons, discussions, ratings, reviews,...
DPGraph Viewer allows you to view and manipulate 2D through 8D graphs created by the full version of DPGraph. DPGraph Viewer comes with hundreds of examples contributed by users from around the world. More: lessons, discussions, ratings, reviews,...
Geomview is an interactive program written by the staff of the Geometry Center for viewing and manipulating geometric objects. It can be used as a standalone viewer for static objects or as a display ... | 677.169 | 1 |
Specific requirements
What will I study?
If you have a grade 3 at GCSE Maths, you will be required to re-sit this during your time at Huddersfield New College.
There is also the opportunity to work towards a grade 5 for those who need to improve beyond grade 4.
Course Content
This one-year programme will follow the AQA Linear specification at foundation level.
The pace of the course is fast and full attendance at the 3 lessons each week is essential for success. A calculator is a necessary item and must be brought to all lessons. There is the possibility of a November entry for students who are likely to get grade 4 at this time.
Assessment
For both the November and May sessions, assessment is by three papers:
• Paper 1 (non-calculator) - 1 hour 30
minutes
• Paper 2 (calculator) - 1 hour 30
minutes
• Paper 3 (calculator) - 1 hour 30
minutes
Assessment
Each FSMQ is assessed by one 60 - 75 minute written paper in late May (calculator) with pre-release material. Results are published in late August. | 677.169 | 1 |
Friday, March 28, 2014
SSC CGL Quantitative Aptitude PDF Study Material
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SSC CGL 2014 Exam is already announced and i am keep on providing study material for the SSC CGL subjects like Logical reasoning, GK and english. So, after all subjects here is the most difficult subject of SSC CGL Exam. Yes, this time Quantitative aptitude, here i am providing some good study notes and topics of Quantitative aptitude subject as per SSC CGL Syllabus in PDF format. You can get all the files easily from below and add some more fuel to your study preparation.
SSC CGL Quantitative Aptitude PDF Study Material
Check out the below mentioned some of the awesome study material for the SSC CGL 2014 Exam, hope you find it helpful. if you want more, you can always request more on below comment box.
1. List of Quantitative Aptitude Formulas - Click here to get it. The above PDF file having list of formulas on arithmetic progression, HCF and LCM, Surds and Indices and many more. You don't have to write it on paper, just get it and save it, reduce your work load. 2. Mental Math with Tricks and Shortcuts PDF- Click here to get it. This file having some tricks and tricks to solve complex sums of Quantitative aptitude section with easy language explanation. you can achieve speed in calculation with this shortcuts. 3. Shortcut for Calculation - Click here to get it. Here is some more useful shortcuts for the arithmetic calculation, more and easy explanation of the sums and more useful shortcuts. 4. System of Speed Maths Techniques - click here to get it. This file have some explanation on how to solve the multiplication sums in easy and fastest way. Do read it, most important for bank exam aspirants. 5. Important Formulas on Quant :- Click here to get it. This file have list of important formulas on time, speed and distance, profit and loss, clocks, problems on trains, which is quite useful in solving sums. 6. Permutation and Combination - click here to get it. This file will teach you how to calculate the sums and SUMMARY OF PERMUTATIONS & COMBINATIONS with easy language explanation, easy to understand. 7. Pipes and cisterns - click here to get it. This file have explained the formula, sums and difficult examples on pipes and cisterns, which you may find it on this PDF. Most important. 8. Notes on Quantitative Aptitude - Click here to get it. A Detailed Discussion On Some New Topics included In Numerical Aptitude Paper Of Ssc 10+2level Data Entry Operator & Ldc Exam, 2011. 9. Maths Funda and Formulas - Click here to get it. Check some important shortcut formulas on number system, Arithmetic, Algebra, Geometry, and modern math. Most important for Bank exam, Government jobs, CAT, XAT & Other MBA Entrance Examinations. 10. Square roots of big numbers - click here to get it. Check how to get the square roots of big numbers in easiest way, with time saving techniques. 11. Quantitative Aptitude Reference Guide - Click here to get it. know basics of real numbers, rational numbers, irrational numbers, integers, common decimal/fractions, terminating, and non-terminating. 12. Total Quantitative Aptitude for SSC CGL - Click here to get it. This file is containing explanation, examples of each and every topics of Quantitative aptitude syllabus of SSC CGL Exam. 13. SSC CGL Solved Question Paper - Click here to get it. This file is contains the Solved question paper of SSC CGL Tier 1, this paper is very helpful to understand the topics and exam pattern of SSC CGL Exam. For now, i found this much files which you can use it as revision material for SSC CGL Examination. if you want more and more study material, then you can use the below comment box and request me anytime. Thank you. Also read:- | 677.169 | 1 |
Algebra 2 Bundle of Unit Investigation Activities
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Here's a fantastic way for you to introduce new units. These investigation activities allow students to work in pairs to discover functions and equations. You can use these lessons to drive your instruction and to thread throughout the units | 677.169 | 1 |
Constructed for students with an advanced understanding of algebra, this course will help them brush up on linear equations, inequalities, graphs, matrices, radicals, functions, and other complex mathematical concepts.
Made for students who have already taken an introductory biology course, this course further explores concepts such as genetics, molecular biology, experimental methods, compounds, parasites, and viruses. | 677.169 | 1 |
Secondary Maths Handbook: Laying the Foundations for Good Mathematics
Description:
This is significantly different to other mathematics books: aiming to help parents, students and teachers with working at home. It aims to explain the underlying reasons behind the theory of mathematics. Mathematics is not a series of un-connected hoops which the reader must jump through whether or not they understand why they are jumping through those hoops. Arranged by topic, the Secondary Maths Handbook can be used sequentially following schoolwork to ensure that: - pupils understand each topic before progressing to the next, ensuring there are no gaps in their knowledge and building confidence that they can be independent learners- parents can revise (or learn!) a topic to help their children learn- teachers can use it to set homework or recommend for pupils with difficulties as an alternative way of learning | 677.169 | 1 |
Topics involving algebra, probability and data analysis are blended into the geometry content. In fact if T < S ∗ then T ⊂ S ∗ and there is a rational number q in S ∗ \T. The second column, "year," gives the year. The elements of a set. 1. can be anything: numbers. Discrete mathematics is a rapidly growing branch of modern mathematics, which includes such fields as combinatorics, graph theory, and operations research. Topics include: prime numbers and related theorems; Euclidean algorithm and quadratic reciprocity; Pythagorean numbers and continued fractions.
Students competing at the regional level have completed and won rigorous local and state level competitions Canada/USA Mathcamp is an intensive 5-week-long summer program for outstandingly talented high school students. Subject: Discrete Mathematics Date: Mon, 14 Jun 1999 13:40:30 -0400 (EDT) Discrete Mathematics Content available to subscribers at: ISSN: 0012-365X AIMS AND SCOPE The aim of Discrete Mathematics is to bring together research papers in different parts of discrete mathematics.
The false, and would affect the answers to a number of exercises. I started reading Isaac Asimov's Foundation. This is Hilbert's program of finitistic reductionism. 28 Although Hilbert did not cite Aristotle, we can imagine that Hilbert would have profited from an examination of Aristotle's distinction between actual and potential infinity. Explore the mysteries of mathematics as the adventure unfolds over 1000s of years. (Of course, the 100s of color images and mathematical milestones are credited and carefully described in the book.
The resource consumption in such algorithms is not only processor cycles on each processor but also the communication overhead between the processors. whereas distributed algorithms utilize multiple machines connected with a network. The concept of set, introduced in the late nineteenth century by Georg Cantor, has had such clarifying power that it occurs everywhere in mathematics. We see that x (3) = P ·x (2) =P· 0.8 0.2 = 0.8 0.2 0.4 0.6 0.8 0.72 · = 0.2 0.28 We find that, on the third day, there is 0.72 probability that it will be sunny and 0.28 probability that it will rain.
When the two versions of the squaring function were discussed, it was necessary to call them S and T in order to say anything about them. For students using this pattern, all grades awarded by the school are averaged in the GPA calculation. Carlitz, L., A Note on Irreducibility of the Bernoulli and Euler Polynomials. Since 50+20=35+25=70, if both players play equal and random strategies, the game gives an even payout to both players. But Mumford's piece is more than a piece on what many now think is an irrelevant debate about whether mathematics is created or discovered.
We must use Fermat's theorem, and we must use our ideas about relatively prime integers. Concentrating on addition and multiplication does not exclude subtraction or division, since subtraction is formally considered to be addition of an additive inverse and division is considered to be multiplication by a multiplicative inverse. Furthersteps need writing or some other system for recording numbers such as tallies or the knotted strings called khipu used by the Inca empire to store numerical data.
Since 2q2 is even, it follows that p2 is even. Cajori, Florian, William Oughtred: A Great Seventeenth-Century Teacher of Mathematics (Chicago: Open Court Publishing, 1916). "The Oughtred Society Slide Rule Homepage"; see I wonder what people think of this form of censorship at the biggest physics forum on the internet? Definition 5.3 Let A be an ordered set and X a subset of A. So any number relatively prime to p raised to the power which is the order of the group will give the identity element, or 1.
Any Miscellaneous Points that Might Help: Comprehensive and profound background of expertise in analysis (advanced courses including: measure and measure-theoretic probability, functional analysis, theory of distributions, analysis on manifolds, differential geometry); reading project in PDE analysis from prospective of distributional solutions; all strong references; regular attendance of regional analysis seminars. In writing a proof you should normally include all these steps: PS.1 Write down what is given, and translate it according to the definitions of the terms involved in the statement of what is given.
Every effort is made to ensure this information is current, but please be aware that some content may have changed. Nor were the Babylonians limited to calculating fractions. It was as a teenager attending university that Gauss discovered (or independently rediscovered) several important theorems. Thus 12 = 2 2 3, 111 = 3 37, and so on. Otherwise not. 1.2 ''And'' and ''Or'' Let A be the statement "Arnold is old." and B be the statement "Arnold is fat." The new statement "A and B" 4 Discrete Mathematics Demystified means that both A is true and B is true. | 677.169 | 1 |
Description
Well paying careers demand skills like problem solving, reasoning, decision making, and applying solid strategies etc. and Algebra provides you with a wonderful grounding in those skills - not to mention that it can prepare you for a wide range of opportunities.
This is a COMPLETE Pre-Algebra guide to well over 325 rules, definitions and examples, including number line, integers, rational numbers, scientific notation, median, like terms, equations, Pythagorean theorem and much more! It will take you step-by-step through the basic building blocks of Algebra giving you a solid foundation for further studies.
Table of Contents
1. Number Line
2. Inequality Symbols
3. Comparing and Ordering
4. Graphs of Real, Integer & Whole Numbers
5. Adding Positive & Negatives
6. Subtracting Numbers & Opposites
7. Multiplying & Dividing Positive & Negatives
8. Properties of Real Numbers
9. Exponents & Properties
10. Order of Operations
11. Divisibility Tests
12. Greatest Common Factor (G.C.F.)
13. Least Common Multiple (L.C.M.)
14. Rational Numbers, Proper, Improper Fractions
15. Reducing Proper & Improper Fractions
16. Adding Fractions
17. Subtracting Fractions
18. Multiplying Fractions
19. Dividing Fractions
20. Adding & Subtracting Decimals
21. Multiplying Decimals
22. Dividing Decimals
23. Fractions to Decimals
24. Decimals to Fractions
25. Rounding Decimals
26. Scientific Notation
27. Percent
28. Percent Problems
29. Averages & Means
30. Medians
31. Mode & Range
32. Variables, Coefficients & Terms, Degrees
33. Like / Unlike Terms
34. Polynomials / Degrees
35. Distributive Property
36. Add/Subtract Polynomials
37. Expression Evaluation
38. Open Sentence / Solutions
39. One-Step Equations
40. Solving ax+b = c Equations
41. Solving ax+b = cx+d Equations
42. Solving a Proportion
43. From Words to Symbols
44. Square Roots / Radical Sign
45. Pythagorean Theorem
Algebra is a very unique discipline. It is very abstract. The abstractness of algebra causes the brain to think in totally new patterns. That thinking process causes the brain to work, much like a muscle. The more that muscle works out, the better it performs on OTHER tasks. In simple terms, algebra builds a better brain! Believe it or not algebra is much easier to learn than many of us think and this guide helps make it easier!
Like all our 'phoneflips', this lightweight application has NO ads, never needs an internet connection and wont take up much space on your iPhone!
For hard copy versions of this and other great products, please visit: Test Review Pre-Algebra covers the pre-algebra that students take in junior high and middle school. Our software is designed to test your understanding and ability to perform basic algebra. Young students who want to master math must know the basic concepts of math backwards and forwards. Our… more
Pre-Algebra One is a complete curriculum for students who need help mastering the skills needed for Algebra and beyond. This teaching tool provides guided instruction, video demonstrations and practice problems.Designed by veteran math teacher Joel Bezaire of the University School of Nashville, Pre… more
Taking pre-algebra? Then you need the Wolfram Pre-Algebra Course Assistant. This definitive app for pre-algebrafrom the world leader in math softwarewill help you work through your homework problems, ace your tests, and learn pre-algebra concepts. Forget canned examples The Wolfram Pre-Algebra… more | 677.169 | 1 |
Learn the concepts and methods of linear algebra, and how to use them to think about computational problems arising in computer science. Coursework includes building on the concepts to write small programs and run them on real data.
Featuring recorded lectures from the Harvard School of Engineering and Applied Sciences course Computer Science 20, this course covers widely applicable mathematical tools for computer science, including topics from logic, set theory, combinatorics, number theory, probability theory, and graph theory. It includes practice in reasoning formally and proving theorems. Students meet twice a week via web conference to solve problems collaboratively.
In recent years calligraphers have felt an increased demand for two unique skills: modern flourishesthat are elegant (without looking like they came out of the eighteenth century) and digitized calligraphythat can be used in print and online in the form of logotypes, advertising, title treatments, printed stationery, and beyond. In this course you will learn my favorite techniques for both skills as you are guided through all the steps of creating your own flourished,
Due to the advancement of video games and game console hardware, the super computer is now a home consumer appliance. Vector Game Math Processors explains to programmers how to write parallel-based integer and floating point based math algorithms for use in video games as well as scientific applications. | 677.169 | 1 |
Course
Exam Board
Pass Rate
Assessment
Entry Requirements
About the Course
If you love number and abstract ideas, then this could be the course for you.
Further Mathematics is a very demanding and challenging course which will develop your Maths at a very fast pace.
This is a suitable course if you are considering a university course in Mathematics, Physics or Engineering.
Course Structure:
This is a double A-level combining A-level Maths with a second qualification in A-level Further Maths. The content for the qualifications are:
A-Level Mathematics is the study of 3 areas of mathematics:
Core
Builds upon the techniques in Algebra, Geometry and Trigonometry developed at GCSE as well as introducing new topics such as calculus, sequences and series, discriminants and logarithms.
Mechanics
Includes topics such as vectors, kinematics, forces, newton laws and moments.
Statistics
Includes topics such as sampling, data presentation and interpretation, probability, statistical distributions and hypothesis testing.
A-level Further Mathematics studies a combination of all areas of mathematics:
Pure
This builds upon the A-level core as well as introducing new topics such as mathematical proof, complex numbers, matrices, polar coordinates, hyperbolic functions and differential equations.
Mechanics
Includes topics such as momentum and collisions, Energy, circular motion, centres of mass and moments.
Statistics
Includes topics such as random variables, expectation, statistical distributions and errors.
Decision
Includes topics such as linear programming, networks, group theory and game theory. | 677.169 | 1 |
Written in a crisp and approachable style, Games and Information uses simple modeling techniques and straightforward explanations to provide students with an understanding of game theory and information economics.
Designed for use in a two-semester course on abstract analysis, Real Analysis: An Introduction to the Theory of Real Functions and Integration illuminates the principle topics that constitute real analysis.
Over the past twenty-five years K-theory has become an integrated part of the study of C*-algebras. This book gives a very elementary introduction to this interesting and rapidly growing area of mathematics. | 677.169 | 1 |
hands-on explanations of topics central to beginning math as related to game development. It differs from books that focus solely on math topics, involving readers in extensive programming activities and focusing on creating anMore...
This book provides hands-on explanations of topics central to beginning math as related to game development. It differs from books that focus solely on math topics, involving readers in extensive programming activities and focusing on creating an understanding of a computer game as a cognitive machine that can be driven almost wholly by table of data generated by math equations. While it does cover some of the math skills conveyed in the typical math book, it also teaches readers how to think about generating mathematical values and then translating them into the coordinate space of the game or into the event model that maps the activities of the game. It addresses both the use of a graphical engine and the creation of intelligence for a game. If you are searching for a gentle introductionto some of the most complex aspects of game development, this book is the ideal guide to get you up and running.
Boris Meltreger graduated from a top mathematics and physics high school in Russia. He went on to earn an advanced degree in optical engineering. After completing a dissertation on acoustics and optics, Boris took up work for the Russian government developing optical computers. He has been the recipient of engineering awards for his work and has owned his own engineering company. Boris has in recent years performed pioneering work in the development of optical technologies for medical applications and currently works as a software engineer. Boris lives in Aurora, Colorado | 677.169 | 1 |
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Online Exclusive! Rick and Morty Comic Bundle
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You . . and it doesn't have to be difficult In just one month, students can gain expertise and ease in all the algebra concepts that often stump students. How? Each lesson gives one small part of the bigger algebra problem, so that every day students build upon what was learned the day before. Fun factoids, catchy memory hooks, and valuable shortcuts make sure that each algebra concept becomes ingrained. With Junior Skill Builders: Algebra in 15 Minutes a Day, before you know it, a struggling student becomes an algebra pro-one step at a time. addition to all the essential practice that kids need to ace classroom tests, pop quizzes, class participation, and standardized exams, Junior Skill Builders: Algebra in 15 Minutes a Day provides parents with an easy and accessible way to help their children exce | 677.169 | 1 |
Math 2A -†This course is the first half of math 2. We will be covering the same material that is covered in Math 2 only we will be going more in-depth since we have more time.
Announcements
Welcome
Welcome to Math 2A. I am Mrs Williams. I am here for extra help for you before school (7am), B lunch and by appointment after school. It is my desire that you will gain both insight and understanding to mathematics this year. I hope that you will begin to see the creativity in mathematics. Mathematics is really not about memorizing a bunch of stuff that is not relevant. | 677.169 | 1 |
Calculating Intervals Using Stokes' Theorem
In this calculating intervals instructional activity, students calculate intervals using Stokes' Theorem. They justify their answers with words and pictures. This two-page instructional activity contains 3 multi-step problems. | 677.169 | 1 |
Stimulating collection of unusual problems dealing with congruence and parallelism, the Pythagorean theorem, circles, area relationships, Ptolemy and the cyclic quadrilateral, collinearity and concurrency, and many other topics. Challenges are arranged in order of difficulty and detailed solutions are included for all. An invaluable supplement to a basic geometry textbook.
Biografía del autor:
Al Posamentier is currently Dean of the School of Education and Professor of Mathematics Education at Mercy College, New York. He is Professor Emeritus of Mathematics Education at The City College of the City University of New York, and former Dean of the School of Education, where he was for 40 years. He is the author and co-author of more than 55 mathematics books for teachers, secondary and elementary school students, and the general readership. Dr. Posamentier is also a frequent commentator in newspapers and journals on topics relating to education. Alfred S. Posamentier: Math's Champion Dr. Alfred S. Posamentier, Professor Emeritus of Mathematics Education at New York's City College and, from 1999 to 2009, the Dean of City College's School of Education, has long been a tireless advocate for the importance of mathematics in education. He is the author or co-author of more than 40 mathematics books for teachers, students, and general readers including The Fascinating Fibonacci Numbers (Prometheus, 2007) and Mathematical Amazements and Surprises: Fascinating Figures and Noteworthy Numbers (Prometheus, 2009) His incisive views on aspects of mathematics education may often be encountered in the Letters columns and on the op-ed pages of The New York Times and other newspapers and periodicals. For Dover he provided, with co-author Charles T. Salkind, something very educational and also fun, two long-lived books of problems: Challenging Problems in Geometry and Challenging Problems in Algebra, both on the Dover list since 1996. Why solve problems? Here's an excerpt from a letter Dr. Posamentier sent to The New York Times following an article about Martin Gardner's career in 2009: "Teachers shouldn't think that textbook exercises provide problem-solving experiences — that's just drill. Genuine problem solving is what Mr. Gardner has been espousing. Genuine problem solving provides a stronger command of mathematics and exhibits its power and beauty. Something sorely lacking in our society." Paperback. Estado de conservación: New. 2nd. 138mm x 14mm x 215mm. Paperback. Stimulating collection of unusual problems dealing with congruence and parallelism, the Pythagorean theorem, circles, area relationships, Ptolemy and the cyclic quadrilateral, collinearity.Shipping may be from multiple locations in the US or from the UK, depending on stock availability. 272 pages. 0.263. Nº de ref. de la librería 9780486691541
Descripción Dover Publications. Paperback. Estado de conservación: New. Paperback. 272 pages. Dimensions: 8.5in. x 5.5in. x 0.6in.Designed for high school students and teachers with an interest in mathematical problem-solving, this volume offers a wealth of nonroutine problems in geometry that stimulate students to explore unfamiliar or little-known aspects of mathematics. Included Invaluable as a supplement to a basic geometry textbook, this volume offers both further explorations on specific topics and practice in developing problem-solving techniques. This item ships from multiple locations. Your book may arrive from Roseburg,OR, La Vergne,TN. Paperback. Nº de ref. de la librería 9780486691541 | 677.169 | 1 |
Calculator Basic Functions Test
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0.02 MB | 2 pages
PRODUCT DESCRIPTION
This is a 10 question quiz/test to see if students know how to use a scientific calculator for basic functions. The questions involve calculations with fractions, decimals, exponents, and order of operations. The computations can be done with the calculator entirely if the student knows how to use it correctly | 677.169 | 1 |
Enrollment Info
Liberal Arts Mathematics 1
Prerequisites:Introductory Algebra Length:Two Semesters
Liberal Arts Mathematics 1 addresses the need for an elective course that focuses on reinforcing, deepening, and extending a student's mathematical understanding. Liberal Arts Mathematics 1 starts with a review of problem-solving skills before moving on to a variety of key algebraic, geometric, and statistical concepts. Throughout the course, students hone their computational skills and extend their knowledge through problem solving and real-world applications.
Course topics include problem solving; real numbers and operations; functions and graphing; systems of linear equations; polynomials and factoring; geometric concepts such as coordinate geometry and properties of geometric shapes; and descriptive statistics.
Within each Liberal Arts Mathematics 1 lesson, students are supplied with a scaffolded note-taking guide, called a Study Sheet, and are given ample opportunity to practice computations in low-stakes Checkup activities before moving on to formal assessment. Additionally, students will have the opportunity to formulate and justify conclusions as they extend and apply concepts through printable exercises and "in-your-own-words" interactive activities.
To assist students for whom language presents a barrier to learning or who are not reading at grade level, Liberal Arts Mathematics 1 includes audio resources in English.
This course is aligned with Florida's Next Generation Sunshine State Standards and Benchmarks | 677.169 | 1 |
Mathematica gives students the power to manipulate interactive graphics and develop complex data models. High-school teacher Abby Brown shares the success she experiences by using Mathematica in her classroom. Includes Spanish audio. | 677.169 | 1 |
Precalculus
About this course
In this interactive pre-Calculus course, you will deepen and extend your knowledge of functions, graphs, and equations from high school algebra and geometry courses so you can successfully work with the concepts in a rigorous university-level calculus course. This course is designed to engage learners in the "doing" of mathematics, emphasizing conceptual understanding of mathematical definitions and student development of logical arguments in support of solutions. The course places major emphasis on why the mathematics topics covered work within the discipline, as opposed to simply the mechanics of the mathematics.
What you'll learn
How to make conjectures, construct logical arguments, and justify your reasoning
The concept of function in mathematics, function characteristics and properties, and rate of change of functions
Modeling of common relations and functions such as linear, power, exponential, and logarithmic functions using statistical regression and matrix methods
An exploration of algebra and geometry, and the connections between the two in analytic geometry
Properties and applications of exponential and logarithmic functions including exponential growth and decay and The Logistic Function
The development of the trigonometric functions and identities along with applications of trigonometry
Limits and rate of change of functions as a precursor to Calculus
Other Coordinates Systems – an investigation of parametrization of the plane and the polar coordinate system along with exploration and use of vectors
Sequences and Series including The Method of Induction
Basic probability and combinatorics used in an investigation and development of the Binomial Theorem and its connections to Pascal's Triangle
CLEP Exam
This course is designed to prepare you for the CLEP Precalculus exam and cover other related topics as well.
Instructors
Mark L. Daniels
Mark L. Daniels is a Clinical Professor of Mathematics and The Associate Director of the UTeach Program in Natural Sciences at the University of Texas at Austin. In addition, he is an Assistant Chair in the Mathematics Department. He has developed the curriculum for and teaches multiple inquiry-based mathematics courses, some specifically designed for pre-service teachers. Dr. Daniels is also the faculty advisor for teaching options in the Mathematics Department as well as the Director of the Discovery Learning Project for the College of Natural Sciences. Moreover, he is the Co-Director of the Inquiry-based Learning Project in Mathematics working with Dr. Michael Starbird. His research interests, of late, involve the preparation of pre-service mathematics teachers and the incorporation of discovery-based instructional methodology in university undergraduate mathematics courses.
Jeremiah W. Lucas
Jeremiah W. Lucas is the Precalculus Course Coordinator with the Center for Teaching and Learning at The University of Texas at Austin. He works with the OnRamps program, a dual-credit and dual-enrollment initiative designed to increase student access to rigorous, high-quality curricula that foster deep conceptual knowledge and academic skills. He earned his BS in Electrical Engineering from the University of Kentucky and his MA in Mathematics from The University of Texas. He taught high school mathematics in diverse and underprivileged schools in central Texas and is currently an Adjunct Professor at Austin Community College.
Karen E. Smid
Karen E. Smid is a Project Manager with the Center for Teaching and Learning at The University of Texas at Austin. She works on a variety of course development and software adoption projects, and has a Ph.D. in Anthropology from the University of Michigan. | 677.169 | 1 |
Student Organizer for Prealgebra
9780321634023
0321634060
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$1.64
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Summary
The Student Organizer is designed to help students develop the study habits they need to be successful. This Organizer guides students through the three main components of studying effectivelynote-taking, practice, and homeworkand helps them develop the habits that will enable them to succeed in future courses. The Student Organizer can be packaged with the text in loose-leaf, notebook-ready format and is also available for download in MyMathLab. | 677.169 | 1 |
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