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Unformatted text preview: 451 ..——'—" 45 Mk+la Ebb HN 1335? SeineHomg 2 1
v’ = Av. Since trial} = 2 and detlfA‘] = —3, the characteristic equation is H—Es—S: ﬂandthechatscteristicrootsareaand —1. Le'th1=[:J
boacharacteristic vectorbelongtngtoﬂ. Thenh+2k= 3h 2h+k= 13. Letv=[:]and.4=[l 2]soﬂtatoursystemisequivalcmto Theseequationsreducetoh=h.sowecan takeh1 = [11 ] Now suppose that lag = [ '2 1 is achsractcrisn’c vector belonging to —1. Then is + 2!: = —h,2h+k=—k.sothath=—k.Hencewecansetb2 = _11 .The getteralsolutiontoﬂtentsoixequauonisv=c1s3‘[: ]+c2e"[ _11 ]. or :t: = e1s3'+ sacﬂy = cps!"t — ego" 5 2
IT. I..L‘.1211=:2 1 tr {.4} = 6 and deth} = 9, the characteristic equation is s2 — {is + 9 = ii
and which has the douhie root 3. Since A as 3!. the matrix A does not have two independent characteristic vectors. Pate: [3].Iliﬂﬂﬂ£}'tﬂ§ﬁﬂmﬂtﬁ{27& —2c;thuscisnota so that our system is equivalent to v’ = Av. Since characteristic vector. Set when ] [3 ]= [i 1 Then at; {t} = ski: and Vg[¢}= eatjtb + c} are independent solutions 2: +1
The general solution is v = elem [2 2 ] + see“ [ 2: ] . Subsn'tuting vfm=[:;].wohavc251+ﬂa=1.2c1=—2Th“scl=1*C2=L Thereforethcsolutionofthe 1Wisv = 4.31:] +e‘3‘[ 2t Elf—1
w[m_2} n+1]: 4 —d.
v; 2 Av. Since REA} = [I and deth} = 15 the characteristic equation is
32 + 16: Iii and the characteristic roots are side. 9. Letv=[:]andﬂ=[4 _3]soﬂ1atoursystemisequivsientto W‘H’L 20]} ~1+N #:l— Embedtuna
4351 conﬂ 44 Ill,.—._ To ﬁnd a characteristic vector 1:: = [ '2 1 {it and it will be omnplen ntu'nhers} 4 —4
or 4h — 3k 2 [4£)h.4h — 4k — Mﬂk. Each of them equations reduces ms = {1+th. 'Ihus wewill puth = [1” 1
complexvalued solution is . 8.3.1:]: = [[1+i}[coa{4t]+isin[4tj}]
coastal] + saints} coslfdt] — unto} J +1. [ einﬂt] + must} holes[gingtos=iii.useneedtosoIIFe[4 _3H::] =[4f}[:], . The corresponding coe[4t] sinHt] The real part and the imaginary part of this solution are themselves solutions of the system, and in fact form a fundamental set of solutions. Therefore. the
general solution is _ cosfnltj— ' {4t} 5' 4t 4t
. — «[ out M ”is?“ J] [ (c; + £2] cosfdt] + [o2 _ solute)
c1 coeﬂt] + cs sinfdt] a I.
1. TheeharacteristicequationofA: [g %]iss3—3s+2=ﬂ;
hence the characteristic roots are 1.2. We can use characteﬁstic vectors ct = [ _11 ] node: = [ i ] belonging to I and 2. respectively. to obtain Eziﬂg. Thus .3513) = independent solutions “IF11:3} = sin; and vﬂt] gt Est
—e' E2, is a fundamental matrix solution.
I
5 X”) = 3:; Z; J is afundamental matrix solution of the associated
homogeneous system.
{a} Sﬂlfp“) = X{i]w[t]. It follows that .'l;'(s}'w’[tj .—. 9* l 1 J ,or
e‘wi + smut; = e‘
hello: + 331w; = e‘. The solution of this equation is uni = D. is; = e“. or of“) =
[33; j . Thus wﬂ} = [ F24 ] , and we have the particular solu sen;5:][_3_.]=[;::]. The general solution is v,[t] + v1.3), where visit} = XEtlc denotes
the general solution of the associated homogeneous equation. ‘trt . W'l'k 'ZﬂD Mia‘si ShiLuHEL’iS Laud— {bl 54'.th = Pt“{t}w(t}. It follows that fitlw’itj = et[ 11 J 1 or
Efw; +Emw; = E: It '21 r
—E'w1+e in? = _et_ 'Ihesoiutionofthisequatinnjsiyi = 1.1”; =U.OIW"EIJ = [3] t
D weir; :iHSHEii "the general solution is split} + wit}, where its“) = {Fit}: denotes
the general solution of the associated homogeneous equation. «new Set wit} 2 Ft'qftlwit]. It follows that Xithr’t't} = te‘ [ l ] ror Tltus wit} = [ J , and we have the particular solution [I
e‘uri + saw; = tst
—e‘w’l + timing. = D.
The solution of this equation is in; = 9’2, = te“,’2, or #{t} =
”2 Thus are) = ‘2“ and h the
iiiifs ' —%E_t[1+t:l ’ we we Far" ticttlar solution _ e' e“ tint _ {sift—t 2—12e'
v13“) — [ _Et E2: J [ _i.e¢{1 + .1} j — [ {—tzf‘ltfﬂ —{X21}et ' 'I'l'tegenera] solution is vpit} + shit}, where whit} = it"{tje denotes
the general solution of the associated homogeneous equation. III}. {a} Lettf = [:1 aotlA =[_11 __11 ] 'I‘heotlren'tetrixforrnoftl'le systemis
t'1
f—
? Fﬁv+[ ﬂ 1 Sinoetrisl}=1+[—1] = ﬂanddetfﬂ}={1][—1]—{1][—l}= I}.
thecharacteristicetntatienofAiss2 = ﬂanditfollowsthats = D is a
double characteristic root. Since A 7E I]  .l'1 the system v“ = Av will have independent solutions
Gfﬂlﬂfﬂfﬂ1V3=ib+CaﬂdV2 = h.wherecisanoneeroseetor that is not characteristic and b = Ac [Ir is a characteristic sector}. Put 1
ti: [a];itiseasytoseedtatAcaéilc.Thenh= [_1].The 1 1+1
fondarnentalrnao'ixis Pitt} = [ _1 _: 1 PipeH, 223]}: HM 4111:) Salk{items
Arc}. [omit Lottk. Set v,{t] = X{t}w{t]. It follows that r't'ﬁtjw'ﬂ} = [ It: ] , or wi+£t+tjw§ = r1
—w{ — m; = o.
The solution of this equation is w{ = —1.w§ = t‘1.or w'itj =
[Ii ].Thus surfs} = [ Int—lil ] ,anci we havethepanieular solution
_ 1 t+1 t _ —t+[t+1}lnt1
‘3'“) ' [—1 —t Hum] _[ t—tlnltf 'Th‘
generai solution is up“) + vhftL where vhﬂ) = .«1’[t}a denotes the
general solution of the associated homogeneous equation. ...
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 Spring '10
 Edgars

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