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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,tl],[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,r32] .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 phaseplane 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
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2 I.
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s 915 44
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a mm ,
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it
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R h = =
1mm m r
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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’ phaseplane trajectoty graph
shown in Fig. 42.16. HINT: Keep in mind that the yoxis 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, LongTerm 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 differen»
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 HigherOwe: 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 pornair, so yan can use those
same nnmberzs :0 identify :lnz curves in {he component solu—
tion graphs 0:: each phaseplane 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 dieoff; 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 mussspring 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. LRCCircuit 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 longterm values of the charge and
current‘? out The EulerCauchy Equation A weli—itnown linear
secondorder 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 EnterCauchy 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. EulerCauchy Equations with Distinct Roots Obtain, fer
t > O, the general solution of the EulerCatich 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 EulerConch)! 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: PhasePlane Trajectories Each of theﬁmcrions
in Problems 7074 is the solution ofo linear secondorder 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 EnterCauchy 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 7679 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 secondorder 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 lefthand 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 ﬁrstorder equation nttlv' + 3013*: [Milﬂlildl + C. which could then he solved by first—order methods (a) Differentiate the righthand 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 secondorder differential equations. to find it one order differential equations with constant coefﬁcients de—
ed must solve a new secondorder 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 twodimensional solution space
for the linear secondorder 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
.. air3 + 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 _,§_ {CREWI'd)!‘ (3) ...
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 Fall '07
 RickRugangYe
 characteristic roots, tlte diﬁ'erenn'ol equa

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