Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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CHAPTER 17 REVIEW 0 355 (b) A boundary-value problem consists. of finding a solution y of a second-order differential equation that al~o satisfies given boundary conditions y(xo) =Yo and y(x1) = Yl· 3. (a) ay" +In/ + CIJ = G{x) where a, b, and care constan ts and G is a continuous function. (b) The complementary equation is the related homogeneous equation ay" +by'+ cy = 0. If we find the general solution Yc of the complementary equation and YP is any particular solution of the original differential equation, then the general solution of the original differential equation is y(x) = Yr>(x) + Yc(x). (c) See Examples 1-5 and the associated discussion in Section 17.2. (d) See the discussion on pages 11 77- 1179 [ ET 1153-11 55]. 4. Second-order linear differential equations can be used to describe the motion of a vibrating spring or to analyze an electric circuit; see the discussion in Section 17 .3. 5. See Example I and the preceding discus.sion in Section 17 .4.. 1. True. See Theorem 17.1.3. TRUE-FALSE QUIZ 3. True. cosh x and sinh x are linearly independent solutions of this linear homogeneous equation. EXERCISES ·1. The auxiliary equation is 4r 2 - 1 = 0 => (2r + 1}(2r - 1) = 0 => r = ±t. Then the general solution 3. The auxiliary equation is r 2 + 3 = 0 => r = ±J3 i. Then the general solution is y = c1 cos ( J3 x) + c 2 sin ( J3 x). 5. r 2 - 4r + 5 = 0 => r = 2 ± i, so Yc (x) = e 2 "'{c 1 cosx + c 2 sin x ). Try YP (x) = Ae 2 "' => y~ = 2Ae 2 "' and y~ = 4Ae 2 "'. Substitution into the differential equation gives 4Ae 2 "' - 8Ae 2 "' + 5Ae 2 "' =-e 2 "' => A = 1 and the general solution is y( x) = e 2 "' ( c1 cos x + c2 sin x) + e 2 "'. 7. r 2 -21·+ l=O => r =landyc(x) = cle"' + c2xe"'. Tryy 1 ,(x)=(Ax +B)cosx + (Cx+D) sin x => y~ = (C- Ax- B) sinx +(A+ Cx +D) cos x andy~= (2C - B- Ax)-cosx + ( -2A- D- Cx) sinx. Substitution gives (-2Cx + 2C- 2A- 2D)cosx + (2Ax - 2A + 2B - 2C}sinx =xcosx => A= 0, B = C = D = - t. The general solution is y( x) = c1 e"' + c2 xe"' - ! cos x - !(x + 1) sin x. 9. r 2 - r- 6 = 0 . => r = -2, r = 3 and Yc(x) = c 1e __:_ 2 "' + C2 e 3 "'. For y" - y'- 6y = 1, try Yr>t (x) = A. Then y~ 1 (x) = y~ 1 (x) = 0 and substitution into the differential equation gives A= -i- For y"- y'- 6y = e- 2 :r try © 2012 Ccngogc Learning. All Righl5 Rescn-.d. May nol be S
356 D CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS yp 2 (x) = Bxe- zx [since y = B e- 2 x satisfies the complementary equation). Then y~ = (B- 2Bx)e- 2 "' and y~ = (4Bx- 4B)e- 2 x, and substitution gives - 5Be- 2 "' = e- 2 "' '* · B = -k· The general solution then is y(x) = c1e- 2 x + Cze 3 "' + yp 1 (x) + yp 2 (x) = c1e- 2 x + cze 3 x - ~- ixe- 2 "'. 11. The auxiliary equation is r 2 + 6r = 0 and the general solution is y(x) = c1 + c 2 e- 6 x = k1 + k 2 e- 6
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- STUDENT SOLUTIONS MANUAL for STEW
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.. BROOKS/COLE ~ I ~~r CENGAGE Lear
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D ABBREVIATIONS AND SYMBOLS CD cu D
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viii o CONTENTS 12.4 The Cross Prod
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10 D PARAMETRIC EQUATIONS AND POLAR
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SECTION 10.1 CURVES DEFINED BY PARA
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SECTION 10.1 CURVES DEFINED BY PARA
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SECTION 10.2 CALCULUS WITH PARAMETR
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SECTION 10.3 POLAR COORDINATES 0 13
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SECTION 10.5 CONIC SECTIONS 0 27 5.
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SECTION 10.5 CONIC SECTIONS 0 29 35
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x2 y2 y2 a:2 _ a2 b 61. ;_2 - - = 1
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SECTION 10.6 CONIC SECTIONS IN POLA
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CHAPTER 10 REVIEW 0 35 the length o
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CHAPTER 10 REVIEW 0 37 EXERCISES 1.
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CHAPTER 10 REVIEW 0 39 25. x = t +
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0 PROBLEMS PLUS l lt sin u dx cost
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11 . D INFINITE SEQUENCES AND SERIE
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SECTION 11.1 SEQUENCES 0 47 35. a,.
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SECTION 11.1 SEQUENCES D 49 71. Sin
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SECTION 11.2 SERIES 0 51 ak+l- a1.:
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. . (1+ c)- 2 scn es and set 1t equ
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SECTION 11.3 THE INTEGRAL TEST AND
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SECTION 11.4 THE COMPARISON TESTS D
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SECTION 11.5 ALTERNATING SERIES 0 6
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SECTION 11.5 ALTERNATING SERIES D 6
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SECTION 11.6 ABSOLUTE CONVERGENCE A
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SECTION 11.6 ABSOLUTE CONVERGENCE A
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17 lim I an+l I= SECTION 11.7 STRAT
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SECTION 11.10 TAYLOR AND MACLAURIN
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61 _ x_ x · sin x - x- tx 3 + 1~
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SECTION 11.11 APPLICATIONS OF TAYLO
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CHAPTER 11 REVIEW 0 97 J'(xn)(xn -
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CHAPTER 11 REVIEW 0 99 oo f (n) (O)
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CHAPTER 11 REVIEW 0 101 l 23. Consi
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49./- 1 - dx = -ln{4- x) + C and 4-
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D PROBLEMS PLUS 1. It would be far
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112 0 CHAPTER 12 VECTORS AND THE GE
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D PROBLEMS PLUS 1. Since three-dime
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l CHAPTER 12 PROBLEMS PLUS 0 149 Eq
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13 D VECTOR FUNCTIONS 13.1 Vector F
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SECTION 13.1 VECTOR FUNCTIONS AND S
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SECTION 13.2 DERIVATIVES AND INTEGR
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2 SECTION 13.4 MOTION IN SPACE: VEL
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CHAPTER 13 REVIEW D 173 43. The tan
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5. f~(t 2 i +tcos 1rtj +sin 1rt k )
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CHAPTER 13 REVIEW 0 177 23. (a) r (
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184 0 CHAPTER 14 PARTIAL DERIVATIVE
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15 D MULTIPLE INTEGRALS 15.1 Double
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61 . Since m :S j(x, y) :S M, ffD m
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-I SECTION 15.6 SURFACE AREA 0 267
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Therefore E = { (x, y, z) I -2 ~ X~
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43. I,. = foL foL foL k(y2 + z2)dz.
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SECTION 15.8 TRIPLE INTEGRALS IN CY
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M xv = I~1f I: I:2 6 - 3 r 2 (zK) r
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SECTION 15.9 TRIPLE INTEGRALS IN SP
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(b) The wedge in question is the sh
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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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