Differential Equations BERNOULLI EQUATIONS. Graham S McDonald. A Tutorial Module for learning how to solve Bernoulli differential equations
|
|
- Derrick Thornton
- 7 years ago
- Views:
Transcription
1 Differential Equations BERNOULLI EQUATIONS Graham S McDonald A Tutorial Module for learning how to solve Bernoulli differential equations Table of contents Begin Tutorial c 2004 g.s.mcdonald@salford.ac.uk
2 Table of contents. Theory 2. Exercises 3. Answers 4. Integrating factor method 5. Standard integrals 6. Tips on using solutions Full worked solutions
3 Section : Theory 3. Theory A Bernoulli differential equation can be written in the following standard form: + P (x)y = Q(x)yn, where n (the equation is thus nonlinear). To find the solution, change the dependent variable from y to z, where z = y n. This gives a differential equation in x and z that is linear, and can be solved using the integrating factor method. Note: Dividing the above standard form by y n gives: ( where we have used y n + P (x)y n = Q(x) + P (x)z = Q(x) ( n) = ( n)y n ).
4 Section 2: Exercises 4 2. Exercises Click on Exercise links for full worked solutions (there are 9 exercises in total) Exercise. The general form of a Bernoulli equation is + P (x)y = Q(x) yn, where P and Q are functions of x, and n is a constant. Show that the transformation to a new dependent variable z = y n reduces the equation to one that is linear in z (and hence solvable using the integrating factor method). Solve the following Bernoulli differential equations: Exercise 2. x y = xy2 Theory Answers IF method Integrals Tips
5 Section 2: Exercises 5 Exercise 3. + y x = y2 Exercise y = ex y 4 Exercise 5. x + y = xy3 Exercise x y = x2 cos x y 2 Theory Answers IF method Integrals Tips
6 Section 2: Exercises 6 Exercise 7. 2 (4x + 5)2 + tan x y = cos x y3 Exercise 8. x + y = y2 x 2 ln x Exercise 9. = y cot x + y3 cosecx Theory Answers IF method Integrals Tips
7 Section 3: Answers 7 3. Answers. HINT: Firstly, divide each term by y n. Then, differentiate z with respect to x to show that 2. y = x2 3 + C x, 3. y = x(c ln x), 4. y 3 = e x (C 3x), 5. y 2 = 2x+Cx 2, 6. y = x2 (sin x + C), 7. y 2 = 2 cos x (4x + 5)3 + C cos x, 8. xy = C + x( ln x), ( n) = y n,
8 Section 3: Answers 8 9. y 2 = sin2 x 2 cos x+c.
9 Section 4: Integrating factor method 9 4. Integrating factor method Consider an ordinary differential equation (o.d.e.) that we wish to solve to find out how the variable z depends on the variable x. If the equation is first order then the highest derivative involved is a first derivative. If it is also a linear equation then this means that each term can involve z either as the derivative OR through a single factor of z. Any such linear first order o.d.e. can be re-arranged to give the following standard form: + P (x)z = Q (x) where P (x) and Q (x) are functions of x, and in some cases may be constants.
10 Section 4: Integrating factor method 0 A linear first order o.d.e. can be solved using the integrating factor method. After writing the equation in standard form, P (x) can be identified. One then multiplies the equation by the following integrating factor : IF= e P (x) This factor is defined so that the equation becomes equivalent to: d (IF z) = IF Q (x), whereby integrating both sides with respect to x, gives: IF z = IF Q (x) Finally, division by the integrating factor (IF) gives z explicitly in terms of x, gives the solution to the equation.
11 Section 5: Standard integrals 5. Standard integrals f (x) x n x f(x) f (x) f(x) xn+ n+ (n ) [g (x)] n g (x) ln x g (x) g(x) [g(x)] n+ n+ (n ) ln g (x) e x e x a x ax ln a (a > 0) sin x cos x sinh x cosh x cos x sin x cosh x sinh x tan x ln cos x tanh x ln cosh x cosec x ln tan x 2 cosech x ln tanh x 2 sec x ln sec x + tan x sech x 2 tan e x sec 2 x tan x sech 2 x tanh x cot x ln sin x coth x ln sinh x sin 2 x cos 2 x x 2 x 2 sin 2x 4 sinh 2 x sinh 2x 4 x 2 + sin 2x 4 cosh 2 x sinh 2x 4 + x 2
12 Section 5: Standard integrals 2 f (x) f (x) f (x) f (x) a 2 +x 2 a tan x a a 2 x 2 2a ln a+x a x (0< x <a) (a > 0) x 2 a 2 2a ln x a x+a ( x > a>0) a 2 x 2 sin x a a 2 +x 2 ( a < x < a) x2 a 2 ln ln x+ a 2 +x 2 a x+ x 2 a 2 a (a > 0) (x>a>0) a2 x 2 a 2 2 [ sin ( ) x a a2 +x 2 a 2 2 ] x2 + x a 2 x 2 a a 2 a [ [ sinh ( x a cosh ( x a ) + x a 2 +x 2 a 2 ] ) + x ] x 2 a 2 a 2
13 Section 6: Tips on using solutions 3 6. Tips on using solutions When looking at the THEORY, ANSWERS, IF METHOD, INTEGRALS or TIPS pages, use the Back button (at the bottom of the page) to return to the exercises. Use the solutions intelligently. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct. Try to make less use of the full solutions as you work your way through the Tutorial.
14 Solutions to exercises 4 Full worked solutions Exercise. + P (x)y = Q(x)yn DIVIDE by y n : y n + P (x)y n = Q(x) SET z = y n : = ( n)y( n ) SUBSTITUTE ( n) = y n ( n) + P (x)z = Q(x) + P (x)z = Q (x) linear in z where P (x) = ( n)p (x) Q (x) = ( n)q(x). Return to Exercise
15 Solutions to exercises 5 Exercise 2. This is of the form + P (x)y = Q(x)yn where DIVIDE by y n : SET z = y n = y : where P (x) = x Q(x) = x and n = 2 y 2 x y = x = y 2 = y 2 x z = x + x z = x
16 Solutions to exercises 6 Integrating factor, IF = e x = e ln x = x x + z = x2 d [x z] = x2 xz = x 2 Use z = y : xz = x3 3 + C x y = x3 3 + C y = x2 3 + C x. Return to Exercise 2
17 Solutions to exercises 7 Exercise 3. This is of the form where P (x) = x, + P (x)y = Q(x)yn Q(x) =, and n = 2 DIVIDE by y n : y 2 + x y = SET z = y n = y : = y 2 = y 2 + x z = x z =
18 Solutions to exercises 8 Integrating factor, IF = e x = e ln x = e ln x = x x x 2 z = x [ d x z] = x x z = x z x = ln x + C Use z = y : yx = C ln x y = x(c ln x). Return to Exercise 3
19 Solutions to exercises 9 Exercise 4. This of the form + P (x)y = Q(x)yn DIVIDE by y n : SET z = y n = y 3 : where P (x) = 3 Q(x) = e x and n = 4 y y 3 = e x = 3y 4 = 3 y z = ex z = 3ex
20 Solutions to exercises 20 Integrating factor, IF = e = e x x e e x z = 3e x e x d [e x z] = 3 e x z = 3 e x z = 3x + C Use z = y : e x 3 y = 3x + C 3 y 3 = ex (C 3x). Return to Exercise 4
21 Solutions to exercises 2 Exercise 5. Bernoulli equation: + y x = y3 with P (x) =, Q(x) =, n = 3 x DIVIDE by y n y 3 : y 3 + x y 2 = SET z = y n z = y 2 : = 2y 3 2 = y x z = 2 x z = 2
22 Solutions to exercises 22 Integrating factor, IF = e 2 x = e 2 ln x = e ln x 2 = x 2 x 2 2 x 3 z = 2 x 2 [ d x z ] = 2 2 x 2 x 2 z = ( 2) ( ) x + C z = 2x + Cx 2 Use z = y 2 : y 2 = 2x+Cx 2. Return to Exercise 5
23 Solutions to exercises 23 Exercise 6. This is of the form + P (x)y = Q(x)yn where where P (x) = 2 x Q(x) = x 2 cos x and n = 2 DIVIDE by y n : y x y = x 2 cos x SET z = y n = y : = y 2 = y x z = x2 cos x 2 x z = x2 cos x
24 Solutions to exercises 24 Integrating factor, IF =e 2 x =e 2 x =e 2 ln x =e ln x 2 = x 2 x 2 2 x 3 z = x2 x 2 cos x [ ] d x 2 z = cos x x 2 z = cos x x 2 z = sin x + C Use z = y : x 2 y = sin x + C y = x2 (sin x + C). Return to Exercise 6
25 Solutions to exercises 25 Exercise 7. Divide by 2 to get standard form: + (4x + 5)2 tan x y = 2 2 cos x y3 This is of the form + P (x)y = Q(x)yn where P (x) = 2 tan x Q(x) = (4x + 5)2 2 cos x and n = 3
26 Solutions to exercises 26 DIVIDE by y n : y tan x (4x + y 2 5)2 = 2 cos x SET z = y n = y 2 : = 2y 3 = 2 y tan x z = (4x + 5)2 2 cos x (4x + 5)2 tan x z = cos x
27 Solutions to exercises 27 ] Integrating factor, IF = e tan x = e sin x cos x [ f (x) e f(x) Use z = y 2 : = e ln cos x = cos x cos x cos x tan x z =cos x(4x+5)2 cos x cos x sin x z = (4x + 5)2 d [cos x z] = (4x + 5)2 cos x z = (4x + 5) 2 ( ) cos x z = 4 3 (4x + 5)3 + C cos x y 2 = 2 (4x + 5)3 + C y 2 = 2 cos x (4x + 5)3 + C cos x. Return to Exercise 7
28 Solutions to exercises 28 Exercise 8. Standard form: + ( ) x y = (x ln x)y 2 P (x) = x, Q(x) = x ln x, n = 2 DIVIDE by y 2 : y 2 + ( x) y = x ln x SET z = y : = y 2 = y 2 + ( x) z = x ln x x z = x ln x
29 Solutions to exercises 29 Integrating factor: IF = e x = e ln x = e ln x = x x x 2 z = ln x d [ x z] = ln x x z = ln x + C [ Use integration by parts: u dv = uv v du, with u = ln x, dv = ] x z = [ x ln x x x ] + C Use z = y : xy = x( ln x) + C. Return to Exercise 8
30 Solutions to exercises 30 Exercise 9. Standard form: (cot x) y = (cosec x) y3 DIVIDE by y 3 : y 3 (cot x) y 2 = cosec x SET z = y 2 : = 2y 3 = 2 y 3 2 cot x z = cosec x + 2 cot x z = 2 cosec x
31 Solutions to exercises 3 Integrating factor: IF = e 2 cos x sin x e 2 f (x) f(x) = e 2 ln(sin x) = sin 2 x. sin 2 x + 2 sin x cos x z = 2 sin x [ d sin 2 x z ] = 2 sin x z sin 2 x = ( 2) ( cos x) + C Use z = y 2 : sin 2 x y 2 = 2 cos x + C y 2 = sin2 x 2 cos x+c. Return to Exercise 9
INTEGRATING FACTOR METHOD
Differential Equations INTEGRATING FACTOR METHOD Graham S McDonald A Tutorial Module for learning to solve 1st order linear differential equations Table of contents Begin Tutorial c 2004 g.s.mcdonald@salford.ac.uk
More informationIntegration ALGEBRAIC FRACTIONS. Graham S McDonald and Silvia C Dalla
Integration ALGEBRAIC FRACTIONS Graham S McDonald and Silvia C Dalla A self-contained Tutorial Module for practising the integration of algebraic fractions Table of contents Begin Tutorial c 2004 g.s.mcdonald@salford.ac.uk
More informationSeries FOURIER SERIES. Graham S McDonald. A self-contained Tutorial Module for learning the technique of Fourier series analysis
Series FOURIER SERIES Graham S McDonald A self-contained Tutorial Module for learning the technique of Fourier series analysis Table of contents Begin Tutorial c 004 g.s.mcdonald@salford.ac.uk 1. Theory.
More informationSUBSTITUTION I.. f(ax + b)
Integrtion SUBSTITUTION I.. f(x + b) Grhm S McDonld nd Silvi C Dll A Tutoril Module for prctising the integrtion of expressions of the form f(x + b) Tble of contents Begin Tutoril c 004 g.s.mcdonld@slford.c.uk
More informationImplicit Differentiation
Implicit Differentiation Sometimes functions are given not in the form y = f(x) but in a more complicated form in which it is difficult or impossible to express y explicitly in terms of x. Such functions
More informationFunction Name Algebra. Parent Function. Characteristics. Harold s Parent Functions Cheat Sheet 28 December 2015
Harold s s Cheat Sheet 8 December 05 Algebra Constant Linear Identity f(x) c f(x) x Range: [c, c] Undefined (asymptote) Restrictions: c is a real number Ay + B 0 g(x) x Restrictions: m 0 General Fms: Ax
More informationVectors VECTOR PRODUCT. Graham S McDonald. A Tutorial Module for learning about the vector product of two vectors. Table of contents Begin Tutorial
Vectors VECTOR PRODUCT Graham S McDonald A Tutorial Module for learning about the vector product of two vectors Table of contents Begin Tutorial c 2004 g.s.mcdonald@salford.ac.uk 1. Theory 2. Exercises
More information14.1. Basic Concepts of Integration. Introduction. Prerequisites. Learning Outcomes. Learning Style
Basic Concepts of Integration 14.1 Introduction When a function f(x) is known we can differentiate it to obtain its derivative df. The reverse dx process is to obtain the function f(x) from knowledge of
More informationUNIT 1: ANALYTICAL METHODS FOR ENGINEERS
UNIT : ANALYTICAL METHODS FOR ENGINEERS Unit code: A/60/40 QCF Level: 4 Credit value: 5 OUTCOME 3 - CALCULUS TUTORIAL DIFFERENTIATION 3 Be able to analyse and model engineering situations and solve problems
More informationMath 432 HW 2.5 Solutions
Math 432 HW 2.5 Solutions Assigned: 1-10, 12, 13, and 14. Selected for Grading: 1 (for five points), 6 (also for five), 9, 12 Solutions: 1. (2y 3 + 2y 2 ) dx + (3y 2 x + 2xy) dy = 0. M/ y = 6y 2 + 4y N/
More informationCHAPTER 2. Eigenvalue Problems (EVP s) for ODE s
A SERIES OF CLASS NOTES FOR 005-006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 4 A COLLECTION OF HANDOUTS ON PARTIAL DIFFERENTIAL EQUATIONS
More information1. First-order Ordinary Differential Equations
Advanced Engineering Mathematics 1. First-order ODEs 1 1. First-order Ordinary Differential Equations 1.1 Basic concept and ideas 1.2 Geometrical meaning of direction fields 1.3 Separable differential
More information2 Integrating Both Sides
2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation
More informationPartial Fractions. p(x) q(x)
Partial Fractions Introduction to Partial Fractions Given a rational function of the form p(x) q(x) where the degree of p(x) is less than the degree of q(x), the method of partial fractions seeks to break
More informationRecognizing Types of First Order Differential Equations E. L. Lady
Recognizing Types of First Order Differential Equations E. L. Lady Every first order differential equation to be considered here can be written can be written in the form P (x, y)+q(x, y)y =0. This means
More informationNotes and questions to aid A-level Mathematics revision
Notes and questions to aid A-level Mathematics revision Robert Bowles University College London October 4, 5 Introduction Introduction There are some students who find the first year s study at UCL and
More informationSeparable First Order Differential Equations
Separable First Order Differential Equations Form of Separable Equations which take the form = gx hy or These are differential equations = gxĥy, where gx is a continuous function of x and hy is a continuously
More informationThe Method of Partial Fractions Math 121 Calculus II Spring 2015
Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method
More information1.7. Partial Fractions. 1.7.1. Rational Functions and Partial Fractions. A rational function is a quotient of two polynomials: R(x) = P (x) Q(x).
.7. PRTIL FRCTIONS 3.7. Partial Fractions.7.. Rational Functions and Partial Fractions. rational function is a quotient of two polynomials: R(x) = P (x) Q(x). Here we discuss how to integrate rational
More informationIntegrals of Rational Functions
Integrals of Rational Functions Scott R. Fulton Overview A rational function has the form where p and q are polynomials. For example, r(x) = p(x) q(x) f(x) = x2 3 x 4 + 3, g(t) = t6 + 4t 2 3, 7t 5 + 3t
More information1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style
Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with
More informationCore Maths C3. Revision Notes
Core Maths C Revision Notes October 0 Core Maths C Algebraic fractions... Cancelling common factors... Multipling and dividing fractions... Adding and subtracting fractions... Equations... 4 Functions...
More informationis identically equal to x 2 +3x +2
Partial fractions 3.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. 4x+7 For example it can be shown that has the same value as 1 + 3
More informationtegrals as General & Particular Solutions
tegrals as General & Particular Solutions dy dx = f(x) General Solution: y(x) = f(x) dx + C Particular Solution: dy dx = f(x), y(x 0) = y 0 Examples: 1) dy dx = (x 2)2 ;y(2) = 1; 2) dy ;y(0) = 0; 3) dx
More informationA Brief Review of Elementary Ordinary Differential Equations
1 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 informationNonhomogeneous Linear Equations
Nonhomogeneous Linear Equations In this section we learn how to solve second-order nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form ay by cy G x where
More informationHomework # 3 Solutions
Homework # 3 Solutions February, 200 Solution (2.3.5). Noting that and ( + 3 x) x 8 = + 3 x) by Equation (2.3.) x 8 x 8 = + 3 8 by Equations (2.3.7) and (2.3.0) =3 x 8 6x2 + x 3 ) = 2 + 6x 2 + x 3 x 8
More informationUsing a table of derivatives
Using a table of derivatives In this unit we construct a Table of Derivatives of commonly occurring functions. This is done using the knowledge gained in previous units on differentiation from first principles.
More informationCalculus. Contents. Paul Sutcliffe. Office: CM212a.
Calculus Paul Sutcliffe Office: CM212a. www.maths.dur.ac.uk/~dma0pms/calc/calc.html Books One and several variables calculus, Salas, Hille & Etgen. Calculus, Spivak. Mathematical methods in the physical
More informationChapter 9. Systems of Linear Equations
Chapter 9. Systems of Linear Equations 9.1. Solve Systems of Linear Equations by Graphing KYOTE Standards: CR 21; CA 13 In this section we discuss how to solve systems of two linear equations in two variables
More informationSect 6.1 - Greatest Common Factor and Factoring by Grouping
Sect 6.1 - Greatest Common Factor and Factoring by Grouping Our goal in this chapter is to solve non-linear equations by breaking them down into a series of linear equations that we can solve. To do this,
More informationRAJALAKSHMI ENGINEERING COLLEGE MA 2161 UNIT I - ORDINARY DIFFERENTIAL EQUATIONS PART A
RAJALAKSHMI ENGINEERING COLLEGE MA 26 UNIT I - ORDINARY DIFFERENTIAL EQUATIONS. Solve (D 2 + D 2)y = 0. 2. Solve (D 2 + 6D + 9)y = 0. PART A 3. Solve (D 4 + 4)x = 0 where D = d dt 4. Find Particular Integral:
More information1 TRIGONOMETRY. 1.0 Introduction. 1.1 Sum and product formulae. Objectives
TRIGONOMETRY Chapter Trigonometry Objectives After studying this chapter you should be able to handle with confidence a wide range of trigonometric identities; be able to express linear combinations of
More information100. In general, we can define this as if b x = a then x = log b
Exponents and Logarithms Review 1. Solving exponential equations: Solve : a)8 x = 4! x! 3 b)3 x+1 + 9 x = 18 c)3x 3 = 1 3. Recall: Terminology of Logarithms If 10 x = 100 then of course, x =. However,
More informationHomework #2 Solutions
MAT Spring Problems Section.:, 8,, 4, 8 Section.5:,,, 4,, 6 Extra Problem # Homework # Solutions... Sketch likely solution curves through the given slope field for dy dx = x + y...8. Sketch likely solution
More informationu dx + y = 0 z x z x = x + y + 2 + 2 = 0 6) 2
DIFFERENTIAL EQUATIONS 6 Many physical problems, when formulated in mathematical forms, lead to differential equations. Differential equations enter naturally as models for many phenomena in economics,
More informationSample Problems. Practice Problems
Lecture Notes Partial Fractions page Sample Problems Compute each of the following integrals.. x dx. x + x (x + ) (x ) (x ) dx 8. x x dx... x (x + ) (x + ) dx x + x x dx x + x x + 6x x dx + x 6. 7. x (x
More information10.3. The Exponential Form of a Complex Number. Introduction. Prerequisites. Learning Outcomes
The Exponential Form of a Complex Number 10.3 Introduction In this Section we introduce a third way of expressing a complex number: the exponential form. We shall discover, through the use of the complex
More informationIntegration by substitution
Integration by substitution There are occasions when it is possible to perform an apparently difficult piece of integration by first making a substitution. This has the effect of changing the variable
More informationCalculus 1: Sample Questions, Final Exam, Solutions
Calculus : Sample Questions, Final Exam, Solutions. Short answer. Put your answer in the blank. NO PARTIAL CREDIT! (a) (b) (c) (d) (e) e 3 e Evaluate dx. Your answer should be in the x form of an integer.
More informationLecture Notes on PDE s: Separation of Variables and Orthogonality
Lecture Notes on PDE s: Separation of Variables and Orthogonality Richard H. Rand Dept. Theoretical & Applied Mechanics Cornell University Ithaca NY 14853 rhr2@cornell.edu http://audiophile.tam.cornell.edu/randdocs/
More information1 Lecture: Integration of rational functions by decomposition
Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.
More information3.1. Solving linear equations. Introduction. Prerequisites. Learning Outcomes. Learning Style
Solving linear equations 3.1 Introduction Many problems in engineering reduce to the solution of an equation or a set of equations. An equation is a type of mathematical expression which contains one or
More information13.5. Click here for answers. Click here for solutions. CURL AND DIVERGENCE. 17. F x, y, z x i y j x k. 2. F x, y, z x 2z i x y z j x 2y k
SECTION CURL AND DIVERGENCE 1 CURL AND DIVERGENCE A Click here for answers. S Click here for solutions. 1 15 Find (a the curl and (b the divergence of the vector field. 1. F x, y, xy i y j x k. F x, y,
More informationTo differentiate logarithmic functions with bases other than e, use
To ifferentiate logarithmic functions with bases other than e, use 1 1 To ifferentiate logarithmic functions with bases other than e, use log b m = ln m ln b 1 To ifferentiate logarithmic functions with
More informationFactoring - Grouping
6.2 Factoring - Grouping Objective: Factor polynomials with four terms using grouping. The first thing we will always do when factoring is try to factor out a GCF. This GCF is often a monomial like in
More informationChapter 7 Outline Math 236 Spring 2001
Chapter 7 Outline Math 236 Spring 2001 Note 1: Be sure to read the Disclaimer on Chapter Outlines! I cannot be responsible for misfortunes that may happen to you if you do not. Note 2: Section 7.9 will
More informationOn closed-form solutions to a class of ordinary differential equations
International Journal of Advanced Mathematical Sciences, 2 (1 (2014 57-70 c Science Publishing Corporation www.sciencepubco.com/index.php/ijams doi: 10.14419/ijams.v2i1.1556 Research Paper On closed-form
More informationThe Exponential Form of a Complex Number
The Exponential Form of a Complex Number 10.3 Introduction In this block we introduce a third way of expressing a complex number: the exponential form. We shall discover, through the use of the complex
More informationChange of Variables in Double Integrals
Change of Variables in Double Integrals Part : Area of the Image of a egion It is often advantageous to evaluate (x; y) da in a coordinate system other than the xy-coordinate system. In this section, we
More informationThis makes sense. t 2 1 + 1/t 2 dt = 1. t t 2 + 1dt = 2 du = 1 3 u3/2 u=5
1. (Line integrals Using parametrization. Two types and the flux integral) Formulas: ds = x (t) dt, d x = x (t)dt and d x = T ds since T = x (t)/ x (t). Another one is Nds = T ds ẑ = (dx, dy) ẑ = (dy,
More information~ EQUIVALENT FORMS ~
~ EQUIVALENT FORMS ~ Critical to understanding mathematics is the concept of equivalent forms. Equivalent forms are used throughout this course. Throughout mathematics one encounters equivalent forms of
More informationClick on the links below to jump directly to the relevant section
Click on the links below to jump directly to the relevant section What is algebra? Operations with algebraic terms Mathematical properties of real numbers Order of operations What is Algebra? Algebra is
More information3.6. Partial Fractions. Introduction. Prerequisites. Learning Outcomes
Partial Fractions 3.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. For 4x + 7 example it can be shown that x 2 + 3x + 2 has the same
More informationy cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx
Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE Evaluate cos 3 x dx.
More informationMerton College Maths for Physics Prelims October 10, 2005 MT I. Calculus. { y(x + δx) y(x)
Merton College Maths for Physics Prelims October 10, 2005 1. From the definition of the derivative, dy = lim δx 0 MT I Calculus { y(x + δx) y(x) evaluate d(x 2 )/. In the same way evaluate d(sin x)/. 2.
More information19.6. Finding a Particular Integral. Introduction. Prerequisites. Learning Outcomes. Learning Style
Finding a Particular Integral 19.6 Introduction We stated in Block 19.5 that the general solution of an inhomogeneous equation is the sum of the complementary function and a particular integral. We have
More informationMath Placement Test Practice Problems
Math Placement Test Practice Problems The following problems cover material that is used on the math placement test to place students into Math 1111 College Algebra, Math 1113 Precalculus, and Math 2211
More informationCollege Algebra - MAT 161 Page: 1 Copyright 2009 Killoran
College Algebra - MAT 6 Page: Copyright 2009 Killoran Zeros and Roots of Polynomial Functions Finding a Root (zero or x-intercept) of a polynomial is identical to the process of factoring a polynomial.
More informationSolutions to Homework 10
Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x
More informationis identically equal to x 2 +3x +2
Partial fractions.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. 4x+7 For example it can be shown that has the same value as + for any
More informationTechniques of Integration
CHPTER 7 Techniques of Integration 7.. Substitution Integration, unlike differentiation, is more of an art-form than a collection of algorithms. Many problems in applied mathematics involve the integration
More information1. (from Stewart, page 586) Solve the initial value problem.
. (from Stewart, page 586) Solve the initial value problem.. (from Stewart, page 586) (a) Solve y = y. du dt = t + sec t u (b) Solve y = y, y(0) = 0., u(0) = 5. (c) Solve y = y, y(0) = if possible. 3.
More informationMATHEMATICAL INSTITUTE UNIVERSITY OF OXFORD. Exploring Mathematics with MuPAD
MATHEMATICAL INSTITUTE UNIVERSITY OF OXFORD Exploring Mathematics with MuPAD Students Guide Michaelmas Term 2011 by Colin MacDonald y 1 0 0.2 0.3 0.4 0.5 0.6 0.7 x 1 2 3 Version 1.0, October 2010, written
More informationSection 4.4. Using the Fundamental Theorem. Difference Equations to Differential Equations
Difference Equations to Differential Equations Section 4.4 Using the Fundamental Theorem As we saw in Section 4.3, using the Fundamental Theorem of Integral Calculus reduces the problem of evaluating a
More information36 CHAPTER 1. LIMITS AND CONTINUITY. Figure 1.17: At which points is f not continuous?
36 CHAPTER 1. LIMITS AND CONTINUITY 1.3 Continuity Before Calculus became clearly de ned, continuity meant that one could draw the graph of a function without having to lift the pen and pencil. While this
More information2.2 Separable Equations
2.2 Separable Equations 73 2.2 Separable Equations An equation y = f(x, y) is called separable provided algebraic operations, usually multiplication, division and factorization, allow it to be written
More informationLagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given.
Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method
More informationEquations, Inequalities & Partial Fractions
Contents Equations, Inequalities & Partial Fractions.1 Solving Linear Equations 2.2 Solving Quadratic Equations 1. Solving Polynomial Equations 1.4 Solving Simultaneous Linear Equations 42.5 Solving Inequalities
More informationDifferential Equations I MATB44H3F. Version September 15, 2011-1949
Differential Equations I MATB44H3F Version September 15, 211-1949 ii Contents 1 Introduction 1 11 Preliminaries 1 12 Sample Application of Differential Equations 2 2 First Order Ordinary Differential Equations
More information20. Product rule, Quotient rule
20. Prouct rule, 20.1. Prouct rule Prouct rule, Prouct rule We have seen that the erivative of a sum is the sum of the erivatives: [f(x) + g(x)] = x x [f(x)] + x [(g(x)]. One might expect from this that
More informationSOLUTIONS. f x = 6x 2 6xy 24x, f y = 3x 2 6y. To find the critical points, we solve
SOLUTIONS Problem. Find the critical points of the function f(x, y = 2x 3 3x 2 y 2x 2 3y 2 and determine their type i.e. local min/local max/saddle point. Are there any global min/max? Partial derivatives
More informationx 2 + y 2 = 1 y 1 = x 2 + 2x y = x 2 + 2x + 1
Implicit Functions Defining Implicit Functions Up until now in this course, we have only talked about functions, which assign to every real number x in their domain exactly one real number f(x). The graphs
More informationMath 113 HW #7 Solutions
Math 3 HW #7 Solutions 35 0 Given find /dx by implicit differentiation y 5 + x 2 y 3 = + ye x2 Answer: Differentiating both sides with respect to x yields 5y 4 dx + 2xy3 + x 2 3y 2 ) dx = dx ex2 + y2x)e
More informationLS.6 Solution Matrices
LS.6 Solution Matrices In the literature, solutions to linear systems often are expressed using square matrices rather than vectors. You need to get used to the terminology. As before, we state the definitions
More informationAnalysis of Stresses and Strains
Chapter 7 Analysis of Stresses and Strains 7.1 Introduction axial load = P / A torsional load in circular shaft = T / I p bending moment and shear force in beam = M y / I = V Q / I b in this chapter, we
More informationTOPIC 4: DERIVATIVES
TOPIC 4: DERIVATIVES 1. The derivative of a function. Differentiation rules 1.1. The slope of a curve. The slope of a curve at a point P is a measure of the steepness of the curve. If Q is a point on the
More informationThe Mathematics Diagnostic Test
The Mathematics iagnostic Test Mock Test and Further Information 010 In welcome week, students will be asked to sit a short test in order to determine the appropriate lecture course, tutorial group, whether
More information(a) We have x = 3 + 2t, y = 2 t, z = 6 so solving for t we get the symmetric equations. x 3 2. = 2 y, z = 6. t 2 2t + 1 = 0,
Name: Solutions to Practice Final. Consider the line r(t) = 3 + t, t, 6. (a) Find symmetric equations for this line. (b) Find the point where the first line r(t) intersects the surface z = x + y. (a) We
More informationy 1 x dx ln x y a x dx 3. y e x dx e x 15. y sinh x dx cosh x y cos x dx sin x y csc 2 x dx cot x 7. y sec 2 x dx tan x 9. y sec x tan x dx sec x
Strateg for Integration As we have seen, integration is more challenging than differentiation. In finding the derivative of a function it is obvious which differentiation formula we should appl. But it
More informationTrigonometric Functions: The Unit Circle
Trigonometric Functions: The Unit Circle This chapter deals with the subject of trigonometry, which likely had its origins in the study of distances and angles by the ancient Greeks. The word trigonometry
More informationcorrect-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were:
Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that
More informationSection 1. Finding Common Terms
Worksheet 2.1 Factors of Algebraic Expressions Section 1 Finding Common Terms In worksheet 1.2 we talked about factors of whole numbers. Remember, if a b = ab then a is a factor of ab and b is a factor
More informationANALYTICAL METHODS FOR ENGINEERS
UNIT 1: Unit code: QCF Level: 4 Credit value: 15 ANALYTICAL METHODS FOR ENGINEERS A/601/1401 OUTCOME - TRIGONOMETRIC METHODS TUTORIAL 1 SINUSOIDAL FUNCTION Be able to analyse and model engineering situations
More informationLinear and quadratic Taylor polynomials for functions of several variables.
ams/econ 11b supplementary notes ucsc Linear quadratic Taylor polynomials for functions of several variables. c 010, Yonatan Katznelson Finding the extreme (minimum or maximum) values of a function, is
More informationSolving DEs by Separation of Variables.
Solving DEs by Separation of Variables. Introduction and procedure Separation of variables allows us to solve differential equations of the form The steps to solving such DEs are as follows: dx = gx).
More informationFactoring Polynomials
Factoring a Polynomial Expression Factoring a polynomial is expressing the polynomial as a product of two or more factors. Simply stated, it is somewhat the reverse process of multiplying. To factor polynomials,
More informationDifferentiation and Integration
This material is a supplement to Appendix G of Stewart. You should read the appendix, except the last section on complex exponentials, before this material. Differentiation and Integration Suppose we have
More information3.1 Solving Systems Using Tables and Graphs
Algebra 2 Chapter 3 3.1 Solve Systems Using Tables & Graphs 3.1 Solving Systems Using Tables and Graphs A solution to a system of linear equations is an that makes all of the equations. To solve a system
More informationIntegrating algebraic fractions
Integrating algebraic fractions Sometimes the integral of an algebraic fraction can be found by first epressing the algebraic fraction as the sum of its partial fractions. In this unit we will illustrate
More informationFactoring Quadratic Expressions
Factoring the trinomial ax 2 + bx + c when a = 1 A trinomial in the form x 2 + bx + c can be factored to equal (x + m)(x + n) when the product of m x n equals c and the sum of m + n equals b. (Note: the
More information1.4. Arithmetic of Algebraic Fractions. Introduction. Prerequisites. Learning Outcomes
Arithmetic of Algebraic Fractions 1.4 Introduction Just as one whole number divided by another is called a numerical fraction, so one algebraic expression divided by another is known as an algebraic fraction.
More informationThe Derivative. Philippe B. Laval Kennesaw State University
The Derivative Philippe B. Laval Kennesaw State University Abstract This handout is a summary of the material students should know regarding the definition and computation of the derivative 1 Definition
More informationSystems of Linear Equations and Inequalities
Systems of Linear Equations and Inequalities Recall that every linear equation in two variables can be identified with a line. When we group two such equations together, we know from geometry what can
More information4.3 Lagrange Approximation
206 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION Lagrange Polynomial Approximation 4.3 Lagrange Approximation Interpolation means to estimate a missing function value by taking a weighted average
More informationAn Introduction to Partial Differential Equations
An Introduction to Partial Differential Equations Andrew J. Bernoff LECTURE 2 Cooling of a Hot Bar: The Diffusion Equation 2.1. Outline of Lecture An Introduction to Heat Flow Derivation of the Diffusion
More informationName: ID: Discussion Section:
Math 28 Midterm 3 Spring 2009 Name: ID: Discussion Section: This exam consists of 6 questions: 4 multiple choice questions worth 5 points each 2 hand-graded questions worth a total of 30 points. INSTRUCTIONS:
More information3.2. Solving quadratic equations. Introduction. Prerequisites. Learning Outcomes. Learning Style
Solving quadratic equations 3.2 Introduction A quadratic equation is one which can be written in the form ax 2 + bx + c = 0 where a, b and c are numbers and x is the unknown whose value(s) we wish to find.
More informationVersion 1.0. General Certificate of Education (A-level) January 2012. Mathematics MPC4. (Specification 6360) Pure Core 4. Final.
Version.0 General Certificate of Education (A-level) January 0 Mathematics MPC (Specification 660) Pure Core Final Mark Scheme Mark schemes are prepared by the Principal Eaminer and considered, together
More informationSolutions for Review Problems
olutions for Review Problems 1. Let be the triangle with vertices A (,, ), B (4,, 1) and C (,, 1). (a) Find the cosine of the angle BAC at vertex A. (b) Find the area of the triangle ABC. (c) Find a vector
More information