Solutions manual of advanced engineering mathematics Peter Neil 8th edition pdf

1. for download complete version of solution (all chapter 1 to 23) click here. ISBN: 978-1-305-63515-9 © 2018, 2012 Cengage Learning WCN: 01-100-101 Cengage Learning 20 Channel Center Street Boston, MA 02210 USA ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher except as may be permitted by the license terms below. Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local office at: international.cengage.com/region. Cengage Learning products are represented in Canada by Nelson Education, Ltd. For your course and learning solutions, visit www.cengage.com/engineering. For product information and technology assistance, contact us at Cengage Learning Academic Resource Center, 1-800-423-0563. Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com. For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions. Further permissions questions can be emailed to permissionrequest@cengage.com. NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED, OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN. READ IMPORTANT LICENSE INFORMATION Dear Professor or Other Supplement Recipient: electronically must be through a password-protected site, with all copy and download functionality disabled, and accessible solely by your students who have purchased the associated textbook for the Course. You may not sell, license, auction, or otherwise redistribute the Supplement in any form. We ask that you take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution. Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement. If you do not accept these conditions, you must return the Supplement unused within 30 days of receipt. Cengage Learning has provided you with this product (the “Supplement”) for your review and, to the extent that you adopt the associated textbook for use in connection with your course (the “Course”), you and your students who purchase the textbook may use the Supplement as described below. Cengage Learning has established these use limitations in response to concerns raised by authors, professors, and other users regarding the pedagogical problems stemming from unlimited distribution of Supplements. All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/or its licensors. The Supplement is furnished by Cengage Learning on an “as is” basis without any warranties, express or implied. This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s conflict of law rules. Cengage Learning hereby grants you a nontransferable license to use the Supplement in connection with the Course, subject to the following conditions. The Supplement is for your personal, noncommercial use only and may not be reproduced, posted electronically or distributed, except that portions of the Supplement may be provided to your students IN PRINT FORM ONLY in connection with your instruction of the Course, so long as such students are advised that they may not copy or distribute any portion of the Supplement to any third party. Test banks and other testing materials may be made available in the classroom and collected at the end of each class session, or posted electronically as described herein. Any material posted Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement. We trust you find the Supplement a useful teaching tool. https://gioumeh.com/product/advanced-engineering-mathematics-solutions/

2. for download complete version of solution (all chapter 1 to 23) click here. INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY ADVANCED ENGINEERNG MATHEMATICS 8th EDITION PETER V.O’NEIL UNIVERSITY OF ALABAMA AT BIRMINGHAM https://gioumeh.com/product/advanced-engineering-mathematics-solutions/

3. for download complete version of solution (all chapter 1 to 23) click here. Contents 1 First-Order Differential Equations 1.1 Terminology and Separable Equations 1.2 The Linear First-Order Equation 1.3 Exact Equations 1.4 Homogeneous, Bernoulli and Riccati Equations Second-Order Differential Equations 2.1 The Linear Second-Order Equation 2.2 The Constant Coefficient Homogeneous Equation 2.3 Particular Solutions of the Nonhomogeneous Equation 2.4 The Euler Differential Equation 2.5 Series Solutions The Laplace Transform 3.1 Definition and Notation 3.2 Solution of Initial Value Problems 3.3 The Heaviside Function and Shifting Theorems 3.4 Convolution 3.5 Impulses and the Dirac Delta Function 3.6 Systems of Linear Differential Equations 1 1 12 19 28 37 37 41 46 53 58 69 69 72 77 86 92 93 2 3 iii https://gioumeh.com/product/advanced-engineering-mathematics-solutions/

4. for download complete version of solution (all chapter 1 to 23) click here. iv CONTENTS 4 Sturm-Liouville Problems and Eigenfunction Expansions 4.1 Eigenvalues and Eigenfunctions and Sturm-Liouville Problems 4.2 Eigenfunction Expansions 4.3 Fourier Series The Heat Equation 5.1 Diffusion Problems on a Bounded Medium 5.2 The Heat Equation With a Forcing Term F(x,t) 5.3 The Heat Equation on the Real Line 5.4 The Heat Equation on a Half-Line 5.5 The Two-Dimensional Heat Equation The Wave Equation 6.1 Wave Motion on a Bounded Interval 6.2 Wave Motion in an Unbounded Medium 6.3 d’Alembert’s Solution and Characteristics 6.4 The Wave Equation With a Forcing Term K(x,t) 6.5 The Wave Equation in Higher Dimensions Laplace’s Equation 7.1 The Dirichlet Problem for a Rectangle 7.2 The Dirichlet Problem for a Disk 7.3 The Poisson Integral Formula 7.4 The Dirichlet Problem for Unbounded Regions 7.5 A Dirichlet Problem in 3 Dimensions 7.6 The Neumann Problem 7.7 Poisson’s Equation Special Functions and Applications 8.1 Legendre Polynomials 8.2 Bessel Functions 8.3 Some Applications of Bessel Functions Transform Methods of Solution 9.1 Laplace Transform Methods 9.2 Fourier Transform Methods 9.3 Fourier Sine and Cosine Transforms Vectors and the Vector Space Rn 10.1 Vectors in the Plane and 3− Space 10.2 The Dot Product 10.3 The Cross Product 10.4 n− Vectors and the Algebraic Structure of Rn 10.5 Orthogonal Sets and Orthogonalization 10.6 Orthogonal Complements and Projections Matrices, Determinants and Linear Systems 11.1 Matrices and Matrix Algebra 11.2. Row Operations and Reduced Matrices 11.3 Solution of Homogeneous Linear Systems 11.4 Nonhomogeneous Systems 11.5 Matrix Inverses 11.6 Determinants 11.7 Cramer’s Rule 11.8 The Matrix Tree Theorem 101 101 107 114 137 137 147 150 153 155 157 157 167 173 190 192 197 197 202 205 205 208 211 217 221 221 235 251 263 263 268 271 275 275 277 278 280 284 287 291 291 295 299 306 313 315 318 320 5 6 7 8 9 10 11 https://gioumeh.com/product/advanced-engineering-mathematics-solutions/

5. for download complete version of solution (all chapter 1 to 23) click here. v 12 Eigenvalues, Diagonalization and Special Matrices 12.1 Eigenvalues and Eigenvectors 12.2 Diagonalization 12.3 Special Matrices and Their Eigenvalues and Eigenvectors 12.4 Quadratic Forms Systems of Linear Differential Equations 13.1 Linear Systems 13.2 Solution of X0= AX When A Is Constant 13.3 Exponential Matrix Solutions 13.4 Solution of X0= AX + G for Constant A 13.5 Solution by Diagonalization Nonlinear Systems and Qualitative Analysis 14.1 Nonlinear Systems and Phase Portraits 14.2 Critical Points and Stability 14.3 Almost Linear Systems 14.4 Linearization Vector Differential Calculus 15.1 Vector Functions of One Variable 15.2 Velocity, Acceleration and Curvature 15.3 The Gradient Field 15.4 Divergence and Curl 15.5 Streamlines of a Vector Field Vector Integral Calculus 16.1 Line Integrals 16.2 Green’s Theorem 16.3 Independence of Path and Potential Theory 16.4 Surface Integrals 16.5 Applications of Surface Integrals 16.6 Gauss’s Divergence Theorem 16.7 Stokes’s Theorem Fourier Series 17.1 Fourier Series on [−L,L] 17.2 Sine and Cosine Series 17.3 Integration and Differentiation of Fourier Series 17.4 Properties of Fourier Coefficients 17.5 Phase Angle Form 17.6 Complex Fourier Series 17.7 Filtering of Signals 323 323 327 332 336 339 339 341 348 350 353 359 359 363 364 369 373 373 376 381 385 387 391 391 393 398 405 408 412 414 419 419 423 428 430 432 435 438 13 14 15 16 17 https://gioumeh.com/product/advanced-engineering-mathematics-solutions/

6. for download complete version of solution (all chapter 1 to 23) click here. vi CONTENTS 18 Fourier Transforms 18.1 The Fourier Transform 18.2 Fourier sine and Cosine Transforms Complex Numbers and Functions 19.1 Geometry and Arithmetic of Complex Numbers 19.2 Complex Functions 19.3 The Exponential and Trigonometric Functions 19.4 The Complex Logarithm 19.5 Powers Complex Integration 20.1 The Integral of a Complex Function 20.2 Cauchy’s Theorem 20.3 Consequences of Cauchy’s Theorem Series Representations of Functions 21.1 Power Series 21.2 The Laurent Expansion Singularities and the Residue Theorem 22.1 Classification of Singularities 22.2 The Residue Theorem 22.3 Evaluation of Real Integrals Conformal Mappings 23.1 The Idea of a Conformal Mapping 23.2 Construction of Conformal Mappings 441 441 448 451 451 455 461 467 468 473 473 477 479 485 485 492 497 497 499 505 515 515 533 19 20 21 22 23 https://gioumeh.com/product/advanced-engineering-mathematics-solutions/

7. for download complete version of solution (all chapter 1 to 23) click here. Chapter 1 First-Order Differential Equations 1.1 Terminology and Separable Equations 1. The differential equation is separable because it can be written 3y2dy dx= 4x, or, in differential form, 3y2dy = 4xdx. Integrate to obtain y3= 2x2+ k. This implicitly defines a general solution, which can be written explicitly as y = (2x2+ k)1/3, with k an arbitrary constant. 2. Write the differential equation as xdy dx= −y, which separates as 1 ydy = −1 xdx if x 6= 0 and y 6= 0. Integrate to get ln|y| = −ln|x| + k. Then ln|xy| = k, so xy = c 1 https://gioumeh.com/product/advanced-engineering-mathematics-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.

8. for download complete version of solution (all chapter 1 to 23) click here. 2 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS with c constant (c = ek). y = 0 is a singular solution, satisfying the original differential equation. 3. If cos(y) 6= 0, the differential equation is dx=sin(x + y) =sin(x)cos(y) + cos(x)sin(y) y cos(y) cos(y) = sin(x) + cos(x)tan(y). There is no way to separate the variables in this equation, so the differen- tial equation is not separable. 4. Write the differential equation as exeydy dx= 3x, which separates in differential form as eydy = 3xe−xdx. Integrate to get ey= −3e−x(x + 1) + c, with c constant. This implicitly defines a general solution. 5. The differential equation can be written xdy dx= y2− y, or y(y − 1)dy =1 1 xdx, and is therefore separable. Separating the variables assumes that y 6= 0 and y 6= 1. We can further write ? Integrate to obtain ? y − 1−1 1 dy =1 xdx. y ln|y − 1| − ln|y| = ln|x| + k. Using properties of the logarithm, this is ???y − 1 ??? = k. ln xy https://gioumeh.com/product/advanced-engineering-mathematics-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.

9. for download complete version of solution (all chapter 1 to 23) click here. 1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 3 Then y − 1 xy = c, with c = ekconstant. Solve this for y to obtain the general solution 1 y = 1 − cx. y = 0 and y = 1 are singular solutions because these satisfy the differential equation, but were excluded in the algebra of separating the variables. 6. The differential equation is not separable. 7. The equation is separable because it can be written in differential form as sin(y) cos(y)dy =1 xdx. This 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 write sec(y) = cx with c constant. The algebra of separating the variables required that cos(y) 6= 0. Now cos(y) = 0 if y = (2n+1)π/2, with n any integer. Now y = (2n+1)π/2 also satisfies the original differential equation, so these are singular solutions. 8. The differential equation itself requires that y 6= 0 and x 6= −1. Write the equation as x y dy dx=2y2+ 1 x and separate the variables to get 1 1 y(2y2+ 1)dy = x(x + 1)dx. Use a partial fractions decomposition to write this as ?1 Integrate to obtain ? ?1 ? 2y 1 y− x− dy = dx. 2y2+ 1 x + 1 ln|y| −1 2ln(1 + 2y2) = ln|x| − ln|x + 1| + c https://gioumeh.com/product/advanced-engineering-mathematics-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.

10. for download complete version of solution (all chapter 1 to 23) click here. 4 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS with c constant. This implicitly defines a general solution. We can go a step further by writing this equation as ! ? ? y x p1 + 2y2 ln = ln + c x + 1 and take the exponential of both sides to get ? ? y x p1 + 2y2= k , x + 1 which also defines a general solution. 9. The differential equation is dy dx= ex− y + sin(y), and this is not separable. It is not possible to separate all terms involving x on one side of the equation and all terms involving y on the other. 10. Substitute sin(x − y) = sin(x)cos(y) − cos(x)sin(y), cos(x + y) = cos(x)cos(y) − sin(x)sin(y), and cos(2x) = cos2(x) − sin2(x) into the differential equation to get the separated differential form (cos(y) − sin(y))dy = (cos(x) − sin(x))dx. Integrate to obtain the implicitly defined general solution cos(y) + sin(y) = cos(x) + sin(x) + c. 11. If y 6= −1 and x 6= 0, we obtain the separated equation y2 y + 1dy =1 xdx. To make the integration easier, write this as ? ? 1 dy =1 y − 1 + xdx. 1 + y Integrate to obtain 1 2y2− y + ln|1 + y| = ln|x| + c. https://gioumeh.com/product/advanced-engineering-mathematics-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.

11. for download complete version of solution (all chapter 1 to 23) click here. 1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 5 This implicitly defines a general solution. The initial condition is y(3e2) = 2, so put y = 2 and x = 3e2to obtain 2 − 2 + ln(3) = ln(3e2) + c. Now ln(3e2) = ln(3) + ln(e2) = ln(3) + 2, so ln(3) = ln(3) + 2 + c. Then c = −2 and the solution of the initial value problem is implicitly defined by 1 2y2− y + ln|1 + y| = ln|x| − 2. 12. Integrate 1 y + 2dy = 3x2dx, assuming that y 6= −2, to obtain ln|2 + y| = x3+ c. This implicitly defines a general solution. To have y(2) = 8, let x = 2 and y = 8 to obtain ln(10) = 8 + c. The solution of the initial value problem is implicitly defined by ln|2 + y| = x3+ ln(10) − 8. We can take this a step further and write ?2 + y ? = x3− 8. ln 10 By taking the exponential of both sides of this equation we obtain the explicit solution y = 10ex3−8− 2. 13. With ln(yx) = xln(y), we obtain the separated equation ln(y) y dy = 3xdx. Integrate to obtain (ln(y))2= 3x2+ c. For y(2) = e3, we need (ln(e3))2= 3(4) + c, https://gioumeh.com/product/advanced-engineering-mathematics-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.

12. for download complete version of solution (all chapter 1 to 23) click here. 6 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS or 9 = 12 + c. Then c = −3 and the solution of the initial value problem is defined by (ln(y))2= 3x2− 3. Solve this to obtain the explicit solution √ 3(x2−1) y = e if |x| > 1. 14. Because ex−y2= exe−y2, the variables can be separated to obtain 2yey2dy = exdx. Integrate to get ey2= ex+ c. To satisfy y(4) = −2 we need e4= e4+ c so c = 0 and the solution of the initial value problem is implicitly defined by ey2= ex, which reduces to the simpler equation x = y2. Because y(4) = −2, the explicit solution is y = −√x for x > 0. 15. Separate the variables to obtain y cos(3y)dy = 2xdx. Integrate to get 1 3y sin(3y) +1 9cos(3y) = x2+ c, which implicitly defines a general solution. For y(2/3) = π/3, we need 1 3 π 3sin(π) +1 9cos(π) =4 9+ c. This reduces to −1 9=4 9+ c, so c = −5/9 and the solution of the initial value problem is implicitly defined by 1 3y sin(3y) +1 or 3y sin(3y) + cos(3y) = 9x2− 1. 9cos(3y) = x2−5 9, https://gioumeh.com/product/advanced-engineering-mathematics-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.

13. for download complete version of solution (all chapter 1 to 23) click here. 1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 7 16. Let T(t) be the temperature function. By Newton’s law of cooling, T0(t) = k(T−60) for some constant k to be determined. This equation is separable and is easily solved to obtain: T(t) = 60 + 30ekt. To determine k, use the fact that T(10) = 88: T(10) = 60 + 30e10k= 88. Then e10k=88 − 60 =14 15, 30 so 1 k = 10ln(14/15). Now we know the temperature function completely: T(t) = 60 + 30ekt= 60 + 30?e10k?t/10 = 60 + 30 15 ?14 ?t/10 . We want to know T(20), so compute ?14 ?2 ≈ 86.13 T(20) = 60 + 30 15 degrees Fahrenheit. To see how long it will take for the object to reach 65 degrees, solve for t in ?14 ?t/10 T(t) = 65 = 60 + 30 . 15 Then ?14 ?t/10 =65 − 60 =1 6, 15 30 so t 10ln(14/15) = ln(1/6) = −ln(6). The object reaches 65 degrees at time t = −10ln(6) ln(14/15)≈ 259.7 minutes. https://gioumeh.com/product/advanced-engineering-mathematics-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.

14. for download complete version of solution (all chapter 1 to 23) click here. 8 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS 17. Suppose the thermometer was removed from the house at time t = 0, and let T(t) be the temperature function. Let A be the ambient temperature outside the house (assumed constant). By Newton’s law, T0(t) = k(t − A). We are also given that T(0) = 70 and T(5) = 60. Further, fifteen minutes after being removed from the house, the thermometer reads 50.4, so T(15) = 50.4. We want to determine A, the constant outside temperature. From the differential equation for T, 1 T − AdT = kdt. Integrate this, as we have done before, to get T(t) = A + cekt. Now, T(0) = 70 = A + c, so c = 70 − A and T(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 e5kto get e5k=60 − A 70 − A. Then ?60 − A ?3 e15k=?e5k?3= . 70 − A Substitute this into the equation T(15) to get ?60 − A ?3 (70 − A) = 50.4 − A. 70 − A Then (60 − A)3= (50.4 − A)(70 − A)2. The cubic terms cancel and this reduces to the quadratic equation 10.4A2− 1156A + 30960 = 0, with roots 45 and (approximately) 66.15385. Clearly the outside temper- ature must be less than 50, and must therefore equal 45 degree. https://gioumeh.com/product/advanced-engineering-mathematics-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.

15. for download complete version of solution (all chapter 1 to 23) click here. 1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 9 18. The amount A(t) of radioactive material at time t is modeled by A0(t) = kA,A(0) = e3 together with the given half-life of the material, A(ln(2)) =1 2e3. Solve this (as in the text) to obtain ?1 ?t/ln(2) e3. A(t) = 2 Then ?1 ?3/ln(2) A(3) = e3 = 1 tonne. 2 19. The problem is like Problem 18, and we find that the amount of Uranium- 235 at time t is ?1 with t in years. Then ?t/(4.5(109)) U(t) = 10 , 2 ?1 ?1/4.5 U(109) = 10 ≈ 8.57 kg. 2 20. At time t there will be A(t) = 12ektgrams, and A(4) = 12e4k= 9.1. Solve this for k to get k =1 4ln The half-life of this element is the time t∗it will take for there to be 6 grams, so A(t∗) = 6 = 12eln(9.1/12)t∗/4. ?9.1 ? . 12 Solve this to get 4ln(1/2) ln(9.1/12)≈ 10.02 minutes. t∗= 21. Let Z∞ e−t2−(x/t)2dt. I(x) = 0 The integral we want is I(3). Compute Z∞ 1 t2e−t2−(x/t)2dt. I0(x) = −2x 0 https://gioumeh.com/product/advanced-engineering-mathematics-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.

16. for download complete version of solution (all chapter 1 to 23) click here. 10 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS Let u = x/t, so t = x/u and dt = −x u2du. Then ?u2 ? Z0 e−(x/u)2−u2 −x I0(x) = −2x = −2I(x). u2du x2 ∞ Then I(x) satisfies the separable differential equation I0= −2I, with general solution of the form I(x) = ce−2x. Now observe that Z∞ in which we used a standard integral that arises often in statistics. Then √π √π 2 e−t2dt = I(0) = = c, 0 e−2x. I(x) = 2 Finally, put x = 3 for the particular integral of interest: Z∞ √π 2 e−t2−(9/t)2dt = e−6. I(3) = 0 22. Begin with the logistic equation P0(t) = aP(t) − bP(t)2, in which a and b are positive constants. Then dP dt = P(a − bP) so 1 P(a − bP)dP = dt and the variables are separated. To make the integration easier, write this equation as ?1 Integrate to obtain ? 1 P+b 1 dP = dt. a − bP a a 1 aln(P) −b aln(a − bP) = t + c. if P(t) > − and a − bP(t) > 0. Using properties of the logarithm, we can write this equation as ? ? P ln = at + k, a − bP https://gioumeh.com/product/advanced-engineering-mathematics-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.

17. for download complete version of solution (all chapter 1 to 23) click here. 1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 11 in which k = ac is still constant. Then P a − bP= eat+k= ekeat= Keat, in which K = ekis a positive constant. Now suppose the initial population (say at time zero) is p0. Then P(0) = p0and p0 = K. a − bp0 We now have P p0 a − bp0eat. a − bP= It is a straightforward algebraic manipulation to solve this for P(t): ap0 a − bp0+ bp0eat. P(t) = This is the solution of the logistic equation with P(0) = p0. Because a − bp0> 0 by assumption, then bp0eat< a − bp0+ bpeat, so P(t) < ap0 bp0eateat=a b. This means that this population function is bounded above. Further, by multiplying the numerator and denominator of P(t) by e(−at, we have ap0 lim t→∞P(t) = lim (a − bp0)e−at+ bp0 ap0 bp0 t→∞ =a = lim t→∞ b. 23. With a and b as given, and p0= 3,929,214 (the population in 1790), the logistic population function for the United States is 123,141.5668 0.03071576577 + 0.0006242342283e0.03134te0.03134t. P(t) = If 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 that Q(10) = 5308483 = 3929214e10k and we can solve for k to get ?5308483 ? 1 ≈ 0.03008667012. k = 10ln 3929214 https://gioumeh.com/product/advanced-engineering-mathematics-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.

18. for download complete version of solution (all chapter 1 to 23) click here. 12 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS year 1790 1800 1810 1820 1830 1840 1850 1860 1870 1880 1890 1900 1910 1920 1930 1940 1950 1960 1970 1980 population 3,929,213 5,308,483 7,239,881 9,638,453 12,886,020 17,169,453 23,191,876 31,443,321 38,558,371 50,189,209 62,979766 76,212,168 92,228,496 106,021,537 123,202,624 132,164,569 151,325,798 179,323,175 203,302,031 226,547,042 P(t) percent error Q(t) percent error 3,929,214 5,336,313 7,228,171 9,757,448 13,110,174 17,507,365 23.193,639 30,414,301 39,374,437 50,180,383 62,772,907 76,873,907 91,976,297 107,398,941 122,401,360 136,329,577 148,679,224 150,231,097 167,943,428 174,940,040 0 3,929,214 5,308,483 7,179,158 7,179,158 13,000,754 17,685,992 23,894,292 32,281,888 43,613,774 58,923,484 79,073,491 107,551,857 145,303,703 196,312,254 0 0 0.52 -0.16 1.23 1.90 2.57 0.008 -3.27 2.12 -0.018 -0.33 0.87 -0.27 1.30 -0.65 3.15 -1.75 -11.2 -17.39 -22.78 -0.94 0.53 1.75 3.61 3.03 2.67 13.11 17.40 26.40 41.12 57.55 83.16 Table 1.1: Census data for Problem 23 The exponential model, using these two data points (1790 and 1800 pop- ulations), is Q(t) = 3929214e0.03008667012t. Table 1.1 uses Q(t) and P(t) to predict later populations from these two initial figures. The logistic model remains quite accurate until about 1960, at which time it loses accuracy quickly. The exponential model becomes quite inaccurate by 1870, after which the error becomes so large that it is not worth computing further. Exponential models do not work well over time with complex populations, such as fish in the ocean or countries throughout the world. 1.2 The Linear First-Order Equation 1. With p(x) = −3/x, and integrating factor is R(−3/x)dx= e−3ln(x)= x−3 e for x > 0. Multiply the differential equation by x−3to get x−3y0− 3x−4= 2x−1. https://gioumeh.com/product/advanced-engineering-mathematics-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.

19. for download complete version of solution (all chapter 1 to 23) click here. 1.2. THE LINEAR FIRST-ORDER EQUATION 13 or dx(x−3y) =2 d x. Integrate to get x−3y = 2ln(x) + c, with c an arbitrary constant. For x > 0 we have a general solution y = 2x3ln(x) + cx3. In the last integration, we can allow x < 0 by replacing ln(x) with ln|x| to derive the solution y = 2x3ln|x| + cx3 for x 6= 0. 2. e exto get y0ex+ yex=1 R dx= exis an integrating factor. Multiply the differential equation by 2(e2x− 1). Then (exy)0=1 2(e2x− 1) and an integration gives us exy =1 4e2x−1 2x + c. Then y =1 4ex−1 2xe−x+ ce−x is a general solution, with c an arbitrary constant. R 2dx= e2xis an integrating factor. Multiply the differential equation by e2x: y0e2x+ 2ye2x= xe2x, 3. e or (e2xy)0= xe2x. Integrate to get e2xy =1 2xe2x−1 4e2x+ c. giving us the general solution y =1 2x −1 4+ ce−2x. https://gioumeh.com/product/advanced-engineering-mathematics-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.

20. for download complete version of solution (all chapter 1 to 23) click here. 14 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS 4. For an integrating factor, compute R Multiply the differential equation by this integrating factor: sec(x)dx= eln|sec(x)+tan(x)|= sec(x) + tan(x). e y0(sec(x) + tan(x)) + sec(x)(sec(x) + tan(x))y = y0(sec(x) + tan(x)) + (sec(x)tan(x) + sec2(x))y = ((sec(x) + tan(x))y)0 = cos(x)(sec(x) + tan(x)) = 1 + sin(x). We therefore have ((sec(x) + tan(x))y)0= 1 + sin(x). Integrate to get y(sec(x) + tan(x)) = x − cos(x) + c. Then x − cos(x) + c sec(x) + tan(x). y = This is a general solution. If we wish, we can also observe that 1 cos(x) 1 + sin(x) sec(x) + tan(x)= to obtain ? ? cos(x) 1 + sin(x) y = (x − cos(x) + c) =xcos(x) − cos2(x) + ccos(x) 1 + sin(x) . 5. First determine the integrating factor R −2dx= e−2x. e Multiply the differential equation by e−2xto get (e−2xy)0= −8x2e−2x. Integrate to get Z e−2xy = −8x2e−2xdx = 4x2e−2x+ 4xe−2x+ 2e−2x+ c. This yields the general solution y = 4x2+ 4x + 2 + ce2x. https://gioumeh.com/product/advanced-engineering-mathematics-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.

21. for download complete version of solution (all chapter 1 to 23) click here. 1.2. THE LINEAR FIRST-ORDER EQUATION 15 R 3dx= e3xis an integrating factor. Multiply the differential equation by e3xto get (e3xy)0= 5e5x− 6e3x. Integrate this equation: 6. e e3xy = e5x− 2e3x+ c. Now we have a general solution y = e2x− 2 + ce−3x. We need y(0) = 2 = 1 − 2 + c, so c = 3. The unique solution of the initial value problem is y = e2x+ 3e−3x− 2. 7. x − 2 is an integrating factor for the differential equation because R(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). e Integrate to get (x − 2)y = x3− 3x2+ c. This gives us the general solution 1 x − 2(x3− 3x2+ c). y = Now we need y(3) = 27 − 27 + c = 4, so c = 4 and the solution of the initial value problem is 1 x − 2(x3− 3x2+ 4). y = R(−1)dx= e−xis an integrating factor. Multiply the differential equation by e−xto get: (ye−x)0= 2e3x. 8. e Integrate to get ye−x=2 3e3x+ c, https://gioumeh.com/product/advanced-engineering-mathematics-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.

22. for download complete version of solution (all chapter 1 to 23) click here. 16 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS and we have the general solution y =2 3e4x+ cex. We need y(0) =2 3+ c = −3, so c = −11/3 and the initial value problem has the solution y =2 3e4x−11 3ex. 9. First derive the integrating factor R(2/(x+1))dx= e2ln(x+1)= eln((x+1)2)= (x + 1)2. Multiply the differential equation by (x + 1)2to obtain ?(x + 1)2y?0= 3(x + 1)2. Integrate to obtain (x + 1)2y = (x + 1)3+ c. e Then c y = x + 1 + (x + 1)2. Now y(0) = 1 + c = 5 so c = 4 and the initial value problem has the solution 4 y = x + 1 + (x + 1)2. 10. An integrating factor is R(5/9x)dx= e(5/9)ln(x)= eln(x5/9)= x5/9. Multiply the differential equation by x5/9to get e (yx5/9)0= 3x32/9+ x14/9. Integrate to get yx5/9=27 41x41/9+9 23x23/9+ c. Then y =27 41x4+9 23x2+ cx−5/9. Finally, we need y(−1) =27 41+9 23− c = 4. Then c = −2782/943, so the initial value problem has the solution y =23 41x4+9 23x2−2782 943x−5/9. https://gioumeh.com/product/advanced-engineering-mathematics-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.

23. for download complete version of solution (all chapter 1 to 23) click here. 1.2. THE LINEAR FIRST-ORDER EQUATION 17 11. Let (x,y) be a point on the curve. The tangent line at (x,y) must pass through (0,2x2), and so has slope y0=y − 2x2 . x This is the linear differential equation y0−1 xy = −2x. An integrating factor is e−R(1/x)dx= e−ln(x)= eln(1/x)=1 x, so multiply the differential equation by 1/x to get 1 xy0− 1 x2y = −2. This is ?1 ?0 = −2. xy Integrate to get 1 xy = −2x + c. Then y = −2x2+ cx, in which c can be any number. 12. Let A(t) be the number of pounds of salt in the tank at time t ≥ 0. Then dA dt ?A(t) We must solve this subject to the initial condition A(0) = 25. The differ- ential equation is A0+ 50 + tA = 6, which is linear with integrating factor = rate salt is added − rate salt is removed ? = 6 − 2 . 50 + t 2 R 2/(50+t)dt= e2ln(50+t)= (50 + t)2. e Multiply the differential equation by (50 + t)2to get (50 + t)2A0+ 2(50 + t)A = 6(50 + t)2. https://gioumeh.com/product/advanced-engineering-mathematics-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.

24. for download complete version of solution (all chapter 1 to 23) click here. 18 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS This is ?(50 + t)2A?0= 6(50 + t)2. Integrate this equation to get (50 + t)2A = 2(50 + t)3+ c, which we will write as c A(t) = 2(50 + t) + (50 + t)2. We need c so that c A(0) = 100 + 2500= 25, so c = 187,500. The number of pounds of salt in the tank at time t is A(t) = 2(50 + t) −187,500 (50 + t)2. 13. Let A1(t) and A2(t) be the number of pounds of salt in tanks 1 and 2, respectively, at time t. Then 1(t) =5 2−5A1(t) A0 ;A1(0) = 20 100 and 2(t) =5A1(t) −5A2(t) 150 A0 ;A2(0) = 90. 100 Solve the linear initial value problem for A1(t) to get A1(t) = 50 − 30e−t/20. Substitute this into the differential equation for A2(t) to get 2+1 30A2=5 2−3 A0 2e−t/20;A2(0) = 90. Solve this linear problem to obtain A2(t) = 75 + 90e−t/20− 75e−t/30. Tank 2 has its minimum when A0 2(t) = 0, and this occurs when 2.5e−t/30− 4.5e−t/20= 0. This occurs when et/60= 9/5, or t = 60ln(9/5). Then A2(t)min= A2(60ln(9/5)) =5450 81 pounds. https://gioumeh.com/product/advanced-engineering-mathematics-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.