Skip to content
Textbook Guides

Solution Manual for Elementary Differential Equations with Boundary Value Problems (Classic Version), 6th Edition

Solution Manual for Elementary Differential Equations with Boundary Value Problems (Classic Version), 6th Edition breaks down difficult textbook problems into simple solutions, making your study time more effective.

Preview 16 of 604 pages

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

  1. Page 1
    Read page 1 as text

    SOLUTIONS MANUAL Elementary DIFFERENTIAL EQUATIONS With BOUNDARY VALUE PROBLEMS Sixth Edition C. Henry Edwards ¢ David E. Penney ‘With the assistance of David Calvis

  2. Page 2
    Read page 2 as text

    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

  3. Page 3
    Read page 3 as text

    ! I ’ CONTENTS 1 FIRST-ORDER DIFFERENTIAL EQUATIONS 1.1 Differential Equations and Mathematical Modeling 1 1.2 Integrals as General and Particular Solutions 9 1.3 Slope Fields and Solution Curves 18 1.4 Separable Equations and Applications 27 1.5 Linear First-Order Equations 44 1.6 Substitution Methods and Exact Equations 52 1.7 Population Models 65 1.8 Acceleration-Velocity Models 78 Chapter 1 Review Problems 87 2 LINEAR EQUATIONS OF HIGHER ORDER 2.1 Introduction: Second-Order Linear Equations 91 2.2 General Solutions of Linear Equations 97 2.3 Homogeneous Equations with Constant Coefficients 105 2.4 Mechanical Vibrations 112 2.5 Nonhomogeneous Equations and the Method of Undetermined Coefficients 122 2.6 Forced Oscillations and Resonance 133 2.7 Electrical Circuits 147 2.8 Endpoint Problems and Eigenvalues 154 3 POWER SERIES METHODS 3.1 Introduction and Review of Power Series 162 3.2 Series Solutions Near Ordinary Points 168 3.3 Regular Singular Points 181 3.4 Method of Frobenius: The Exceptional Cases 194 | | -

  4. Page 4
    Read page 4 as text

    i I BN h 3.5 Bessel's Equation 203 3.6 Applications of Bessel Functions 211 4 LAPLACE TRANSFORM METHODS 4.1 Laplace Transforms and Inverse Transforms 216 4.2 Transformation of Initial Value Problems 221 4.3 Translation and Partial Fractions 230 4.4 Derivatives, Integrals, and Products of Transforms 238 4.5 Periodic and Piecewise Continuous Input Functions 245 4.6 Impulses and Delta Functions 258 5 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 5.1 First-Order Systems and Applications 267 5.2 The Method of Elimination 276 53 Linear Systems and Matrices 297 5.4 The Eigenvalue Method for Homogeneous Linear Systems 305 5.5 Second-Order Systems and Mechanical Applications 335 5.6 Multiple Eigenvalue Solutions 348 5.7 Matrix Exponentials and Linear Systems 362 5.8 Nonhomogeneous Linear Systems 371 6 NUMERICAL METHODS 6.1 Numerical Approximation: Euler's Method 380 6.2 A Closer Look at the Euler Method 387 6.3 The Runge-Kutta Method 397 6.4 Numerical Methods for Systems 407 | | -

  5. Page 5
    Read page 5 as text

    7 NONLINEAR SYSTEMS AND PHENOMENA 7.1 Equilibrium Solutions and Stability 415 7.2 Stability and the Phase Plane 428 7.3 Linear and Almost Linear Systems 437 7.4 Ecological Applications: Predators and Competitors 454 7.5 Nonlinear Mechanical Systems 469 7.6 Chaos in Dynamical Systems 481 8 FOURIER SERIES METHODS 8.1 Periodic Functions and Trigonometric Series 486 8.2 General Fourier Series and Convergence 496 8.3 Fourier Sine and Cosine Series 510 8.4 Applications of Fourier Series 524 8.5 Heat Conduction and Separation of Variables 530 8.6 Vibrating Strings and the One-Dimensional Wave Equation 536 8.7 Steady-State Temperature and Laplace's Equation 543 9 EIGENVALUES AND BOUNDARY VALUE PROBLEMS 9.1 Sturm-Liouville Problems and Eigenfunction Expansions 554 9.2 Applications of Eigenfunction Series 563 9.3 Steady Periodic Solutions and Natural Frequencies 575 9.4 Cylindrical Coordinate Problems 588 9.5 Higher-Dimensional Phenomena 596 APPENDIX Existence and Uniqueness of Solutions 600 | | | ©

  6. Page 6
    Read page 6 as text

    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 differential 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 y,=cos2x and y,=sin2x, then y{=-2sin2x and y;=2cos2x so YW = —4cos2x = —4y, and y; = —4sin2x = —4y, Thus y/+4y, = 0 and y)+4y, = 0. 4. If y,=¢” and y,=e¢™, then y, =3¢ and y,=-3e™ so yy =9¢" =9y and py =9e™ =9y, 5. If y=e¢*—e™, then y'=e*+e* so y-y = (er +e) ~(e* -e™) = 2e”*. Thus y= y+2e”. 6. If yy=e™ and y,=xe™, then y]=-2¢™, yi=4e™, y,=¢™ -2xe™, and Vy=—4e +4xe™. Hence Viray ay, = (46) +a(-2e7)+4(e>) = 0 and Vi+dy,+dy, = ( —4e> +hxe™)+4(e™ -2xe™)+4(xe™) =0. mr + stay

  7. Page 7
    Read page 7 as text

    8. If y,=cosx—cos2x and y,=sinx—cos2x, then y|=-sinx+2sin2x, y/=—cosx+4cos2x, and yj, =cosx+2sin2x, yj =—sinx+4cos2x. Hence +» = (-cosx+4cos2x)+(cosx —cos2x) = 3cos2x and 3+, = (-sinx+4cos2x) + (sinx —cos2x) = 3cos2x. 11. If y=y =x7 then y'=—-2x7 and 3"=6x", so X*Y"+5xy' +4y = x? (6x7*)+ Sx(-2x7)+4(x?) = 0. If y=y,=x"Inx then y'=x7-2x7Inx and 3" =-5x"+6x*Inx, so XY" +5xy +4y = x? (-5x +6x7 Inx)+ Sx (x -2x7 Inx)+4 (x Inx) = (-5x7 +552) + (6x7 ~10x +42) Inx = 0. 13. Substitution of y=e™ into 3y'=2y gives the equation 3re™ = 2¢™ that simplifies to 3r=2. Thus r=2/3. 14. Substitution of y=e™ into 4y”"=y gives the equation 4r>e™ = e™ that simplifies to 472 =1. Thus r=+1/2. 15. Substitution of y=e™ into y"+3'-2y = 0 gives the equation r’e™ +re™ —2¢™ =0 that simplifiesto 72 +7—-2 = (r+2)(*—1) = 0. Thus r=-2 or r=1. 16. Substitution of y=e™ into 3y"+3y'—4y = 0 gives the equation 3r%e™ +3re™ —4e™ =0 that simplifies to 37° +3r—4 = 0. The quadratic formula then gives the solutions » = (-3 +57 ) 6. 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. 2 Chapter 1 — StudyXY 1 1 1 a

  8. Page 8
    Read page 8 as text

    17. C=2 18. C=3 0 S—r : \ | \ \ / / 03) ©2) \ / / / No ) ) J 1 | \ / | | 19. If y(x) = Ce" ~1 then y(0)=5 gives C~1 = 5, so C = 6. The figure is on the left below. | / | | \ ss / / 09 \ BREE Pe / \ 7 | JERI . 20. If y(x) = Ce™+x~1 then »(0)=10 gives C~1 = 10, so C = 11. The figure is on the right above. 21. C = 7. The figure is on the left at the top of the next page. i stuancr 1 1 1 n

  9. Page 9
    Read page 9 as text

    oT s LN | \ \ ———— Jo =—— 22. If y(x) = In(x+C) then y(0)=0 gives InC = 0, so C = 1. The figure is on the right above. 23. If y(x) = $x°+Cx7 then y(2)=1 gives the equation %-32+C-% = 1 with solution C=-56. The figure is on the left below. —~ | / © 5 / 3 DN 24. C= 17. The figure is on the right above. 4 Chapter 1 | | | i v

  10. Page 10
    Read page 10 as text

    25. If y(x) = tan(x”+C) then ¥(0)=1 gives the equation tan C = 1. Hence one value of C is C=m/4 (asis this value plus any integral multiple of 7). ht) “1 [] 1 2 26. Substitution of x=7 and y=0 into y = (x+C)cosx yields the equation 0=@=+C)-1),s0 C = —7. : LAA N 27. y =x+y Section 1.1 5 | 1 | | .

  11. Page 11
    Read page 11 as text

    28. The slope of the line through (x,y) and (x/2,0) is »' = (y=0)/(x—x/2) = 2y/x, so the differential equationis xy’ = 2y. 29. If m=" is the slope of the tangent line and nm’ is the slope of the normal line at (x,), then the relation mm'=—1 yields m' = 1/y" = (y~1)/(x~0). Solution for y' then gives the differential equation (1-y)y' = x. 30. Here m=) and m' =D, (x’ +k) = 2x, so the orthogonality relation mm'=—1 gives the differential equation 2x)’ = —1. 31. The slope of the line through (x,y) and (-y,x) is y' = (x-y)/(=y—x), so the differential equationis (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, the word "is" with the = sign, the phrase "proportional to" with %, and finally translate the remainder of the given sentence into symbols. 32. dP/dt = KP 33. dv/dt = kV? 34. dv/dt = k(250-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 y(x) =1 or p(x) = x ofthe differential equation y"=0. 38. A function whose derivative equals itself, and hence a solution of the differential equation y'=y is y(x) = €*. 39. Wereason thatif y= kx’, then each term in the differential equation is a multiple of x. The choice k=1 balances the equation, and provides the solution y(x) = x*. 40. If y isa constant, then »'=0 so the differential equation reduces to y? =1. This gives the two constant-valued solutions y(x) = 1 and y(x) = —1. 6 Chapter 1 — StudyXY | | | r 1

  12. Page 12
    Read page 12 as text

    41. Wereason that if y =e”, then each term in the differential equation is a multiple of e*. The choice k=1 balances the equation, and provides the solution y(x) = Le*. 42. Two functions, each equaling the negative of its own second derivative, are the two solutions y(x) = cosx and y(x) = sinx of the differential equation y"=-y. 43. (a) We need only substitute x(t) =1/(C —kf) in both sides of the differential equation x'= kx? for a routine verification. (b) The zero-valued function x(#)=0 obviously satisfies the initial value problem x’ =kx?, x(0)=0. 44. (a) The figure on the left below shows typical graphs of solutions of the differential equation x'=1x’. - (b) The figure on the right above shows typical graphs of solutions of the differential equation x'=—1x’. We see that — whereas the graphs with k =1 appear to "diverge to infinity" — each solution with k =—1 appears to approach 0 as ¢— co. Indeed, we see from the Problem 43(a) solution x(r)=1/(C—4r) that x(f) > w as — 2C. However, with k =—1 it is clear from the resulting solution x(t) =1/(C + 11) that x(#) remains bounded on any finite interval, but x(f) => 0 as f —> +o. 45. Substitution of P'=1 and P=10 into the differential equation P'=kP? gives k =<, so Problem 43(a) yields a solution of the form P(f)=1/(C ~#/100). The initial condition P(0)=2 now yields C=1, so we get the solution Section 1.1 7 — StudyXY | | | i H

  13. Page 13
    Read page 13 as text

    1 100 P(t) = —— = ——, © 1_1t 50-¢ 2 100 We now find readily that P=100 when t=49, and that P=1000 when ¢=49.9. It appears that P grows without bound (and thus "explodes") as ¢ approaches 50. 46. Substitution of v'=-1 and v=5 into the differential equation v'=kv? gives k ==, so Problem 43(a) yields a solution of the form v(f) = 1/(C +¢/25). The initial condition v(0) =10 now yields C =, so we get the solution 1 50 1)= —- = ——. 0 =T LT Sex 10 25 We now find readily that v=1 when #=22.5, and that v=0.1 when ¢=247.5. It appears that v approaches 0 as ¢ increases without bound. Thus the boat gradually slows, but never comes to a "full stop" in a finite period of time. 47. (a) y(10)=10 yields 10=1/(C-10), so C=101/10. (b) There is no such value of C, but the constant function y(x)=0 satisfies the conditions y' =»? and y(0)=0. (©) 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 y'=3?, y(a)=bh. 48. (b) Obviously the functions u(x)=—x* and v(x)=+x* both satisfy the differential equation xy’ = 4. But their derivatives #'(x)=-4x" and v(x) =+4x’ match at x = 0, 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 x<0 and x 20, respectively). © If a> 0 (for instance), choose C, fixed so that C,a* = b. Then the function x) = C xt if x<0, 7 Cx* if x20 satisfies the given differential equation for every real number value of C._. 8 Chapter 1 — StudyXY | | | [ i

  14. Page 14
    Read page 14 as text

    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 concepts of velocity and acceleration, as well as the fps and mks unit systems. 1. Integration of y'=2x+1 yields y(x) = Jex +1 dx = x*+x+C. Then substitution of x=0, y=3 gives 3 =0+0+C = C, so y(x) = x +x+3. 2. Integration of y'=(x-2)* yields y(x) = fox —2)%dx = 1(x-2)’+C. Then substitution of x=2, y=1gives 1 = 0+C = C, so y(x) = T(x-2)° +1. 3. Integration of y'=+x yields y(x) = | Vxdx = 2x +C. Then substitution of x=4, y=0 gives 0=%+C, so y(x) = 3(x**-8). 4. Integration of y'=x7 yields y(x) = fx? dx = —1/x+C. Then substitution of x=1, y=5gives 5=-1+C, so y(x) = -1/x+6. S. Integration of y= (x+2)™? yields y(x) = fx +2)? dx = 2Jx+2+C. Then substitution of x=2, y=-1 gives —1=2-2+C, so p(x) = 2/x+2-5. 6. Integration of y'=x(x*+9)"? yields y(x) = fre? +9) dx = L(x? +9)? +C. Then substitution of x=-4, y=0 gives 0= 15’ +C, so yx) = Hx? +9)y"2~ 125]. 7. Integration of »'=10/(x" +1) yields p(x) = fron? +1)dx = 10tan™ x+C. Then substitution of x=0, y=0 gives 0=10-0+C, so p(x) = 10tan' x. 8. Integration of y'=cos2x yields y(x) = feos2xax = 1sin2x+C. Then substitution of x=0, y=1 gives 1=0+C, so y(x) = 1sin2x +1. 9. Integration of y'=1/v1-x yields y(x) = fur V1-x*dx = sin" x+C. Then substitution of x=0, y=0 gives 0=0+C, so p(x) = sin” x. Section 1.2 9 I~ StudyXY | | | ! i

  15. Page 15
    Read page 15 as text

    10. Integration of y'=xe™ yields y(x) = [xe ax = Jue* au = (u-De" = —(x+De*+C (when we substitute #=—x and apply Formula #46 inside the back cover of the textbook). Then substitution of x=0, y=1 gives 1=-1+C, so y(x) = —(x+1)e +2. 11. If a(t) = 50 then v(t) = Js0a = 50t+v, = 50¢+10. Hence x(t) = J(50¢+10)dr = 2512 +101 +x, = 257 +101 +20. 12. If a(f) = -20 then v(f) = f(=20)ar = ~20f+v, = —20r-15. Hence X(t) = [(2201-15)dr= ~107 ~15t +x, = —10£~15¢ +5. 13. If a(t) = 31 then v(¢) = [3tdt = $£+v, = 3° +5. Hence x(t) = [Ge +5)dr= 18 4514x, = 1 +5¢. 4. If a(n) = 2r+1then v(t) = [Qr+1)dr = +1+v, = +17. Hence ©) = [(+t-Ndt= LF +11 -Tr+x, = LF +11-Tt +4. 15. If af) = 4(+3). then v(t) = [4(+3) dr = 4(+3)°+C = 4(1+3) =37 (taking C=-37 so that v(0)=-1). Hence x(t) = [$437 -37]dr= 2(+3)' -37+C = L(t+3)' ~37r-26. 16. If a() = 1/i+4 then v() = [1/i+d di = 2Ji+4+C = 2Ji+4-5 (taking C=-5 so that v(0)=-1). Hence X(t) = [@VI+A-5)dr= $0+4y7 =51+C = $(t +4)? —51-2 (taking C =-29/3 so that x(0)=1). 10 Chapter 1 | | | [ :

  16. Page 16
    Read page 16 as text

    17. If a(t) = (t+1)7 then v(r) = f+? dt = 1+)? +C = L(+)? +4 (taking C=1 so that v(0) = 0). Hence x0) = [[-3e+D)? +4]dr= Le+)' +41+C = a+ +e-1] (taking C=—1 so that x(0)=0). 18. If a(t) = 50sin5t then v(t) = [50sinst dt = —10cos5t+C = —10cos5t (taking C=0 so that v(0)=-10). Hence X(t) = [(-10cos5t)dr= ~2sinS5t+C = —2sin5t+10 (taking C=-10 so that x(0)=8). 19. Note that v(r)=5 for 0<¢<5 and that v(r)=10—¢ for 5<¢<10. Hence x(@)=5t+C for 0<r<5 and x(r)=10r-4r*+C, for 5<¢<10. Now C,=0 because x(0) = 0, and continuity of x() requires that x(f)=>5¢ and x(t) = 106 -1¢ +C, agree when ¢=35. This implies that C,=-%, and we get the following graph. a > y % 2 4 6 8 10 t Section 1.2 11 — StudyXY | | | ! I

588 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.