Exercícios Selecionados Livro Calc2 Pt2-2021 2

138 e CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS systems are also able, by means of their dsolve commands, to provide explicit solu- tions of homogeneous linear constant-coefficient differential equations. In the classic text Differential Equations by Ralph Palmer Agnew’ (used by the author as a student) the following statement is made: It is not reasonable to expect students in this course to have computing skill and equipment necessary for efficient solving of equations such as d*y d*y d’y dy 4.317 — + 2.179 — + 1.416 — + 1.295 — + 3.169y = 0. 13 dx! dx? dx? dx y (13) Although it is debatable whether computing skills have improved in the intervening years, it is a certainty that technology has. If one has access to a computer algebra sys- tem, equation (13) could now be considered reasonable. After simplification and some relabeling of output, Mathematica yields the (approximate) general solution y = cye0778852* c9s(0.618605x) + cre °778852* sin(0.618605x) + c,¢9476478* Cos(0.75908 Lx) + cye°4778* sin(0.75908 Lx). Finally, if we are faced with an initial-value problem consisting of, say, a fourth-order equation, then to fit the general solution of the DE to the four initial conditions, we must solve four linear equations in four unknowns (the c1, C2, ¢3, C4 in the general solution). Using a CAS to solve the system can save lots of time. See Problems 59 and 60 in Exercises 4.3 and Problem 35 in Chapter 4 in Review. “McGraw-Hill, New York, 1960. EXE RC | S E S 4. 3 Answers to selected odd-numbered problems begin on page ANS-4. In Problems(—14 find the general solution of the given d’x dx second-order differential equation. 20. de de 4x = 0 1. 4y" + y' =0 2. y" — 36y =0 21. y" + 3y" + 3y' ty =0 3. y"—y’ — 6y =0 4, y" — 3y' + 2y =0 22. y — 6y + 12y' — 8y =0 23. (4) 4 ym ” 5. y" + 8y' + l6y =0 6. y" — 10y’ + 25y =0 Sty yo 24. y — 2y"+ y=0 7. 12y" — S5y’ — 2y =0 8. y" + 4y’-y=0 25 16% 4 142 4 9 =0 9, y"+ 9y =0 10. 3y” + y=0 aor re dx y 11. y" — 4y' + 5y =0 12. 2y"+ 2y’+,y=0 ° ° , ° se er en 13. 3y" + 2y’ +y=0 14. 2y" — 3y’ + 4y =0 dx dx 5 4 3 2 In Problems 15—28 find the general solution of the given 27. au +5 au 2 au 10 du +4 du +5u=0 higher-order differential equation. dr dr’ dr’ dr dr ax d*x d°x dx 15. y" — 4y" — Sy’ = 0 28. 2—= - 7 + 12-5 + 8-5 = 0 ds? ds‘ ds? ds? 16. y”"-y=0 17. y" — Sy" + 3y' + 9y =0 In Problems29- 36) solve the given initial-value problem. 18. y” + 3y” — 4y’ — 12y = 0 29, y" + loy=0, y(0) = 2, y'(0) = —2 du du d’y (2) (2) 19. —~ + —> - 2u= 30. —+y=0, —}=0,y|-] =2 4.3 HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS e 139 d’y dy 45. y 31. —, —-4— — 5y = 0, 1)=0,y'(1) =2 We i yQ) y'() 32. dy" — 4y’ — 3y=0, yO) = 1, y'(0) = 5 x 33. y" + y' + 2y=0, yO) = yO) = 0 34. y”— 2y'’+y=0, yO) =5,y'(0) = 10 35. y" + 12y" + 36y’ =0, yO) = 0, y'(0) = 1, y"O) = —7 FIGURE 4.3.4 Graph for Problem 45 36. y” + 2y"” — Sy’ — 6y =0, yO) = y’(0) = 0, y"(0) = 1 y In Problem(7—4)solve the given boundary-value problem. 46. 37. y”— 10y' + 25y =0, y(0) = 1, y\1) = 0 38. y" + 4y=0, (0) = 0, (7) = 0 * 7 39. y"+y=0, yO)= oy(2) =0 2 FIGURE 4.3.5 Graph for Problem 46 40. y’— 2y' + 2y=0, yO) =1,y(7) = 1 . . 47. y In Problems 41 and 42 solve the given problem first using the form of the general solution given in (10). Solve again, this time using the form given in (11). 41. y"—3y=0, y@O)=1,y'(0) =5 ae 42. y"-y=0, yO)=1, yd) =0 In Problems43—48)each figure represents the graph of a particular solution of one of the following differential FIGURE 4.3.6 Graph for Problem 47 equations: (a) y" — 3y’ — 4y =0 (b) y" + 4y =0 48 y (c) y" + 2y' +y=0 (d) y"+y=0 , (e) y” + 2y' + 2y=0 (f) y" — 3y’ + 2y =0 Match a solution curve with one of the differential equa- tions. Explain your reasoning. TT x 43. y x FIGURE 4.3.7 Graph for Problem 48 FIGURE 4.3.2 Graph for Problem 43 Discussion Problems y 49. The roots of a cubic auxiliary equation are m, = 4 and 44. m2 = m3 = —5. What is the corresponding homogeneous linear differential equation? Discuss: Is your answer unique? x 50. Two roots of a cubic auxiliary equation with real coef- ficients are m, = —3 and m2 = 3 + i. What is the corre- FIGURE 4.3.3 Graph for Problem 44 sponding homogeneous linear differential equation? 4.6 VARIATION OF PARAMETERS e 161 The first 1 — 1 equations in this system, like y;uj + y.u5 = 0 in (4), are assumptions that are made to simplify the resulting equation after y, = u)(x)y)(x) + +++ + Un(xX)y,(x) is substituted in (9). In this case Cramer’s rule gives Ww m=, k= 1,2,....0, W where W is the Wronskian of y;, y2,..., y, and W; is the determinant obtained by replacing the Ath column of the Wronskian by the column consisting of the right- hand side of (10)—that is, the column consisting of (0, 0,..., f(x)). When n = 2, we get (5). When n = 3, the particular solution is y, = uyjy,; + uzy2 + u3y3, Where y1, Y2, and y3 constitute a linearly independent set of solutions of the associated homogeneous DE and uy, u2, uz are determined from ’ Ww, , Ww, , W, dd 1) ui =—, uy =—, uy, =—, ' Ww Ww > WwW 0 yy ¥3 yy O- ys vy yw O Yi 2 3 Wi=|0 ys ys), Wo= | 90 ys], We= yi yp 0 |, and W=|yy ys ys). FO) ys y3 yi f@ ys yi ys f@) yi Y2 y3 See Problems 25 and 26 in Exercises 4.6. REMARKS (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 141. 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 yy. Depending on how the antiderivatives of uj and wu} 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, = 5 sinx — 5 x cos x and y, = j sinx — } x cos x are valid answers. In either case the general solution y = y, + y, simplifies to y =c, cos x + cz sin x — +x cos x. Why? EX E RC | S E S 4 . 6 Answers to selected odd-numbered problems begin on page ANS-5. . . . . 1 In Problem kolve each differential equation by varia- 11. y" + 3y’ + 2y = —— tion of parameters. 1+ n” ' — e 1. y"+ y = secx 2. y" +y =tanx 12. y ay +Y=Ty_ 3. y" + y = sinx 4. y” + y =sec 6 tan 0 13. y" + 3y' + 2y = sin e* no ’ — _,t 5. y" + y = cos2x 6. y" ty =sec2x 14. y" — 2y’ + y =e‘ arctant 15. y"+2y’+y=e'lnt 16. 2y"+ 2y’ +y=4 7. y" — y =coshx 8. y" — y = sinh 2x y vow yen y yoy va 17. 3y" — 6y' + 6y = e* sec x n” e* n” 9x 9. y" — 4y = —— 10. y" — 99 = 18. 4y"—4y' ty =ePV1- 9 162 e CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS In Problem(19-22 olve each differential equation by 30. Find the general solution of xty" + xy! —A xy =] variation of parameters, subject to the initial conditions given that y, = x2 is a solution of the associated homo- y(0) = 1, y’(0) = 0. geneous equation. 19. 4y" — y = xe*? @) Suppose y,(x) = ui(x)yi(x) + u2(x)y2(x), where u; and 20. 2y"+ y'-y=x4+1 uz are defined by (5) is a particular solution of (2) on an Ly" + 2y" — 95-2 -_ interval J for which P, Q, and f are continuous. Show yi + dy’ — By = 2e" —e that y, can be written as 22. y" — 4y’ + 4y = (12x? — 6x)e” In Problems 23 and 24 the indicated functions are known lin- Yp(r) = [ Gx, Of dt, (12) early independent solutions of the associated homogeneous ” differential equation on (0, ©). Find the general solution of where x and xo are in J, the given nonhomogeneous equation. yiDyax) — yir)yo(t) n” , t SOOO 1 xy" + xy’ + ( - 1) = 3; G(x, 1) Wit) , (13) y, =x !? cosx, yo =x !? sin x and W(t) = W(yi(0), yo(t)) is the Wronskian. The func- xy" + xy’ + y = sec(In x); tion G(x, f) in (13) is called the Green’s function for the y; = cos(In x), y2 = sin(In x) differential equation (2). In Problems 25 and 26 solve the given third-order differen- 32. Use (13) to construct the Green’s function for the differ- tial equation by variation of parameters. ential equation in Example 3. Express the general solu- 25, y" +y' =tanx 26. y" + 4y! = sec 2x tion given in (8) in terms of the particular solution (12). 33. Verify that (12) is a solution of the initial-value problem Discussion Problems ; dy , dy In Problems 27 and 28 discuss how the methods of unde- We + P dx + Oy = f(x), yA) = 9, y'(X%) = O. termined coefficients and variation of parameters can be combined to solve the given differential equation. Carry out on the interval I. [Hint: Look up Leibniz’s Rule for your ideas. differentiation under an integral sign.] 27. 3y" — 6y’ + 30y = 15 sinx + e* tan 3x 34. Use the results of Problems 31 and 33 and the Green’s 28. y" — 2y' +y =4x27-3 4x let function found in Problem 32 to find a solution of the initial-value problem 29. What are the intervals of definition of the general solu- tions in Problems 1, 7, 9, and 18? Discuss why the inter- y"-y=e, y(0)=0, y'(0)=0 val of definition of the general solution in Problem 24 is not (0, ©). using (12). Evaluate the integral. | 4.7) CAUCHY-EULER EQUATION REVIEW MATERIAL e Review the concept of the auxiliary equation in Section 4.3. INTRODUCTION The same relative ease with which we were able to find explicit solutions of higher-order linear differential equations with constant coefficients in the preceding sections does not, in general, carry over to linear equations with variable coefficients. We shall see in Chapter 6 that when a linear DE has variable coefficients, the best that we can usually expect is to find a solution in the form of an infinite series. However, the type of differential equation that we consider in this section is an exception to this rule; it is a linear equation with variable coefficients whose general solution can always be expressed in terms of powers of x, sines, cosines, and logarithmic functions. Moreover, its method of solution is quite similar to that for constant-coefficient equations in that an auxiliary equation must be solved. ANS-4 e ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS EXERCISES 3.3 (PAGE 110) EXERCISES 4.1 (PAGE 128) 1. x(t) = xe" ly= Se - te — MAL a at 3. y = 3x — 4xInx 0) Ay ~ rN (e c ) 9. (-®, 2) e sinh x 11. = >— (e- e* b) y = — = »(1- 2 fo ear AL gm] (a) y 1! e") (b) y sinh | 2 | 2 | 13. (a) y = e* cos x — e* sinx 3. 5, 20, 147 days. The time when y(f) and z(t) are the (b) no solution same makes sense because most of A and half of B (c) y=e*cosx + e77e* sin x are gone, so half of C should have been formed. (d) y = cpe* sin x, where c2 is arbitrary dx 15. dependent 17. dependent = 5. 7 =6- zx + HX 19. dependent 21. independent ow ' 23. The functions satisfy the DE and are linearly independent tH dx, =2,—2, on the interval since W(e~>*, e+”) = Je* # 0; es dt 25 “1 25 2 y= cje * + oe, a d 25. The functions satisfy the DE and are linearly independent UO 7. (a) 1 3 a 2—1— on the interval since W(e* cos 2.x, e* sin 2x) = 2e?* # 0; dt 100 — ¢ 100 + ¢ = .o% xo e y = cye’ cos 2x + cre” sin 2x. n dx, =? ML 3 %2 27. The functions satisfy the DE and are linearly > dt 100 + ¢ 100 — ¢ independent on the interval since W(x3, x4) = x° #0; 4 (b) x1(2) + x2(t) = 150; x2(30) ~ 47.4 lb y = cx? + cox". O di 29. The functions satisfy the DE and are linearly independent 13. Li + (R; + R3)in + Riis = E(t) on the interval since W(x, x7,x 7 Inx) = 9x ° # 0; a t y=axt ox? + 03x 7 Inx. di — 42 2x, —_ — 6y — 1423 = Le + Ryiy + (R, 4 R;) i, = E(t) 35. (b) Vp =x + 3x + 3e°*; Vp = 2x? 6x 3 e Lu co > 15. i(0) = ig, s(0) =n — ip, r(0) = 0 5 EXERCISES 4.2 (PAGE 132) a 1. y> = xe”* 3. yo = sin 4x a CHAPTER 3 IN REVIEW (PAGE 113) 5. y) = sinh x 7. yo = xe2"3 OQ iL. dP/dt = 0.15P 9. yo = x" In|x| Hl. yx = 1 Lu 3. P(45) = 8.99 billion 13. y2 = x cos (In x) 15. yo=x°t+x+2 5 Vion 17. y, = e”, yy = —} 19. y, = e”, yy = > ee uw 10 + V100 — JS 5x5 191n( #2 VO) ~ Vi00-¥ n y a ° 7. (a) BT, + T, BT, + T, EXERCISES 4.3 (PAGE 138) re 1+B 1+B 1. y=c, + oe *!4 3. y= cje* + coe >* S (b) T(t) = BT, + T 4 T, — Th ek(+B)r 5. y= ce + coxe y= ce2*!3 + e,e7*4 a 1+B 1+B 9. y = c,cos 3x + cosin 3x Z Lo 11. y = e?*(c,cos x + eosin x) < 9. w= {ho O=1< 10 13. y = e*(c,cos ! V2 x + cy sin V2 x) 20, 1= 10 15. y=c, to2e *+ c3e* ace! 17. y=cje *+ ce’* + c3xe** 11. x(t) = Ts com’ y(t) = (1 + cet) h!h 19. u=cyje' +e ' (cocos t + c3sin f) ' 21. y=cye * + oxe *+ axe * 13. x=—y+1+ce” 23. y=cy toxter”? (c;cos 5V3x + c,sin} V3 x) I 25. y = c, cos} V3x + cy sin} V3x 15. (a) p(X) = pte» +e | q(x) ax) + cx cost V3x + cyx sink V3x _ r r —r -r —5r (b) The ratio is increasing; the ratio is constant. aa, y _ se eae 1 fe + care" ¥ c5e “y= 73 __ Kp _ |___ kp BL y= Fe) $i eED (8) 0) = Ty + fama ~ \2CK— Be) 93, y = 0 ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS e ANS-5 — 5 5. 6x 1 —6x 35. y= ag ~ ase * + Gxe® 35. y=cje * + c9e* — 6 — x Xx 37. y= e-*— xe 37. y=c, t+ coe * + 3x 39. y=0 39. y= ce + oxe*+3x41 41. y = (1 _ =) e738 4 (1 +4 =) eV3e: 41. y=c, +Ox+coe* + = x4 — Sx + 8x? 2 V3 2 V3 43. y= cje* + ce" + sxet 5 _ y = cosh V3x + 7 sinh V3x 45. y=cye * + ce — e* +3 47. y =c,cos5x + cy) sin 5x + 7 sin x 49, y=cye * + Exe * — pre + sae _ —x 1 1 x Lx EXERCISES 4.4 (PAGE 148) Sl. y= ce * + ce + xe’ — gre’ + gxe’ — 5 L. y=cye® + pe 2* +3 53. y = e*(c,cos 2x + cysin 2x) + fe sin x ~yrel 2 3. y=ce* + cxxe* + Sx +3 55. y = c,cos 5x + cysin 5x — 2x cos 5x + _— —2x —2x 2 7 V3 V3 oc 5. y=cje *+ axe *+x 4x + 5 57. y= (6, cos—— x + c) sin 3.) tu 7. y = c, cos V3x + c) sin V3x + (—4x2 + 4x - 4) 038 2 2 [au + sinx + 2 cos x — x cos x < 9. y=c, + c9e* + 3x s no 7g 4 = 11. y = cye? + coxet? + 12 + Fx? er? 59. y = cy + eax + c3e + g5gX° + aX” — 76% Uv _ x 1 13. y = c, cos 2x + cy sin 2x — ux cos 2x 61. y = cyet + cyxe* + exes + Grek + x — 13 o _ X Xx 1 Xx 1 15. y = c,cosx + c)sinx — $x° cosx + $x sinx 63. y = c) + ex + ce + cyxe* + yx e+ 5° > _ 5-81 4 58% _ 1 17. y = cye*cos 2x + cye*sin 2x + } xe" sin 2x 65. y =e + ge — | = — _ 41 4 41 sy _ 1 9 19, y=cye* + Exe* — + cos x 67. y = ~y55 + jag" — GX t+ 95% O + 2 sin 2x — 2 cos 2x 69. y = —mcosx — tsinx — cos 2x + 2x cos x ie = x 3 2x oj 1 3 3 21. y =, tox + ce — 42 — Scosx + Lsinx 71. y = 2e** cos 2x — Ge* sin2x + gx + Fx? + 35K a 23. y = cye" + cyxe* + cyxe* — x — 3 — Exe ai 25. y = cy cosx + co sin x + c3x cos x + c4x sin x EXERCISES 4.6 (PAGE 161) S Fat = 2x = 3 1. y =c, cosx + cy sinx + xsinx + cos x In| cos x| 5 27. y= V2 sin 2x ~ 3 3 = o cosx + ysinx — ! cos x QO 29, y = —200 + 200e*/5 — 3x? + 30x ea zee a 31. y = —10e72* cos x + 9e~2* sin x + Te 5. y = c, cOSx + c)sinx + 5 — ¢cos 2x O _ yg lig 33 _ _ Fy 7.y= ce + oe* + 5x sinh x a 7e Tp net 3 tcos wt At kK wo o 9. y= ce +t oe ¢ fer] rs nae O _ : else -y=ce coe ** + gf e?* In|x| — e dt}, uw 35. y = 11 — lle* + 9xe* + 2x — 12x e + 5e xy ot a 37. y = 6 cos x — 6(cot 1) sinx + x7 — 1 Xo > O o 39 —4 sin V3x 4 11. y= ce + ce 2% + ( + e7?*) In(1 + e*) O . = _ —2x —x 72x go x y sin V3 + V3 cos V3 x 13. y=cje “+ c2e e ‘sine an 5 1 15. y=cje'+ ote! + iPe! Int — +Pet co cos 2x + ?sin2x + ;sinx, OSx< 7/2 . ' : — 41. y=}, dx + Sein? > a/2 17. y = cye*sinx + c,e* cos x + 4xe* sin x = 5 cos 2x + 2 sin 2x, x> 7 n ° ° + 4e*cos x In| cos x| Zz 19. y= 1 oad 4 3 pri2 4 12 oxl2 _ 1, x2 . 4 4 8 4 EXERCISES 4.5 (PAGE 156) 21. y = tet + Ber — fer 4 te 1. GD — 2)3D + 2)y = sin x 23. y=cyx ! cosx + cox? sin x + x71? : yD 5) + ay =x—6 25. y = c, + c) cosx + c3sin. x — In|cos x| . y=e ; 7. (D —1)(D —2)(D + 5)y = xe? — sin x In|sec x + tan x| 9. D(D + 2)(D? — 2D + 4)y =4 15. D* 17. D(D — 2) 19. D244 21. D3(D? + 16) EXERCISES 4.7 (PAGE 168) 23. (D + 1)(D — 1)3 25. D(D* — 2D + 5) 1. y=cyx ! + epx? 27. 1, x, x7, x3, x4 29, 0%, e 3x2 3. y=c, +coInx 31. cos V5x, sin V5x 33. 1, e, xe* 5. y = c, cos(2 In x) + c2 sin(2 In x)