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4.3 - 2.38 Chaptei 4 33:3 Higher—Order Linear...

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Unformatted text preview: 2.38 Chaptei 4 33:3: Higher—Order Linear Differential Equations j Going On and On Consider the fifth~order equation (15)) d4): d3}; d2): (115+ dr" + (1:3 + ctr? The characteristic equation is 5 +3f‘4+3r3+f2 at), which factors into (r + 1)3r2 :2 0 with a tripie root r 2 —i and a double root r 2 O. The sointion is 3’ m (C1 "i- Czr + C3f2)8~r ~§~ (:4 + c5: . for triple root for éoubie root Repeated Complex Roots Consider 32(4) "i“ 8y” + 16)) m 0.: The characteiistic equation is r4 + 8;"2 + 16 x 0, which can be factored: (r2 + 4)?— = 0, yielding repeated compiex conjugate roots r = :iz2i, £211 The solution is y =(c1+ (:31) cos 2: + (53 + at) sin 2:. We have completed the solution of the second—order linear homegeneous differential equation with constant 1‘ coefficients for all cases of the characteristic roots (or eigenvalues): real and distinct real and equal or complex _:;_: conjugates The general solution of the equation generates a two— dimensional vector space in all thiee cases __ The resuits are applied to overdamped, criticain damped, and underdamped vibrations foi‘ the damped harmonic osciliatot. 4.3 Problems Solutions in Genera! For Problems 1—10, determine the gen— 7. y” w my —|— 26y m D 8. 332" + 432’ + 9y = 0 am! solution and give the basis B w big, y;) for the satiation space 9. y” — y’ + y m 0 10. y” + y’ ~i~ 2y = 0 L y” + 9y = 0 2. y” + y’ + y = 0 Initial-Value Problems Solve the NP; in Problems 11—16. 3. y” —— 4y’ + 5y m G 4. y” + 23” + 8): m 0 11. ya: + 43: m 9‘ M0) = 1' fat) m S. y” + 232’ «1» 4y = 0 6“ y” _ 432’ + 7y = 0 12. y” _ 4y’ +13), = G, y(0)=1, y’(0) m Section 4 3 Complex Characteristic Roots 239 13. y” + 23” + 2y m 0. MO) = I, y’t0) = 0 14. y” - y” + 3' = 0, MO} = 0, y’tO} = l 15. y” w 434’ + 7y = 0, MO) m 0, WW) m —-E 16- y” + 2.)“ + 5y m 0, 31(0) = 1: y’tfll = -1 Working Backwards Write the standard form (equation (16) with leading coefi‘icient (1,, = I) of the nth order linear ho— mogeneous [lifter-timid! equation with real eoefiiciems whose roots are given in Problems 17—20 17. 3rd order, r m i, 1, l 18. 3rd ordet, two of the roots are r m 4, l — i 19, 3rd order, two of the roots are r = —2 + i, 2 +i 20. 4th order, three of the roots are 2, —2, 4 -§- 1‘ Matching Problem For Problems 21—28, determine which graph of the particular solution shown in Fig 439 matches each difi‘erentiai equation. 21. y”-—y’=0 22. y”+y’=0 23. y”+3y’+2y=0 24. y”—5y’+6y=0 25 27 29 v” + 31’ -i- y 2 0 26 y” + y = 0 y” +4y’ + 4y a: 0 28“ y" — .V’ "i“ y m 0 Euler’s Formula You can use the following process to justify Euler’s formuia a“? = cost? + i sin& to) Write out explicitly the first dozen or so terms of the Maciautin series (the Taylor expansion about the oxi- gin) given by n! ' "m0 (E3) The series is valid for both real and complex numbers Replace x by i9 and write the expression for 2‘9 to) Simplify the results by using the periodicity of'powers oi'i: jezi‘lmzsz =1, llm15=i9_ ml: ‘) . l 1'21621m— =wl, 13mi7=ili_ m—i (d) Collect the real and imaginary terms. (e) Obtain Euler’s formula by recognizing the two Maolaurin series that appear in part (d) (G) (H) F! G U RE 4 , 3 . 9 Particular solutions that match the differential equations in Problems 21w28 The dotted curves in (G) and (H) give the envelopes for the solutions, TICSTMTM * x_x_ t... H A __.. MT :5:..~‘.*n.'.i.‘::;sv?“. if?” m‘fl' : LEW-m: MA am. . rm 249 Chapter 4 Higher—Order Linear Differential Equations Long~Term Behavior of Solutions Suppose that rl and r; are the characteristic rootsfor ay” 4: by’ + cy = 0, so the solu- tion is y(!) 2 6:6” ~+ czar“ For Problems 30«-35, discuss the long-term solution behaviorsforihe given r1, r2 combinations Assume fl # O, 30» r1< 0,f'2 <0 31.1“1<0,r3 =0 32. r=o::|:5i .33ii";=0,i‘2=0 .34.r1>0,r‘2<0 .35.!‘mifii 36. Linear Independence Verify that e‘" cost}; and 2“” sin ,6: are lineariy independent on any interval. 37. Real Coefficients Suppose the roots of the characteris tic equation fo; (1) are complex conjugates or i iii, which gives rise to the general soiution y m k;elu+i5)’ -i~ [gem—'73”, Where In and k; are any constants (even com~ piex), Show {hat in order for the solution y(t) to be real, in and 162 must be complex conjugates 38. Solving d"y/dt" a 0 (a) Solve the equation day/d!“ m 0 (18) by successive integration, getting d3‘y/dt3 2 k3 and day/tit2 = k3! +k2 to obtain 320'). (b) Determine the characteristic equation for (18) and use its roots to find the generai solution. Compare this solution with the soiution you found in (a). (c) Generalize the process in (a) to solve d" y Mt" = 0. Higher-Order DES Find the solutions for the following higher-order equalions Remember thaifor each repetition of a root, a term with an additional factor of! must be included [fa factorization of the characteristic equation f0} m 0 is not obvious, look fora small iniegerq that satisfies f{q) = 0. Divide the characteristic equation by r .... an 39. y‘” — 4):“) + 4y” m 0 40. y'” + 4y” — 7)?’ —- 10)) = O 41. yo) _ y: 2 0 42. y’” —- 4y" + Sy’ —— 2y = 0 43. y'" + 6y”+12y’+ 8y = O 44. y”) — y 2 0 Linking Graphs For the sets ofry, ty’, and yy’ graphs in Problems 45 and 46. match the corresponding trajectories. They are numbered on the phase portrait. so you can use those some numbers to identify the curves in the component solution graphs. On each phaseuplane trajectory, mark the point where t m 0 and add arrowheads to show the direcn‘on of motion as t geis larger: 45. Fig. 4.310 46. Fig 4 3.11 FIG U R E 4 n 3.1 1 Graphs to be linked for Problem 46‘ Section 4.3 El Complex Characteristic Roots 241 (b) F i G U R E 4 . 3.1 2 Solutions and phase-piane trajectories for Problem 47. 47» Changing the Damping Consider the mass-spring system .‘C' + bit +1: m 0, MO) m 4, x(0) = 0. For damping coefficient 15 2 O, 0.5, 1, 2, 4, the corre- sponding solutions are plotted together in Fig. 4.312(a). Their phase-plane trajectories are plotted in Fig. 4.312(b). Make a trace of'both graphs and iabel each curve with the appropriate value of b. 48. Changing the Spring Consider the mass-spring system s+s+kx ma. x(0) at, are) :0 (a) For spring constant k m 0. 25, 0‘5, 1, 2, 4, the cor- responding solutions are plotted together in Fig. 4. 3. 13(a). Their phase-plane trajectories are plot- ted in Fig, 4.3.13Cn). Make a trace of both graphs and label each curve with the appropriate value of k. (13) From your observations of the graphs in Fig. 4.3.13, do the oscillations increase in frequency and ampli- tude as the spring constant is increased (Le, as the spring becomes “stiffer”)? Explain. 49“ Changing the Mess A mass—spring system has a mass m attached in standard fashion with a damping factor b ..—. O and a spring constant k = 16 (a) (b) (a) Discuss how the vaiue of m affects the motion. (b) How wouid the frequency of the motion be affected if the mass were doubled? (c) Discuss how much damping would he required for the critical damping if the mass were increased. 50» Finding the Maximum (21) For the mass-spring system for which at = l, b m 2, k = 3, and x(0) m 1, 4'10) = 0, find the maximum displacement attained. (HINT: Differentiate the solu- tion x(t) and set $0) an O to find the critical point.) (b) Do the same thing for m = 1, b m 2, k = 10, and x(0) = 0, 3(0) m 2 (underdampedl (c) Do the same thing for m m 1, b = 4, k 2 4E, and x(0) = 0, HO) z 2 (criticaiiy damped). Oscillating Euler-Cauchy E‘uler— Cauchy equations were in- troduced in Sec. 4.2 Problem 60; Problems 51—54 consider Euler-Cauchy equations with named characteristic roots. The solutions than have the final form 310‘) =t"{c1cos(;3int)+0;sin(filnt)}. (19) 51. Verify the solution (19). HINT: Use the reiation Iaifli m smasher F [G U RE 4. 3 .1 3 Solutions and phase-plane trajectories for Problem 48 242 Chapter 4 El i-liglier~0rcler Linear Differential Equations 52. Solve :2 ii” ~§~ Zty’ + y = l} (a) Choosing g/ L e: I, what is the motion of the inverted a pendulum for ..\'(0) = l} and .ilG) = l? 53" 3‘3”“? + 3’3’ “My =0 (in) Are thereinitialconditions that will make tiieinveited . , ._ '. m r) 54" Solve [2in + 170,, + 16y m 0 pendulum approach A _ .1 Gas 1 w-t‘ oo. 59. Pendulum and Inverted Pendulum For nnall displace— ments, whete sinx m x, the pendulum and inverted pen« dulum of Fig. 4.3 14 are modeled (setting g/L = i) by SS. Third-Order Euler-«Cauchy Use the substitution y = r’ for: >- 0 to obtain the characteristic equation for the fol« lowing third—order EuienCauchy equation: may," + btzy” «2» cty' + dy = 0 for r >- 0 :6 ~l~w .x = 0 (iinearized pendulum equation), .."-sm 1' ‘d' td -dl t' Third-Order Euler-Cauchy Problems Use the results from I Y O (meanze invere pen uum equaion) Problem 55 to solve the specific Euler-Cauchy equations of Let us examing [liege models in the language oi'linear Problems 56 and 57, algebra 56. 13y” +12y” —— Zty’ + 2y = 0 (a) Show that fundamental solutions of' the pendulum equation are e” and e‘” , while those of the inverted 57¢ r3y’” + 3(2):” -t~ Sty’ = D pendulum are a' and a"? , _ . . (b) Show that fundamental solutions of the pendulum 58. Inverted Pendulum Since the genezai SOlUEiDfl oi the ‘un- equation are cos I and sin r while those of the inverted earized pendulum equation it +(8/L)x = O is the class of pendulum equation are cosh: and sinhr (hyperbolic sinusoidal osciilations, for small displacements, the pen“ cosine and hyperbolic sine). dulom oscillates back and forth about its equilibrium point. The equation 'i W (g/le = 0 describes the inverted pendulum, (See Fig 4.3 1403)) (c) Are the solutions to both equations teal? Explain. 60. Finding the Damped Oscillation Determine the con— stants for the clamped oscillation x0) = e”'(c1cost + c; sinr), subject to the initial conditions x(0) = 1 and M0} = 1. Graph the solution. were {223v 61. Extremes of Damped Oscillations Show that the max- ima and minima of 3.x: .\ llama-w, a; . ._,_..i =. i .xtt} 2 emu, cos mt + C3 sin cut) for a < 0 occur at equidistant values of 1‘, adjacent values diffeiing by arr/co .— - q-n 'x'fll‘uil" g..‘ in: r‘ . .. Ill: use “131m“. .. 1.. q, “ n r-nAnT'M ear-r m: (:1) Stable equilibrium point 62. Underdamped Mass-Spring System Find and graph for t Halli-m0 the motion of a damped mass—spring system with mass m m 0 25 slugs, spring constant k = 4 Wit, and damping constant b = 1 lb secfft The mass is initially pulled to the right, stretching the spring by 1 ft, and then released .- . ‘_~1 634 Damped Mass-Spring System The motion of a mass— spring system obeys x‘ + bi + 64x 2 0, MG) = l. 5c(0) = 0 Determine Mr) and sketch the motion for (a) b = 10; (b) b = 16; (c) b = 20, Series Circuit Visualize the solutions for LRC—cit‘cuits (b) Unstable equilibrium point for if —- (gills : {l 64. LRCvCircuit I A sezies LRC—ciicuit (Fig. 4.3 15) has a Fl (3 URE 43.14 Pendulum and resistor of 8 ohms, an inductor of‘ l henry and a capaci— inverted pendulum for Problem 59 to: of 0.04 farads. The initial charge on the capacitor is v__.w._~..‘_mmmww, . Section 4 3 £21 Compiex Characteristic Roots 243 FIGURE 4.3.15 Aseries LRC‘mcircuit; the arrow for 10') indicates the positive direction for the current 1 coulomb, and there is initially no current in the circuit. Assume Wt) = Dior-r > 0: (a) Formulate the IVP for the charge across the capacitor: (b) Determine the charge across the capacitor fort > O. (c) Find the current in the circuit for: > 0. (d) What are the long-term values of charge and current in the circuit? 65. LRC-Circuit II A series LRC-circuit has a resistor of 1 ohm. an inductor of 025 hensies. and a capacitor of 0125 farads The initial charge on the capacitor is l coulomb, and there is initially no current in the circuit. Assume VU) m 0 fort > O. (a) Formulate the IV? for the charge across the capacitor: (b) Determine the charge across the capacitor fort > 0. (c) Find the current in the circuit for: >2 0. (cl) What are the long-term values of charge and current in the circuit? 66. Computer Lab: Damped Free Vibrations Improve your understanding of damped oscillations by working through Lab 9 of the 101?. software package, skipping over parts 37 and 2 5 on Energy Linear Oscillators: free Response Lab 9 provides a simple visual and visceral introduction Effects of Nonconstant Coefficients In our study of the clamped mass—spring system with mass in, spring constant k, and damping constant I), we have used as our model the second- ora’cr linear dijfierenrial equation mx‘ + bi + k3: = 0 having constant coefiicr‘enrs. When coefi‘icienrs change with time, the analytic solutions we have found for constant coejjficients do not work Explain why, Then, for Problems 67MB, consider" some DB in which m, h, and k change with time. (it) Use your intuition and/or a computer to describe the motion of the system under these changing conditions. (b) Use a computer to draw a solution xtr) for r > G, x(0) m 2. 13(0) 2 t). HINT: When you need to avoid t =2 0, try a trick like starting your plot or I m 0 it (c) Discuss what followed your intuition, t-r-lurr did nor, and whorfiu‘rher' questions you might now ask i 1 67¢ t “it“ “II E G 63. 't “i“ 7.1: +11.“ m 0 69. tuiE+.r=0 791 .t-t—(xzmltic-i-xmo 1 71¢ fic‘wi»(siur)5:+x=D 72.1“+}-5c+rx20 73. .t‘ + (sin2t)x = O Boundary—Value Problems Two boundary conditions flat} = b: and ytag) = b; can be used to specify the solution to what is now called a boundary—value problem, provided that the two conditions do not lead to a contradiction Find all solutions ofy” + y = O satisfying the boundary conditions in Problems 74-77. if the given hourulary condition leads to a contradiction, stare this fact explicitly and show that it is so 74- 1(0) 2 0. Mfr/'2) = 0 75« M0) = 0, Mil/2) m i 76. 31(0) m i, y(n) =1 77. 32(0) m 0, )l{JI/2} = 0 Exact Second-Order Differential Equations The difieren- rial equation 3/” + ptru’ t qmy = 0 is called an exact second—order“ equation if it can be tin-irrcn in aform that can be integrated directly, An example is y” + lgtoyi’ = 0. where glr) is determined from p(t) and q{t) This DE can be integrated directly to get afirsruorder linear equation, which in turn can be integrated using the integrating factor method For Problems 78430, solve the given exact equations. 1 1 78. y” + :31’ m fly = o (HINt:g(r)=1/r) 79 H 2' 2 »y+?)’—t—g)’ il 0 80“ (r3 — 20y” +4rr w my +2); m 0 (Hun-z Let gm = r3 m 2: and show that the left—hand side equals (3)21”) 81. Suggested journal Entry The subsection entitled “S urn- marizing Solutions for Real and Complex Characteristic Roots” and Problems 30—35 on longwterrn behavior specify certain types ot'charactetistic roots and tell how the solu- tion evolves, Summarize these outcomes using the various behaviors as categories That is, answer such questions as the following (a) When do solutions tend to zero as r —> 00? (b) When do solutions remain hounded but not tend to zero? (c) When do solutions osciilate in an unbounded manner"? ...
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