Exercícios Selecionados Livro Calc2 Pt4-2021 2

7.1. DEFINITION OF THE LAPLACE TRANSFORM e 261 REMARKS (i) Throughout this chapter we shall be concerned primarily with functions that are both piecewise continuous and of exponential order. We note, however, that these two conditions are sufficient but not necessary for the existence of a Laplace transform. The function f(t) = t~'/? is not piecewise continuous on the interval [0, ©), but its Laplace transform exists. See Problem 42 in Exercises 7.1. (ii) As a consequence of Theorem 7.1.3 we can say that functions of s such as F\(s) = 1 and F(s) = s/(s + 1) are not the Laplace transforms of piecewise continuous functions of exponential order, since F)\(s) #0 and F>(s) #0 as s — ©, But you should not conclude from this that F\(s) and F2(s) are not Laplace transforms. There are other kinds of functions. EX E RC | S E S 7 . 1 Answers to selected odd-numbered problems begin on page ANS-10. In Problems 1-18 use Definition 7.1.1 to find F{f(p}. fo @) -1, Ost<l c — t = | | fo) | (p= _ t 5 44, Ost<2 a? »fO= 0, t=2 FIGURE 7.1.9 Graph for Problem 10 t Ost<l va 3. f(t) = { (= 1 11. f(t) =e" 12. fi) =e > - 13. f(t) = te 14, f(t) = Pe 2+1, O<1<1 KO = te I= Fe 4.10 = 45 f=] 15. f(t) =e 'sint 16. f(t) = e' cost sint, O<t<a7 17. f(t) = tcost 18. f() =tsint (3) fo = 0, t=7 In Problem(19—36)ise Theorem 7.1.1 to find Z{f()}. 0, Ost<7a/2 6 FO = {° t= a2 19. fd) = 214 20. 7 =8 7, f(t) 21. f(t) = 4t — 10 22. f(t) = 7t +3 (2, 2) : J 23. f(t) = 0? + 6t — 3 24. f(t) = —40? + 16t + 9 ! 25. f(t) = (t+ 19 26. f(t) = (2t — 1) 1 } 27. f(t) = 1+ e* 28. fX =P -—e +5 FIGURE 7.1.6 Graph for Problem 7 29. f(t) = (1 +e?" 30. f(t) =(e'-—e'y 8. fo) 2.2) 31. f(t) = 40° — 5 sin 3t 32. f(t) = cos St + sin 2t » 33. f(t) = sinh kt 34. f(t) = cosh kt 35. f(t) = e' sinh t 36. f(t) = e ‘cosht t 1 In Problem(37—40)find L{f()} by first using a trigono- FIGURE 7.1.7 Graph for Problem 8 metric identity, 9. fo 37. f(t) = sin 2t cos 21 38. f(t) = cos?r 1 ry 39. f(t) = sin(4t + 5) 40. f(t) = 10 cos| t — 6 t ' One definition of the gamma function is given by the FIGURE 7.1.8 Graph for Problem 9 improper integral [(a) = [9 t* 'e ‘dt, a > 0. 262 e CHAPTER 7 THE LAPLACE TRANSFORM (a) Show that [(a@ + 1) = al(a). the observation that 7 >In M + ct, for M> 0 and t T(a + 1) sufficiently large, show that e” > Me“ for any c? (b) Show that #{t*} = ——-, a> -l. se 46. Use part (c) of Theorem 7.1.1 to show that Use the fact that P'(+) = Vand Problem 41 to find the L{earibyny = Saat where a and D are real Laplace transform of (s— ay +b _ 12 _ A _ 3/2 and i = —1. Show how Euler’s formula (page 134) can (a) fort (b) = 1 (©) fOr. then be used to deduce the results Discussion Problems Fe" cos bt} = Wy 43. Make up a function F(f) that is of exponential order but (s — ay + where f(t) = F(t) is not of exponential order. Make up le gin bt = b a function f that is not of exponential order but whose tes'sin bt} = (s — a)? + b? Laplace transform exists. 47. Under what conditions is a linear function 44, Suppose that S{f,()} = F\(s) for s>c, and that f(x) = mx + b, m # 0, a linear transform? Lif} = F,(s) for s > cz. When does 48. The proof of part (b) of Theorem 7.1.1 requires LAAAO + f(D} = Fi(s) + Fy(s)? the use of mathematical induction. Show that if 45. Figure 7.1.4 suggests, but does not prove, that the func- L{t" |} =(n— 1)!/s" is assumed to be true, then tion f(t) = e” is not of exponential order. How does Lit") = n!/s"*! follows. 7.2 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES REVIEW MATERIAL e Partial fraction decomposition e See the Student Resource and Solutions Manual INTRODUCTION In this section we take a few small steps into an investigation of how the Laplace transform can be used to solve certain types of equations for an unknown function. We begin the discussion with the concept of the inverse Laplace transform or, more precisely, the inverse of a Laplace transform F(s). After some important preliminary background material on the Laplace transform of derivatives f’(4), f"(f), ..., we then illustrate how both the Laplace transform and the inverse Laplace transform come into play in solving some simple ordinary differential equations. 7.2.1 INVERSE TRANSFORMS THE INVERSE PROBLEM If F(s) represents the Laplace transform of a function f(d, that is, L{f()} = F(s), we then say f(t) is the inverse Laplace transform of F(s) and write f(t) = L7'{F(s)}. For example, from Examples 1, 2, and 3 of Section 7.1 we have, respectively, Transform Inverse Transform 1 1 SF{ly=- i= 294] Ss S 1 1 L{th=s t=L > ine4 {| 1 1 L —3ty = -3t = fo! te") s+3 . { + +} 7.2. INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES e 269 The desired decomposition (15) is given in (4). This special technique for determining coefficients is naturally known as the cover-up method. (iii) In this remark we continue our introduction to the terminology of dynamical systems. Because of (9) and (10) the Laplace trans- form is well adapted to linear dynamical systems. The polynomial P(s) = a,s" + a,_,s" ! + +++ + agin (11) is the total coefficient of Y(s) in (10) and is simply the left-hand side of the DE with the derivatives d‘y/dt* replaced by powers s*, k = 0, 1, ..., n. It is usual practice to call the recipro- cal of P(s)—namely, W(s) = 1/P(s)—the transfer function of the system and write (11) as Y(s) = W(s)Q(s) + W(s)G(s). (16) In this manner we have separated, in an additive sense, the effects on the response that are due to the initial conditions (that is, W(s)Q(s)) from those due to the input function g (that is, W(s)G(s)). See (13) and (14). Hence the response y(f) of the system is a superposition of two responses: y(t) = L'{W(s)Q(s)} + L'{W(s)G(s)} = yo(t) + yi). If the input is g(f)=0, then the solution of the problem is yo(t) = L-'{W(s)O(s)}. This solution is called the zero-input response of the system. On the other hand, the function y,(t) = £Y~'{ W(s)G(s)} is the output due to the input g(f). Now if the initial state of the system is the zero state (all the initial conditions are zero), then Q(s) = 0, and so the only solution of the initial- value problem is y(t). The latter solution is called the zero-state response of the system. Both yo(t) and y;(#) are particular solutions: yo(f) is a solution of the IVP consisting of the associated homogeneous equation with the given initial condi- tions, and y;() is a solution of the IVP consisting of the nonhomogeneous equa- tion with zero initial conditions. In Example 5 we see from (14) that the transfer function is W(s) = 1/(s* — 3s + 2), the zero-input response is ae t) = £1, ———__; = —3e! + 4e”!, pe ls = 16 = 5] “i and the zero-state response is 1 1 1 1 t) = Bre pA —_— = —- Ht t+ et 4+ Ke", a enaaaer al 5° "6° * 30° Verify that the sum of yo(f) and yj(f) is the solution y(t) in Example 5 and that yo(0) = 1, yo(0) = 5, whereas y(0) = 0, y;(0) = 0. EX E RC | S E S 7 . 2 Answers to selected odd-numbered problems begin on page ANS-10. 7.2.1 INVERSE TRANSFORMS 1 1 1 4 6 1 ———_F yO 7, $'y= - - + — 8 Ly - + = - — ss gs-2 s 9 st+8 In Problen{d 30 de appropriate algebra and Theorem 7.2.1 to find the given inverse Laplace transform. 9, £-! ! 10. ¢! ! 4s +1 , 5s —2 1 1 LNA LN 5 1 1g {| 2. of {3 . £4} 12, g-4 Os s* + 49 s* + 16 1 48 2 1/ 3. els Zz + 4. eA(2 _ 5) 1 -1 4s 14 =I ! ss sos 3 Flas +1 7S aes + 1) + 2) 2s — 6 +1 5. £! (+I 6. £7! (s + 2y 1b. £4 16. £-')- st Ss se +9 se +2 270 e CHAPTER 7 THE LAPLACE TRANSFORM 7. ¢ 4 13. gt 39, 2y" + 3y" — 3y'— 2y=e%, y(0)=0, yO) =0, , s? + 35 , s? — 4s y"(0) = 1 s 1 40. y" + 2y"—y’ — 2y = sin3t, y(0)=0, y’(0) =0, 19. £71; ————_ 20. L's ———_ "0) = {=~ —| {=| yO=l The inverse forms of the results in Problem 46 in _ 0.9s . 21. #-!§ ——————<Joq Exercises 7.1 are (s — 0.1)(s + 0.2) sa _ L!; —————_¢ = e“ cos bt 22, gf 3-3 la — ay + aI a. (s — V3)(s + V3) L! __ 9 \y_ e” sin bt. 23. vf 8 (s-arv +h (s — 2)(s — 3)(s — 6) ; In Problems@q_and 4Duse the Laplace transform and these 24, £-! | stl | inverses to solve the given initial-value problem. , — 1)(s + 1s — 2 s(s — Ds + fs ~ 2) 41. y’ +y =e cos 2t, y(0) =0 1 Ss ” , ’ 25. £71} 26. £-1; 42. y"— 2y' + 5y=0, yO)=1, y'(O)=3 {= + =| {= + 2)(s? + ;| Discussion Problems 2s—4 1 27. £ —_<*_*__ 28. L-'5 7 . . . . . (s? + s)(s* + 1) si — 9 43. (a) With a slight change in notation the transform in (6) 1 6s +3 is the same as S —] —1 — , 29. £ lass Shes ;| 30. L {3 a + LUf(O} = s£LUFO} — fO). With f(t) = te“’, discuss how this result in conjunc- 7.2.2 TRANSFORMS OF DERIVATIVES tion with (c) of Theorem 7.1.1 can be used to evalu- 66h Oa seer ate P{te""}. In Probtem(31-40) se the Laplace transform to solve the (b) Proceed as in part (a), but this time discuss how to given initial-value problem. use (7) with f(t) = f sin kt in conjunction with (d) and (e) of Theorem 7.1.1 to evaluate ¥{t sin kr}. 31. dy _ y=1, y@)=0 44, Make up two functions f; and fp that have the same dt Laplace transform. Do not think profound thoughts. 32. 2 dy +y=0, y(0)=-3 45. Reread Remark (iii) on page 269. Find the zero-input dt and the zero-state response for the [VP in Problem 36. 33. y' + 6y =e", y(0) =2 46. Suppose f(t) is a function for which f’(f) is piecewise 34. y’ —y=2cos5t, y(0) =0 continuous and of exponential order c. Use results in 35. y" +5y' +4y=0, y(0)=1, y(0)=0 this section and Section 7.1 to justify 36. y” — 4y’ = 6e* —3e, y(0)=1, y(0)=-1 f(0) = lim sF(s), 37. y' ty = V2sin V21, (0) = 10, y"(0) = 0 where F(s)=L{f()}. Verify this result with 38. y" + 9y =e, yO) =0, y'(0)=0 f(t) = cos kt. | 7.3) OPERATIONAL PROPERTIES | REVIEW MATERIAL e Keep practicing partial fraction decomposition e Completion of the square INTRODUCTION _ Itis not convenient to use Definition 7.1.1 each time we wish to find the Laplace transform of a function f(#). For example, the integration by parts involved in evaluating, say, L{e'r sin 3t} is formidable, to say the least. In this section and the next we present several labor- saving operational properties of the Laplace transform that enable us to build up a more extensive list of transforms (see the table in Appendix III) without having to resort to the basic definition and integration. 278 e CHAPTER 7 THE LAPLACE TRANSFORM SOLUTION Recall that because the beam is embedded at both ends, the boundary conditions are y(0) = 0, y’(0) = 0, y(L) = 0, y'(L) = 0. Now by (10) we can express w(x) in terms of the unit step function: waofi-2) of 2) ob w(x) = Wo ha Wo 1" x— 5 2wo| L L L = “0 fx 4(r—2)a(x—4) . L {2 2 2 Transforming (19) with respect to the variable x gives 2wy| L/2 1 1 El(s*¥(s) _ s°y(0) _ sy'(0) — sy"(0) — y’(0)) = Z| Hl? -s+ Lew] L Ss Ss Ss 2wo|L/2 1 1 ay _ ”" _ ym — “0 Jey —Ls/2 . or s*Y(s) — sy"(O) — y"(0) al , 2 2° | If we let c; = y"(O) and cp = y'"(0), then Cy C2 2Wo L/2 1 1 —~Ls/2 Ys) =—3+44+—7|)—~7-a+7e" (s) ssf mia S56 soe , and consequently C1 -p_,| 2! Cr _,_,} 3! 2wo}| L/2 __,|4! 1 tS! 1. JS! _ =O gels S gels AS pel gab 4 = gp 2 YO) = 5) {= 3! {5 EIL| 4! sf 5 sf * 5! soe L L\? L = 2124234 "0 Lu s+ (c-4) ax - 5) . 2 6 60 EIL| 2 2 2 Applying the conditions y(L) = 0 and y'(L) = 0 to the last result yields a system of equations for c; and c2: LV? L> 49w 4 q—to—+—~=0 2 6 1920ET Lv 85wyL3 cL+oa—+—— =0. 2 960ET Solving, we find c) = 23woL?/(960ED and c, = —9woL/(40ED). Thus the deflec- tion is given by 23wo Ll? 3WoL SL L\ L yx) = OE 7p — Os 4 Mo Las +(r-4) n(x 5) . a 1920EI 80EI 60EIL | 2 2 2 EX E RC | S E S 7 . 3 Answers to selected odd-numbered problems begin on page ANS-11. 7.3.1 TRANSLATION ON THE s-AXIS 1 1 ry a 11. Ly; — 12. £1, ——_ (s + 29% (s — 1)4 In Problem: (1-29) nd either F(s) or f(t), as indicated. 1 1 13. £-!; _—___~ 14, £-'!} _—__ 1. L{te™} 2. L{te} 3 {3 — 6s + 5 | {3 +25 + +} 3. Lite >} 4, Lite ™) Iw+5 ref} wef ts 5. L{tle' + e')} 6. L{er(t — 1)?} s+4s+5 s’ + 65 + 34 7. L{e'sin 3t} 8. L{e~'cos 4t} 7. | s | 18. | 5s | 9. L{(1 — e + 3e-*) cos St} (s + 1) (s — 2) _,| 25-1 _{(s + 1? 319 — we 19, £1; 20. £1; —— 10. ac (o 4t + 10 sin 5 As + DS 0 (s + 2) 7.3, OPERATIONAL PROPERTIES | e 279 In problems) the Laplace transform to solve the given initial-vatue problem. Ey) = L R 21. y' +4y=e%, yO) =2 22. y’ -y=1+ te’, y0)=0 C 23. y" + 2y'+y=0, y(0)= 1 y'(0)=1 FIGURE 7.3.9 — Series circuit in Problem 35 24. y" — 4y’ + 4y = re, yO) = 0, y'(0) = 0 > * * » _ “ , > _ 36. Use the Laplace transform to find the charge q(t) 5. yi — by + 9y = 4, yO) = 0, y'O) = 1 in an RC series circuit when q(0)=0O and 26. y"—4y' + 4y =P, yO) =1,y'0) =0 E(t) = Eoe ™, k > 0. Consider two cases: k # 1/RC 27. y" — 6y' + 13y =0, y(0) = 0, y'(0) = —3 and k = 1/RC. 28. 2y" + 20y’ + Sly =0, y(O) = 2, y'(0) =0 7.3.2. TRANSLATION ON THE ¢-AXIS 29. y"—y' =e'cost, y(0) =0,y'(0) =0 30. y’ — 2y'+5y=14+4, y(0)=0,y'(0) =4 In Problems7—48)ind either F(s) or f(f), as indicated. In Problems 31 and 32 use the Laplace transform and 37. Lit — Ut — 1} 38. L{e™! Ut — 2)} the procedure outlined in Example 9 to solve the given 39. LS{t Ut — 2)} 40. L{Bt+ 1) Ut — 1)} boundary-value problem. 41. L{cos2+Ut— m} 42 sin (1 — =) 31. y" + 2y'+y=0, y'(O) =2,yC) =2 2 ” , _ _ ’ _ —2s 1+ —2s)2 32. y" + 8y’ + 20y=0, y(0)=0,y'(a) = 0 43. oe} 44, vf\ o ) | 33. A 4-pound weight stretches a spring 2 feet. The weight . . is released from rest 18 inches above the equilibrium je” “| se 72 position, and the resulting motion takes place in a 45. £ etl 46. £ ec +4 medium offering a damping force numerically equal to - 5 i times the instantaneous velocity. Use the Laplace 47. | e° | 48. | e* | transform to find the equation of motion x(2). s(s + 1) s(s — 1) 4. Recall that the differential tion for the instanta- 3 vce ah the cablerental equalon 108 fae insane In Problems(@9—54) match the given graph with one of neous charge qg(f) on the capacitor in an LRC series . Lo, . er the functions in (a)-(f). The graph of f(f) is given in circuit is given by . Figure 7.3.10. d* dq 1 Loa t Rot oa = Bl. (20) (a) f — SUE — a) ' he Lal ' find (b) f(t — b) U(t — b) See Section 5.1. Use the Lap ace transform to fin q(t) (c) fi) Ut — a) when L = 1h, R= 20Q, C = 0.005 f, E(t) = 150 V, t > 0, (0) = 0, and i(0) = 0. What is the current i(t)? (d) f() — FU — b) e) f(t) Ut — a) — f(t) Ut — b 35. Consider a battery of constant voltage Ep that charges Ce) FO UE a) — FOU ) the capacitor shown in Figure 7.3.9. Divide equa- (£) f(t — a) Ut — a) — ft — a) UE — B) tion (20) by L and define 2A = R/L and w? = 1/LC. fi) Use the Laplace transform to show that the solution q(t) of g" + 2Aq' + w*q = Eo/L subject to (0) = 0, i(0) = Ois a b t — eA A\/)2 — 42 B¢| e “cos A ot FIGURE 7.3.10 Graph for Problems 49-54 ert yt G2 + Vata ae sinh Xr w } A> a, 49. fl) q(t) = ( E,Ctl — e (1 + Ad], A= a, f\ | B¢|| - e™ (cos Vw" — rt | Xr Xr <wo. a b f + —=——— sin Va? — A’t) P Var — ) FIGURE 7.3.11. Graph for Problem 49 280 e CHAPTER 7 THE LAPLACE TRANSFORM 50. FO 0 0=t< 3/2 58. fi) =4.7 m sin t, t= 3/2 | | t O=t<2 |_| 59. f(t) = ab t 0, t=2 FIGURE 7.3.12 Graph for Problem 50 60. f() = snt, OSt<27 0, t227 51. fO 61. fO 1 [1 NY A | | | | a b ? t a b rectangular pulse FIGURE 7.3.13 Graph for Problem 51 FIGURE 7.3.17 Graph for Problem 61 62. FO | 52. fO 3 — | | S\ ° a | boy | | IF ors ab t bot tt t FIGURE 7.3.14 Graph for Problem 52 ' 2 34 staircase function fo FIGURE 7.3.18 Graph for Problem 62 53. t In Problems@3-79) use the Laplace transform to solve the given initial-value problem. | 0, Ost<i1 b t 63. y +y=f(), yO) =0, wh N=). a yi ty =f, yO) =0, where fio {° =| FIGURE 7.3.15 Graph for Problem 53 64. y’ + y =f, (0) = 0, where Jl ost<l 54. iO) FO) _) t=1 65. y' + 2y =f(t), y(O) = 0, where ft, Ost<l a b t SO ~~ 0, t=1 FIGURE 7.3.16 Graph for Problem 54 66. y" + 4y = f(), + y(O) = 0, y(0) = —1, where 1 O0Ost<il fO= jo >] In Problem 63)erite each function in terms of unit step > = functions. Find the Laplace transform of the given function. 67. y" + 4y = sint U(t — 27m), yO) = 1, y'(0) =0 55 {* 0=t<3 68. y” — 5y' + 6y = Ur— D, yO) = 0, y'(O) = 1 . n= fo —2, t=3 69. y" +y =f, yO) = 0, yO) = 1, where 1 Osr<4 0, Ost<7 56. f() = )0, 45t<5 fO= 91, wst<2a7 I, t=5 0, t227 57. f(t) = (" Osr<l 70. y" + 4y' +3y=1-U(t— 2) -— Ut — 4) + M(t — ©), r r= y(0) = 0, y'(0) = 0 7.4 OPERATIONAL PROPERTIES II e 289 1 L/R L/R From a L/R _ / s(s + R/L) s s+ R/L we can then rewrite (13) as ws) =4(-— Ja - Ts —2s _ —35 4 ) R\s s+ RIL con ee 1 (2 eo? es es 1 ( 1 1 _ es es = —(-— — — 4 — — —— +... . J — =| — © — ss 4 — Os ssSsSFse.e) R\s os s s R\s + R/L s+ R/L s+ R/L s+R/L By applying the form of the second translation theorem to each term of both series, we obtain . 1 iD = 70 — Ut -1)+ Ut - 2)- Ut-3)+-+-°) _ 1 g-rm = eH RUDIL a(t = 1) + eWRO-DL YE = 2) — eH REIL Qt — 3) + +) R or, equivalently, , 1 _ ly —Re- i(t) = —(1 — e Ry) + — SD (- 1)" ee RO), MU = 0). R Ry=1 To interpret the solution, let us assume for the sake of illustration that R = 1, L = 1, and 0 = t < 4. In this case i)=1-et—-d-e'uer-)t+d-e”) Ue - 2) - d -— ee») UE — 3); i 2 in other words, 1.5 1 1-e", O0Osr<l 0.5 vn. pre tte, 1st<2 t Os l-ett+te)—e, 2=1t<3 1 2 3 4 -—ette Ve Ie) Z54< 4 FIGURE 7.4.5. Graph of current i() in The graph of i(t) for 0 =r < 4, given in Figure 7.4.5, was obtained with the help Example 8 of a CAS. = EXE RC | S E S 7.4 Answers to selected odd-numbered problems begin on page ANS-11. 7.4.1. DERIVATIVES OF A TRANSFORM 11. y" + 9y = cos 3t, y(0)=2, y'(0) =5 In Problem{I —8 ise Theorem 7.4.1 to evaluate the given 12, yi ty=sint yO)=1, yO)=—1 Laplace transform. 13. y’ + loy=f(), y(O)=0, y’(0O) = 1, where L. L{te“ 2. L{Pe'} fi) = {OSE OS tS Tt 0, t=7 3. L{t cos 2t} 4. L{t sinh 32} 14. y"+y=f(, 0)=1, y'() =0, wh 5. L{Psinh t} 6. L{ cos t} tye IO, yO) y'©) = 0, where st< 7. L{ te" sin 61} 8. L{ te cos 31} f) = {\ O=t< a/2 sin t, t= 7/2 In Problem 9-19) use the Laplace transform to solve the given initial-value problem. Use the table of Laplace In Problems 15 and 16 use a graphing utility to graph the transforms in Appendix III as needed. indicated solution. 9. y'+y=rsint, y(0)=0 15. y(t) of Problem 13 for0 =t < 27 10. y' —y=te'sint, y(0)=0 16. y(t) of Problem 14 for 0 < t < 37 ‘Lhe inverse form of (/), ‘ F [ f(r) dr = gf Pa) (8) 290 e CHAPTER 7 THE LAPLACE TRANSFORM ° ° In some instances the Laplace transform can be used to solve In Problems 37-46 use the Laplace transform to solve the linear differential equations with variable monomial coeffi- given integral equation or integrodifferential equation. cients. In Problems 17 and 18 use Theorem 7.4.1 to reduce the given differential equation to a linear first-order DE ‘ in the transformed function Y(s) = L{y(t)}. Solve the first- 37. f(t) + F (@— )f@dr=t order DE for Y(s) and then find y(t) = L'{ Y(s)}. t 17. ty"—y' =2%, y) =0 38. f() = 2t -— | sin T f(t — 7) dt 18. 2y" + ty’ —2y=10, yO) =y'(0) =0 t 39. fi) = te’ + | tf(t — adr 0 t 7.4.2. TRANSFORMS OF INTEGRALS 40. f(r) + 2| (9) cos (t — 1) dr = 4e + sint convolugao 0 In Problems@—30use Theorem 7.4.2 to evaluate the given ' Laplace transform. Do not evaluate the integral before 41. f(t) + [ f()dt=1 transforming. 0 t 19. L{1*P} 20. L{P* te} 42. f(t) = cost + [err — dt 21. Lf{e"*e' cos t} 22. L{e?' * sin r} 8 ft 43. f=1+t v3 (7 — tP f(a dr t t 0 23. | ear] 24. al cos rar} , ° ° 44. 1 — 2f(t) = [ (e7 — e-") f(t — ddr t t 0 25. A] e "cos var} 26. || 7 sin rar} 1 o o 45. y(t) = 1 — sint — [ y(t) dt, yO) =0 0 t t 27. A] rerar} 28. | sin T cos (t — 7) ar| dy t 0 0 46. Ut + 6yt) +9] yadr=1, yO)=0 0 t t 29. £ i sint dt 30. £ i te dt . . . 0 0 In Problems 47 and 48 solve equation (10) subject to i(0) = 0 /' with L, R, C, and E(#) as given. Use a graphing utility to graph In Problem@i —34)use (8) to evaluate the given inverse the solution forO = tS 3. transform. 47. L=0.1h,R=3 0, C = 0.05 f, 1 1 = —])- _ 31. -! 32. £4 E(t) = 100[%(t — 1) — Ut — 2)] s(s — 1) s(s — 1) 48. L=0.005h,R=10,C = 0.02 f, _ 1 x 1 E(t) = 100[t — (¢ — 1)%€(t — 1)] 33. £71; ——— 34, £1}; ——— (a - 5| lia — a 35. The table in Appendix III does not contain an entry for 7.4.3. TRANSFORM OF A PERIODIC gl 8k3s FUNCTION (2 + RPP) In Problems9 —5D)use Theorem 7.4.3 to find the Laplace (a) Use (4) along with the results in (5) to evaluate this transform of the given periodic function. inverse transform. Use a CAS as an aid in evaluating the convolution integral. 49. fo (b) Reexamine your answer to part (a). Could you have 1 i i obtained the result in a different manner? bot ot a, 2a\ 3a, 4a) t 36. Use the Laplace transform and the results of Problem 35 i | | | | to solve the initial-value problem y’+y=sint+rsin¢, yO)=0, y'(0)=0. meander function Use a graphing utility to graph the solution. FIGURE 7.4.6 Graph for Problem 49 7.4 OPERATIONAL PROPERTIES II e 291 50. fO 57. m= $s B=1, k=5, f is the meander function in l i Problem 49 with amplitude 10, anda = 7,0 =t< 27. bot 7; 58. m= 1, B = 2,k = 1, fis the square wave in Problem 50 aq 2a 3a 4a with amplitude 5, and a = 7,0 <1 < 4m. square Wave FIGURE 7.4.7. Graph for Problem 50 Discussion Problems 59. Discuss how Theorem 7.4.1 can be used to find 51. f(t) a eli =) |4 | |4 | s+] b 2b 3b 4b ‘ 60. In Section 6.3 we saw that ty” + y’ + ty = Ois Bessel’s h funct equation of order v = 0. In view of (22) of that section Sawtooth Function and Table 6.1 a solution of the initial-value problem FIGURE 7.4.8 Graph for Problem 51 ty” + y’ + ty = 0, y(0) = 1, y’(0) = 0, is y = Jo(t). Use this result and the procedure outlined in the instructions 52. SO to Problems 17 and 18 to show that I 1 L{I(o)} = =. , { of )} Ve +1 1 2 3 4 [Hint: You might need to use Problem 46 in triangular wave Exercises 7.2.] FIGURE 7.4.9 Graph for Problem 52 61. (a) Laguerre’s differential equation 53. fo) ty” + (1—dy' + ny =0 1 is known to possess polynomial solutions when n is a nonnegative integer. These solutions are naturally tn 3n 4n t called Laguerre polynomials and are denoted by L,(t). Find y = L,(0), for n = 0, 1, 2, 3, 4 if it is full-wave rectification of sin t known that L,,(0) = 1. FIGURE 7.4.10 Graph for Problem 53 (b) Show that t qd" 54. fF n-th — . Ae a e | Y(s), where Y(s) = L{y} and y = L,(f) is a polynomial 1X on 3n 4n t solution of the DE in part (a). Conclude that half-wave rectification of sin t L,(t) = od no-t n=0.1.2 n n! dt" ? ato mar ree FIGURE 7.4.11 Graph for Problem 54 This last relation for generating the Laguerre poly- In Problems 55 and 56 solve equation (12) subject to nomials is the analogue of Rodrigues’ formula for i(0) = 0 with E(t) as given. Use a graphing utility to graph the Legendre polynomials. See (30) in Section 6.3. the solution for 0 = t < 4 in the case when L = | andR = 1. 55. E(t) is the meander function in Problem 49 with Computer Lab Assignments amplitude | anda = 1. 62. In this problem you are led through the commands in 56. E(t) is the sawtooth function in Problem 51 with Mathematica that enable you to obtain the symbolic amplitude 1 and b = 1. Laplace transform of a differential equation and the so- lution of the initial-value problem by finding the inverse In Problems 57 and 58 solve the model for a driven transform. In Mathematica the Laplace transform of spring/mass system with damping a function y(t) is obtained using LaplaceTransform ax dx [y[t], t, s]. In line two of the syntax we replace me + B tt +kx =f), x(0)=0, x'(0) = 0, LaplaceTransform [y[t], t, s] by the symbol Y. Uf you f i do not have Mathematica, then adapt the given proce- where the driving function fis as specified. Use a graphing dure by finding the corresponding syntax for the CAS utility to graph x(¢) for the indicated values of t. you have on hand.) 7.6 SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS e 295 EX E RC | S E S 7 . 5 Answers to selected odd-numbered problems begin on page ANS-12. In Problems C12 }s0 the Laplace transform to solve the is free at its right end. Use the Laplace transform to given initial-value problem. determine the deflection y(x) from , _ _ d*y 1 1. y'—3y=6(t— 2), yO) =0 El = wo d(x — 4L), x 2. y' ty =S(t- 1), yO) =2 where y(0) = 0, y’(0) = 0, y"(L) = 0, and y"(L) = 0. 3. y"+ y= 6(t— 27), yO) =0,y'(0) = 1 ; ; . . 14. Solve the differential equation in Problem 13 subject to 4, y" + loy = 6(t— 27), y(0) =0,y'(0) = 0 y(0) = 0, y'(0) = 0, y(L) = 0, y'(L) = 0. In this case 5 y"+ty= 8(t _ 7) + 8(t _ 3x7) the beam is embedded at both ends. See Figure 7.5.5. . 2 27) y(0) = 0, y'(0) = 0 > 7 0 n 6. y" + y= 82m) + Br 4m), yO) = 1, yO) = 0 b E 7. y" + 2y’ = 6-1), yO) =0,y'(0) =1 | i Ao a is a L tL: 8. y"—2y’ = 148-2), yO) =0,y') = 1 dy t 9. y" + dy’ + Sy = 6(t— 277), yO) = 0, y'(0) = 0 FIGURE 7.5.5 Beam in Problem 14 10. y" + 2y'+y=6(t— 1), yO) =0,y'(0) = 0 Discussion Problems 11. y” + 4y' + 13y = 6(¢ — 7) + 8(t — 37), y(0) = 1, y'(0) = 0 15. Someone tells you that the solutions of the two [VPs 12. y"—7y' + 6y =e + S(t — 2) + S(t — 4), y” + 2y' + 10y = 0, y(0) =0, yO)=1 y(0) = 0, y'(0) = 0 y” + 2y’ + 10y = 6(A), y(0) =0, y(0) =0 13. A uniform beam of length L carries a concentrated load are exactly the same. Do you agree or disagree? Defend Wo atx = SL. The beam is embedded at its left end and your answer. 7.6 SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS REVIEW MATERIAL e Solving systems of two equations in two unknowns INTRODUCTION When initial conditions are specified, the Laplace transform of each equation in a system of linear differential equations with constant coefficients reduces the system of DEs to a set of simultaneous algebraic equations in the transformed functions. We solve the system of algebraic equations for each of the transformed functions and then find the inverse Laplace trans- forms in the usual manner. COUPLED SPRINGS Two masses m and m2 are connected to two springs A and B of negligible mass having spring constants k; and k2, respectively. In turn the two springs are attached as shown in Figure 7.6.1. Let x(t) and x2(f) denote the vertical displacements of the masses from their equilibrium positions. When the system is in motion, spring B is subject to both an elongation and a compression; hence its net elongation is x2 — x;. Therefore it follows from Hooke’s law that springs A and B exert forces —k,x; and k2(x2 — x1), respectively, on mj. If no external force is impressed on the system and if no damping force is present, then the net force on m; is —k,x, + k2(x2 — x1). By Newton’s second law we can write d’x, mp = hx, + ky(x, — x). ANS-10 e ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS es ‘ 1 1 1 (—1)" sin(VAr) = —~ = 34,6 _ =~ _ 9 4... 33, (b) y(t) _ >» Gna pila _ ar 21. y(x) Co 1 3° + 32 , a 33 , 31" + 1 1 S (-1)" cos(V/At + ay 7 y(t) =r! > a (Vxr)2" = cos( VA) ) “ |: ay 4-7" n=0 (2n)! t Vr Wr — tgs + [bebe (c) y= Cx sin(~) + Cyx cos () 4°7- 10 2 3 x Xx 1, iv. + x6 — PP + +s 37-2) 37-3! EXERCISES 6.3 (PAGE 250) EXERCISES 7.1 (PAGE 261) 1. y = ey Jij3(x) + c2J-1/3(%) i 3. y = e1Jsi2(a) + €2J—s/2(X) test Ps 11, aa 5. y = cyJo(x) + c2Yo(x) “'s s “ss? - 7. y = cyJ2(3x) + c2¥2(3x) l+e™ 1 1. <x 9. y = €1J2/3(5X) + €2J—2/3(5x) 3 Sq 7 eo + oe" 5 11. y= Cy Tylax) + cox7 7? I-p(ax) 1 1 1 el ; 13. y= x? [ey I) (4x7) + cr ¥1(4x")] 9.-- s+ 5e° 1. —— 15. y = x[eWJi(x) + 2100] sos os s— $ 17. y = x'? [ey J3o(x) + €2¥3/2(x) 13. 1 15. — SF 19, y= x Yewip(bx) + oJ-1(42)] (s — 4) st+2s+2 a O 23. y= x"? [eyJia(x) + C2J-12)] 17 sa 19 48 eo = C; sinx + Cy cos x “(52 + 1p * 55 a 25. y= xl? [elt x?) + CJ -4(42°)] 21. e _ 10 23. 2 + 6 _3 i = C,x73?sin(! x?) + C)x73cos(# x2) 6 ; 3 1 ° 1 ° wat+atst- .-+— = 35. y= cx! Ty3(2 ax*?) + cx! 4(2 ax?!) . st sss? 27 s s—4 Zz 45. P2(x), P3(x), P4(x), and Ps(x) are given in the text, 29 1 4 2 + 1 31 8 5 A Po(x) = 7g (231x° — 315x4 + 105x° — 5), “s s-2 s-4 “ss 9 +9 O P,(x) = £ (429x7 — 693x5 + 315x3 — 35x) . okt — okt O 47. Ay = 2, Ao = 12, A3 = 30 33. Use sinh kt = —z show that - . k 1 L{sinh kt} “2 ke ul CHAPTER 6 IN REVIEW (PAGE 253) I I 2 35. ———-—- - — 37. =—— 5 1. False 2s —2) 2s s+ 16 il — C po 4cos5 + (sin5)s we 7. x-(x—- ly" +y' +y=0 39. 2+ 16 Ww 1 = 9, y= >> n= 0 nan =C x2], —-ly + 1irP-1 8+... Z yi) = Cx! | 4 . 630% EXERCISES 7.2 (PAGE 269) < yx) = C[L — x + Gx? — arte | 1. iP 3. t— 244 IL. yi) = o[l + pe + pt gate] 5.14+3r+3P4+!P 74-1 t er y2(x) = cy [x + vy + ix +: | 9. ye ti4 11. 2 sin Tt = = t 13. 1. = 3, =0 13. cos = 15. 2 cos 3f — 2 sin 3¢ yx) = C3 [1 t+ixt grr aete: | 2 1 _ 1-3 347-3 L y,(x) = C1 txt 12] 17. ;-—3e 7" 19. je" + ge’ 15, y(x) = 31 — 2 tit — ba te] 21. 0.3e°" + 0.6e°°7 23, 56" — e + 50% —2[x-te4is—datt.-- 25. £— cos V5t 27. —4+ 3e'+ cost+3sint 17. ba 29, isint — isin 2r 31. y=-l1+e' 19. x = Ois an ordinary point 33. y= pete 35. y=te'—ter ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS e ANS-11 37. y = 10 cost + 2 sint — V2 sin V2t 73. q(t) = 2 Ut — 3) — ze 5-3) U(t — 3) 39. y = eth tte t+ 3etiet ' ' 0 41. y= je'— fe *cos 2t + fe *sin 2r 75. (a) i(t) = i’ ~ 191 °° t+ Tor sit! _ 10 100-3712) ult — 3a EXERCISES 7.3 (PAGE 278) 101 2 1 6 10 37 37 1. —— 3. ——j 7 — 2 yo" (s— 10) (s + 2) + 101 cos 7 a 2 1 2 1 3 1 5 $, 2 1, 1,3 + stasin( 22) a 22) (s-2P (s-3P (s- 4) (s — 1? +9 101 2 2 9. = So s—l1 43 s+4 (b) imax ~ 0.1 at t ~ 1.7, imin ~ —O.1 at t ~ 4.7 N st+25 (s— 1) + 25 (s + 4) + 25 wol2 woL w co 12,728 Bt gi 77. y(x) = ee 3 yp = 11. ae 13. e°' sint 16EI 12EI QAEI e 15. e-?'cost — 2e7?' sint 17. e'— te"! } <t 19.5 —t—5e'—4tet — sPet A we (s = ‘) as _ ‘) 5 21. y=te"+2e" 23. y=e'+ 2te' ie 25. y=ptts- Ret t+ Pte™ 27. y= —Se*sin 2r Wol2 WoL n — Moh 2 Wor 13 29. y=} -elcost + te'sint 79. y(x) 48EI. 94 EI* = 31. y=(e+ Ite +(e- Det WW EB uy L = + —|—x-xw+lx-= a(x — 5)] 3 V15 7V 15 V15 33. x(t) = —=e77"2cos ———t — ——e-7" sin t 60EIL | 2 2 2 ie 2 2 10 2 a —s —2s —2s dT _ Lu 37. <— 39, — 4 9o 81. (a) a k(T — 70 — 57.5t — (230 — 57.51) U(t — 4)) a Ss s Ss co 5 —7s 1 _ 2 _— Waa? 43. 30 — 2)" Ur — 2) EXERCISES 7.4 (PAGE 289) Z 45. —sint U(t — 7) 47. Ut -— 1) -e~) Ut -— 1) 1 1 3 ys 4 2 49. (c) 51. (f) . (s + 10) ° (s? +4 Ay O 53. 3. @) 657 + 2 12s — 24 a 55 = — — . = 2 — 4 —3s 5. 7 aN 7. UND pad -E - f(t) =2—-4U(t — 3); LF} = 5 ee (s? — 1) [(s — 2)? + 36] UO uw e° e’ es 9. y = —te' + hcost—ttcost + 4tsint rr 57. f(t) = PUG — 1); L{f()} = 25 + 25 + — ° ° ” f . . 11. y = 2cos 3¢ + 3sin 3¢ + Zr sin 3¢ 5 1 es es nme 59. f(t) = t— 1 Ut — 2); AfO} = 3 - — - 2— 13. y = jsin4¢ + frsin 4¢ n S S Ss oc eras pbs — 3(t— msin4(t — mUC— 7m) S 61. fi) = Ut — a) — Ut — b); L{f(} = — — — 6 nv . . y= Gh +P 19. = Z 63. y = [5 — Se] Mr — 1) s < 65. y= —4+5t+4e%—-iUt—1) 21 s—1l 23 1 eo _— 7 _ S(t _ 1) U(t _ 1) + Fel) Ut _ 1) (s + KG 1) + 1] s(s 1) 67. y = cos 2t — ésin 2(t — 27) U(t — 27) 25. stl 27. a 1. s[(s + 1 + IJ s*(s — 1) + 3sin (tf — 277) U(t — 277) . 387 + 1 : 69. y = sint + [1 — cos(t — m]U(t — wm) 29. 4+ 1p 31. e- 1 — [1 — cos(t — 27)] U(t — 27) - 33. e —5f -—t-1 37. f(t) = sint 71. x(t) = 24 — Zsin 4t — 2 — 5) U(r — 5) LO +4 Zsin A(t — 5) U(t — 5) — 25 U(t — 5) 39. f(t) = ie + ce! + + tel + ie! 41. fo =e + » cos 4(t — 5) W(t — 5) 43. f(t) = ser + gen + 5COs 2t+ ;sin 2t ANS-12 e ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS . = sint — tsi 1 2V6 2 2 45. y() = sins ~ ptsint 13. x, = =sint + 2V6 V6t + =cost — =cos V6t 47. i(t) = 100[e 1") = e UG = 1) 5 va 5 p ~100— —20(1— 2 6 4 1 — 100[e7 19 — eM) Ut — 2) XxX) = =sint — —sin V6t + cost + cos V6t Lo ee 5 15 5 5 49. sd te) 15. (b) iy _ ” _ 100 e901 all 1 i; = > — Pe 51. 2(2 - =) (c) i) = 20 — 20e 2 53 coth (7rs/2) 17. i, = —Ber + Bel + Bcost + S sine “st +1 is = Pe + Bee — Meosr + sine 1 2 N 55. i(t) = Rl! - eT Rill) 19, i, = : - oem cosh 50V2t — OV? sm sinh 50V2t a E +2 Yop = ek UE ~ n) i, = 2 $10 cosh 50-V31 — 2? o-0 ginh 50-V3 4 xt n=1 = O 57. x(t) = 2(1 — e-'cos 3¢ — 4e~'sin 37) ; vo CHAPTER 7 IN REVIEW (PAGE 300) a +4 > (-)"[1 — e" cos 3¢ — nm) 12. Ss n=l 1. 5>- se 3. false Mo — Fe" sin 3(t — nm |Ut — nz) “¢ ' co 5. t 7. — O me s+7 G& —_—_EXERCISES 7.5 (PAGE 295) » As 9. =— 11. ——\} QO nyse MU — 2) e+4 (s? + 4 a ud 3. y = sine + sint Ut — 27) 13. 26 15. $e 5. y= —cost Ut — 7) + cost Ut — 32 17. e*'cos 2t + 2e°'sin 2t y 2 2 2 5 Ty= 4 _ se! +4 [} _ Le-2-)] U(t — 1) 19. cos w(t — 1)%(t — 1) + sin w(t — 1)%(t — 1) U —k(s—a a 9. y = e222 sin t U(t — 277) 21. —5 23. e *°-9F(s — a) O 11. y = e*'cos 3t + =e" sin 3t 25. f()U(t — 0) 27. f(t — t)U(t — t) a + Le Mm gin 3¢¢— m) Ut — 2) 29. f() =t-—(@— UE - 1) — Ur — 4); . 1 1 1 O + be2t-3) sin 3(¢ — 30) U(t — 37) LF) =Z- Get ~ es a Po(L 1s) o<xvel ven <2 be eo EI\4 6 2 tefO} = (—IP Gb O BM \pen 1) 4 Lu ro (hy _ «) -~<;<Lb _ ! oe 46-D YN 4ET \2 12 2 s— 1 a = 31. ff) =2 + (t — 2) U(t — 2); n 2 1 < 1. x = —}e + tel 3. x = —cos 3¢ — 3 sin 3¢ > 1 y= te + ze! y = 2cos 3t — i sin 3t Lle'f(o} = s-1 + (s — 12 ere) 5.x = —2e% + 3e%— $F 7. x= —3t-7V2sin V2t 33. y = Ste’ + te! — 8 31 5 2t 1 _— 1 3 : y=7e" — sel — = y=—it+iV2sinV2t ° ‘ , a 35. y= —2£ tir t3et— Bes — 4 ar — 2) %x=8t oP toe — ta - 2) Wr - 2) + fe? Ut — 2) ‘ ‘ — 2-5-2 Qi — 2 e t ye 2 paste 100 ( ) 3! 4) 37. y=14+4+5P Wx =5P+tt+1l-et 39. x= —1 + Set g Let y= + }e't+itet y=ttre%*—1em