4.2 - 223 Chapter 4 E Higher—Orde: Linear Differential...

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Unformatted text preview: 223 Chapter 4 E Higher—Orde: Linear Differential Equations 4.2 Problems Rea! Characteristic Roots Determine the general solution: for the dtfierenttol equations in Problems ]~l-’l 1. y” = 0 2" y” — )t’ = 0 3t 3!" w 9y m D 4. y" w y = 0 S. y” — 33" + 2y 2 0 6. y” — y’ —— 2v = 0 '7. y” + 2)” + )1 r: (J 8. 4y" — 4y’ + y m {I 9. 2y“ - 3)” + y = 0 10- t” - 6}" + 9y w 0 11, y” -~ Sy' + 16y : O 12» y" — y' — 6)r = O 1.3. y” + 2)" — y = 14“ 9y” —§- 6.)” + y u 0 Initial Vaiues Specified For Problems 15—22, solve the initial—value problem. 15. y” m 25)! = 0, )'(U) m 1, )"(D) = 0 16. y” + y’ — 2y 2 0‘ y(0) = 1, y’lO) m 0 17» y” + 2)” + "v = 0, #0) m 0, H0) 2 i 13. 3'" - 9y 3 0, MO) 2 — E, y’(0} = 0 19 y" u 5}" “11* 9y = 0» NO) = 0, WC) = —1 20. y” + y’ — 6y 2 0, y(0) = 1, M0) : 1 21- )r” - t’ w 0. NO} x 2, J"(0) 2 —I 22. y” -~ 43" — 12y m 0, MO) 2 I, y’(0) = w} Buses and Solution Spaces For each of tlte difi'erenn'ol equa- tions in Problems 2.3«26, give a basis and a solution space in terms oftlte be its 23. y" — 4y’ 2 0 24. y” —» 10y'+ ZSy m 0 FIG U RE 4 .2 ~3 Graphs to son for Problem :32, + 55: + 6x : 0. For eoc . (a) M 25. 5y"~—10y’— 15ym O . t _ tlt 26“ y” + 2J3)“ + 2y = 0 ‘ at (In) W y: Other Bases Use the Solution Space Theorem to show that (c) D the sets given in Problems 2 7 and 28 are each o basis for the DE 16 27. y’—4ym0; {crate—2’], [cosh 2t,sinh 2t], [ez’mositZt] 33 WHO: {1.t},[t+1,t-l],[21.3t—l] The Wt‘onskian Test USE the Wronskt'on Test in Prob— lems 29—31 to determine if the set of solutions it ct bow": for the given DE 29. 3th mo, {1 +1,r*1.r3+r,t3) ym_10yu m 15),: m ’ {IEMSt’ 65:, 285! ~ FEGURE 3L 3'”) =0. {r+1.t3~+2r,r3-2] .32“ Sorting Graphs For the DE, + 52% + 61' m 0 of Ex~ ample i, Fig 42.8 adds to Fig 4 2.} the Hoked solution graph for it t) Label the phase-plane ttajectories from left to tight as A, B, C, D, E. Then attach the same iabets to the appropriate linked solutions xlt) and in), Relating Graphs Problems 33—35 give lt'nkedsolutt'ott graph; and a phase portrait for a single particular“ solution to + 53': + 6.x = Ofi'om Example 1. Problems 36—39 give linked solution graphs and a phase p01“!t'flllf0r‘€l single particular solution to - i‘ —— 6x m 0 (See Problem 12 ) FIGURE l l l I J y l n ‘.._--m_ FiGURE s 2 w 7“ 1W... 2 I. l l S r. 2 t‘ m .x 4 m m U 135 r“. .. 1] .m mum. Kw 22 lat ,( s 915 44 .H Hm New.” e Wm. 0 “Fr... t rin 1 . . C a hm 2 r d. a mm , h 5v. .u. C mm F. it t I 9".va a a.” 03 .. 6 [Mary 0 R h = = 1mm m r a .ne «24 cs 9. 6%.}... a.” 1W2 _22 2 EM 5,4 $.44 4 l + g _ gnu d .. .1 . Hi n (\ .JF .X F 0 ma.“ mfiom H 51 F33 . 0 0.. C O z m. e __ S t = 0) for the trajectories shown, and odd onowsfor iicit formtiiafrom part (b) the direction 5; of each trajectory Write down the initial conditions “15(0), ti:(0)for‘ the phase—plane trajectory. your initial condition to solve the IVF: (c} Describe how the graph for the solution x( t) relates {a} Mark on the phase portrait the starting points“ (where {[3} Write the explicit sohuionsforxfl) andifi), then use For each Problem, relate the given graphs oxfih'lom ' FIG U RE 42. T 0 Graphs to relate for Problem 34. + 53': + 6x FIG U RE 4 .2 "1 1 Graphs to relate for Problem 35, 4» Si «k 6x FIG U RE 4. 2 .9 Graphs to relate for Problem 33. 2% + if + 6x W e r“. t i 0 t {0 mm re 1. hum flmfit .m: 4.115 1 mm M. mi {Mmd mm. 5 .1 MW m pa, ) emumw m; 1:; fl ll... mama m Hm 3.1 a». 0.1 S I. .w 0 5 need I "are If. fr 9.5.“. 11er Ffllhfi flew 0M m” mee PPgd . {E 9 .r........... Jud“! C Ctrlx 10 BE D v, 4 7. 4 b1 0.U.H f y B n e.m a H,” n 03h n Dumx.$ m 9.“ _ WW 8 h“ I .u m 4‘ H; m hm r‘ E Co 1.AW r. U 11 B “5.! fl 5 A“ W.C I . n t .L r. l n (u w G MWMMm 2 M fl 4 4 H P.4flfl v W 4 M . H n” . H 5 fl p . . I, p 4 n .I. H .x u n . .I. H H o x. T , - - t -T . H- HWHH :H w- -- “H H H -- “RUNWJ Flu” H H EHHH H” H“- EHHHHH H :HH KHHHLH” HHHHHH ,HHHH r---------- lwrrr--.--- ---- -p-------.; - .-;r,- -r.-----.., ---.‘r-.--- --,‘,,r- ==0 fferential Equat 6x -—6x =0. .x—x— ,J'éwx near Di —).—6x=0. _ "x 1 Order L H ig h e r- s to relate for Problem 39 1 5 Graph Chapter4 a 2. F l G U R E 4 V2 "1 .7. Graphs to relate for Problem 36 F l G U RE 4.2 41 3 Graphs to relate for Problem 37 FIG U RE 4“ 2 .1 4 Graphs to relate for Problem 38, FIGURE 4. 224 Secrion 4 (A) (B) F l G U RE 4 .. 2 .1 6 Phase portraits to match to ?rohlerns 40—43 Phase Portraits Match the ty solution graphs in Problems 40—43 to the corresponding yy’ phase-plane trajectoty graph shown in Fig. 42.16. HINT: Keep in mind that the y-oxis is vertical in one graph and horizontal in the other. You may also want to think about what the ty' graph would look like. 44. 45. 46" Independent Soiutions Show that if n and r2 are distinct real characteristic roots of equation (1). then the soiotions e’” and 9’“ are lineariy independent. Second Solution Verify that if the discriminant of' equa- tion (1) as given by A m b?- —«~ 4m: is zero, so that h2 40c and the characteristic root is r ~—~%, then substi— tuting y m nun—“W” into (1) leads to the condition v”{t) u 0 Independence Again In the “repeated roots" case oi equation (1), where A = b2 w 4a.: = O and r 2: F234,, show that the solutions e’wml‘ and te‘lb/Zal’ are linearly independent 47. Repeated Roots, Long-Term Behavior Show that in the “repeated roots" case of equation (1), the solution, which is given by .r(t) = cle"(b/2“)’ «t» Cate—WM“, for 5% > 0, tends toward zero as 1 becomes large. HINT: You may need l‘I—lopital’s Rule: If, as .x approaches a, both fix} (C) 48. 49. 50. 51. 52.. Real Characteristic Roots 22,5 and g(.x) approach zero, then the hints,“ _f(x)/g(,x) is indeterminate But Marquis l'i—iopital (1661~1704) came to the rescue by pubiishing a result of Johann chouili (1667—1748}, that we can find the iimit of the quotient by . W) M . f"(x) if; g(..r} _.l{§z11 g'(.x}' providing, of course, that both derivatives exist in a neigh» horhood of n and approach nonzero limits. Negative Roots Verify that in the overdarnped mass‘ spring system, for which A w. bl —~ 4:71]: > 0, both char— acteristic roots are negative. Circuits and Springs (a) What conditions on the resistance R, the capacitance C and the inductance L in equation (1 1) correspoad to overdarnping and critical damping in the mass~spring system? ($3) Show that these conditions are directly analogous to 13> «Mink for overdarnping and bmwtmk for critical damping for the mass—spring system. Use Tabieélifl A Test of Your intuition We have two curves. The first starts at y(0) = i and its rate of increase equals its height; that is, it satisfies y’ : y. The second curve also starts at y(0) = 1 with the same slope, and its second deriva— tive, measuring upward curvature, equais its height; that is, it satisfies y” = y Which curve lies above the other? Make an educated guess before resolving the question analytically. An Overdamped Spring The solution of the differ-en» tiai equation for an overdamped vibration has the form fit) = o'er" + oge’“. with both :1 and c; nonzero. (a) Show that x(t) is zero at most once (b) Show that aim is zero at most once A Critically Damped Spring The soiution of the differ- entiai equation for a criticaliy damped Vibration has the form xlt) m (5'; -§- £205”, with both cl and C3 nonzero (a) Show that xlt) is zero at most once. (13) Show that :‘c(t) is zero at most once. 226 Chapter 4 B Higher-Owe: Linear Differential FIGURE 4,2.17 Graphs to iink for i’roblem 53 FIG U RE 4 .2 .1 9 Graphs to link for Problem 55 Linking Graphs For the sets of Iy, :3”, and )Iy’ graphs in 56. Problems 53—55, march the corresponding Irajccrorics They are nmnbered on Ilia phase porn-air, so yan can use those same nnmberzs :0 identify :lnz curves in {he component solu— tion graphs 0:: each phase-plane Imjecmry, mark the point where I z 0 and add crn'mrrhends :0 show (In! direcrion of motion as r gem larger: 53“ Fi ([9 42:7 54. Fig 4213 55. Fig 4219 Dampecf Vibration A smail objecl of mass 1 slug rests on a frictionless table and is attached, via a sailing, to the wall The damping consiani is b m 2 lb sec/ft and the spring consmnz is k = I lb/ft A: timer = 0, the object is pnliecl .3 in to ihe rigiil ané i‘cieascd Show :hat the mass times not overshoot the equilibrium pcséiion at .x m 0 Surge Functions The lunctionxfl} m Are‘“ can be used {0 model events for which there is a surge and die-off; for am: hi ghly i the soil Assure: and 5:{( the soil 59. [RC— age 51 and a capac (a) B c: (l3) 5 fr (C) E 3111:: Euler (6 g , fly”) Section 4 2 Real Chaiactetistic Roots 2.27 to: example, the sales of a “hot” toy or the incidence of a highly infectious disease. This function can be obtained as the solution of a muss-spring system, mft? + bi: + hr = 0. Assume m = 1 Find b and k and initial coaditions 37(0) and nit-(0} in terms of parameters A and i" that Would yieid the solution shown in Fig‘ 4.2 20 FIG URE liuZAZG A particular solution to the WP of Prohlem 57 58. LRC-Circuit I A series LRC~circuit in a power grid has no input voltage, a resistor of 101 ohms, an inductor of 2 henries and a capacitorof'O 02 farads Initialiy, the charge on the capacitor is 99 coulomhs, and there is no current (See Fig. 4 2.6) (:1) Determine the IV? for the charge across the capacitoi (b) Solve the IVEJ in (a) for the charge across the capacitor fort > {it {c) Determine the current in the circuit for t > 0 (d) What are the long~tcrm values of the charge and current? Series Circuits This tool provides heir) in visualizing the resuiting cuwent and Iong«terrn behavior in such circuits 59. LRCT‘Circuit II A series LRCmcircuit with no input voib- age has a resistor of 15 ohms, an inductor of 1 henry, and a capacitor of 0 GP. farads Initially. the charge on the capacitor is 5 coulombs, and there is no current (a) Determine the {VP for the charge across the capacitor (b) Solve the WP in {a} lot the charge across the capacitor for: >- 0 (c) Determine the current in the circuit for t > O (d) What are the long-term values of the charge and current‘? out The Euler-Cauchy Equation A weli—itnown linear second-order equation with variable coefficients is the EuierbCauchy Equation3 may” + bty’ + or = 0i I > 01 (14) where o. b. c E R and a 3i 0. Show by substituting y :1" that solutions of this form are obtained when r is a solution of the Enter-Cauchy characteristicgquaiion or(r-~1)+br+c m0 (5) Then verify that if rl and r; are distinct solutions of ($5), the general solution of (14) is given by y{t) m :1!" +C3r’3. r > 0, int arbitrary ct, Cg E R. Euler-Cauchy Equations with Distinct Roots Obtain, fer t > O, the general solution of the Euler-Catich equations in Problems 613454 61. 12y" + Zty’ — 12y = 0 62¢ Lirzy” + Sty' — By M O 63“ t2)!” + 4ty’ +— 2}! = 0 64. ley" + 3ty’ — y = O 65. Repeated Enter—Cauchy Roots Verify that if the charac— teristic equation (15) for the Euler—Cauchy equation (14) has a ['B§€fii6d teai root r, a second soiution is given by t’ lot and that r’ and r’ in: are linearly indegendent. Solutions for Repeated Euier‘~Cauchy Roots Obtain, for t> O, the general solution of the Euler-Conch)! equations in Problems 66—69 66“ t3)!” + Sty’ + 4y = 0 68" 9I1ylr+ 3iy’ + y = a 67. :33)” n 3iy' i 4y = o 69. atrzy” + 8ry’ + y m 0 Computer: Phase-Plane Trajectories Each of thefimcrions in Problems 70-74 is the solution ofo linear second-order dif- ferential equation with constant coefiicients in each case. do thefoiiowr‘ngi {a} Determine the 13.6 (b) Calculate the derivative y’ and the initial condition M0}. Viol (c) Plot the trajectory [310), y’(t)} on the vectorfieid in the yy’—plnne JThe Enter-Cauchy equation can be recognized in standard form because the power oft is the same as the order of the derivative in each term (a g , 13 y”) 228 78. 71. 72” 7.3, 74. 75. Chapter 4 y“) = 2e" + 9'“ Vi!) = e" + 2*“ W) m e’ + e“r W) = e" + re" y{!) m 3 + 283' Reduction of Order4 For a solution y] of y" + p(x)v’ + thlv = 0 (I6) on interval 1 , such that yl is not the zero function on I, use the following steps to find the conditions on a function u 0er such that .)’2 = 11.))! is a solution to equation (16) that is linearly independent from v; on 1. (a) Determine yé and y; and substitute them into equa- tion (l6). Regroup and use the fact that y] is a solution of (16) to obtain nu” + (23’; + pmv' = 0- 0)} Set 12’ = to. Solve the resulting first~order DE to obtain 6—] p(.x)dx u m :i: / —-—————2—d.x (t7) . .3’1 so that ew p(.r}d: y2 = M / —3w—dxr . .yl Establish the fact that {y} , 372} is a linearly independent set by showing that 1) cannot be a constant function on I HINT: Show that u' cannot be identically zero on I . (c) Reduction of OrdergSeeond Solution Use the steps or fire formula for ya developed in Problem 7.5 to find a second linearly independent solution to the secondmm‘er difibi‘fl!“ u'al equations of Problems 76-79 for which y; is a known solution. HINT: Put the DE in standard form before using the formula. ‘ 76. y" — 6)" + 9y = 0. y] = 83’ 77. y" m 4y + 4y 2 0. y, = 81’ 78. 79. Higher—Order Linear Differential Equations rzy" - 1y' + .v = 0, .n m I (r2 +133’H" 3r)” + 2y = 0. 3’1 L“! Classical Equations The equations in Problems 80—82 are some of the most famous dificerential equations in plrysicr.5 Use 517 ’Alemberr '5 reduction of order method described in Prob- lem 7.5 along with the given solution 3!; to find a second solution 3’20). HINT: Be prepared for integrals that you cannot eval- uate." Those answers should be left in terms of unevoluated integrals. 89. 81. 82. 83. y” — Zry’ + 4y = 0, (Hermite’s equation) MU) w E d 29 (1 w Ills” — W‘ + v = 0, 3’10) W! (Chebyshev’s equation) = Chehyshev’s Equation ‘ Graphical solutions give more insight than more formulas Iv"+(1~!)y’+vmfi, 3’10) 2 l “*3 (Laguer're’s equation) Lagrange’s Adjoth Equation The integrating factor method, which was an effective method for solving first- order differential equations, is not a viable approach for solving second-order equations. To see what happens, even for the simplest equation. consider the differential equation yn+3yr+2y : Lagrange sought a function Mr) such that if one multiv plied the left-hand side of (18) by Mr), one would get (13) d m tutthwstthtl. (19) d: where gm is to be determined. In this way, the given dif— ferential equation would he converted to mmfl+f+fl= d Elutrlv’ + girls] = WM“). which could he integrated, giving the first-order equation nttlv' + 3013*: [Milfllildl + C. which could then he solved by first—order methods (a) Differentiate the right-hand side of (19) and set the coefficients of y, y’, and 3'" equal to each other to find gm. “The reduction of order method of solving second~order DES is of long standing. and is attributed to French mathematician lean to Rond d’AIembert ([717w1783) SThese classical equations ofphysics were named for Charles Hermite {1 822—4 90] ) and Edmond Nicolas baguette (1834—1886). aiso Frenchmen. and {or Russian mathematician Pafnuti Lvovich Chebyshev{1821~1894). Section 4.3 Complex Characteristic Roots 2.2.9 . .g ____.nm__#. 9—4 (b) Show that the integrating factor pm satisfies the (c) Show that the adjoint equation of'the general second- second—order homogeneous equation order linear equation y” + plrly’ wt atrly 7» it!) re ll” _ 1L" ‘l‘ n” = 0: is the homogeneous equation is 14”“ 52min” + [rim -~ p’trilu m 04 lb. called the adjoint equation of (18) In other words, on aithought it is possible to find an “integrating factor” 84. Suggested Journal Entry The theory of linear second» 31. for second-order differential equations. to find it one order differential equations with constant coefficients de— ed must solve a new second-order equation for the inte~ pends on the nature of the solutions of aquadratic equation. grating factor Mt), which might he every hit as hard Give other examples from precaicuius or calculus where as the original equation. (In Sections 4 4 and 4 5. we you have met a similar classification based on the sign of m) will develop other methods ) the discriminant of a quadratic. m) 4.3 Complex Characteristic Roots S YN OPS! S: We complete the description of the two-dimensional solution space for the linear second-order homogeneous differential equation with constant coeffi- cients for the case where the roots of the characteristic equation are complex num- bers with imaginary terms These solutions exhibit a variety of long—term behaviors, including periodic motions and damped oscillations on) Real and Complex Solutions ctor :_ In solving the linear second~order homogeneous differential equation with con- ‘nst— .7 stant coefficients, 2 for ' i 5- ens, ny" + by’ + cy = O, (l) atial in the case of complex characteristic roots, we will encounter (nonreal) complex» valued solutions. It turns out, however, that the real and imaginary parts of these (18) .. objects are also solutions and are, in fact, the actual real solutions that we want. .ulti- :_. Of course, we can aiways just verify directly, by substitution, that the real parts or at .' the imaginary putts satisfy ( 1). But there is a general principle that can be checked too. If u(r) + i v(t) is a solution 013(1), then u(t) and utt) are individual solutions (19) as well, because ‘élf' a(u —§— iv)" ~l~ Mu + iu)’ + C(u + iv) = (au” ~§- 1311' + car) + itav" -l— bu' + cu), and a complex number is zero if and only if both its real and imaginary parts are . zero. anon '55 'x In the previous section, we studied solutions of equation (1) for the cases in 3" which the discriminant A m £22 —— 4nc is positive or zero. We now complete the discussion with Case 3, supposing that A < 0 h Case 3: Complex Characteristic Roots (A<O) at t e 0 find When the discriminant A = b2 w Linc is negative, the characteristic equation .. air-3 + br + c = O has the complex conjugate solutions Int/E b.J:E Road I r1=“:2"£+l 2a marl—rig and T'2="“‘2—(:—£ 2a The general solution can be written zawifi, (2) :hmeni y ______ kle(a’+lflll _,§_ {CREW-I'd)!‘ (3) ...
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This note was uploaded on 04/07/2008 for the course MATH 5A taught by Professor Rickrugangye during the Fall '07 term at UCSB.

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4.2 - 223 Chapter 4 E Higher—Orde: Linear Differential...

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