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HW 2 Problems - E g'4:11:1 ‘ “tines[Nation licalivc...

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Unformatted text preview: E. - g:'4:11:1:\ ‘- “tines [Nation licalivc condi- Iou can ll value (28} 22). we .rr equal 1d the ion in 1'6 [116 tions of Linear Homogeneous Equations; the W/ronshian 155 PROBLEMS In each of Problems 1 through 6 find the Wronskian of the given pair of functions. # 1. 92’, 3’3”” 2. cost, sint 3. 93“”, 384' 4. x, xex 5. e’sint, e'cost 6. cosze, 1+c0329 In each of Problems 7 through 12 determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution. 7. tr” + 3y = I: 32(1): 1, 3111): 2 8. (t — 1)y” — 302’ + 4y = sin I, y(——2) = 2, y’{—2) = 1 9. I0 - 4))!” + 3Iy’ + 4y z 2» M3) = 0, y’(3) = —1 10. y” + (cos t)y' + 3(ln |t|)y = 0, y(2) : 3, y’(2) = 1 11- (x e 3)y" + xy’ + {1n leh} = 0, y(1) = 0. y’(1)= 1 12. (x W 2))!” + y' + (x a 2) (tanx)y = o, y(3) .= 1, y'{3) = 2 13. Verify that y1(r) : t2 and y2(t) : 1’1 are two solutions of the differential equation :2)!” — 2y : 0 for t > 0. Then show that y : c112 + Qt” is also a solution of this equa- tion for any c1 and 62. 14. Verify that y1(r) :: 1 and y2(t) = 1112 are solutions of the differential equation yy” + (y’)2 = 0 for r > 0. Then show that y 2 c1 + (321?“2 is not, in general, a solution of this equation. Explain why this result does not contradict Theorem 3.2.2. 15. Show that if y = ¢(t) is a solution of the differential equation y" +p(t)y' + q(r)y : g(r), where g0?) is'not always zero, then y : c4501), where c is any constant other than 1, is not a solution. Explain why this result does not contradict the remark following Theorem 3.2.2. 16. CE y : siI1(t2) be a solution on an interval containing 1‘ = 0 of an equation y” + p(t)y’ + q(t)y = U with continuous coefficients? Explain your answer. 17. If the Wronskian W off and g is 33‘”, and iff(t) = 32‘, find go). 18. If the Wronskian W off and g is tze‘, and iff(r) = I, find g0). 19. If WQ‘Ngr) is the Wronskian off and g, and if it 2 2f — g, u = f + 2g, find the Wronskian W(u,,u) of u and v in terms of W(f,g). 20. If theWronskian of f and gistcost — sin Land if u = f + 3g, 1) = f — g,find theWronskian of u and v. 21. Assume that y} and y; are a fundamental set of solutions of y” + p(t)y’ + q(t)y 2 O and let ya, = alyl + azyz and y4 :: blyl + b2y2,where ahag. b1, and b; are any constants. Show that W03ay4) = (flibz — azbill’VOJLyzl- Are y3 and 314 also a fundamental set of solutions? Why or why not? In each of Problems 22 and 23 find the fundamental set of solutions specified by Theorem 3.2.5 for the given differential equation and initial point. 22.y"-l—y'—2y=0, 1‘020 23- y”+4y'+3y=0, 10:1 In each of Problems 24 through 27 verify that the functions y; and yg are solutions of the given differential equation. Do they constitute a fundamental set of solutions? ~ 24' J’” + 4y 2 0; y1(r) : cos 2t, y2(t) = sin2t " " ' 25 y" - 2)" + y = 0; 3W) 2 8‘. 3220) = te’ i a‘ I I i . l E E l E 26- xzy” —x(x+2)y’ + (X+2)y : 0. x > 0; y1(x):x, yztx) Ixe" 27. (i _xcotx)y” —xy’ +y 2: 0, 0 < x < Jr; y1(x) =x, y2(x) = sinx 28. Consider-the equation y” — y’ — 2); m 0. 41. Chapter 3. Second Order Linear Equafi - - 0'13 (a) Show that yd!) = e" and 3120:) = 22‘ form a fundamental set of solutions. (b) hit 3130) = —2€2’, MU) = 3’1 (1‘) “I” 25bit), and J’s“) = 2M!) * Zyslfl- Are YBU} y and 3250‘) also solutions of the given differential equation? ' all (c) Determine whether each of the following pairs forms a fundamental set of 1(t),ys(l‘)l; U2(I)J3(f)]; [Yi(t),y4(f)}; {314(r),y5(r)]. In each of Problems 29 through 32 find the Wronskian of two solutions oft differential equation without solving the equation. 29. ray” —~ :0? + 2)y’ + (I + 2)y = 0 30. (cos 0y” + (sin r)y’ — ty 2 0 31. xzy” 4— xy’ -i— (x2 — v2)y = 0, Bessel’s equation 32. (1 — x2)y” — ny’ + ot(ct + 1) y 2 0, Legendre’s equation soiutin' he give 33. Show that if p is differentiable and 170‘) > 0, then the Wronskian WU) of two solutio- of [p(t)y’]’ + q(t)y : 0 is WU) = c/p(t), where c is a constant. 34. if yl and y; are a fundamental set of solutions of ty”+2y’+te‘y=0 and 5f W(yl,y2)(i) = 2, find the value of W(y1,y2)(5). 35. If yl and y2 are a fundamental set of solutions of tzy” —2y’ + (3 -§- 0y = D and iii WUI,}’2)(2) = 3,find the value of W01,yg)(4). g 36. It the Wronskian of any two solutions of y” +p(t)y’ + q(t)y = 0 is constantwhatdocs: I this imply about the coefficients p and q? ' 37. If f, g, and h are differentiable functions, show that W(fg,fh) : fZW(g,h). In Problems 38 through 40 assume that p and q are continuous and that the functions yl an yg are solutions of the differential equation y” + p(t)y’ + q(t) y = O on an open interval 1'. ' 38. Prove that ify1 and y; are zero at the same point in I, then they cannot be a fundamenta set of solutions on that interval. 39. Prove that if y1 and y; have maxima or minima at the same point in I , then they canno be a fundamental set of solutions on that interval. 40. Prove that if y; and y; have a common point of inflection to in I, then they cannotb a fundamental set of solutions on I unless both 1) and q are zero at to. Exact Equations. The equation P(x)y” + Q(x)y’ + R (x)y : O is said to be exact if it ca be written in the form [P{x)y’]’ + [f(x)y}’ : 0, where f (x) is to be determined in termSO P05), Q05), and Jim. The latter equation can be integrated once immediately, refillltl'fl in a first order linear equation for y that can be solved as in Section 2.1. By equating ”1 coefficients of the preceding equations and then eliminating f (x), show that a necessary condition for exactness is P”(x) — Q’ (x) + R05) 2 0. It can be shown that this is also sufficient condition. In each of Problems 42 through 45 use the result of Problem 41 to determine whether th given equation is exact. If so, solve the equation. 42. y”+xy’+y=0 43. y”+3x2y’+xy=0 44. xy” -— (cosx)y’ + {sinx)y = 0, x > 0 45. xzy” +xy’ —y = 0, x > 0 46. The Adjoint Equation. If a second order linear homogeneous equation is not exact! can be made exact by multiplying by an appropriate integrating factor MU)- “will require that JMic) be such that ,u.(x)P(x)y” + u(x)Q(x)y’ + M(.?C)R(x)y = 0 can be “mm 3.3 Complex 3 3 Complex Roots of the Characteristic! Equation , .1 W: note that it the real part of the roots is zero, as in this example, then there is no exponential factor in the solution. Figure 3.3.3 shows the graph of two typical solutions of Eq. (28). In each case the solution is a pure oscillation whose amplitude is determined by the initial conditions. Since there is no exponential factor in the solution (29), the amplitude of each OScillation remains constant in time. In each of Problems 1 through 6 use Euler’s formula to write the given expression in the form M a + ib. 1. exp(1 + 21') 2. exp(2 —— 3i) 3. ebr 4 eZ—(n/ZH 5. 21—t‘ 6 ”Al-+2! In each of Problems 7 through 16 find the general solution of the given differential equation. 7.y”m2y’+2y=0 8.y”—2y'+6y:0 9. y"+2y’m8y=0 10. y”+2y’+2y:0 11. y"-£—6y’+13y=0 12. 4y”+9y=0 13. y” -i- Zy’ + 1.25); z 0 14. 9y” + 9y’ — 4y : 0 15. y" +y’ + 1.25)) = 0 16. y” + 452’ + 6.25); z 0 In each of Problems 17 through 22 find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing 1. 17. y” + 4y = O, y(0) = 0, y’{0) : 1 18. y” + 4)” + 532 :x 0, y(0) = 1, y’(0) = 0 19. y” — 2y’ + 531 = 0, yOr/Z) ..—_ O, for/2) z 2 20. y” +y = 0, yer/3) = 2, yer/3) = —4 21. y” +y’ + 1.25): = 0, y(0) = 3, y'(0) m 1 22. y” + 2y’ + 2y 2 O, y{:rr/4) = 2, y’(7r/4) : ——2 $52, 23. Consider the initial value problem 3M” —— u’ + 2n = 0, MO) 2 2, u’(0) = 0. (a) Find the solution um of this problem. (b) Fort > 0 find the first time at which |u(t)i = 10. $52, 24. Consider the initial value problem 5H" + Zu’ + Tu = 0, MO) 2 2, u’(0) = 1. (a) Find the solution MU) of this problem. (b) Find the smallest T such that |u(r)| 5 0.1 for all t > T. $2 25. Consider the initial value probiem y” + 2)}! + 6y = 0, (3) Find the solution y(t) of this problem. (b) Find at so that y = 0 when t 2 1. (c) Find, as a function of a, the smallest p (d) Determine the limit of the expression W3) = 2, y’(0) = 0t 3 0. ositive value of I for which y :0 found in part (c) as or —> oo. 3_4 Repeated Roots; Reduction of Order 1 73 Then think of r1 as fixed and use L’I—Iospital’s rule to evaluate the limit of ¢(t;r1,r2) as r; —> r1, thereby obtaining the second solution in the case of equal roots. 22. (a) If er2 + br + c = O has equal roots r1, show that 31. 32. L[er1} = (1(8’1)” + bk”), + CE" = [10" _ “Fen. G) Since the right side of Eq. (i) is zero when r 2 n, it follows that exp(r1 t) is a solution of LEy] = ay” + by’ + cy = 0. (b) Differentiate Eq.’ (i) with respect to r and interchange differentiation with respect to r and with respect to I, thus showing that 51L[e”] : L [32"] = L[te”] : ate"(r — ml + Zae"(r - 1'1). (ii) Br Br Since the right side of Eq. (ii) is zero when r = r1, conclude that t expmt) is also a solution of L[y} : 0. In each of Problems 23 through 30 use the method of reduction of order to find a second solution of the given differential equation. 23. 24. 25. 26. 27. 2,8. 29. 30. tzy” w 4ty’ + 6y = O, t > 0; 321(1) 2 t2 tly” + 2t)" — 2y = 0, t > 0; yitt) = 32y” + 3a’ +y = 0. r> 0; rim = r1 fly” — t(r+2)y' + (r + 2))! = 0, r > 0; 3210):: Jay” —— y’ + 4x3}; 2 O, x > 0; y1(x) 2 Shut:2 (x—1)y”—xy'+y=0, X>1; y1IX)=e‘ xzy” — (x — 01875)}: 2 0, x > 0; 3210:) = imam x2)!” + xy’ + (i:2 w 0.25))! = 0, x > 0; y1(x) 2 15’1” sinx The differential equation xy” — (x + N)y’ + Ny = 0, where N is a nonnegative integer, has been discussed by several authors.6 One reason why it is interesting is that it has an exponential solution and a polynomial solution. (a) Verify that one solution is y1(x) : e". (b) Show that a second solution has the form y; (x) 2 oz" f xNe’x dx. Calculate yg(x) for N = 1 and N = 2; convince yourself that, with c = —1/NI, 2 N _1 x x x y2(x)- +E+Z+.H+“lel Note thatyzor) is exactly the first N + 1 terms in the Taylor series about x = 0 for e", that is, for y1(x). The differential equation y” + 5(xy’ +y) = 0 arises in the study of the turbulent flow of a uniform stream past a circular cylinder. Verify that y1 (x) = exp{—6x2/2) is one solution and then find the general solution in the form of an integral. . ,. .32; f:’ 6T. A. Newton, “On Using a Differential Equation to Generate Polynomials,” American Mathematical Monthly 8] (1974), pp. 592—601. Also see the references given there. 1 f4 . Chapter 3. Second Order Linear E 33. The method of Problem 20 can be extended to second order equations with Vtr' fecoeflicients. If yl is a known nonvanishing solution of y” + p(t)y’ + q,(:)y : 0’ shtiwi _—_a second solution y; satisfies (y; /y1)’ = Wtyt,y2)/y§, Wham WO’iJz) is the Wl‘Onski yl and 312. Then use Abel‘s formula [Eq. (22) of Section 3.2] to determine y2_ shit that iln at In each of Problems 34 through 37 use the method of Problem 33 to find a second inde solution of the given equation. 34. fly” + 3ty’ + y = 0, t > 0; y1(t) : F] 35. ty” —- y’ + 4t3y : 0, t > 0; y1(t) = sin(t‘2) 36. (x—l)y”—xy’+y:0, x>1; y1(x):e" 37. xzy” + xy’ + (x2 — 0.25)y : 0, x > 0; y1(x) 2 16‘1"2 sinx pendent Behavior of Soiutions as t ~—> oo. Probierns 38 through 40 are concerned with the beam-i"E of solutions as t ~+ oo. ' 38. If a, b, and c are positive constants, show that all solutions of fly” + by’ + cy = 0 aPPmach zero as t w+ oo. 39-. (a) If a > 0 and c > 0, but .5 : 0,show that the result of Problem 38 is no longer truchut that all solutions are bounded as t —+ 00. (b) Ifa > 0 and b > 0, but c = 0, show that the result of Problem 38 is no longer truehug that all soiutions approach a constant that depends on the initial conditions as t ‘9 x Determine this constant for the initial conditions y(0) : yo, y’(0) : Yir 40. Show that y : sint is a solution of y” + (k sinz t)y’ + (1 w it cos tsin 0y 2 0 for any value of the constant k. If 0 < k < 2,show that 1 — k cos t 5th > 0 and k sinlt 3 (t, Thus observe that even though the coefficients of this variable-coefficient differential equa- tion are nonnegative (and the coefficient of y’ is zero only at the points t = 0,a,2;r, . H). it has a solution that does not approach zero as t e 00. Compare this situation with the result of Problem 38. Thus we observe a not unusual situation in the study of differential equations: equations that are apparently very similar can have quite different properties Euler Equations. In each of Problems 41 through 46 use the substitution introduced in Proh- lem 34 in Section 3.3 to solve the given differential equation. 41. fly” —3ty’+4y=0, t> 0 42. fly" + Zty’ + 0.25y = 0, :> 0 43. 2:23;" — Sty’ + 5y 2 0, r> 0 44. ray” + 3ty’ +y = 0. r > 0 4S. 4t2y” — Sty’ + 9y = 0, t > 0 46. tzy”+51y’+13y20, z> 0 We now return to the nonhomogeueous equation My] = y” +p(t)y’ + (10))! = $0), where p, q, and g are given (continuous) functions on the open interval 1. fl“ equation (7) (ll Lty] = y” +190)? + q(t)y = 0, “:5. 4 l l E 3_ 5 Nonhomogeneous Equations; Method of Undetermined Coefficients 183 and we should choose Y0) = e(a+ifl)’(AUr" + - - - + An) + €(d_ifi)r(Botn + - - . + B"), or, equivalently, ’ Ye) = rampant” + - - - + A”) cos B: + e‘“(Bor” + - - - + 3,.) sin er. Usually, the latter form is preferred. It or :I: to satisfy the characteristic equation corresponding to the homogeneous equation, we must, of course, multiply each of the polynomials by t to increase their degrees by one. If the nonhornogeneous function involves both cos ,Bt and sin fit, it is usually con" venient to treat these terms together, since each one individually may give rise to the same form for a particular solution. For example, if g(t) : tsint + 2 cost, the form for YO) would be YO?) = (A0? + 141) Shit + (Bot “l" Bl) cos r, provided that sin: and cos 3‘ are not solutions of the homogeneous equation. PROBLEMS In each of Problems 1 through 12 find the general solution of the given differential equation. 1- 12”»- 232’ — 3y = 362' 2. y” + 2y' + 5y = 3sin2t 3- y” e 232’ — 3y = —3te" 4. y” + Zy’ = 3 + 4 sin2r 5- y” + 952 = 963’ + 6 6. y" + Zy’ + y = 2e“! 7. 2y”+3y'+y=t2+35int 8. y”+y=35in2x+icosZt 9. u” + cage = cos cut, a? gé mg 10. u.” «1- age = cos mo: 11. y” + y’ + 4y : 28inht Hint: sinht = {e‘ — e")/2 12. y” — y’ — 2y = costh Hint: cosht 2 (e[ + e“)/2 In each of Problems 13 through 18 find the solution of the given initial value problem. 13. y” +y’,w 2y : 21, 32(0) : O, y’(0) = 1 14. y" + 4y :12 + 38‘, 32(0) 2 0, y’(0) = 2 15. y” — 232’ +J2 = te’ + 4, y(0) = 1, y’(0) = 1 16. y” — 232’ — 3y = 3:69”, 32(0) 2 1, y'{0) = 0 17. y" + 4): z 35inZt, 32(0) := 2, y’(0) = +1 18. y” + 232’ -1— 5y 2 497‘ cos2r, 32(0) = 1, y'{0) : 0 _ In each of Problems 19 through 26: (a) Determine a suitable form for YU) if the method of undetermined coefficients is to be used. (b) Use a computer algebra system to find a particular solution of the given equation. 9% 19. y" + 3y’ = 2:4 +1484“ + sin 3: 9:52, 20. y” +y : at + sin :) $2 21. y” ~5y’+6yze‘cos21+e2‘(3t+4)sinr ,. $8 22. y" + Zy’ + 232 2: 3e~r ~t— 26” cost + 45":2 sin: _ _:____:___ ' $2 23. y” _ 452 + 4y = 2:2 + 4152” + {sin 2; $3 24. y” + 4)! z :2 sin 2: + (6: + 7) cos 21: 655% 25. y” + 332’ + 2y : e'(r2 + 1) sinZI + 36" cost + 4e“ 3 5 Variation of Parameters PROBLEMS 189 . where IQ is any cenveniently chesen point in I. The general Solution is __ . y": emit) +-c'z'y.:3cr) +‘ Ya); ' ' ' ' _ -_ "(29) I as prescribed by Theorem-3.5.2. - I. - By examining the expression (28) and reviewing the process by which we derived it, we can see that there may be two major difficulties in using the method of variation of parameters. As we have mentioned earlier, one is the determination of y1 (t) and y2(t); a fundamental set of solutions of the homogeneous equation (18), when the coefficients in that equation are not constants. The other possible difficulty lies in the evaluation of the integrals appearing in Eq. (28). This depends entirely on the nature of the functions y1, 32;, and g. In using Eq. (28), be sure that the differential equation is exactly in the form (16); otherwise, the nonhomogeneous term g(t) will not be correctly identified. A major advantage of the method of variation of parameters is that Eq. (28) pro- vides an expression for the particular solution Y(t) in terms of an arbitrary forcing function go). This expression is a good starting point if you wish to investigate the effect of variations in the forcing function, or if you wish to analyze the response of a system to a number of different forcing functions. In each of Problems 1 through 4 use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answer by using the method of undetermined coefficients. 1- y" — 5y' + 6y = 26' 2- y” w- y’ — 2y = 26“ 3. y” + Zy’ + y = 33” 4- 4y” W 4)” + y = 166’” In each of Problems 5 through 12 find the general solution of the given differential equation. In Problems 11 and 12, g is an arbitrary continuous function. ' 5. y”+y=tant, 0<t<rrf2 6. y”-l—9y=9sec23t, 0<t<rr/6 7. y” + 4y“ + 4y = file—2‘, I> 0 8. y” + 4y = 3csc2t, 0 < t < 3172 9. 4y” +y = Zsec(t/2), —-J‘I < t < 7: 10. y" — 2y' +y = e’/(1 + t2) 11. y" — Sy' + 6y = g(1) 12. y” + 4y 2 g(t) In each of Problems 13 through 20 verify that the given functions y1 and 312 satisfy the corre- sponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and 20; g is an arbitrary continuous function. 13. t2)!” — 2y 2 3:2 — 1, t > 0; y1(t) =12, ygU) = r"1 14. tzy” — t(t + 2)y' + (t +2)y : 2:3, r> 0; y1(t) = r, ygu) x 123‘ 15. ty” — (1 + t)y' +y = 1262’, t > 0; y1(t) = 1 + t, 3120‘) x e' 16. (1 — t)y” + [32' —y = 20‘ —— D22“, 0 < t < 1; 311(3): e’, yg(t) = t 17. xzy” — Sxy’ + 4)} = 2lnx, x > 0; 3210:) :13“, y2(x) =x11nx 18. xzy” +xy’ + (2c2 — 0.25))! = 3x312 sinx, x > O; " - - y1(x) = if”2 sin 3:; y2(x) = if”2 cosx .. 19. (1 —x)y” +xy' my = g(x), 0 < x < 1; y1(x) 2 ex, y2(x) = x 20. xzy” +xy’ + {x2 a 0.25)); = g(x), x > O; y1(x) 2 x‘”2 sinx, y2(x) = fl” cosac l l l i 3 physically tcurrent in ne form as 1d example :ond order 1 in terms of i at leads to l :pression in [1. 1e pesition oosition? _ y of 2 NS. itional3ia. nass at an)’ ii irium POSi'- ontracling I " rmine ”19 i Q10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 3 Mechanical and Electrical Vibrations 203 W A mass weighing 16 1b stretches a spring 3 in The mass is attached to a viscous damper with a damping constant of 2 lb- sift. If the mass is set in motion from its equilibrium position with a downward velocity of 3 in/s, find its position u at any time 1‘. Plot u versus 1. Determine when the mass first returns to its equilibrium position. Also find the time r such that |u(t)i < 0.01 in for all t > r. A spring is stretched 10 cm by a force of 3 N. A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 N when the velocity of the mass is 5 1111's. It the mass is pulled down 5 cm beiow its equilibrium position and given an initial downward velocity of 10 Cfl'i/S, determine its position it at any time t. Find the quasi frequency pi and the ratio of p, to the natural frequency of the corresponding undamped motion. A series circuit has a capacitor of 10‘5 F, a resistor of 3 x 102 Q, and an inductor of 0.2 H. The initial charge on the capacitor is 10‘6 C and there is no initial current. Find the charge Q on the capacitor at any time t. A certain vibrating system satisfies the equation it” + yu’ + u 2 0. Find the value of the damping coefficient 3/ for which the quasi pe...
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