ADVANCED ENGINEERING MATHEMATICS, 8TH EDITION
A Student’s Solutions Manual to AccompanyADVANCED ENGINEERING MATHEMATICS,8TH EDITIONPETER V. O’NEIL
STUDENT'S SOLUTIONS MANUALTO ACCOMPANYAdvanced EngineeringMathematics8th EDITIONPETER V. O’NEIL
Contents1First-Order Differential Equations1.1 Terminology and Separable Equations1.2 The Linear First-Order Equation1.3 Exact Equations1.4 Homogeneous, Bernoulli and Riccati Equations2 Second-Order Differential Equations2.1 The Linear Second-Order Equation2.2 The Constant Coefficient Homogeneous Equation2.3 Particular Solutions of the Nonhomogeneous Equation2.4 The Euler Differential Equation2.5 Series Solutions3 The Laplace Transform3.1 Definition and Notation3.2 Solution of Initial Value Problems3.3 The Heaviside Function and Shifting Theorems3.4 Convolution3.5 Impulses and the Dirac Delta Function3.6 Systems of Linear Differential Equationsiii118111519192124272935353740444848
ivCONTENTS4Sturm-Liouville Problems and Eigenfunction Expansions4.1 Eigenvalues and Eigenfunctions and Sturm-Liouville Problems4.2 Eigenfunction Expansions4.3 Fourier Series5 The Heat Equation5.1 Diffusion Problems on a Bounded Medium5.2 The Heat Equation With a Forcing Term F (x, t)5.3 The Heat Equation on the Real Line5.4 The Heat Equation on a Half-Line5.5 The Two-Dimensional Heat Equation6The Wave Equation6.1 Wave Motion on a Bounded Interval6.2 Wave Motion in an Unbounded Medium6.3 d’Alembert’s Solution and Characteristics6.4 The Wave Equation With a Forcing Term K(x, t)6.5 The Wave Equation in Higher Dimensions7Laplace’s Equation7.1 The Dirichlet Problem for a Rectangle7.2 The Dirichlet Problem for a Disk7.3 The Poisson Integral Formula7.4 The Dirichlet Problem for Unbounded Regions7.5 A Dirichlet Problem in 3 Dimensions7.6 The Neumann Problem7.7 Poisson’s Equation8Special Functions and Applications8.1 Legendre Polynomials8.2 Bessel Functions8.3 Some Applications of Bessel Functions9Transform Methods of Solution9.1 Laplace Transform Methods9.2 Fourier Transform Methods9.3 Fourier Sine and Cosine Transforms10Vectors and the Vector Space Rn10.1 Vectors in the Plane and 3 Space10.2 The Dot Product10.3 The Cross Product10.4 n Vectors and the Algebraic Structure of Rn10.5 Orthogonal Sets and Orthogonalization10.6 Orthogonal Complements and Projections11Matrices, Determinants and Linear Systems11.1 Matrices and Matrix Algebra11.2. Row Operations and Reduced Matrices11.3 Solution of Homogeneous Linear Systems11.4 Nonhomogeneous Systems11.5 Matrix Inverses11.6 Determinants11.7 Cramer’s Rule11.8 The Matrix Tree 5156158160163163165167171175176178179
v12Eigenvalues, Diagonalization and Special Matrices12.1 Eigenvalues and Eigenvectors12.2 Diagonalization12.3 Special Matrices and Their Eigenvalues and Eigenvectors12.4 Quadratic Forms13Systems of Linear Differential Equations13.1 Linear Systems13.2 Solution of X0 AX When A Is Constant13.3 Exponential Matrix Solutions13.4 Solution of X0 AX G for Constant A13.5 Solution by Diagonalization14Nonlinear Systems and Qualitative Analysis14.1 Nonlinear Systems and Phase Portraits14.2 Critical Points and Stability14.3 Almost Linear Systems14.4 Linearization15Vector Differential Calculus15.1 Vector Functions of One Variable15.2 Velocity, Acceleration and Curvature15.3 The Gradient Field15.4 Divergence and Curl15.5 Streamlines of a Vector Field16Vector Integral Calculus16.1 Line Integrals16.2 Green’s Theorem16.3 Independence of Path and Potential Theory16.4 Surface Integrals16.5 Applications of Surface Integrals16.6 Gauss’s Divergence Theorem16.7 Stokes’s Theorem17Fourier Series17.1 Fourier Series on [ L, L]17.2 Sine and Cosine Series17.3 Integration and Differentiation of Fourier Series17.4 Properties of Fourier Coefficients17.5 Phase Angle Form17.6 Complex Fourier Series17.7 Filtering of 237239242244245247248
viCONTENTS18Fourier Transforms18.1 The Fourier Transform18.2 Fourier sine and Cosine Transforms19Complex Numbers and Functions19.1 Geometry and Arithmetic of Complex Numbers19.2 Complex Functions19.3 The Exponential and Trigonometric Functions19.4 The Complex Logarithm19.5 Powers20Complex Integration20.1 The Integral of a Complex Function20.2 Cauchy’s Theorem20.3 Consequences of Cauchy’s Theorem21Series Representations of Functions21.1 Power Series21.2 The Laurent Expansion22Singularities and the Residue Theorem22.1 Classification of Singularities22.2 The Residue Theorem22.3 Evaluation of Real Integrals23Conformal Mappings23.1 The Idea of a Conformal Mapping23.2 Construction of Conformal 275279283283284287293293303
Chapter 1First-Order DifferentialEquations1.1Terminology and Separable Equations1. The differential equation is separable because it can be written3y 2dy 4x,dxor, in differential form,3y 2 dy 4x dx.Integrate to obtainy 3 2x2 k.This implicitly defines a general solution, which can be written explicitlyasy (2x2 k)1/3 ,with k an arbitrary constant.3. If cos(y) 6 0, the differential equation isysin(x y) dxcos(y)sin(x) cos(y) cos(x) sin(y) cos(y) sin(x) cos(x) tan(y).There is no way to separate the variables in this equation, so the differential equation is not separable.1 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS5. The differential equation can be writtenxordy y 2 y,dx11dy dx,y(y 1)xand is therefore separable. Separating the variables assumes that y 6 0and y 6 1. We can further write 111 dy dx.y 1 yxIntegrate to obtainln y 1 ln y ln x k.Using properties of the logarithm, this islnTheny 1 k.xyy 1 c,xywith c ek constant. Solve this for y to obtain the general solutiony 1.1 cxy 0 and y 1 are singular solutions because these satisfy the differentialequation, but were excluded in the algebra of separating the variables.7. The equation is separable because it can be written in differential form assin(y)1dy dx.cos(y)xThis assumes that x 6 0 and cos(y) 6 0. Integrate this equation to obtain ln cos(y) ln x k.This implicitly defines a general solution. From this we can also writesec(y) cxwith c constant.The algebra of separating the variables required that cos(y) 6 0. Nowcos(y) 0 if y (2n 1)π/2, with n any integer. Now y (2n 1)π/2 alsosatisfies the original differential equation, so these are singular solutions. 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1.1. TERMINOLOGY AND SEPARABLE EQUATIONS39. The differential equation isdy ex y sin(y),dxand this is not separable. It is not possible to separate all terms involvingx on one side of the equation and all terms involving y on the other.11. If y 6 1 and x 6 0, we obtain the separated equation1y2dy dx.y 1xTo make the integration easier, write this as 11dy dx.y 1 1 yxIntegrate to obtain1 2y y ln 1 y ln x c.2This implicitly defines a general solution. The initial condition is y(3e2 ) 2, so put y 2 and x 3e2 to obtain2 2 ln(3) ln(3e2 ) c.Nowln(3e2 ) ln(3) ln(e2 ) ln(3) 2,soln(3) ln(3) 2 c.Then c 2 and the solution of the initial value problem is implicitlydefined by1 2y y ln 1 y ln x 2.213. With ln(y x ) x ln(y), we obtain the separated equationln(y)dy 3x dx.yIntegrate to obtain(ln(y))2 3x2 c.For y(2) e3 , we need(ln(e3 ))2 3(4) c,or 9 12 c. Then c 3 and the solution of the initial value problemis defined by(ln(y))2 3x2 3. 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
4CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONSSolve this to obtain the explicit solution 2y e 3(x 1)if x 1.15. Separate the variables to obtainy cos(3y) dy 2x dx.Integrate to get11y sin(3y) cos(3y) x2 c,39which implicitly defines a general solution. For y(2/3) π/3, we need141πsin(π) cos(π) c.3399This reduces to41 c,99so c 5/9 and the solution of the initial value problem is implicitlydefined by115y sin(3y) cos(3y) x2 ,399or3y sin(3y) cos(3y) 9x2 1. 17. Suppose the thermometer was removed from the house at time t 0, andlet T (t) be the temperature function. Let A be the ambient temperatureoutside the house (assumed constant). By Newton’s law,T 0 (t) k(t A).We are also given that T (0) 70 and T (5) 60. Further, fifteen minutesafter being removed from the house, the thermometer reads 50.4, soT (15) 50.4.We want to determine A, the constant outside temperature. From thedifferential equation for T ,1dT kdt.T AIntegrate this, as we have done before, to getT (t) A cekt .Now,T (0) 70 A c, 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1.1. TERMINOLOGY AND SEPARABLE EQUATIONS5so c 70 A andT (t) A (70 A)ekt .Now use the other two conditions:T (5) A (70 A)e5k 15.5 and T (15) A (70 A)e15k 50.4.From the equation for T (5), solve for e5k to gete5k 60 A.70 AThene15k e 5k 3 60 A70 A 3.Substitute this into the equation T (15) to get 360 A(70 A) 50.4 A.70 AThen(60 A)3 (50.4 A)(70 A)2 .The cubic terms cancel and this reduces to the quadratic equation10.4A2 1156A 30960 0,with roots 45 and (approximately) 66.15385. Clearly the outside temperature must be less than 50, and must therefore equal 45 degree.19. The problem is like Problem 18, and we find that the amount of Uranium235 at time t is t/(4.5(109 ))1U (t) 10,2with t in years. Then 1/4.51 8.57 kg.U (10 ) 102921. LetZI(x) 2e t (x/t)2dt.0The integral we want is I(3). ComputeZ 1 t2 (x/t)20I (x) 2xedt.t20Let u x/t, so t x/u anddt xdu.u2 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONSThen0Z0I (x) 2x u2x2 2e (x/u) u2 xduu2 2I(x).Then I(x) satisfies the separable differential equation I 0 2I, withgeneral solution of the form I(x) ce 2x . Now observe that Z 2πI(0) e t dt c,20in which we used a standard integral that arises often in statistics. Then π 2xe.I(x) 2Finally, put x 3 for the particular integral of interest: Z 22π 6e .e t (9/t) dt I(3) 2023. With a and b as given, and p0 3, 929, 214 (the population in 1790), thelogistic population function for the United States isP (t) 123, 141.5668e0.03134t .0.03071576577 0.0006242342283e0.03134tIf we attempt an exponential model Q(t) Aekt , then take A Q(0) 3, 929, 214, the population in 1790. To find k, use the fact thatQ(10) 5308483 3929214e10kand we can solve for k to get 15308483k ln 0.03008667012.103929214The exponential model, using these two data points (1790 and 1800 populations), isQ(t) 3929214e0.03008667012t .Table 1.1 uses Q(t) and P (t) to predict later populations from these twoinitial figures. The logistic model remains quite accurate until about 1960,at which time it loses accuracy quickly. The exponential model becomesquite inaccurate by 1870, after which the error becomes so large that itis not worth computing further. Exponential models do not work wellover time with complex populations, such as fish in the ocean or countriesthroughout the world. 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1.1. TERMINOLOGY AND SEPARABLE 7,042P 40,040percent 73,491107,551,857145,303,703196,312,2547percent 1257.5583.16Table 1.1: Census data for Problem 23 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
8CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS1.2The Linear First-Order Equation1. With p(x) 3/x, and integrating factor iseR( 3/x) dx e 3 ln(x) x 3for x 0. Multiply the differential equation by x 3 to getx 3 y 0 3x 4 2x 1 .or2d 3(x y) .dxxIntegrate to getx 3 y 2 ln(x) c,with c an arbitrary constant. For x 0 we have a general solutiony 2x3 ln(x) cx3 .In the last integration, we can allow x 0 by replacing ln(x) with ln x to derive the solutiony 2x3 ln x cx3for x 6 0.R3. e 2 dx e2x is an integrating factor. Multiply the differential equation bye2x :y 0 e2x 2ye2x xe2x ,or(e2x y)0 xe2x .Integrate to get1 2x 1 2xxe e c.24e2x y giving us the general solution11x ce 2x .24y 5. First determine the integrating factorRe 2 dx e 2x .Multiply the differential equation by e 2x to get(e 2x y)0 8x2 e 2x .Integrate to gete 2xZy 8x2 e 2x dx 4x2 e 2x 4xe 2x 2e 2x c.This yields the general solutiony 4x2 4x 2 ce2x . 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1.2. THE LINEAR FIRST-ORDER EQUATION97. x 2 is an integrating factor for the differential equation becauseeR(1/(x 2)) dx eln(x 2) x 2.Multiply the differential equation by x 2 to get((x 2)y)0 3x(x 2).Integrate to get(x 2)y x3 3x2 c.This gives us the general solutiony 1(x3 3x2 c).x 2Now we needy(3) 27 27 c 4,so c 4 and the solution of the initial value problem isy 1(x3 3x2 4).x 29. First derive the integrating factoreR(2/(x 1)) dx2 e2 ln(x 1) eln((x 1)) (x 1)2 .Multiply the differential equation by (x 1)2 to obtain(x 1)2 y 0 3(x 1)2 .Integrate to obtain(x 1)2 y (x 1)3 c.Theny x 1 c.(x 1)2Nowy(0) 1 c 5so c 4 and the initial value problem has the solutiony x 1 4.(x 1)211. Let (x, y) be a point on the curve. The tangent line at (x, y) must passthrough (0, 2x2 ), and so has slopey0 y 2x2.x 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONSThis is the linear differential equationy0 1y 2x.xAn integrating factor ise R(1/x) dx e ln(x) eln(1/x) 1,xso multiply the differential equation by 1/x to get11 0y 2 y 2.xxThis is 1yx 0 2.Integrate to get1y 2x c.xTheny 2x2 cx,in which c can be any number.13. Let A1 (t) and A2 (t) be the number of pounds of salt in tanks 1 and 2,respectively, at time t. ThenA01 (t) 5 5A1 (t) ; A1 (0) 202100and5A1 (t) 5A2 (t) ; A2 (0) 90.100150Solve the linear initial value problem for A1 (t) to getA02 (t) A1 (t) 50 30e t/20 .Substitute this into the differential equation for A2 (t) to get15 3A2 e t/20 ; A2 (0) 90.302 2Solve this linear problem to obtainA02 A2 (t) 75 90e t/20 75e t/30 .Tank 2 has its minimum when A02 (t) 0, and this occurs when2.5e t/30 4.5e t/20 0.This occurs when et/60 9/5, or t 60 ln(9/5). ThenA2 (t)min A2 (60 ln(9/5)) 545081pounds. 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1.3. EXACT EQUATIONS1.311Exact EquationsIn these problems it is assumed that the differential equation has the formM (x, y) N (x, y)y 0 0, or, in differential form, M (x, y) dx N (x, y) dy 0.1. With M (x, y) 2y 2 yexy and N (x, y) 4xy xexy 2y. Then N M 4y exy xyexy x yfor all (x, y), so the differential equation is exact on the entire plane. Apotential function ϕ(x, y) must satisfy ϕ M (x, y) 2y 2 yexy xand ϕ N (x, y) 4xy xexy 2y. yChoose one to integrate. If we begin with ϕ/ x M , then integratewith respect to x to getϕ(x, y) 2xy 2 exy α(y),with α(y) the “constant” of integration with respect to x. Then we musthave ϕ 4xy xexy α0 (y) 4xy xexy 2y. yThis requires that α0 (y) 2y, so we can choose α(y) y 2 to obtain thepotential functionϕ(x, y) 2xy 2 exy y 2 .The general solution is defined implicitly by the equation2xy 2 exy y 2 c, ,with c an arbitrary constant.3. M/ y 4x 2x2 and N/ x 4x, so this equation is not exact (onany rectangle).5. M/ y 1 N/ x, for x 6 0, so this equation is exact on the planeexcept at points (0, y). Integrate ϕ/ x M or ϕ/ y N to find thepotential functionϕ(x, y) ln x xy y 3for x 6 0. The general solution is defined by an equationln x xy y 3 k. 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS7. For this equation to be exact, we need M N 6xy 2 3 3 2αxy 2 . y xThis will be true if α 3. By integrating, we find a potential functionϕ(x, y) x2 y 3 3xy 3y 2and a general solution is defined implicitly byx2 y 3 3xy 3y 2 k.9. Because M/ y 12y 2 N/ x, this equation is exact for all (x, y).Straightforward integrations yield the potential functionϕ(x, y) 3xy 4 x.A general solution is defined implicitly by3xy 4 x k.To satisfy the condition y(1) 2, we must choose k so that48 1 k,so k 47 and the solution of the initial value problem is specified by theequation3xy 4 x 47.In this case we can actually write this solution explicitly with y in termsof x.11. First, M N 2x sin(2y x) 2 cos(2y x) , y xso the differential equation is exact for all (x, y). For a potential function,integrate ϕ 2x cos(2y x) ywith respect to y to getϕ(x, y) x sin(2y x) c(x).Then we must have ϕ x cos(2y x) sin(2y x) x x cos(2y x) sin(2y x) c0 (x). 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1.3. EXACT EQUATIONS13Then c0 (x) 0 and we can take c(x) to be any constant. Choosing c(x) 0yieldsϕ(x, y) x sin(2y x).The general solution is defined implicitly by x sin(2y x) k.To satisfy y(π/12) π/8, we need πsin(π/6) k,12so choose k π/24 to obtain the solution defined by x sin(2y x) π24which of course is the same asx sin(2y x) We can also writey π.24 π 1 x arcsin224xfor x 6 0.13. ϕ c is also a potential function if ϕ is because (ϕ c) ϕ x xand ϕ (ϕ c) . y yThe function defined implicitly byϕ(x, y) kis the same as that defined byϕ(x, y) c kif k is arbitrary.15. First,3 N M x y 5/2 and 2x. y2 xand these are not equal on any rectangle in the plane.In differential form, the differential equation is(xy y 3/2 ) dx x2 dy 0. 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONSMultiply this equation by xa y b to get(xa 1 y b 1 xa y b 3/2 ) dx xa 2 y b dy 0
A Student’s Solutions Manual to Accompany ADVANCED ENGINEERING MATHEMATICS, 8TH EDITION PETER V. O’NEIL. STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY Advanced Engineering Mathematics 8th EDITION PETER V. O’NEIL. Contents 1 First-Order Di erential Equations 1 1.1 Terminology and Separable Equations 1 1.2 The Linear First-Order Equation 8
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ADVANCED ENGINEERING MATHEMATICS By ERWIN KREYSZIG 9TH EDITION This is Downloaded From www.mechanical.tk Visit www.mechanical.tk For More Solution Manuals Hand Books And Much Much More. INSTRUCTOR’S MANUAL FOR ADVANCED ENGINEERING MATHEMATICS imfm.qxd 9/15/05 12:06 PM Page i. imfm.qxd 9/15/05 12:06 PM Page ii. INSTRUCTOR’S MANUAL FOR ADVANCED ENGINEERING MATHEMATICS NINTH EDITION ERWIN .
Erwin Kreyszig, Advanced Engineering Mathematics, 10 th Edition, Wiley-India 2. Micheael Greenberg, Advanced Engineering Mathematics, 9 th edition, Pearson edn 3. Dean G. Duffy, Advanced engineering mathematics with MATLAB, CRC Press 4. Peter O’neil, Advanced Engineering Mathematics, Cengage Learning. 5.
Erwin Kreyszig, Advanced Engineering Mathematics, 10 th Edition, Wiley-India 2. Micheael Greenberg, Advanced Engineering Mathematics, 9 th edition, Pearson edn 3. Dean G. Duffy, Advanced engineering mathematics with MATLAB, CRC Press 4. Peter O’neil, Advanced Engineering Mathematics, Cengage Learning. 5.
End of the Year Celebrations June 3rd: 8th Grade Breakfast (OLA Gym; 8:30 AM); Grades 1-4 Field Trip to the Bronx Zoo June 4th: 8th Grade Dinner-Dance (7 PM) June 5th: Altar Server/SCOLA Trip June 6th: 8th Grade Trip to Philadelphia June 7th: 8th Grade Honors (Morning Announcements, 8:05 AM); Graduation Mass (OLA Church 8:30 AM) June 8th: 8th Grade Graduation (OLA Church, 1 PM)
Engineering Mechanics Statics 8th Edition meriam Solutions Manual Author: meriam Subject: Engineering Mechanics Statics 8th Edition meriam Solutions ManualInstant Download Keywords: Engineering Mechanics Statics
The purpose of this document is to provide keystone fundamentals of MLA 8th Edition. This document is not intended to serve or function as an end-all/be-all MLA resource or a fully instructive guide. It seeks to describe and exemplify the process of MLA 8th Edition. For more information about MLA 8th Edition, please visit https://style.mla.org.
2 Annual Book of ASTM Standards, Vol 01.06. 3 Annual Book of ASTM Standards, Vol 01.01. 4 Annual Book of ASTM Standards, Vol 15.08. 5 Annual Book of ASTM Standards, Vol 03.02. 6 Annual Book of ASTM Standards, Vol 02.05. 7 Annual Book of ASTM Standards, Vol 01.08. 8 Available from Standardization Documents Order Desk, Bldg. 4 Section D, 700 Robbins Ave., Philadelphia, PA 19111-5094, Attn: NPODS .
