Lista 2 Exercícios Livro com Resposta-2021 2

4.4 UNDETERMINED COEFFICIENTS—SUPERPOSITION APPROACH e 147 enim MARKS oo smnnnnnnnnnnnnsnnsse (i) In Problems 27—36 in Exercises 4.4 you are asked to solve initial-value problems, and in Problems 37-40 you are asked to solve boundary-value problems. As illustrated in Example 8, be sure to apply the initial conditions or the boundary conditions to the general solution y = y, + y,. Students often make the mistake of applying these conditions only to the complementary function y. because it is that part of the solution that contains the constants Ci, C2, ++ +5 Cn. (ii) From the “Form Rule for Case I” on page 144 of this section you see why the method of undetermined coefficients is not well suited to nonhomogeneous linear DEs when the input function g(x) is something other than one of the four basic types highlighted in color on page 140. For example, if P(x) is a polyno- mial, then continued differentiation of P(x)e** sin Bx will generate an indepen- dent set containing only a finit number of functions—all of the same type, namely, a polynomial times e** sin Bx or a polynomial times e** cos Bx. On the other hand, repeated differentiation of input functions such as g(x) = In x or g(x) = tan” !x generates an independent set containing an infinit number of functions: derivatives of \n pal? lerivatives of In x: ss > am derivati earl! 1 —2x —2 + 6x7 an 2 —, ——__, ———_> : : + + erivatives of x l+e O+xyp +e EX E RC | S E S 4 . 4 Answers to selected odd-numbered problems begin on page ANS-5. In Problems 1—26 solve the given differential equation by 16. y" — Sy’ = 2x3 — 4x2 —x +6 undetermined coefficients ” , - 17. y" — 2y' + 5y = e* cos 2x 1. y" + 3y’ + 2y =6 18. y” — 2y’ + 2y = e?*(cos x — 3 sin x) 2. 4y" + 9y = 15 19, y” + 2y’ + y = sinx + 3 cos 2x 3. y" = 10y' + 25y = 30x + 3 20. y” + 2y’ — 24y = 16 — (x + 2)e** 4. y"+ y’ — 6y = 2x 2. y" — 6y" =3 —cosx 1 m ” ' 5. ree ty ty=x2-2x 22, y" — 2y" — dy’ + 8y = 6xe?* 23, y” — 3y"” + 3y’ ~y =x — 4e* 6. y” — 8y’ + 20y = 100x? — 26xe* , » yy 5 24, y" —y" — 4y' + 4y =5—e* + e** 7. y" + 3y = —48x763* y y y y e e (4) ” >=(x— 1) 8. dy" — dy’ — 3y = cos 2x 25. yO + 2y" + y =(x - 1) 9, y"-y=-3 26. y — y" = 4x + 2xe™% Aa ym _— p72x 10. y" + 2y'=2x+5—e In Problems 27—36 solve the given initial-value problem. 1 mo pe x/2 » , WM. y"— yi t7ya3te 27. y" + 4y = 2, y(m/8) = 1 y'(a/8) = 2 Dy" + 3y! — 2y = 142 — 4 — 0) = 0, y'(0) = 12. y" — 16y = 20% 28. 2y" + 3y’ — 2y = 14x° -4x—- 11, y(0) =0, y'(0) =0 13. y" + 4y =3 sin 2x 29. Sy" + y' = —6x, y(0) = 0, y'(0) = —10 14. y" — 4y = (x? — 3) sin 2x 30. y" + 4y' +4y=34xJe*, y0)=2, yO) =5 15. y" + y = 2x sinx 31. y” + 4y’ + 5y =35e*, (0) = -3, y'(0) = 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 148 ° CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS 32. y’—y=coshx, y(0) = 2, y'(0) = 12 45. Without solving, match a solution curve of y” + y = f(x) ax shown in the figure with one of the following functions: 33. We + wx = Fysinwt, x(0) = 0,x'(0) = 0 (i) f(x) = 1, (ii) f(x) =e, ‘ (iii) f(x) = e*, (iv) f(x) = sin 2x, d. =p ; =g 34, + wx = Fycos t, x(0) = 0,x'(0) =0 (v) f(x) = e* sin x, (vi) f(x) = sin x. dt Briefly discuss your reasoning 35. y" — 2y" + y! =2 — 24e* + 40e*, yO) = 5, yO) =3y"O = 3 (a) » 36. y" + 8y =2x—5+8e*, y(0) = —5, y'(0) = 3, y"@)=-4 In Problems 37—40 solve the given boundary-value problem. 37. y"ty="V 41, y(0)=5, yr) =0 38. y" — 2y’ + 2y =2x-2, y(0)=0, y(7) = 7 39. y" + 3y = 6x, y(0) =0,y(1) + y'(I) =0 FIGURE 4.4.1 Solution curve 40. y" + 3y = 6x, yO) + yO) = 0, y(1) = 0 y In Problems 41 and 42 solve the given initial-value problem (b) in which the input function g(x) is discontinuous. [Hint: Solve each problem on two intervals, and then find a solu- tion so that y and y’ are continuous at x = 7 /2 (Problem 41) and at x = 7 (Problem 42).] 41. y” + 4y = g(x), y(0) = 1, y'(0) = 2, where x sinx, 0<x< 7/2 FIGURE 4.4.2 Solution curve 8@) = 0, x > 1/2 (©) , 42. y" — 2y’ + 10y = g(x), y(0) = 0, y'(0) = 0, where x (x) 20, OSxS7 y= § 0, x>7 FIGURE 4.4.3 — Solution curve Discussion Problems 43. Consider the differential equation ay” + by’ + cy = e*, (d) y where a, b,c, and k are constants. The auxiliary equation of the associated homogeneous equation is an? +bm+c=0. (a) If k is not a root of the auxiliary equation, show x that we can find a particular solution of the form Yp = Ae, where A = 1/(ak? + bk +c). FIGURE 4.4.4 Solution curve (b) Ifk is a root of the auxiliary equation of multiplicity one, show that we can find a particular solution of the form y, = Axe, where A= 1/(2ak + b). Computer Lab Assignments Explain how we know that ke ~/ (2a). . In Problems 46 and 47 find a particular solution of the given (c) Ifk is a root of the auxiliary equation of multiplicity differential equation. Use a CAS as an aid in carrying out two, show that we can finda particular solution of the differentiations, simplifications, and algebra form y = Axe, where A = 1/(2a). . . 46. y" — 4y’ + 8y = (2x? — 3x)e”* cos 2 44. Discuss how the method of this section can be used y 9 y= Ox 2 ye cos a : . ” : + (10x° — x — le sin 2x to find a particular solution of y” + y = sin x cos 2x. Carry out your idea. 47, yO + 2y"+ y =2 cosx — 3xsinx Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4.6 VARIATION OF PARAMETERS e 161 where the uw;, k = 1, 2,...,n are determined by the n equations yup toll Fr tb YM, = 0 Yi + yay Hes + Vpn = 0 : : (14) yf a + yPPu be + Pa = FD. The first — 1 equations in this system, like y,u; + y.u5 = 0 in (8), are assumptions that are made to simplify the resulting equation after y, = u)(x)y\(x) +++ + + un(X)yn(x) is substituted in (13). In this case Cramer’s Rule gives W, up=—~ k=1,2,...,n, W where W is the Wronskian of y,, y2,..., y, and W;, is the determinant obtained by replacing the kth column of the Wronskian by the column consisting of the right- hand side of (14)—that is, the column consisting of (0, 0,..., f(x)). When n = 2, we get (9). When n = 3, the particular solution is y, = u;y, + u2y2 + u3y3, where Y1, Y2, and y3 constitute a linearly independent set of solutions of the associated homogeneous DE and wy, u2, u3 are determined from Ww W, , _ W3 u=—, uy ==, u,=—, 15 (=ap May way (15) 0 yw Ys v 0 ys yy OO Yi Y2 3 Wi=} 0 ys ysl, Wo=}yi 0 yi), Wa= Jy yp 0 |, and W=|y, yoy). SQ) yo y3 yi fQ) 3 yi ys fQ@) yi yo See Problems 25-28 in Exercises 4.6. enn EMARKS oc csuunnsnnnnnnnnne (i) Variation of parameters has a distinct advantage over the method of undetermined coefficients in that it will always yield a particular solution y, provided that the associated homogeneous equation can be solved. The pre- sent method is not limited to a function f(x) that is a combination of the four types listed on page 140. As we shall see in the next section, variation of parameters, unlike undetermined coefficients, is applicable to linear DEs with variable coefficients (ii) In the problems that follow, do not hesitate to simplify the form of yp. Depending on how the antiderivatives of u; and w5 are found, you might not obtain the same y, as given in the answer section. For example, in Problem 3 in Exercises 4.6 both y, = } sin x — } x cos x and y, = j sin x — } x cos x are valid answers. In either case the general solution y = y, + yp simplifies to y =c, cos x + cy sinx — }.x cos x. Why? EX E RC | S E S 4 . 6 Answers to selected odd-numbered problems begin on page ANS-6. In Problems 1—18 solve each differential equation by varia- 5. y’ty= cos2x 6.y"ty= sec2x tion of parameters. 7. y" — y =coshx 8. y” — y = sinh 2x 1. y"+y=secx 2. y" +y = tanx ” — e ” — 9x 3. y+ y = sinx 4. y" + y =sec 6 tan 6 9. y Aya 10. y — OY = Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 162. ¢ CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS 1. y" + 3y’ + 2y = i 24, x?y" + xy’ + y = sec(In x); ‘. e y, = cos(In x), y2 = sin(In x) 12. y"-2y' +y= Tee In Problems 25—28 solve the given third-order differential equation by variation of parameters. 13. y" + 3y’ + 2y = sin e* 14, y" — 2y’ + y = e‘ arctan t 25. y" + y' = tanx 15. y" + 2y' ty =etlne 26. y"” + 4y’ = sec 2x 16. 2y" + 2y’ + y =4Vx 27. y" — 2y"— yl + 2y =e 2x 17. 3y" — 6y’ + 6y = e* sec x 28. y" — 3y" + 2y' = ss o 18. 4y"— 4y' ty = e2VT = 2 In Problems 19-22 solve each differential equation by Discussion Problems vO et of b eS subject to the initial conditions In Problems 29 and 30 discuss how the methods of unde- yO) = Ly'@) = 0. termined coefficients and variation of parameters can be 19, 4y" — y = xe"? combined to solve the given differential equation. Carry out 20. 2y"+y'’ -~y=xt1 your ideas. 21. y" + 2y’ — 8y =2e7* —e* 29. 3y" — oy’ + 30y = 15 sinx + e* tan 3x 22. y" — 4y' + 4y = (12x? — 6x)e* 30. y" = 2y! ty = 4x7 —3 + x let In Problems 23 and 24 the indicated functions are known lin- 31. What are the intervals of definition of the general solu- early independent solutions of the associated homogeneous tions in Problems 1, 7, 9, and 18? Discuss why the inter- differential equation on (0, ). Find the general solution of val of definition of the general solution in Problem 24 is the given nonhomogeneous equation. not (0, ). , 32. Find the general solution of xty" + x3y! — 4x?y = 1 23, xy" + xy! + (2 - yy = 5 given that y; = x? is a solution of the associated homo- yp = x7! cos x, yy = x7"? sin x geneous equation. = Cauchy-Euler Equation A linear differential equation of the form d’y dy dy Gy,X" —— + yx"! + + tax + ay = g(x), dx" 1 ax! x oY = B(x) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS e ANS-5 EXERCISES 4.2 (PAGE 131) 21. y =, + ex + Ce% — $x? — Scosx t+ $sinx 1. yo = xe2* 3. yo = sin 4x 23. y = ye’ + Exe + xe — x — 3 - Fe 5. y2 = sinh x Te y2= xe2X/3 25. y = ci cosx + c2 sinx + c3x cos x + Cax sin x a ya = xi ble ie yal 5 + 2-2 —3 y= Sy exer txt . 17 ” = *c08t ~ 19 * = 2 a 5 3x 27. y = V2sin2x — 4 92 OY = 9 *% yp 3 29. y = —200 + 2007/5 — 3x? + 30x 31. y = —10e7* cos x + 9e~?* sinx + Je 4* 33. x= si Fo EXERCISES 4.3 (PAGE 137) XS Tp Sin oF ~ Ff Os wt lyse toe" 3. y= ce + oe? 35. y = 11 — lle’ + 9xe" + 2x — 12x e* + Se 5S. y=cje *+exe* y= ce 3 + oe" 37. y = 6 cos x — 6(cot 1) sinx +.x* - 1 = i : v 9. y cycos 3x + c2sin 3x —4 sin V3x ~ 11. y = e**(c, cos x + c2sin x) 39. Y= Gave V3008V3 7 2% ww . sin cos 13. y= e*3(c, cos+V2x + csin4 V2 x) E 15. y=c, +c9e* + c30* 41. y= cos 2x + ?sin2x + }sin x, OSx5 7/2 = 17. y=cye* + ce** + c3xe>* ‘> 2 cos 2x + ° sin 2x, x> 7/2 O 19. u=cye' +e '(cycost + c3sIin f) e 21. y =cje* + Exe * + 63x72 n 23. y = c) + x +e? (cyc08 }V3.x + cysin} V3x) EXERCISES 4.5 (PAGE 155) = = 1 “od — 25. y = c,cos}V3x + c,sin} V3x 1. 3D — 2)(3D + 2)y =sinx ma + C3X cos} V3.x + cyx sing V3x 3. (D— 6)(D + 2)y=x-6 9 27. u = cye’ + core’ + c3e°" + cyre” + cose 5. D(D + 5)*y = e* 29. y = 2cos 4x —4sin4x 7. (D— 1)(D — 2)(D + S)y = xe a 31. y = be) + 1 SUD 9. D(D + 2)(D* — 2D + 4)y =4 ~ 3. y=0 3 5 15. D4 17. D(D - 2) co "y / / 19. D2 +4 21. D3(D? + 16) = 35, y= 2 — 5 em 6x 4 Lye 6x 3 3 Ss » Y 36 ~ 36 6 23. (D + 1)(D — 1) 25. D(D? — 2D + 5) z 37. y =e — xe* 27. 1, x, x2, x3, x4 29, e&*, e 3%? a 39. y=0 31. cos V5x, sin V5x 33. 1, e*, xe* a 1 5 1 5 x 35. y=cie * + cre** — 6 Oo y= — ) e-V3x 4 = — ) ev3x- 41. y (1 Saye eg lie Saye ; 37. y=c) tere? + 3x 2 5 39. y = ce + xe +4x41 5 y = cosh V3x + 7 sinhV3x 4. y=c +x t ce +283 + 82 o += eee 4x 4 ly ody ws 49. y" — 6y' +5y=0 51. y’ —2y' =0 43. y= cye** + ce + 5 xe nA 45. y=cye* + cre** — ee +3 x 53. y" + 9y =0 55. y" + 2y' +2y=0 ye ere oN oe @ . O 57 8 F 47. y = c, cos Sx + cy) sin 5x + 3 sinx ou _ yl — gy = A y y 49. y = ce + cove ® — hve + e™ o 51. y=cye*+oe+ gxet - per + pre —5 = EXERCISES 4.4 (PAGE 147) 53. y = e*(c,cos 2x + cysin 2x) + te sin x Z Ll y=cje*+ ce %* +3 55. y = cicos 5x + c2sin 5x — 2x cos 5x = 3. y= ce + coxe® + 2x +3 V3 _ V3 y ' > 5 5 7 57. y = e*| ec, cos— x + cy sin —x 5. y= ce + xe * +x — 4x45 2 2 7. y = c, cos V3x + cy sin V3x + (-42 + 4x - ‘Jes + sin.x + 2 cos. x — x cos x 9. y=c) + me" + 3x 9. y=c toxtaqe*+2rt+ire—-ie VW. y = cye*? + eyxe? + 12 + ger? 61. y = ce + Exe + xe" + ive +x—- 13 13. y = c, cos 2x + cy sin 2x — 3x cos 2x 63. y =c, + Cox + 03° + Cyxe® + sxe + $x 15. y = c, cosx + c sinx — $x? cos x + 4x sinx 65. y = fe + fe —F 17. y = cye*cos 2x + coe*sin 2x + txe* sin 2x 67. y= —h + Be -— te + fx 19. y = ce" + Exe — 5 cos x 69. y = —7cos x — 4 sin x — ¥ cos 2x + 2x cos x 12: 9 >= 702" — 3 B2x Gi 1.3 43 3 + 53 sin 2x — 5; cos 2x 71. y = 2e* cos2x — Ge*sin2x + gx + gx + 5x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. ANS-6 e ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS EXERCISES 4.6 (PAGE 161) EXERCISES 4.8 (PAGE 179) 1. y =c, cosx + cy sinx + xsinx + cos x In|cos x| iy’. . 1 1. y,(x) = g] sinh 4(@x — Af(dt 3. y =c, cosx + cy sinx — 5.x cos x Xo 5. y = c, cosx + cy, sinx + } — i cos 2x x (9 . 3. y,(x) = | (x — De~S-%f(Hdt Ty=ce+ome*+ ix sinh x Yel) . ) AO) x et x = 6 ek -2x 4 1{ 22x -2x all. 9. y = ce + coe + 5 ( e* In|x| — e 7 at > 5. y,(x) = 3] sin3(x — Of(ddt Xo Xo xy > 0 if = een 4 4x pl] ge _ ~21 MW. y=ce%+ coe 2 + (eo + e2) In(l + e*) 7.y=ce + oe + 3 _ Sinh4 (x t)te~ “dt 13. y=cje * + oe *—e > sine* rh st 15. y= ce! + tet + $Pe Int —3 Pe 9. y = cje* + Exe + I (x — Ne~*~%e~'dt . : x ea 17. y = cye‘sinx + c,e* cos x + ixe* sin x ° &E . ‘. . =z + 3e*cos x In| cos x| ll. y = cycos3x + c,sin3x + { sin3(x — A(t + sint)dt Xo L149, y= ter? + 32 4 12 or2 — Ly gr/2 OU yr4 4 8 4 13. y,(x) = lye _ eer + ke 2. y = Sends 4 2x — pax pg e yr 9 36 4 ° 15. y,(x) = 1y2e5t Q 23. y=cyx 7? cos x + cox! sin x + x7!” 7. . . => . 17. y,(x) = —cosx + Ssinx — xsinx — cosx In| sinx| wy 25. y =c¢, + ce cosx + cysinx — In|cos x| 2 5 — sin x In|sec x + tan x| 19. y = Be-® — Fe + txe* S27. y= cye + ce * + cye* + yo 21. y = —e® + 6xe® + Fre a 23. y = —xsinx — cosxIn|sinx| ws oc 25. y = (cos! — 2)e~* + (1 + sinl — cosl)e~* — e~**sine* uw EXERCISES 4.7 (PAGE 168) y= ) ( ) S “4 2 27. y = 4x — 2x? — xInx Ll. y=cyx + Cox 3 3. y=cy t+c)Inx 29. y= Bx —dr724+ 5 - Linx a 5. y = c, cos(2 In x) + cz sin(2 In x) a Tey = cyx2-V9 + cy x2+VO) 31. y(x) = Se’ + 3e7* + y,(x), Q 2% y=c, cos (: In x) + c) sin (: In x) where y,(x) = 1 — coshx,x <0 5 5 yp —1+coshx,x 0 O VW. y=cix * + cox“ Inx wi . 33. y = cosx — sinx + y,(x), q By= x "Ac, cos(t V3 In x) +c sin(t V3 In x)| » »@) n 0, x<0 5 15. y=cye+ c,cos(V2In x) + 63 sin( V2 In x) where y,(x) = 410 — 10cosx, Ox <3a 17. y=cy t+eoox + c3x2 + cax 3 —20cos x, x> 37 Yn & 19% y= ce, + Ox + $5 Inx * | 35. y,(x) = («— LD] tfdt + x] (¢- lIf(dadt = 21. y =cyx + cox Inx + x(In x)? 9 * Z 23. y=cyx | +e.x—Inx 37. y,(x) = 5x2 - bx < 25. y=2-2x? in( 1) : sin(x — sinx 27. y = cos(In x) + 2 sin(In x) 39. y,(x) = : -—— +1 . sinl sin] 29. y= 3 —Inx+ ge 41. y,(x) = —e*cosx — e*sinx + e* -1 24 1 BL. y = cyx7 + cnx? 43. y,(x) = 3(Inx) + 3Inx 33. y= ex t+ ex 8+ px . 43 EXERCISES 4.9 (PAGE 184) 35. y = x°[c, cos(3 Inx) + c, sin(3 Inx)] + 4 + 7x 1. x =cye' + cote! 37. y = 2(—x)!?2 — 5(—x)!? In(—x), x < 0 y = 2(-2)'? = 5(-a)!? In(—x) yal epe! + epte! 39. y = ex + 3P + e(x + 3)’ 3. x=c,cost+cosint+tt+1 41. y = c,cos[In(x + 2)] + cosin[In(@ + 2)] y=e,sint— c,cost+t— 1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.