First Order Linear Differential Equations
|
|
- Laurence Wilkerson
- 5 years ago
- Views:
Transcription
1 LECTURE 6 First Orer Linear Differential Equations A linear first orer orinary ifferential equation is a ifferential equation of the form ( a(xy + b(xy = c(x. Here y represents the unknown function, y its erivative with respect to the variable x, an a(x, b(c an c(x are certain prescribe functions of x (the precise functional form of a(x, b(x, an c(x will fixe in any specific example. So long as a(x 0, this equation is equivalent to a ifferential equation of the form (2 y + p(xy = g(x where p(x = b(x a(x, g(x = c(x a(x. We shall refer to a ifferential equation (2 as the stanar form of ifferential equation (. (In general, we shall say that an orinary linear ifferential equation is in stanar form when the coefficient of the highest erivative is. Our goal now is to evelop a formula for the general solution of (2. To accomplish this goal, we shall first construct solutions for several special cases. Then with the knowlege gaine from these simpler examples, we will evelop a general formula for the solution of any ifferential equation of the form (2.. The Homogeneous Case Let s begin with the case when the function g (x on the right han sie of (2 is ientically 0. ((3 y + p (x y = 0. The solution of this special case will actually play a funamental role in our solution of the general case later on. A similar phenomenon will happen when we look at secon orer linear ifferential equations. Because of this, we have a special nomenclature for the two basic possibilities for y-inepenent terms of a linear ifferential equation. We say that y + p (x y = 0 is a homogeneous st orer linear ifferential equation an y + p (x y = g (x is a nonhomogeneous st orer linear ifferential equation whenever g (x 0. OK. Let s see if we can solve (3. This is in fact relatively easy since (3 can be written in separable form: y + p (x y = 0 y 24 y = p (x x
2 . THE HOMOGENEOUS CASE 25 Applying the technique we evelope for Separable Equations: y = p (x x y y y = p (x x + C ln y = p (x x + C ( y = exp p (x x + C ( y = exp p (x x exp (C ( y = A exp p (x x where A = exp (C is an arbitrary constant Example 6.. Integrating both sies of the latter equation (the left han sie with respect to y an the right han sie with respect to x yiels ln(y = x p(x x + C or, exponentiating both sies [ x y = exp p(x x + C [ x y = exp p(x x + C [ x = e C exp p(x x [ x = A exp p(x x In the last step we have simply replace the constant e C, which is arbitrary since C is arbitrary, by another arbitrary constant A. There is nothing tricky here; the point is that in the general solution the numerical factor in front of the exponential function is arbitrary an so rather than writing this factor as e C we use the simpler form A. Thus, the general solution of is Example 6.2. y + 3 x y = 0 y + p(x = 0 [ x y = A exp p(x x First, we recognize this ifferential equation as a separable ifferential equation: y + 3 x y = 0 y. y x = 3 x We next apply the technique for solving separable equations: y = 3xx y y y = 3 x x + C ln y = 3 ln x + C y = exp ( 3 ln x + C y = exp ( 3 ln x exp (C y = A exp ( 3 ln x where A = exp (C is an arbitrary constant Digression: (a very common an useful ientity Let me begin by recalling three basic facts:
3 2. A SIMPLE NONHOMOGENOUS CASE 26 (i The exponential function exp (x e x where e = is the natural logarithmic base. (ii The logarithm function is the functional inverse of the exponential function: exp (ln x = x = ln (exp (x. (iii Powers of powers of a number obey ( A b c = A bc. Now consier the ientity Thus, we have an ientity x = exp (ln x by (ii x = e ln x by (i ( x a = e ln(x a = e a ln x exp (a ln x = x a by (i (4 exp (a ln x = x a by (iii Using this ientity, we can simply our expression for the general solution of y + 3 x y = 0: y = A exp ( 3 ln x y (x = Ax 3 Example 6.3. (Example Reloae In the preceing example, we solve y + 3 xy = 0 by repeating the steps by which we solve the general case y +p (x y = 0. However, the main reason for treating the general case, was to get a general formula for that case. Inee, ropping the intermeiary steps we have from our initial iscussion (5 y + p (x y = 0 y (x = A exp ( p (x x So another way of attacking y + 3 xy = 0 is to recognize it as a particular instance of the case governe by the formula (5. Thus, we can also procee as follows: y + 3 x y = 0 is of the form y + p (x y = 0 when we take p (x = 3 x. Plugging p (x = 3 x into the formula (5 for the solution of y + p (x y = 0 yiels ( 3 y (x = A exp x x = A exp ( 3 ln x = Ax 3 (applying ientify (4 Thus, recovering the same answer but a little more efficiently.. 2. A simple nonhomogenous case We are still not prepare to attack ifferential equations of the form y + p (x y = g (x x when g (x 0. However, we ll continue to pursue our strategy of eveloping solution techniques by introucing complications in a very controlle way. Having hanle the case when g (x = 0, the next simple case we might consier is when g (x 0 but at least the function p (x is only a constant; say p (x = a for all x. So now we ll try to solve (6 y + ay = g (x x where a is a constant.
4 3. THE GENERAL CASE 27 To solve this equation we ll employ a trick. Let s multiply both sies of this equation by e ax : e ax y + ae ax y = e ax g(x. Noticing that the right han sie is x (eax y (via the prouct rule for ifferentiation, we have x (eax y = e ax g(x. We now take anti-erivatives of both sies to get e ax y = x e ax g(x x + C or Example 6.4. y(x = x e ax e ax g(x x + Ce ax. y 2y = x 2 e 2x This equation is of type (iii with p = 2 g(x = x 2 e 2x. So we multiply both sies by e 2x to get ( e 2x y = e 2x (y 2y = e 2x ( x 2 e 2x = x 2 x Integrating both sies with respect to x, an employing the Funamental Theorem of Calculus on the left yiels or Let us now confirm that this is a solution e 2x y = 3 x3 + C y = 3 x3 e 2x + Ce 2x. y = x 2 e 2x x3 e 2x + 2Ce 2x so 2y = 2 3 x3 e 2x 2Ce 2x y 2y = x 2 e 2x 3. The General Case We are now prepare to hanle the case of a general first orer linear ifferential equation; i.e., ifferential equations of the form (7 y + p(xy = g(x with p(x an g(x are arbitrary functions of x. Note: This case inclues all the preceing cases. Accoring to Richar Feynman, if you employ a trick more than once, it becomes a metho. Inee, shortly, the trick introuce here will lea to a bone-fie metho - the integrating factors metho.
5 3. THE GENERAL CASE 28 We shall construct a solution of this equation in a manner similar to case when p(x is a constant; that is, we will try fin a function µ(x satisfying (8 µ(x (y + p(xy = x (µ(xy Multiplying (7 by µ(x, we coul then obtain which when integrate yiels or (µ(xy = µ(xg(x x µ(xy = x µ(x g(x x + C (9 y = x µ(x g(x x + C µ(x µ(x It thus remains to fin a suitable function µ(x; i.e., we nee to fin a function µ(x so that This will certainly be true if (0 For then x (µ(xy = µ(xy + p (x µ(xy µ(x = p(xµ(x. x ( x (µ(xy = µ(xy + x µ (x y = µ(xy + p (x µ(xy Thus, we have to solve another first orer, linear, ifferential equation of type (iii. As before we re-write (0 in terms of ifferentials to get µ µ = p(xx, an then integrate both sies; yieling ln(µ = x p(x x + A. Exponentiating both sies of this relation yiels ( x µ = exp p(x x + A ( x µ = exp p(x x + A ( x = A exp p(x x So a suitable function µ(x is ( x µ(x = A exp p(x x Inserting this expression for µ(x into our formula (9 for y yiels [ ( x x y(x = A exp ( x p(x x A exp p(x x g(x x + C
6 3. THE GENERAL CASE 29 It is easily see that the constant A in the enominator is irrelevant to the final answer. This is because it can be cancele out by the A within the integral over x, an it can be absorbe into the arbitrary constant C in the secon term. Thus, the general solution to a first orer linear equation is given by (a y(x = µ(x where (b Example 6.5. y + p(xy = g(x x µ(x g(x x + ( x µ(x = exp p(x x (2 xy + 2y = sin(x C µ(x Putting this equation in stanar form requires we set Now so Therefore, p(x x = p(x g(x = 2 x = sin(x x 2 x x = 2 ln(x = ln ( x 2, µ(x = exp [ x p(x x = exp [ ln ( x 2 = x 2 x y(x = µ(x g(x x + C µ(x µ(x = x x 2 (x 2 sin(x x x + C x 2 = x 2 x x sin(x x + C x 2 Now x sin(x x can be integrate by parts. Set Then u = x, v = sin(xx u = x, v = v = cos(x an the integration by parts formula, uv = uv vu, tells us that x sin(x x = x cos(x + cos(x x = x cos(x + sin(x.
7 3. THE GENERAL CASE 30 Therefore, we have as a general solution of (2, y(x = x 2 ( x cos(x + sin(x + C x 2 = x 2 sin(x x cos(x + C x 2.
First Order Linear Differential Equations
LECTURE 8 First Orer Linear Differential Equations We now turn our attention to the problem of constructing analytic solutions of ifferential equations; that is to say,solutions that can be epresse in
More informationLinear First-Order Equations
5 Linear First-Orer Equations Linear first-orer ifferential equations make up another important class of ifferential equations that commonly arise in applications an are relatively easy to solve (in theory)
More information2 ODEs Integrating Factors and Homogeneous Equations
2 ODEs Integrating Factors an Homogeneous Equations We begin with a slightly ifferent type of equation: 2.1 Exact Equations These are ODEs whose general solution can be obtaine by simply integrating both
More informationLINEAR DIFFERENTIAL EQUATIONS OF ORDER 1. where a(x) and b(x) are functions. Observe that this class of equations includes equations of the form
LINEAR DIFFERENTIAL EQUATIONS OF ORDER 1 We consier ifferential equations of the form y + a()y = b(), (1) y( 0 ) = y 0, where a() an b() are functions. Observe that this class of equations inclues equations
More informationImplicit Differentiation
Implicit Differentiation Thus far, the functions we have been concerne with have been efine explicitly. A function is efine explicitly if the output is given irectly in terms of the input. For instance,
More informationReview of Differentiation and Integration for Ordinary Differential Equations
Schreyer Fall 208 Review of Differentiation an Integration for Orinary Differential Equations In this course you will be expecte to be able to ifferentiate an integrate quickly an accurately. Many stuents
More informationLectures - Week 10 Introduction to Ordinary Differential Equations (ODES) First Order Linear ODEs
Lectures - Week 10 Introuction to Orinary Differential Equations (ODES) First Orer Linear ODEs When stuying ODEs we are consiering functions of one inepenent variable, e.g., f(x), where x is the inepenent
More informationDifferentiation ( , 9.5)
Chapter 2 Differentiation (8.1 8.3, 9.5) 2.1 Rate of Change (8.2.1 5) Recall that the equation of a straight line can be written as y = mx + c, where m is the slope or graient of the line, an c is the
More informationOrdinary Differential Equations
Orinary Differential Equations Example: Harmonic Oscillator For a perfect Hooke s-law spring,force as a function of isplacement is F = kx Combine with Newton s Secon Law: F = ma with v = a = v = 2 x 2
More informationIntegration Review. May 11, 2013
Integration Review May 11, 2013 Goals: Review the funamental theorem of calculus. Review u-substitution. Review integration by parts. Do lots of integration eamples. 1 Funamental Theorem of Calculus In
More informationThe Exact Form and General Integrating Factors
7 The Exact Form an General Integrating Factors In the previous chapters, we ve seen how separable an linear ifferential equations can be solve using methos for converting them to forms that can be easily
More informationThe derivative of a function f(x) is another function, defined in terms of a limiting expression: f(x + δx) f(x)
Y. D. Chong (2016) MH2801: Complex Methos for the Sciences 1. Derivatives The erivative of a function f(x) is another function, efine in terms of a limiting expression: f (x) f (x) lim x δx 0 f(x + δx)
More informationChapter 7. Integrals and Transcendental Functions
7. The Logarithm Define as an Integral Chapter 7. Integrals an Transcenental Functions 7.. The Logarithm Define as an Integral Note. In this section, we introuce the natural logarithm function using efinite
More informationIntegration by Parts
Integration by Parts 6-3-207 If u an v are functions of, the Prouct Rule says that (uv) = uv +vu Integrate both sies: (uv) = uv = uv + u v + uv = uv vu, vu v u, I ve written u an v as shorthan for u an
More informationSturm-Liouville Theory
LECTURE 5 Sturm-Liouville Theory In the three preceing lectures I emonstrate the utility of Fourier series in solving PDE/BVPs. As we ll now see, Fourier series are just the tip of the iceberg of the theory
More informationDifferentiability, Computing Derivatives, Trig Review. Goals:
Secants vs. Derivatives - Unit #3 : Goals: Differentiability, Computing Derivatives, Trig Review Determine when a function is ifferentiable at a point Relate the erivative graph to the the graph of an
More informationQF101: Quantitative Finance September 5, Week 3: Derivatives. Facilitator: Christopher Ting AY 2017/2018. f ( x + ) f(x) f(x) = lim
QF101: Quantitative Finance September 5, 2017 Week 3: Derivatives Facilitator: Christopher Ting AY 2017/2018 I recoil with ismay an horror at this lamentable plague of functions which o not have erivatives.
More informationMath 251 Notes. Part I.
Math 251 Notes. Part I. F. Patricia Meina May 6, 2013 Growth Moel.Consumer price inex. [Problem 20, page 172] The U.S. consumer price inex (CPI) measures the cost of living base on a value of 100 in the
More informationDifferentiability, Computing Derivatives, Trig Review
Unit #3 : Differentiability, Computing Derivatives, Trig Review Goals: Determine when a function is ifferentiable at a point Relate the erivative graph to the the graph of an original function Compute
More informationA. Incorrect! The letter t does not appear in the expression of the given integral
AP Physics C - Problem Drill 1: The Funamental Theorem of Calculus Question No. 1 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question
More informationd dx [xn ] = nx n 1. (1) dy dx = 4x4 1 = 4x 3. Theorem 1.3 (Derivative of a constant function). If f(x) = k and k is a constant, then f (x) = 0.
Calculus refresher Disclaimer: I claim no original content on this ocument, which is mostly a summary-rewrite of what any stanar college calculus book offers. (Here I ve use Calculus by Dennis Zill.) I
More informationFall 2016: Calculus I Final
Answer the questions in the spaces provie on the question sheets. If you run out of room for an answer, continue on the back of the page. NO calculators or other electronic evices, books or notes are allowe
More informationx 2 2x 8 (x 4)(x + 2)
Problems With Notation Mathematical notation is very precise. This contrasts with both oral communication an some written English. Correct mathematical notation: x 2 2x 8 (x 4)(x + 2) lim x 4 = lim x 4
More informationSection 7.1: Integration by Parts
Section 7.1: Integration by Parts 1. Introuction to Integration Techniques Unlike ifferentiation where there are a large number of rules which allow you (in principle) to ifferentiate any function, the
More informationMath 1B, lecture 8: Integration by parts
Math B, lecture 8: Integration by parts Nathan Pflueger 23 September 2 Introuction Integration by parts, similarly to integration by substitution, reverses a well-known technique of ifferentiation an explores
More informationFinal Exam Study Guide and Practice Problems Solutions
Final Exam Stuy Guie an Practice Problems Solutions Note: These problems are just some of the types of problems that might appear on the exam. However, to fully prepare for the exam, in aition to making
More informationFebruary 21 Math 1190 sec. 63 Spring 2017
February 21 Math 1190 sec. 63 Spring 2017 Chapter 2: Derivatives Let s recall the efinitions an erivative rules we have so far: Let s assume that y = f (x) is a function with c in it s omain. The erivative
More informationd dx But have you ever seen a derivation of these results? We ll prove the first result below. cos h 1
Lecture 5 Some ifferentiation rules Trigonometric functions (Relevant section from Stewart, Seventh Eition: Section 3.3) You all know that sin = cos cos = sin. () But have you ever seen a erivation of
More informationf(x) f(a) Limit definition of the at a point in slope notation.
Lesson 9: Orinary Derivatives Review Hanout Reference: Brigg s Calculus: Early Transcenentals, Secon Eition Topics: Chapter 3: Derivatives, p. 126-235 Definition. Limit Definition of Derivatives at a point
More informationJUST THE MATHS UNIT NUMBER DIFFERENTIATION 2 (Rates of change) A.J.Hobson
JUST THE MATHS UNIT NUMBER 10.2 DIFFERENTIATION 2 (Rates of change) by A.J.Hobson 10.2.1 Introuction 10.2.2 Average rates of change 10.2.3 Instantaneous rates of change 10.2.4 Derivatives 10.2.5 Exercises
More information1 dx. where is a large constant, i.e., 1, (7.6) and Px is of the order of unity. Indeed, if px is given by (7.5), the inequality (7.
Lectures Nine an Ten The WKB Approximation The WKB metho is a powerful tool to obtain solutions for many physical problems It is generally applicable to problems of wave propagation in which the frequency
More information3 Algebraic Methods. we can differentiate both sides implicitly to obtain a differential equation involving x and y:
3 Algebraic Methods b The first appearance of the equation E Mc 2 in Einstein s handwritten notes. So far, the only general class of differential equations that we know how to solve are directly integrable
More informationDifferentiation Rules Derivatives of Polynomials and Exponential Functions
Derivatives of Polynomials an Exponential Functions Differentiation Rules Derivatives of Polynomials an Exponential Functions Let s start with the simplest of all functions, the constant function f(x)
More informationChapter 6: Integration: partial fractions and improper integrals
Chapter 6: Integration: partial fractions an improper integrals Course S3, 006 07 April 5, 007 These are just summaries of the lecture notes, an few etails are inclue. Most of what we inclue here is to
More informationSolutions to Math 41 Second Exam November 4, 2010
Solutions to Math 41 Secon Exam November 4, 2010 1. (13 points) Differentiate, using the metho of your choice. (a) p(t) = ln(sec t + tan t) + log 2 (2 + t) (4 points) Using the rule for the erivative of
More informationcosh x sinh x So writing t = tan(x/2) we have 6.4 Integration using tan(x/2) = 2 2t 1 + t 2 cos x = 1 t2 We will revisit the double angle identities:
6.4 Integration using tanx/) We will revisit the ouble angle ientities: sin x = sinx/) cosx/) = tanx/) sec x/) = tanx/) + tan x/) cos x = cos x/) sin x/) tan x = = tan x/) sec x/) tanx/) tan x/). = tan
More informationSome functions and their derivatives
Chapter Some functions an their erivatives. Derivative of x n for integer n Recall, from eqn (.6), for y = f (x), Also recall that, for integer n, Hence, if y = x n then y x = lim δx 0 (a + b) n = a n
More information1 Lecture 20: Implicit differentiation
Lecture 20: Implicit ifferentiation. Outline The technique of implicit ifferentiation Tangent lines to a circle Derivatives of inverse functions by implicit ifferentiation Examples.2 Implicit ifferentiation
More informationCenter of Gravity and Center of Mass
Center of Gravity an Center of Mass 1 Introuction. Center of mass an center of gravity closely parallel each other: they both work the same way. Center of mass is the more important, but center of gravity
More informationImplicit Differentiation. Lecture 16.
Implicit Differentiation. Lecture 16. We are use to working only with functions that are efine explicitly. That is, ones like f(x) = 5x 3 + 7x x 2 + 1 or s(t) = e t5 3, in which the function is escribe
More information0.1 The Chain Rule. db dt = db
0. The Chain Rule A basic illustration of the chain rules comes in thinking about runners in a race. Suppose two brothers, Mark an Brian, hol an annual race to see who is the fastest. Last year Mark won
More information1 Lecture 18: The chain rule
1 Lecture 18: The chain rule 1.1 Outline Comparing the graphs of sin(x) an sin(2x). The chain rule. The erivative of a x. Some examples. 1.2 Comparing the graphs of sin(x) an sin(2x) We graph f(x) = sin(x)
More informationIntroduction to Differential Equations Math 286 X1 Fall 2009 Homework 2 Solutions
Introuction to Differential Equations Math 286 X1 Fall 2009 Homewk 2 Solutions 1. Solve each of the following ifferential equations: (a) y + 3xy = 0 (b) y + 3y = 3x (c) y t = cos(t)y () x 2 y x y = 3 Solution:
More informationDesigning Information Devices and Systems II Fall 2017 Note Theorem: Existence and Uniqueness of Solutions to Differential Equations
EECS 6B Designing Information Devices an Systems II Fall 07 Note 3 Secon Orer Differential Equations Secon orer ifferential equations appear everywhere in the real worl. In this note, we will walk through
More informationChapter 2. Exponential and Log functions. Contents
Chapter. Exponential an Log functions This material is in Chapter 6 of Anton Calculus. The basic iea here is mainly to a to the list of functions we know about (for calculus) an the ones we will stu all
More informationSection The Chain Rule and Implicit Differentiation with Application on Derivative of Logarithm Functions
Section 3.4-3.6 The Chain Rule an Implicit Differentiation with Application on Derivative of Logarithm Functions Ruipeng Shen September 3r, 5th Ruipeng Shen MATH 1ZA3 September 3r, 5th 1 / 3 The Chain
More informationSYDE 112, LECTURE 1: Review & Antidifferentiation
SYDE 112, LECTURE 1: Review & Antiifferentiation 1 Course Information For a etaile breakown of the course content an available resources, see the Course Outline. Other relevant information for this section
More informationRules of Differentiation
LECTURE 2 Rules of Differentiation At te en of Capter 2, we finally arrive at te following efinition of te erivative of a function f f x + f x x := x 0 oing so only after an extene iscussion as wat te
More informationImplicit Differentiation
Implicit Differentiation Implicit Differentiation Using the Chain Rule In the previous section we focuse on the erivatives of composites an saw that THEOREM 20 (Chain Rule) Suppose that u = g(x) is ifferentiable
More informationLecture 16: The chain rule
Lecture 6: The chain rule Nathan Pflueger 6 October 03 Introuction Toay we will a one more rule to our toolbo. This rule concerns functions that are epresse as compositions of functions. The iea of a composition
More informationEuler equations for multiple integrals
Euler equations for multiple integrals January 22, 2013 Contents 1 Reminer of multivariable calculus 2 1.1 Vector ifferentiation......................... 2 1.2 Matrix ifferentiation........................
More informationThe Principle of Least Action
Chapter 7. The Principle of Least Action 7.1 Force Methos vs. Energy Methos We have so far stuie two istinct ways of analyzing physics problems: force methos, basically consisting of the application of
More information2.5 The Chain Rule Brian E. Veitch
2.5 The Chain Rule This is our last ifferentiation rule for this course. It s also one of the most use. The best way to memorize this (along with the other rules) is just by practicing until you can o
More informationMath 2163, Practice Exam II, Solution
Math 63, Practice Exam II, Solution. (a) f =< f s, f t >=< s e t, s e t >, an v v = , so D v f(, ) =< ()e, e > =< 4, 4 > = 4. (b) f =< xy 3, 3x y 4y 3 > an v =< cos π, sin π >=, so
More information1 Definition of the derivative
Math 20A - Calculus by Jon Rogawski Chapter 3 - Differentiation Prepare by Jason Gais Definition of the erivative Remark.. Recall our iscussion of tangent lines from way back. We now rephrase this in terms
More informationComputing Derivatives
Chapter 2 Computing Derivatives 2.1 Elementary erivative rules Motivating Questions In this section, we strive to unerstan the ieas generate by the following important questions: What are alternate notations
More informationTable of Contents Derivatives of Logarithms
Derivatives of Logarithms- Table of Contents Derivatives of Logarithms Arithmetic Properties of Logarithms Derivatives of Logarithms Example Example 2 Example 3 Example 4 Logarithmic Differentiation Example
More informationCalculus of Variations
16.323 Lecture 5 Calculus of Variations Calculus of Variations Most books cover this material well, but Kirk Chapter 4 oes a particularly nice job. x(t) x* x*+ αδx (1) x*- αδx (1) αδx (1) αδx (1) t f t
More informationApplications of the Wronskian to ordinary linear differential equations
Physics 116C Fall 2011 Applications of the Wronskian to orinary linear ifferential equations Consier a of n continuous functions y i (x) [i = 1,2,3,...,n], each of which is ifferentiable at least n times.
More informationA Brief Review of Elementary Ordinary Differential Equations
A A Brief Review of Elementary Ordinary Differential Equations At various points in the material we will be covering, we will need to recall and use material normally covered in an elementary course on
More informationQuestion Instructions Read today's Notes and Learning Goals.
63 Proucts an Quotients (13051836) Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 Instructions Rea toay's Notes an Learning Goals. 1. Question Details fa15 62 chain 1 [3420817] Fin all
More informationDiagonalization of Matrices Dr. E. Jacobs
Diagonalization of Matrices Dr. E. Jacobs One of the very interesting lessons in this course is how certain algebraic techniques can be use to solve ifferential equations. The purpose of these notes is
More information8/28/2017 Assignment Previewer
Proucts an Quotients (10862446) Due: Mon Sep 25 2017 07:31 AM MDT Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 Instructions Rea toay's Notes an Learning Goals. 1. Question Details
More informationMA 2232 Lecture 08 - Review of Log and Exponential Functions and Exponential Growth
MA 2232 Lecture 08 - Review of Log an Exponential Functions an Exponential Growth Friay, February 2, 2018. Objectives: Review log an exponential functions, their erivative an integration formulas. Exponential
More information1. Find the equation of a line passing through point (5, -2) with slope ¾. (State your answer in slope-int. form)
INTRO TO CALCULUS REVIEW FINAL EXAM NAME: DATE: A. Equations of Lines (Review Chapter) y = m + b (Slope-Intercept Form) A + By = C (Stanar Form) y y = m( ) (Point-Slope Form). Fin the equation of a line
More information1 Heisenberg Representation
1 Heisenberg Representation What we have been ealing with so far is calle the Schröinger representation. In this representation, operators are constants an all the time epenence is carrie by the states.
More informationP(x) = 1 + x n. (20.11) n n φ n(x) = exp(x) = lim φ (x) (20.8) Our first task for the chain rule is to find the derivative of the exponential
20. Derivatives of compositions: the chain rule At the en of the last lecture we iscovere a nee for the erivative of a composition. In this lecture we show how to calculate it. Accoringly, let P have omain
More information11.7. Implicit Differentiation. Introduction. Prerequisites. Learning Outcomes
Implicit Differentiation 11.7 Introuction This Section introuces implicit ifferentiation which is use to ifferentiate functions expresse in implicit form (where the variables are foun together). Examples
More informationMake graph of g by adding c to the y-values. on the graph of f by c. multiplying the y-values. even-degree polynomial. graph goes up on both sides
Reference 1: Transformations of Graphs an En Behavior of Polynomial Graphs Transformations of graphs aitive constant constant on the outsie g(x) = + c Make graph of g by aing c to the y-values on the graph
More informationMath 115 Section 018 Course Note
Course Note 1 General Functions Definition 1.1. A function is a rule that takes certain numbers as inputs an assigns to each a efinite output number. The set of all input numbers is calle the omain of
More informationUnit #6 - Families of Functions, Taylor Polynomials, l Hopital s Rule
Unit # - Families of Functions, Taylor Polynomials, l Hopital s Rule Some problems an solutions selecte or aapte from Hughes-Hallett Calculus. Critical Points. Consier the function f) = 54 +. b) a) Fin
More informationHigher. Further Calculus 149
hsn.uk.net Higher Mathematics UNIT 3 OUTCOME 2 Further Calculus Contents Further Calculus 49 Differentiating sinx an cosx 49 2 Integrating sinx an cosx 50 3 The Chain Rule 5 4 Special Cases of the Chain
More informationSeries Solutions of Differential Equations
Chapter 6 Series Solutions of Differential Equations In this chapter we consider methods for solving differential equations using power series. Sequences and infinite series are also involved in this treatment.
More informationSolving the Schrödinger Equation for the 1 Electron Atom (Hydrogen-Like)
Stockton Univeristy Chemistry Program, School of Natural Sciences an Mathematics 101 Vera King Farris Dr, Galloway, NJ CHEM 340: Physical Chemistry II Solving the Schröinger Equation for the 1 Electron
More informationMathematical Review Problems
Fall 6 Louis Scuiero Mathematical Review Problems I. Polynomial Equations an Graphs (Barrante--Chap. ). First egree equation an graph y f() x mx b where m is the slope of the line an b is the line's intercept
More information7.3 Singular points and the method of Frobenius
284 CHAPTER 7. POWER SERIES METHODS 7.3 Singular points and the method of Frobenius Note: or.5 lectures, 8.4 and 8.5 in [EP], 5.4 5.7 in [BD] While behaviour of ODEs at singular points is more complicated,
More informationMath 1271 Solutions for Fall 2005 Final Exam
Math 7 Solutions for Fall 5 Final Eam ) Since the equation + y = e y cannot be rearrange algebraically in orer to write y as an eplicit function of, we must instea ifferentiate this relation implicitly
More informationPartial Differential Equations
Chapter Partial Differential Equations. Introuction Have solve orinary ifferential equations, i.e. ones where there is one inepenent an one epenent variable. Only orinary ifferentiation is therefore involve.
More informationMATH 120 Theorem List
December 11, 2016 Disclaimer: Many of the theorems covere in class were not name, so most of the names on this sheet are not efinitive (they are escriptive names rather than given names). Lecture Theorems
More informationCalculus of Variations
Calculus of Variations Lagrangian formalism is the main tool of theoretical classical mechanics. Calculus of Variations is a part of Mathematics which Lagrangian formalism is base on. In this section,
More informationExam 2 Review Solutions
Exam Review Solutions 1. True or False, an explain: (a) There exists a function f with continuous secon partial erivatives such that f x (x, y) = x + y f y = x y False. If the function has continuous secon
More informationPhysics 251 Results for Matrix Exponentials Spring 2017
Physics 25 Results for Matrix Exponentials Spring 27. Properties of the Matrix Exponential Let A be a real or complex n n matrix. The exponential of A is efine via its Taylor series, e A A n = I + n!,
More informationcosh x sinh x So writing t = tan(x/2) we have 6.4 Integration using tan(x/2) 2t 1 + t 2 cos x = 1 t2 sin x =
6.4 Integration using tan/ We will revisit the ouble angle ientities: sin = sin/ cos/ = tan/ sec / = tan/ + tan / cos = cos / sin / tan = = tan / sec / tan/ tan /. = tan / + tan / So writing t = tan/ we
More information16.30/31, Fall 2010 Recitation # 1
6./, Fall Recitation # September, In this recitation we consiere the following problem. Given a plant with open-loop transfer function.569s +.5 G p (s) = s +.7s +.97, esign a feeback control system such
More informationARAB ACADEMY FOR SCIENCE TECHNOLOGY AND MARITIME TRANSPORT
ARAB ACADEMY FOR SCIENCE TECHNOLOGY AND MARITIME TRANSPORT Course: Math For Engineering Winter 8 Lecture Notes By Dr. Mostafa Elogail Page Lecture [ Functions / Graphs of Rational Functions] Functions
More information1 Applications of the Chain Rule
November 7, 08 MAT86 Week 6 Justin Ko Applications of the Chain Rule We go over several eamples of applications of the chain rule to compute erivatives of more complicate functions. Chain Rule: If z =
More informationKramers Relation. Douglas H. Laurence. Department of Physical Sciences, Broward College, Davie, FL 33314
Kramers Relation Douglas H. Laurence Department of Physical Sciences, Browar College, Davie, FL 333 Introuction Kramers relation, name after the Dutch physicist Hans Kramers, is a relationship between
More informationII. First variation of functionals
II. First variation of functionals The erivative of a function being zero is a necessary conition for the etremum of that function in orinary calculus. Let us now tackle the question of the equivalent
More information4.2 First Differentiation Rules; Leibniz Notation
.. FIRST DIFFERENTIATION RULES; LEIBNIZ NOTATION 307. First Differentiation Rules; Leibniz Notation In this section we erive rules which let us quickly compute the erivative function f (x) for any polynomial
More informationHyperbolic Functions. Notice: this material must not be used as a substitute for attending. the lectures
Hyperbolic Functions Notice: this material must not be use as a substitute for attening the lectures 0. Hyperbolic functions sinh an cosh The hyperbolic functions sinh (pronounce shine ) an cosh are efine
More informationUNDERSTANDING INTEGRATION
UNDERSTANDING INTEGRATION Dear Reaer The concept of Integration, mathematically speaking, is the "Inverse" of the concept of result, the integration of, woul give us back the function f(). This, in a way,
More information6.0 INTRODUCTION TO DIFFERENTIAL EQUATIONS
6.0 Introduction to Differential Equations Contemporary Calculus 1 6.0 INTRODUCTION TO DIFFERENTIAL EQUATIONS This chapter is an introduction to differential equations, a major field in applied and theoretical
More informationCalculus I Announcements
Slie 1 Calculus I Announcements Office Hours: Amos Eaton 309, Monays 12:50-2:50 Exam 2 is Thursay, October 22n. The stuy guie is now on the course web page. Start stuying now, an make a plan to succee.
More information016A Homework 10 Solution
016A Homework 10 Solution Jae-young Park November 2, 2008 4.1 #14 Write each expression in the form of 2 kx or 3 kx, for a suitable constant k; (3 x 3 x/5 ) 5, (16 1/4 16 3/4 ) 3x Solution (3 x 3 x/5 )
More information23 Implicit differentiation
23 Implicit ifferentiation 23.1 Statement The equation y = x 2 + 3x + 1 expresses a relationship between the quantities x an y. If a value of x is given, then a corresponing value of y is etermine. For
More informationYear 11 Matrices Semester 2. Yuk
Year 11 Matrices Semester 2 Chapter 5A input/output Yuk 1 Chapter 5B Gaussian Elimination an Systems of Linear Equations This is an extension of solving simultaneous equations. What oes a System of Linear
More informationOptimization Notes. Note: Any material in red you will need to have memorized verbatim (more or less) for tests, quizzes, and the final exam.
MATH 2250 Calculus I Date: October 5, 2017 Eric Perkerson Optimization Notes 1 Chapter 4 Note: Any material in re you will nee to have memorize verbatim (more or less) for tests, quizzes, an the final
More informationLECTURE 4-1 INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS
130 LECTURE 4-1 INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS DIFFERENTIAL EQUATIONS: A differential equation (DE) is an equation involving an unknown function and one or more of its derivatives. A differential
More informationMAP 2302 Midterm 1 Review Solutions 1
MAP 2302 Midterm 1 Review Solutions 1 The exam will cover sections 1.2, 1.3, 2.2, 2.3, 2.4, and 2.6. All topics from this review sheet or from the suggested exercises are fair game. 1 Give explicit solutions
More informationIntegration via a Change of Variables
LECTURE 33 Integration via a Change of Variables In Lectures an 3 where we evelope a general technique for computing erivatives that was base on two ifferent kins of results. First, we ha a table that
More information