MTH 3311 Test #1. order 3, linear. The highest order of derivative of y is 2. Furthermore, y and its derivatives are all raised to the
|
|
- Colleen Sparks
- 5 years ago
- Views:
Transcription
1 MTH 3311 Test #1 F 018 Pat Rossi Name Show CLEARLY how you arrive at your answers. 1. Classify the following according to order and linearity. If an equation is not linear, eplain why. (a) y + y y = 4 order 3, linear. The highest order of derivative of y is 3. Furthermore, y and its derivatives are all raised to the 1 st power, no derivative of y is a co-factor of y or any other derivative of y, and neither y nor any of its derivatives are the inner function of a composite function, so the equation is linear. (b) y (5) + yy = sin () order 5, non-linear. The highest order of derivative of y is 5. (y (5) is the fifth derivative of y it is NOT y 5.) Since y is a co-factor of y, the equation is non-linear. (c) y (4) + y + y = 6 6 order 4, non-linear. The highest order of derivative of y is 4. (y (4) is the fourth derivative of y it is NOT y 4.) Since y is raised to a power other than 1, the equation is non-linear. (d) sin () y 3y + 3 y = e + sin () order 3, linear. The highest order of derivative of y is 3. Furthermore, y and its derivatives are all raised to the 1 st power, no derivative of y is a co-factor of y or any other derivative of y, and neither y nor any of its derivatives are the inner function of a composite function, so the equation is linear. (e) y y + 4y = +1 order, linear. The highest order of derivative of y is. Furthermore, y and its derivatives are all raised to the 1 st power, no derivative of y is a co-factor of y or any other derivative of y, and neither y nor any of its derivatives are the inner function of a composite function, so the equation is linear.. Show that the function y = c 1 e 3 + c e is a solution of the differential equation y 9y = Observe: y = c 1 e 3 + c e y = 3c 1 e 3 + 3c e y = 9c 1 e 3 + 9c e Plugging into the left side of the equation, we have: y 9y = ( 9c 1 e 3 + 9c e ) 9 ( c 1 e 3 + c e ) = (9 9) c 1 e 3 + (9 9) c e 3 + ( ) = i.e., y 9y = Hence, y = c 1 e 3 + c e is a solution of the differential equation: y 9y =
2 3. Solve: dy = y + y + + ; subject to the initial condition y (0) = 1 (Assume that, y 0) Solve the equation by separating the variables, and then solve the equation using the integrating factor method. The variables can be separated. = ( + ) y + + = ( + ) y + ( + ) = ( + ) (y + 1) 1 y+1dy = ( + ) 1 y+1 dy = ( + ) ln (y + 1) = C e ln(y+1) = e 1 ++C = e 1 + e C = e 1 + C 1 = C 1 e 1 + i.e., y + 1 = C 1 e 1 + y = C 1 e Recall: y (0) = 1 1 = C 1 e 1 (0) +(0) 1 = C 1 1 = C 1 y = e Alternative Solution on the net page
3 Alternatively: We can solve this equation using the Integrating Factor Method i) Re-epress the equation in the form: y + p () y = Q () dy = y + y + + y = y + y + + y = ( + ) y + + y ( + ) y = ( + ) y + ( ) y = ( + ) p() ii) Compute the integrating factor: e p() = e ( ) = e 1 iii) Multiply both sides by the integrating factor e 1 y + ( ) y = ( + ) e 1 iv) Epress the left side as the derivative of a product [ ] e 1 y = ( + ) e 1 v) Integrate! [ ] d e 1 y = ( + ) e 1 i.e., [ ] d e 1 y = e 1 ( + ) = e u ( du) = e u du = e u + C = e 1 + C e u [ ] e 1 y = e 1 + C vi) Solve for y y = 1 + Ce 1 + i.e., y = Ce du Incorporating the initial condition y (0) = 1, we have: 1 = Ce 1 (0) +(0) 1 = C 1 i.e., 1 = C 1 C = y = e
4 4. Solve: y y = 1 +1 sin () ; using the integrating factor method. (Assume that, y > 0.) i) Epress the equation in the form: y + p () y = Q () y + 1 y = 1 sin () p() Q() ii) Compute the integrating factor: e p() = e ( +1) 1 = e } ln +1 = {{ e ln(+1) } = ( + 1) because > 0 iii) Multiply both sides by the integrating factor ( + 1) y 1 + ( + 1) +1 y = ( + 1) 1 +1 sin () i.e., ( + 1) y + y = sin () iv) Epress the left side as the derivative of a product d [( + 1) y] = sin () v) Integrate! d [( + 1) y] = sin () i.e., ( + 1) y = cos () + C vi) Solve for y y = cos() +1 + C +1 = cos()+c1 +1 4
5 5. Determine whether or not the equation is eact. If the equation is eact, solve it. ( 3 y + y cos () + y 3 + e ) + ( 3 + 6y + sin () + cos (y) e sin(y)) dy = 0 M N The equation will be eact if M y = N Check: M y = [ y 3 y + y cos () + y 3 + e ] = cos () y N = [ 3 + 6y + sin () + cos (y) e sin(y)] = cos () y i.e., M y = cos () y = N Hence, the original equation is eact. The solution to the Differential Equation is of the form: U (, y) = C, where U (, y) = M = Ndy. U (, y) = M = [ 3 y + y cos () + y 3 + e ] = 3 y + y sin () + y 3 + e + f (y) = C U (, y) = N = [ 3 + 6y + sin () + cos (y) e sin(y)] = 3 y + y 3 + y sin () + e sin(y) + g () = C Comparing terms, we see that f (y) = e sin(y) and g () = e Thus, U = 3 y + y 3 + sin () y + e + e sin(y) = C Our solution y is given implicitly by the equation: 3 y + y 3 + sin () y + e + e sin(y) = C 5
6 6. Solve: y dy = 3 + 4y using the substitution v = y. (Assume that, y > 0) dy i) Re-write in the form: = f ( ) y y dy = 3 + 4y y dy = ( y ) = 3 1 ( y ( y + ) ) f( ) y ii) Make the following substitutions: v = y ; dy = v + dv v + dv dv = 3 1 = 3 1 i.e. dv = +3 iii) Separate! +3 dv = 1 iv) Integrate: +3 dv = 1 v + v + v = 3 + v = 3 (Eq. 1) + = +3 ***************************************************************** Scratchwork: +3 dv = 1 dv u = 1 u du ( 1 du) = 1 1 u du = 1 ln u = 1 ln + 3 = 1 ln ( + 3 ) i.e., +3 dv = 1 ln ( + 3 ) ***************************************************************** Substituting this into Eq. 1, we have: 1 ln ( + 3 ) = ln () + C ln ( + 3 ) = ln () + C 1 ln ( + 3 ) = ln ( ) + C 1 e ln( +3) = e ln( )+C 1 = e ln( ) e C 1 = C e ln( ) = C i.e., + 3 = C ( ) y + 3 = C ( ) y + 3 = C 6
7 y + 3 = C 4 Our solution is given implicitly by the equation: y + 3 = C 4 7
Practice Midterm Solutions
Practice Midterm Solutions Math 4B: Ordinary Differential Equations Winter 20 University of California, Santa Barbara TA: Victoria Kala DO NOT LOOK AT THESE SOLUTIONS UNTIL YOU HAVE ATTEMPTED EVERY PROBLEM
More information(a) x cos 3x dx We apply integration by parts. Take u = x, so that dv = cos 3x dx, v = 1 sin 3x, du = dx. Thus
Math 128 Midterm Examination 2 October 21, 28 Name 6 problems, 112 (oops) points. Instructions: Show all work partial credit will be given, and Answers without work are worth credit without points. You
More information(a) Show that there is a root α of f (x) = 0 in the interval [1.2, 1.3]. (2)
. f() = 4 cosec 4 +, where is in radians. (a) Show that there is a root α of f () = 0 in the interval [.,.3]. Show that the equation f() = 0 can be written in the form = + sin 4 Use the iterative formula
More informationDIFFERENTIATION RULES
3 DIFFERENTIATION RULES DIFFERENTIATION RULES 3.6 Derivatives of Logarithmic Functions In this section, we: use implicit differentiation to find the derivatives of the logarithmic functions and, in particular,
More informationMath 123, Week 9: Separable, First-Order Linear, and Substitution Methods. Section 1: Separable DEs
Math 123, Week 9: Separable, First-Order Linear, and Substitution Methods Section 1: Separable DEs We are finally to the point in the course where we can consider how to find solutions to differential
More information2.13 Linearization and Differentials
Linearization and Differentials Section Notes Page Sometimes we can approimate more complicated functions with simpler ones These would give us enough accuracy for certain situations and are easier to
More informationUnit #11 - Integration by Parts, Average of a Function
Unit # - Integration by Parts, Average of a Function Some problems and solutions selected or adapted from Hughes-Hallett Calculus. Integration by Parts. For each of the following integrals, indicate whether
More informationMath 180, Exam 2, Spring 2013 Problem 1 Solution
Math 80, Eam, Spring 0 Problem Solution. Find the derivative of each function below. You do not need to simplify your answers. (a) tan ( + cos ) (b) / (logarithmic differentiation may be useful) (c) +
More informationINTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS. Introduction It is possible to integrate any rational function, constructed as the ratio of polynomials by epressing it as a sum of simpler fractions
More informationMATHEMATICS 200 April 2010 Final Exam Solutions
MATHEMATICS April Final Eam Solutions. (a) A surface z(, y) is defined by zy y + ln(yz). (i) Compute z, z y (ii) Evaluate z and z y in terms of, y, z. at (, y, z) (,, /). (b) A surface z f(, y) has derivatives
More informationMATH MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS Calculus, Fall 2017 Professor: Jared Speck. Problem 1. Approximate the integral
MATH 8. - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS 8. Calculus, Fall 7 Professor: Jared Speck Problem. Approimate the integral 4 d using first Simpson s rule with two equal intervals and then the
More informationM151B Practice Problems for Exam 1
M151B Practice Problems for Eam 1 Calculators will not be allowed on the eam. Unjustified answers will not receive credit. 1. Compute each of the following its: 1a. 1b. 1c. 1d. 1e. 1 3 4. 3. sin 7 0. +
More informationPreface. Computing Definite Integrals. 3 cos( x) dx. x 3
Preface Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have more or less detail than other solutions. The level of detail in each solution will depend up on
More informationReview Problems for the Final
Review Problems for the Final Math 6-3/6 3 7 These problems are intended to help you study for the final However, you shouldn t assume that each problem on this handout corresponds to a problem on the
More informationFirst Midterm Examination
Çankaya University Department of Mathematics 016-017 Fall Semester MATH 155 - Calculus for Engineering I First Midterm Eamination 1) Find the domain and range of the following functions. Eplain your solution.
More informationMath 307 E - Summer 2011 Pactice Mid-Term Exam June 18, Total 60
Math 307 E - Summer 011 Pactice Mid-Term Exam June 18, 011 Name: Student number: 1 10 10 3 10 4 10 5 10 6 10 Total 60 Complete all questions. You may use a scientific calculator during this examination.
More information2. Jan 2010 qu June 2009 qu.8
C3 Functions. June 200 qu.9 The functions f and g are defined for all real values of b f() = 4 2 2 and g() = a + b, where a and b are non-zero constants. (i) Find the range of f. [3] Eplain wh the function
More informationLesson 50 Integration by Parts
5/3/07 Lesson 50 Integration by Parts Lesson Objectives Use the method of integration by parts to integrate simple power, eponential, and trigonometric functions both in a mathematical contet and in a
More informationChapter 9. Derivatives. Josef Leydold Mathematical Methods WS 2018/19 9 Derivatives 1 / 51. f x. (x 0, f (x 0 ))
Chapter 9 Derivatives Josef Leydold Mathematical Methods WS 208/9 9 Derivatives / 5 Difference Quotient Let f : R R be some function. The the ratio f = f ( 0 + ) f ( 0 ) = f ( 0) 0 is called difference
More informationMath 112 Section 10 Lecture notes, 1/7/04
Math 11 Section 10 Lecture notes, 1/7/04 Section 7. Integration by parts To integrate the product of two functions, integration by parts is used when simpler methods such as substitution or simplifying
More informationChapter 3: Transcendental Functions
Chapter 3: Transcendental Functions Spring 2018 Department of Mathematics Hong Kong Baptist University 1 / 32 Except for the power functions, the other basic elementary functions are also called the transcendental
More informationMATH 307: Problem Set #3 Solutions
: Problem Set #3 Solutions Due on: May 3, 2015 Problem 1 Autonomous Equations Recall that an equilibrium solution of an autonomous equation is called stable if solutions lying on both sides of it tend
More information2.2 Separable Equations
2.2 Separable Equations Definition A first-order differential equation that can be written in the form Is said to be separable. Note: the variables of a separable equation can be written as Examples Solve
More informationDT7: Implicit Differentiation
Differentiation Techniques 7: Implicit Differentiation 143 DT7: Implicit Differentiation Moel 1: Solving for y Most of the functions we have seen in this course are like those in Table 1 (an the first
More informationDaily Lessons and Assessments for AP* Calculus AB, A Complete Course Page 584 Mark Sparks 2012
The Second Fundamental Theorem of Calculus Functions Defined by Integrals Given the functions, f(t), below, use F( ) f ( t) dt to find F() and F () in terms of.. f(t) = 4t t. f(t) = cos t Given the functions,
More informationMath 180, Final Exam, Spring 2008 Problem 1 Solution. 1. For each of the following limits, determine whether the limit exists and, if so, evaluate it.
Math 80, Final Eam, Spring 008 Problem Solution. For each of the following limits, determine whether the limit eists and, if so, evaluate it. + (a) lim 0 (b) lim ( ) 3 (c) lim Solution: (a) Upon substituting
More informationThe Fundamental Theorem of Calculus Part 3
The Fundamental Theorem of Calculus Part FTC Part Worksheet 5: Basic Rules, Initial Value Problems, Rewriting Integrands A. It s time to find anti-derivatives algebraically. Instead of saying the anti-derivative
More informationSpring 2011 solutions. We solve this via integration by parts with u = x 2 du = 2xdx. This is another integration by parts with u = x du = dx and
Math - 8 Rahman Final Eam Practice Problems () We use disks to solve this, Spring solutions V π (e ) d π e d. We solve this via integration by parts with u du d and dv e d v e /, V π e π e d. This is another
More informationLesson 53 Integration by Parts
5/0/05 Lesson 53 Integration by Parts Lesson Objectives Use the method of integration by parts to integrate simple power, eponential, and trigonometric functions both in a mathematical contet and in a
More informationThe Explicit Form of a Function
Section 3 5 Implicit Differentiation The Eplicit Form of a Function The normal way we see function notation has f () on one sie of an equation an an epression in terms of on the other sie. We know the
More informationReview for Ma 221 Final Exam
Review for Ma 22 Final Exam The Ma 22 Final Exam from December 995.a) Solve the initial value problem 2xcosy 3x2 y dx x 3 x 2 sin y y dy 0 y 0 2 The equation is first order, for which we have techniques
More informationCalculus 1: Sample Questions, Final Exam
Calculus : Sample Questions, Final Eam. Evaluate the following integrals. Show your work and simplify your answers if asked. (a) Evaluate integer. Solution: e 3 e (b) Evaluate integer. Solution: π π (c)
More informationMa 221 Homework Solutions Due Date: January 24, 2012
Ma Homewk Solutions Due Date: January, 0. pg. 3 #, 3, 6,, 5, 7 9,, 3;.3 p.5-55 #, 3, 5, 7, 0, 7, 9, (Underlined problems are handed in) In problems, and 5, determine whether the given differential equation
More informationIntegration by Parts
Calculus 2 Lia Vas Integration by Parts Using integration by parts one transforms an integral of a product of two functions into a simpler integral. Divide the initial function into two parts called u
More informationPart Two. Diagnostic Test
Part Two Diagnostic Test AP Calculus AB and BC Diagnostic Tests Take a moment to gauge your readiness for the AP Calculus eam by taking either the AB diagnostic test or the BC diagnostic test, depending
More informationAPPM 1360 Final Exam Spring 2016
APPM 36 Final Eam Spring 6. 8 points) State whether each of the following quantities converge or diverge. Eplain your reasoning. a) The sequence a, a, a 3,... where a n ln8n) lnn + ) n!) b) ln d c) arctan
More informationFinal Exam 2011 Winter Term 2 Solutions
. (a Find the radius of convergence of the series: ( k k+ x k. Solution: Using the Ratio Test, we get: L = lim a k+ a k = lim ( k+ k+ x k+ ( k k+ x k = lim x = x. Note that the series converges for L
More informationPractice Midterm 1 Solutions Written by Victoria Kala July 10, 2017
Practice Midterm 1 Solutions Written by Victoria Kala July 10, 2017 1. Use the slope field plotter link in Gauchospace to check your solution. 2. (a) Not linear because of the y 2 sin x term (b) Not linear
More informationMA 113 Calculus I Fall 2009 Exam 4 December 15, 2009
MA 3 Calculus I Fall 009 Eam December 5, 009 Answer all of the questions - 7 and two of the questions 8-0. Please indicate which problem is not to be graded b crossing through its number in the table below.
More informationSection 7.4 #1, 5, 6, 8, 12, 13, 44, 53; Section 7.5 #7, 10, 11, 20, 22; Section 7.7 #1, 4, 10, 15, 22, 44
Math B Prof. Audrey Terras HW #4 Solutions Due Tuesday, Oct. 9 Section 7.4 #, 5, 6, 8,, 3, 44, 53; Section 7.5 #7,,,, ; Section 7.7 #, 4,, 5,, 44 7.4. Since 5 = 5 )5 + ), start with So, 5 = A 5 + B 5 +.
More informationFirst Order ODEs, Part I
Craig J. Sutton craig.j.sutton@dartmouth.edu Department of Mathematics Dartmouth College Math 23 Differential Equations Winter 2013 Outline 1 2 in General 3 The Definition & Technique Example Test for
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 information3 x 2 / 3 2. PhysicsAndMathsTutor.com. Question Answer Marks Guidance 1 5x(x + 1) 3(2x + 1) = (2x + 1)(x + 1) M1*
Question Answer Marks Guidance 5( + ) 3( + ) ( + )( + ) * 3 4 4 0 dep* Multiplying throughout by ( + )( + ) or combining fractions and multiplying up oe (eg can retain denominator throughout) Condone a
More informationENGI 3424 First Order ODEs Page 1-01
ENGI 344 First Order ODEs Page 1-01 1. Ordinary Differential Equations Equations involving only one independent variable and one or more dependent variables, together with their derivatives with respect
More informationVII. Techniques of Integration
VII. Techniques of Integration Integration, unlike differentiation, is more of an art-form than a collection of algorithms. Many problems in applied mathematics involve the integration of functions given
More information0.1 Problems to solve
0.1 Problems to solve Homework Set No. NEEP 547 Due September 0, 013 DLH Nonlinear Eqs. reducible to first order: 1. 5pts) Find the general solution to the differential equation: y = [ 1 + y ) ] 3/. 5pts)
More informationUnit 3. Integration. 3A. Differentials, indefinite integration. y x. c) Method 1 (slow way) Substitute: u = 8 + 9x, du = 9dx.
Unit 3. Integration 3A. Differentials, indefinite integration 3A- a) 7 6 d. (d(sin ) = because sin is a constant.) b) (/) / d c) ( 9 8)d d) (3e 3 sin + e 3 cos)d e) (/ )d + (/ y)dy = implies dy = / d /
More informationMidterm Exam #1. (y 2, y) (y + 2, y) (1, 1)
Math 5B Integral Calculus March 7, 7 Midterm Eam # Name: Answer Key David Arnold Instructions. points) This eam is open notes, open book. This includes any supplementary tets or online documents. You are
More information(a) During what time intervals on [0, 4] is the particle traveling to the left?
Chapter 5. (AB/BC, calculator) A particle travels along the -ais for times 0 t 4. The velocity of the particle is given by 5 () sin. At time t = 0, the particle is units to the right of the origin. t /
More informationMTH 3311 Test #2 Solutions
Pat Rossi MTH 3311 Test #2 Solutions S 2018 Name Directions: Do two of the three exercises. 1. A paratrooper and parachute weigh 160 lb. At the instant the parachute opens, she is traveling vertically
More informationM343 Homework 3 Enrique Areyan May 17, 2013
M343 Homework 3 Enrique Areyan May 17, 013 Section.6 3. Consider the equation: (3x xy + )dx + (6y x + 3)dy = 0. Let M(x, y) = 3x xy + and N(x, y) = 6y x + 3. Since: y = x = N We can conclude that this
More informationSolutions to Math 41 Final Exam December 9, 2013
Solutions to Math 4 Final Eam December 9,. points In each part below, use the method of your choice, but show the steps in your computations. a Find f if: f = arctane csc 5 + log 5 points Using the Chain
More informationLEARN ABOUT the Math
1.5 Inverse Relations YOU WILL NEED graph paper graphing calculator GOAL Determine the equation of an inverse relation and the conditions for an inverse relation to be a function. LEARN ABOUT the Math
More informationRadical Expressions and Functions What is a square root of 25? How many square roots does 25 have? Do the following square roots exist?
Topic 4 1 Radical Epressions and Functions What is a square root of 25? How many square roots does 25 have? Do the following square roots eist? 4 4 Definition: X is a square root of a if X² = a. 0 Symbolically,
More informationVector fields, line integrals, and Green's Theorem
Vector fields, line integrals, and Green's Theorem Line integral strategy suggested problems solutions Bunch of assorted line integral of vector filed problems here. In the solutions, I ll show the way
More information7.3 Adding and Subtracting Rational Expressions
7.3 Adding and Subtracting Rational Epressions LEARNING OBJECTIVES. Add and subtract rational epressions with common denominators. 2. Add and subtract rational epressions with unlike denominators. 3. Add
More informationUBC-SFU-UVic-UNBC Calculus Exam Solutions 7 June 2007
This eamination has 15 pages including this cover. UBC-SFU-UVic-UNBC Calculus Eam Solutions 7 June 007 Name: School: Signature: Candidate Number: Rules and Instructions 1. Show all your work! Full marks
More informationEXACT EQUATIONS AND INTEGRATING FACTORS
MAP- EXACT EQUATIONS AND INTEGRATING FACTORS First-order Differential Equations for Which We Can Find Eact Solutions Stu the patterns carefully. The first step of any solution is correct identification
More informationCalculus 2 - Examination
Calculus - Eamination Concepts that you need to know: Two methods for showing that a function is : a) Showing the function is monotonic. b) Assuming that f( ) = f( ) and showing =. Horizontal Line Test:
More informationC6-2 Differentiation 3
chain, product and quotient rules C6- Differentiation Pre-requisites: C6- Estimate Time: 8 hours Summary Learn Solve Revise Answers Summary The chain rule is used to differentiate a function of a function.
More informationChapter1. Ordinary Differential Equations
Chapter1. Ordinary Differential Equations In the sciences and engineering, mathematical models are developed to aid in the understanding of physical phenomena. These models often yield an equation that
More informationMath 115 Second Midterm November 12, 2018
EXAM SOLUTIONS Math 5 Second Midterm November, 08. Do not open this eam until you are told to do so.. Do not write your name anywhere on this eam. 3. This eam has 3 pages including this cover. There are
More informationChapter 3: Topics in Differentiation
Chapter 3: Topics in Differentiation Summary: Having investigated the derivatives of common functions in Chapter (i.e., polynomials, rational functions, trigonometric functions, and their combinations),
More informationAnswer Key 1973 BC 1969 BC 24. A 14. A 24. C 25. A 26. C 27. C 28. D 29. C 30. D 31. C 13. C 12. D 12. E 3. A 32. B 27. E 34. C 14. D 25. B 26.
Answer Key 969 BC 97 BC. C. E. B. D 5. E 6. B 7. D 8. C 9. D. A. B. E. C. D 5. B 6. B 7. B 8. E 9. C. A. B. E. D. C 5. A 6. C 7. C 8. D 9. C. D. C. B. A. D 5. A 6. B 7. D 8. A 9. D. E. D. B. E. E 5. E.
More informationProblem Set 9 Solutions
8.4 Problem Set 9 Solutions Total: 4 points Problem : Integrate (a) (b) d. ( 4 + 4)( 4 + 5) d 4. Solution (4 points) (a) We use the method of partial fractions to write A B (C + D) = + +. ( ) ( 4 + 5)
More informationMath 1B Final Exam, Solution. Prof. Mina Aganagic Lecture 2, Spring (6 points) Use substitution and integration by parts to find:
Math B Final Eam, Solution Prof. Mina Aganagic Lecture 2, Spring 20 The eam is closed book, apart from a sheet of notes 8. Calculators are not allowed. It is your responsibility to write your answers clearly..
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 informationMath 226 Calculus Spring 2016 Practice Exam 1. (1) (10 Points) Let the differentiable function y = f(x) have inverse function x = f 1 (y).
Math 6 Calculus Spring 016 Practice Exam 1 1) 10 Points) Let the differentiable function y = fx) have inverse function x = f 1 y). a) Write down the formula relating the derivatives f x) and f 1 ) y).
More informationDiff. Eq. App.( ) Midterm 1 Solutions
Diff. Eq. App.(110.302) Midterm 1 Solutions Johns Hopkins University February 28, 2011 Problem 1.[3 15 = 45 points] Solve the following differential equations. (Hint: Identify the types of the equations
More informationStudy 5.5, # 1 5, 9, 13 27, 35, 39, 49 59, 63, 69, 71, 81. Class Notes: Prof. G. Battaly, Westchester Community College, NY Homework.
Goals: 1. Recognize an integrand that is the derivative of a composite function. 2. Generalize the Basic Integration Rules to include composite functions. 3. Use substitution to simplify the process of
More informationCalculus Integration
Calculus Integration By Norhafizah Md Sarif & Norazaliza Mohd Jamil Faculty of Instrial Science & Technology norhafizah@ump.e.my, norazaliza@ump.e.my Description Aims This chapter is aimed to : 1. introce
More informationSection 3.7: Solving Radical Equations
Objective: Solve equations with radicals and check for extraneous solutions. In this section, we solve equations that have roots in the problem. As you might expect, to clear a root we can raise both sides
More informationExact Differential Equations. The general solution of the equation is f x, y C. If f has continuous second partials, then M y 2 f
APPENDIX C Additional Topics in Differential Equations APPENDIX C. Eact First-Order Equations Eact Differential Equations Integrating Factors Eact Differential Equations In Chapter 6, ou studied applications
More informationRegent College Maths Department. Core Mathematics 4 Trapezium Rule. C4 Integration Page 1
Regent College Maths Department Core Mathematics Trapezium Rule C Integration Page Integration It might appear to be a bit obvious but you must remember all of your C work on differentiation if you are
More informationAnswer Key. ( 1) n (2x+3) n. n n=1. (2x+3) n. = lim. < 1 or 2x+3 < 4. ( 1) ( 1) 2n n
Math Midterm Eam #3 December, 3 Answer Key. [5 Points] Find the Interval and Radius of Convergence for the following power series. Analyze carefully and with full justification. Use Ratio Test. L lim a
More informationUnit 9: Symmetric Functions
Haberman MTH 111 Section I: Functions and Their Graphs Unit 9: Symmetric Functions Some functions have graphs with special types of symmetries, and we can use the reflections we just studied to analyze
More informationa x a y = a x+y a x a = y ax y (a x ) r = a rx and log a (xy) = log a (x) + log a (y) log a ( x y ) = log a(x) log a (y) log a (x r ) = r log a (x).
You should prepare the following topics for our final exam. () Pre-calculus. (2) Inverses. (3) Algebra of Limits. (4) Derivative Formulas and Rules. (5) Graphing Techniques. (6) Optimization (Maxima and
More informationMT410 EXAM 1 SAMPLE 1 İLKER S. YÜCE DECEMBER 13, 2010 QUESTION 1. SOLUTIONS OF SOME DIFFERENTIAL EQUATIONS. dy dt = 4y 5, y(0) = y 0 (1) dy 4y 5 =
MT EXAM SAMPLE İLKER S. YÜCE DECEMBER, SURNAME, NAME: QUESTION. SOLUTIONS OF SOME DIFFERENTIAL EQUATIONS where t. (A) Classify the given equation in (). = y, y() = y () (B) Solve the initial value problem.
More informationThe region enclosed by the curve of f and the x-axis is rotated 360 about the x-axis. Find the volume of the solid formed.
Section A ln. Let g() =, for > 0. ln Use the quotient rule to show that g ( ). 3 (b) The graph of g has a maimum point at A. Find the -coordinate of A. (Total 7 marks) 6. Let h() =. Find h (0). cos 3.
More informationBe sure this exam has 8 pages including the cover The University of British Columbia MATH 103 Midterm Exam II Mar 14, 2012
Be sure this exam has 8 pages including the cover The University of British Columbia MATH Midterm Exam II Mar 4, 22 Family Name Student Number Given Name Signature Section Number This exam consists of
More informationDifferential calculus. Background mathematics review
Differential calculus Background mathematics review David Miller Differential calculus First derivative Background mathematics review David Miller First derivative For some function y The (first) derivative
More informationQuestion 1. (8 points) The following diagram shows the graphs of eight equations.
MAC 2233/-6 Business Calculus, Spring 2 Final Eam Name: Date: 5/3/2 Time: :am-2:nn Section: Show ALL steps. One hundred points equal % Question. (8 points) The following diagram shows the graphs of eight
More informationSubstituting this into the differential equation and solving for u, we have. + u2. du = 3x + c. ( )
Worked Solutions 27 Chapter 6: Simplifying Through Substitution 6 a Let u = 3 + 3y + 2 Solving for y computing y we then see that 3y = u 3 2 y = u 3 2 3 = d [ u 3 2 3 = 3 Substituting this into the differential
More informationTaylor Series 6B. lim s x. 1 a We can evaluate the limit directly since there are no singularities: b Again, there are no singularities, so:
Taylor Series 6B a We can evaluate the it directly since there are no singularities: 7+ 7+ 7 5 5 5 b Again, there are no singularities, so: + + c Here we should divide through by in the numerator and denominator
More informationAP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals
AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals 8. Basic Integration Rules In this section we will review various integration strategies. Strategies: I. Separate
More informationMath 181, Exam 2, Fall 2014 Problem 1 Solution. sin 3 (x) cos(x) dx.
Math 8, Eam 2, Fall 24 Problem Solution. Integrals, Part I (Trigonometric integrals: 6 points). Evaluate the integral: sin 3 () cos() d. Solution: We begin by rewriting sin 3 () as Then, after using the
More informationy x is symmetric with respect to which of the following?
AP Calculus Summer Assignment Name: Note: Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers for which f () is a real number. Part : Multiple Choice Solve
More informationHonours Advanced Algebra Unit 2: Polynomial Functions What s Your Identity? Learning Task (Task 8) Date: Period:
Honours Advanced Algebra Name: Unit : Polynomial Functions What s Your Identity? Learning Task (Task 8) Date: Period: Introduction Equivalent algebraic epressions, also called algebraic identities, give
More informationSolutions to Exam 1, Math Solution. Because f(x) is one-to-one, we know the inverse function exists. Recall that (f 1 ) (a) =
Solutions to Exam, Math 56 The function f(x) e x + x 3 + x is one-to-one (there is no need to check this) What is (f ) ( + e )? Solution Because f(x) is one-to-one, we know the inverse function exists
More informationCalculus for Engineers II - Sample Problems on Integrals Manuela Kulaxizi
Calculus for Engineers II - Sample Problems on Integrals Manuela Kulaxizi Question : Solve the following integrals:. π sin x. x 4 3. 4. sinh 8 x cosh x sin x cos 7 x 5. x 5 ln x 6. 8x + 6 3x + x 7. 8..
More informationPre-Calculus Mathematics Limit Process Calculus
NOTES : LIMITS AND DERIVATIVES Name: Date: Period: Mrs. Nguyen s Initial: LESSON.1 THE TANGENT AND VELOCITY PROBLEMS Pre-Calculus Mathematics Limit Process Calculus The type of it that is used to find
More informationMath 222 Spring 2013 Exam 3 Review Problem Answers
. (a) By the Chain ule, Math Spring 3 Exam 3 eview Problem Answers w s w x x s + w y y s (y xy)() + (xy x )( ) (( s + 4t) (s 3t)( s + 4t)) ((s 3t)( s + 4t) (s 3t) ) 8s 94st + 3t (b) By the Chain ule, w
More informationSolution to Review Problems for Midterm II
Soluion o Review Problems for Miderm II MATH 3860 001 Correcion: (i) should be y () ( + )y () + ( + )y() = e (1 + ). Given ha () = e is a soluion of y () ( + )y () + ( + )y() = 0. You should do problems
More informationThe Explicit Form of a Function
Section 3 5 Implicit Differentiation The Eplicit Form of a Function Function Notation requires that we state a function with f () on one sie of an equation an an epression in terms of on the other sie
More informationMath Reading assignment for Chapter 1: Study Sections 1.1 and 1.2.
Math 3350 1 Chapter 1 Reading assignment for Chapter 1: Study Sections 1.1 and 1.2. 1.1 Material for Section 1.1 An Ordinary Differential Equation (ODE) is a relation between an independent variable x
More informationSOLUTIONS to Problems for Review Chapter 15 McCallum, Hughes, Gleason, et al. ISBN by Vladimir A. Dobrushkin
SOLUTIONS to Problems for Review Chapter 1 McCallum, Hughes, Gleason, et al. ISBN 978-0470-118-9 by Vladimir A. Dobrushkin For Exercises 1, find the critical points of the given function and classify them
More informationA-LEVEL Mathematics MPC3
A-LEVEL Mathematics MPC UNIT: Pure Core Mark scheme 660 June 07 Version:.0 Final MARK SCHEME A LEVEL MATHEMATICS MPC JUNE 07 Mark schemes are prepared by the Lead Assessment Writer and considered, together
More informationThe stationary points will be the solutions of quadratic equation x
Calculus 1 171 Review In Problems (1) (4) consider the function f ( ) ( ) e. 1. Find the critical (stationary) points; establish their character (relative minimum, relative maimum, or neither); find intervals
More informationMath 312 Lecture Notes Linearization
Math 3 Lecture Notes Linearization Warren Weckesser Department of Mathematics Colgate University 3 March 005 These notes discuss linearization, in which a linear system is used to approximate the behavior
More informationOne Solution Two Solutions Three Solutions Four Solutions. Since both equations equal y we can set them equal Combine like terms Factor Solve for x
Algebra Notes Quadratic Systems Name: Block: Date: Last class we discussed linear systems. The only possibilities we had we 1 solution, no solution or infinite solutions. With quadratic systems we have
More information