Skip to content
Textbook Guides

Solution Manual for Differential Equations: Computing and Modeling (Tech Update), 5th Edition

Solution Manual for Differential Equations: Computing and Modeling (Tech Update), 5th Edition provides the perfect textbook solutions, giving you the help you need to succeed in your studies.

Preview 16 of 651 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

     S OLUTIONS M ANUAL D IFFERENTIAL E QUATIONS C OMPUTING AND M ODELING D IFFERENTIAL E QUATIONS AND B OUNDARY V ALUE P ROBLEMS C OMPUTING AND M ODELING F IFTH E DITION C. Henry Edwards David E. Penney The University of Georgia David Calvis Baldwin-Wallace University

  2. Page 2
  3. Page 3
    Read page 3 as text

    Contents 1 First-Order Differential Equations 1.1 Differential Equations and Mathematical Models 1 1.2 Integrals as General and Particular Solutions 8 1.3 Slope Fields and Solution Curves 16 1.4 Separable Equations and Applications 27 1.5 Linear First-Order Equations 44 1.6 Substitution Methods and Exact Equations 62 Chapter 1 Review Problems 86 2 Mathematical Models and Numerical Methods 2.1 Population Models 100 2.2 Equilibrium Solutions and Stability 116 2.3 Acceleration-Velocity Models 127 2.4 Numerical Approximation: Euler's Method 137 2.5 A Closer Look at the Euler Method 144 2.6 The Runge-Kutta Method 155 3 Linear Equations of Higher Order 3.1 Introduction: Second-Order Linear Equations 167 3.2 General Solutions of Linear Equations 174 3.3 Homogeneous Equations with Constant Coefficients 182 3.4 Mechanical Vibrations 190 3.5 Nonhomogeneous Equations and Undetermined Coefficients 201 3.6 Forced Oscillations and Resonance 214 3.7 Electrical Circuits 227 3.8 Endpoint Problems and Eigenvalues 234 4 Introduction to Systems of Differential Equations 4.1 First-Order Systems and Applications 241 4.2 The Method of Elimination 250 4.3 Numerical Methods for Systems 270 5 Linear Systems of Differential Equations 5.1 Matrices and Linear Systems 280 5.2 The Eigenvalue Method for Homogeneous Linear Systems 288 5.3 Solution Curves of Linear Systems 313 5.4 Second-Order Systems and Mechanical Applications 319 5.5 Multiple Eigenvalue Solutions 331 5.6 Matrix Exponentials and Linear Systems 345 5.7 Nonhomogeneous Linear Systems 355 iii 

  4. Page 4
    Read page 4 as text

    6 Nonlinear Systems and Phenomena 6.1 Stability and the Phase Plane 363 6.2 Linear and Almost Linear Systems 372 6.3 Ecological Applications: Predators and Competitors 389 6.4 Nonlinear Mechanical Systems 404 6.5 Chaos in Dynamical Systems 415 7 Laplace Transform Methods 7.1 Laplace Transforms and Inverse Transforms 422 7.2 Transformation of Initial Value Problems 427 7.3 Translation and Partial Fractions 436 7.4 Derivatives, Integrals, and Products of Transforms 444 7.5 Periodic and Piecewise Continuous Input Functions 451 7.6 Impulses and Delta Functions 464 8 Power Series Methods 8.1 Introduction and Review of Power Series 473 8.2 Series Solutions Near Ordinary Points 479 8.3 Regular Singular Points 492 8.4 Method of Frobenius—The Exceptional Cases 505 8.5 Bessel’s Equation 514 8.6 Applications of Bessel Functions 522 9 Fourier Series Methods and Partial Differential Equations 9.1 Periodic Functions and Trigonometric Series 527 9.2 General Fourier Series and Convergence 537 9.3 Fourier Sine and Cosine Series 551 9.4 Applications of Fourier Series 565 9.5 Heat Conduction and Separation of Variables 571 9.6 Vibrating Strings and the One-Dimensional Wave Equation 577 9.7 Steady-State Temperature and Laplace’s Equation 585 10 Eigenvalue Methods and Boundary Value Problems 10.1 Sturm-Liouville Problems and Eigenfunction Expansions 596 10.2 Applications of Eigenfunction Series 607 10.3 Steady Periodic Solutions and Natural Frequencies 619 10.4 Cylindrical Coordinate Problems 631 10.5 Higher-Dimensional Phenomena 643 Appendix Existence and Uniqueness of Solutions 644   iv 

  5. Page 5
    Read page 5 as text

    1 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS SECTION 1.1 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS The main purpose of Section 1.1 is simply to introduce the basic notation and terminology of dif- ferential equations, and to show the student what is meant by a solution of a differential equation. Also, the use of differential equations in the mathematical modeling of real-world phenomena is outlined. Problems 1-12 are routine verifications by direct substitution of the suggested solutions into the given differential equations. We include here just some typical examples of such verifications. 3. If 1 cos 2 y x  and 2 sin 2 y x  , then 1 2sin 2 y x    2 2 cos 2 y x   , so 1 1 4 cos 2 4 y x y      and 2 2 4sin 2 4 y x y      . Thus 1 1 4 0 y y   and 2 2 4 0 y y    . 4. If 3 1 x y e  and 3 2 x y e   , then 3 1 3 x y e  and 3 2 3 x y e    , so 3 1 1 9 9 x y e y    and 3 2 2 9 9 x y e y     . 5. If x x y e e    , then x x y e e     , so     2 . x x x x x y y e e e e e           Thus 2 . x y y e     6. If 2 1 x y e   and 2 2 x y x e   , then 2 1 2 x y e     , 2 1 4 x y e    , 2 2 2 2 x x y e x e      , and 2 2 2 4 4 . x x y e x e       Hence       2 2 2 1 1 1 4 4 4 4 2 4 0 x x x y y y e e e             and       2 2 2 2 2 2 2 2 4 4 4 4 4 2 4 0. x x x x x y y y e x e e x e x e                 8. If 1 cos cos 2 y x x   and 2 sin cos 2 y x x   , then 1 sin 2sin 2 , y x x     1 cos 4 cos 2 , y x x     2 cos 2sin 2 y x x    , and 2 sin 4 cos 2 . y x x     Hence     1 1 cos 4cos 2 cos cos 2 3cos 2 y y x x x x x        and     2 2 sin 4cos 2 sin cos 2 3cos 2 . y y x x x x x        

  6. Page 6
    Read page 6 as text

    2 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS 11. If 2 1 y y x    , then 3 2 y x     and 4 6 , y x    so       2 2 4 3 2 5 4 6 5 2 4 0. x y x y y x x x x x             If 2 2 ln y y x x    , then 3 3 2 ln y x x x      and 4 4 5 6 ln y x x x       , so           2 2 4 4 3 3 2 2 2 2 2 2 5 4 5 6 ln 5 2 ln 4 ln 5 5 6 10 4 ln 0. x y x y y x x x x x x x x x x x x x x x x                            13. Substitution of rx y e  into 3 2 y y   gives the equation 3 2 rx rx r e e  , which simplifies to 3 2. r  Thus 2 / 3 r  . 14. Substitution of rx y e  into 4 y y   gives the equation 2 4 rx rx r e e  , which simplifies to 2 4 1. r  Thus 1 / 2 r   . 15. Substitution of rx y e  into 2 0 y y y      gives the equation 2 2 0 rx rx rx r e r e e    , which simplifies to 2 2 ( 2)( 1) 0. r r r r       Thus 2 r   or 1 r  . 16. Substitution of rx y e  into 3 3 4 0 y y y      gives the equation 2 3 3 4 0 rx rx rx r e r e e    , which simplifies to 2 3 3 4 0 r r    . The quadratic formula then gives the solutions   3 57 6 r    . The verifications of the suggested solutions in Problems 17-26 are similar to those in Problems 1-12. We illustrate the determination of the value of C only in some typical cases. However, we illustrate typical solution curves for each of these problems. 17. 2 C  18. 3 C 

  7. Page 7
    Read page 7 as text

    Section 1.1 3 19. If   1 x y x Ce   , then   0 5 y  gives 1 5 C   , so 6 C  . 20. If   1 x y x C e x     , then   0 10 y  gives 1 10 C   , or 11 C  . 21. 7 C  . 22. If   ( ) ln y x x C   , then   0 0 y  gives ln 0 C  , so 1 C  . −4 0 4 −4 0 4 x y (0, 2) Problem 17 −5 0 5 −5 0 5 x y (0, 3) Problem 18 −5 0 5 −10 −5 0 5 10 x y (0, 5) Problem 19 −10 −5 0 5 10 −20 0 20 x y (0, 10) Problem 20

  8. Page 8
    Read page 8 as text

    4 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS 23. If 5 2 1 4 ( ) y x x C x    , then   2 1 y  gives 1 1 4 8 32 1 C     , or 56 C   . 24. 17 C  . 25. If   3 tan y x C   , then   0 1 y  gives the equation tan 1 C  . Hence one value of C is / 4 C   , as is this value plus any integral multiple of  . −2 −1 0 1 2 −10 −5 0 5 10 x y (0, 7) Problem 21 −20 −10 0 10 20 −5 0 5 x y (0, 0) Problem 22 0 1 2 3 −30 −20 −10 0 10 20 30 x y (2, 1) Problem 23 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −30 −20 −10 0 10 20 30 x y (1, 17) Problem 24

  9. Page 9
    Read page 9 as text

    Section 1.1 5 26. Substitution of x   and 0 y  into   cos y x C x   yields    0 1 C     , so C    . 27. y x y    28. The slope of the line through   , x y and   2, 0 x is 0 2 / 2 y y y x x x      , so the differen- tial equation is 2 x y y   . 29. If m y   is the slope of the tangent line and m  is the slope of the normal line at ( , ), x y then the relation 1 m m    yields     1 1 0 m y y x        . Solving for y  then gives the differential equation   1 y y x    . 30. Here m y   and 2 ( ) 2 x m D x k x     , so the orthogonality relation 1 m m    gives the differential equation 2 1. xy    31. The slope of the line through   , x y and ( , ) y x  is     y x y y x      , so the differen- tial equation is ( ) . x y y y x     In Problems 32-36 we get the desired differential equation when we replace the “time rate of change” of the dependent variable with its derivative with respect to time t , the word “is” with the = sign, the phrase “proportional to” with k , and finally translate the remainder of the given sentence into symbols. 32. dP dt k P  33. 2 dv dt kv  −2 −1 0 1 2 −4 −2 0 2 4 x y (0, 1) Problem 25 0 5 10 −10 −5 0 5 10 x y (  , 0) Problem 26

  10. Page 10
    Read page 10 as text

    6 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS 34.   250 dv dt k v   35.   dN dt k P N   36.   dN dt kN P N   37. The second derivative of any linear function is zero, so we spot the two solutions   1 y x  and ( ) y x x  of the differential equation 0 y   . 38. A function whose derivative equals itself, and is hence a solution of the differential equa- tion y y   , is ( ) x y x e  . 39. We reason that if 2 y kx  , then each term in the differential equation is a multiple of 2 x . The choice 1 k  balances the equation and provides the solution 2 ( ) y x x  . 40. If y is a constant, then 0 y   , so the differential equation reduces to 2 1 y  . This gives the two constant-valued solutions ( ) 1 y x  and ( ) 1 y x   . 41. We reason that if x y ke  , then each term in the differential equation is a multiple of x e . The choice 1 2 k  balances the equation and provides the solution 1 2 ( ) x y x e  . 42. Two functions, each equaling the negative of its own second derivative, are the two solu- tions   cos y x x  and ( ) sin y x x  of the differential equation y y    . 43. (a) We need only substitute   ( ) 1 x t C kt   in both sides of the differential equation 2 x kx   for a routine verification. (b) The zero-valued function ( ) 0 x t  obviously satisfies the initial value problem 2 x kx   , (0) 0 x  . 44. (a) The figure shows typical graphs of solutions of the differential equation 2 1 2 x x   . (b) The figure shows typical graphs of solutions of the differential equation 2 1 2 . x x    We see that—whereas the graphs with 1 2 k  appear to “diverge to infinity”—each solu- tion with 1 2 k   appears to approach 0 as . t   Indeed, we see from the Problem 43(a) solution   1 2 ( ) 1 x t C t   that ( ) x t   as 2 t C  . However, with 1 2 k   it is clear from the resulting solution   1 2 ( ) 1 x t C t   that ( ) x t remains bounded on any bounded interval, but ( ) 0 x t  as t   .

  11. Page 11
    Read page 11 as text

    Section 1.1 7 45. Substitution of 1 P   and 10 P  into the differential equation 2 P kP   gives 1 100 , k  so Problem 43(a) yields a solution of the form   1 100 ( ) 1 P t C t   . The initial condition (0) 2 P  now yields 1 2 , C  so we get the solution 1 100 ( ) 1 50 2 100 P t t t     . We now find readily that 100 P  when 49 t  and that 1000 P  when 49.9 t  . It ap- pears that P grows without bound (and thus “explodes”) as t approaches 50. 46. Substitution of 1 v    and 5 v  into the differential equation 2 v kv   gives 1 25 , k   so Problem 43(a) yields a solution of the form   ( ) 1 25 v t C t   . The initial condition (0) 10 v  now yields 1 10 , C  so we get the solution 1 50 ( ) 1 5 2 10 25 v t t t     . We now find readily that 1 v  when 22.5 t  and that 0.1 v  when 247.5 t  . It ap- pears that v approaches 0 as t increases without bound. Thus the boat gradually slows, but never comes to a “full stop” in a finite period of time. 47. (a) (10) 10 y  yields   10 1 10 C   , so 101 10 C  . (b) There is no such value of C , but the constant function ( ) 0 y x  satisfies the condi- tions 2 y y   and (0) 0 y  . 0 1 2 3 4 0 1 2 3 4 5 t x Problem 44a 0 1 2 3 4 0 1 2 3 4 5 6 t x Problem 44b

  12. Page 12
    Read page 12 as text

    8 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELS (c) It is obvious visually (in Fig. 1.1.8 of the text) that one and only one solution curve passes through each point ( , ) a b of the xy -plane, so it follows that there exists a unique solution to the initial value problem 2 y y   , ( ) y a b  . 48. (b) Obviously the functions 4 ( ) u x x   and 4 ( ) v x x   both satisfy the differential equa- tion 4 . x y y   But their derivatives 3 ( ) 4 u x x    and 3 ( ) 4 v x x    match at 0 x  , where both are zero. Hence the given piecewise-defined function   y x is differentiable, and therefore satisfies the differential equation because   u x and   v x do so (for 0 x  and 0 x  , respectively). (c) If 0 a  (for instance), then choose C  fixed so that 4 C a b   . Then the function   4 4 if 0 if 0 C x x y x C x x         satisfies the given differential equation for every real number value of C  . SECTION 1.2 INTEGRALS AS GENERAL AND PARTICULAR SOLUTIONS This section introduces general solutions and particular solutions in the very simplest situation — a differential equation of the form   y f x   — where only direct integration and evaluation of the constant of integration are involved. Students should review carefully the elementary con- cepts of velocity and acceleration, as well as the fps and mks unit systems. 1. Integration of 2 1 y x    yields   2 ( ) 2 1 y x x dx x x C       . Then substitution of 0 x  , 3 y  gives 3 0 0 C C     , so   2 3 y x x x    . 2. Integration of   2 2 y x    yields       2 3 1 3 2 2 y x x dx x C       . Then substitution of 2 x  , 1 y  gives 1 0 C C    , so     3 1 3 2 1 y x x    . 3. Integration of y x   yields   3/2 2 3 y x x dx x C     . Then substitution of 4 x  , 0 y  gives 16 3 0 C   , so     3/2 2 3 8 y x x   . 4. Integration of 2 y x    yields   2 1 y x x dx x C       . Then substitution of 1 x  , 5 y  gives 5 1 C    , so   1 6 y x x    .

  13. Page 13
    Read page 13 as text

    Section 1.2 9 5. Integration of   1 2 2 y x     yields     1 2 2 2 2 y x x dx x C        . Then substitu- tion of 2 x  , 1 y   gives 1 2 2 C     , so   2 2 5 y x x    . 6. Integration of   1 2 2 9 y x x    yields       1 2 3 2 2 2 1 3 9 9 y x x x dx x C       . Then substitution of 4 x   , 0 y  gives 3 1 3 0 (5) C   , so     3/2 2 1 3 9 125 y x x          . 7. Integration of 2 10 1 y x    yields   1 2 10 10 tan 1 y x dx x C x       . Then substitution of 0 x  , 0 y  gives 0 10 0 C    , so   1 10 tan y x x   . 8. Integration of cos 2 y x   yields   1 2 cos 2 sin 2 y x x dx x C     . Then substitution of 0 x  , 1 y  gives 1 0 C   , so   1 2 sin 2 1 y x x   . 9. Integration of 2 1 1 y x    yields 1 2 1 ( ) sin 1 y x dx x C x       . Then substitution of 0 x  , 0 y  gives 0 0 C   , so   1 sin y x x   . 10. Integration of x y xe    yields       1 1 x u u x y x xe dx ue du u e x e C             , using the substitution u x   together with Formula #46 inside the back cover of the textbook. Then substituting 0 x  , 1 y  gives 1 1 , C    so ( ) ( 1) 2. x y x x e      11. If   50 a t  , then   0 50 50 50 10 v t dt t v t       . Hence     2 2 0 50 10 25 10 25 10 20 x t t dt t t x t t          . 12. If   20 a t   , then     0 20 20 20 15 v t dt t v t          . Hence     2 2 0 20 15 10 15 10 15 5 x t t dt t t x t t             . 13. If   3 a t t  , then   2 2 3 3 0 2 2 3 5 v t t dt t v t       . Hence     2 3 3 3 1 1 0 2 2 2 5 5 5 x t t dt t t x t t         . 14. If   2 1 a t t   , then     2 2 0 2 1 7 v t t dt t t v t t          . Hence     2 3 3 1 1 1 1 0 3 2 3 2 7 7 7 4 x t t t dt t t t x t t t             .

  14. Page 14
    Read page 14 as text

    10 INTEGRALS AS GENERAL AND PARTICULAR SOLUTIONS 15. If     2 4 3 a t t   , then         2 3 3 4 4 3 3 4 3 3 3 37 v t t dt t C t          (taking 37 C   so that   0 1 v   ). Hence         3 4 4 4 1 1 3 3 3 3 37 3 37 3 37 26 x t t dt t t C t t             . 16. If   1 4 a t t   , then   1 2 4 2 4 5 4 v t dt t C t t          (taking 5 C   so that   0 1 v   ). Hence         3/2 3/2 29 4 4 3 3 3 2 4 5 4 5 4 5 x t t dt t t C t t             (taking 29 3 C   so that   0 1 x  ). 17. If     3 1 a t t    , then         3 2 2 1 1 1 2 2 2 1 1 1 v t t dt t C t               (taking 1 2 C  so that   0 0 v  ). Hence         2 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 1 x t t dt t t C t t                     (taking 1 2 C   so that   0 0 x  ). 18. If   50sin 5 a t t  , then   50 sin 5 10 cos 5 10 cos 5 v t t dt t C t        (taking 0 C  so that   0 10 v   ). Hence   10 cos 5 2 sin 5 2 sin 5 10 x t t dt t C t          (taking 10 C   so that   0 8 x  ). Students should understand that Problems 19-22, though different at first glance, are solved in the same way as the preceding ones, that is, by means of the fundamental theorem of calculus in the form       0 0 t t x t x t v s ds    cited in the text. Actually in these problems     0 t x t v s ds   , since 0 t and   0 x t are each given to be zero. 19. The graph of   v t shows that   5 if 0 5 10 if 5 10 t v t t t          , so that   1 2 1 2 2 5 if 0 5 10 if 5 10 t C t x t t t C t            . Now 1 0 C  because   0 0 x  , and continuity of   x t requires that   5 x t t  and   2 1 2 2 10 x t t t C    agree when 5 t  . This implies that 25 2 2 C   , leading to the graph of   x t shown.

  15. Page 15
    Read page 15 as text

    Section 1.2 11 Alternate solution for Problem 19 (and similar for 20-22): The graph of   v t shows that   5 if 0 5 10 if 5 10 t v t t t          . Thus for 0 5 t   ,     0 t x t v s ds   is given by 0 5 5 t ds t   , whereas for 5 10 t   we have     5 0 0 5 2 2 2 5 5 10 75 25 25 10 25 10 10 . 2 2 2 2 2 t t s t s x t v s ds ds s ds s t t s t t                            The graph of   x t is shown. 20. The graph of   v t shows that   if 0 5 5 if 5 10 t t v t t         , so that   2 1 1 2 2 if 0 5 5 if 5 10 t C t x t t C t           . Now 1 0 C  because   0 0 x  , and continuity of   x t requires that   2 1 2 x t t  and   2 5 x t t C   agree when 5 t  . This implies that 25 2 2 C   , leading to the graph of   x t shown. 21. The graph of   v t shows that   if 0 5 10 if 5 10 t t v t t t          , so that   2 1 1 2 2 1 2 2 if 0 5 10 if 5 10 t C t x t t t C t            . Now 1 0 C  because   0 0 x  , and continuity of 0 2 4 6 8 10 0 10 20 30 40 t x (5, 25) Problem 19 0 2 4 6 8 10 0 10 20 30 40 t x (5, 12.5) Problem 20

  16. Page 16
    Read page 16 as text

    12 INTEGRALS AS GENERAL AND PARTICULAR SOLUTIONS   x t requires that   2 1 2 x t t  and   2 1 2 2 10 x t t t C    agree when 5 t  . This implies that 2 25 C   , leading to the graph of   x t shown. 22. For 0 3 t   , 5 3 ( ) v t t  , so   2 5 1 6 x t t C   . Now 1 0 C  because   0 0 x  , so   2 5 6 x t t  on this first interval, and its right-endpoint value is   15 2 3 x  . For 3 7 t   ,   5 v t  , so   2 5 x t t C   Now 15 2 (3) x  implies that 15 2 2 C   , so   15 2 5 x t t   on this second interval, and its right-endpoint value is   55 2 7 x  . For 7 10 t   ,   5 3 5 7 v t     , so   5 50 3 3 v t t    . Hence   2 5 50 3 6 3 x t t t C     , and 55 2 (7) x  implies that 290 3 6 C   . Finally,   2 1 6 ( 5 100 290) x t t t     on this third inter- val, leading to the graph of   x t shown. 23.   9.8 49 v t t    , so the ball reaches its maximum height ( 0 v  ) after 5 t  seconds. Its maximum height then is       2 5 4.9 5 49 5 122.5 meters y     . 24. 32 v t   and 2 16 400 y t    , so the ball hits the ground ( 0 y  ) when 5 sec t  , and then   32 5 160 ft/sec v     . 25. 2 10 m/s a   and 0 100 km/h 27.78 m/s v   , so 10 27.78 v t    , and hence   2 5 27.78 x t t t    . The car stops when 0 v  , that is 2.78 s t  , and thus the distance traveled before stopping is   2.78 38.59 meters x  . 26. 9.8 100 v t    and 2 4.9 100 20 y t t     . 0 2 4 6 8 10 0 10 20 30 40 t x (5, 12.5) Problem 21 0 2 4 6 8 10 0 10 20 30 40 t x (3, 7.5) (7, 27.5) Problem 22

635 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Mathematics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.