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# HW9soln - Mot-H1 2-5315 “N 44:51 Shim-item‘s E 4...

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Unformatted text preview: Mot-H1 2-5315 “N 44:51 Shim-item‘s E 4. Letﬁtftt} =y"+3y’+2y.Theoharuo‘teristicmtsofﬁ[y}=[lure-+1 and -2. Let us start by ﬁnding a particular solution to“) of ﬂy] = c“. Put m = Aeﬂ. Then ﬁlls-1] = Awe-3* + tiemt + 2:”; = Atlseﬂ} = a”? soﬁ=ﬁaudy1=rlfcﬂ. A particular solution ygft} of my} = —e‘* will have the form 3;; = Etc“. what: ﬁle} = sum — 22"} + atntr‘ + a") + 2m": = Rte-*3 = —e-1. 1husB=—1mdy2=—tc“.1herefore 1 it -rt 3'}: HI F2 12 1'. Let ﬂy} 2 y" — y. The characteristic roots of Ely) = t] are i]. A particular solution ypﬁtj of ﬂy} = sinﬁﬂt} + 2cocl2t) will have the form yp = Asiatic} + B cos{2t}. ctypi = -4A uinEEt] —4.s cos{2t}-—Asin{2t)—-B match = —5A sin[2£}— somtct} = sintct} + 2cou{2t] Thus A = —§, = is and y? = —%[sin{2£] + Ecosﬂtl} 10. Let Ely} = y” + y. The characteristic roots of ﬂy} 2 D are :l:-i. Thus the general solution of £{y} = I] is yh = Cl oosft] + {3'2 sin[t]. Let yptt} = A212 + Alt + an be a particular solution. Theo y; = 2A2, and lhﬂrﬂfﬂl'ﬂ Slim.) = 2112 + flat: + A1! + AOL: t2 + [3+ i]. Comparing Eh: I (‘12 = 1 coefﬁcients of the powers of t, we have A; = ﬂ Thus A; = 1T 2A: + do = El. A1 = Li, anti An = -2. It follows that yPIIt} = t2 — 2T and the general solution is y = t2 — 2-i- 01 £05“) + 03 sinitj. 13. Lot ﬂy] = y” + 43:“ + 43;. The homogeneous equation Em) = t] has a double characteristic tom. —2. Thus the general solution of My} = ﬂ is lift- : 5-2t|:01 + Cat]. Let yptjt} = Atac'2t be a particular solution. Then 3;; = lie-W2: — 2:”), y; = Ate-mm — st 4- 415’), and therefore My?) = Ede—3‘. Thus A = 1t It follows that yr“) = éeze‘a‘, out! the general solution is y = étzﬂ- + c‘mE-ﬂl + Car}, Hence if = or" — tze‘m + e'ml[—2C?1 + —2€2t + Cg}. Substituting ytfﬂ} = U, and If [U] = 1. weohtain C" =EI {—1201+02=1 mucl=ﬂsmd02=LThesolutioois 1 y = Ecze'ﬁ + rte-2* 5‘5 l._,_.— 5. mm 2.01} HM \$01 Eblmﬁcuﬁ weput nit] = smith-1+ wafﬂt‘n, and the system becom In?) I..—II 4. T. 11. wi{t}t'1+w3{t]t"2 = o -—wi[t}t'2 — Ewiﬂﬁ'a Thus mi = Int, w; = —t int. Integrating these expi'eSsions yields w] = tInt— tnndurg =¥-':%Then 2 2 tir[t]-[tlnt-tjt_1 + (a- J Edit-3 1 3 — ilnt— 1 "so — 2} 1:3 — 2:112": aft 3 “Inf [t-E'Je—“ctt B-Ioo \$31me —-{t-2]dc‘“ — Iim[—1[t—2)e"'3+fﬁic'“dt] _ B—«m s g u s 1 .. =Blimf—3I13—23em- :Eefi} _ 3+1 " s s” [a] ﬁt} is continuous on [[1, co}. Since limtqm girl“ = D. for all 31 :> ll, we conclude that ﬁt} has n Laplace transform to} ﬁrm.“ size?” = limb“. Etti-ni = on. for all 51 we conclude that ﬁt} is not is not of exponential order as t —I- oo, thus it has no Laplace transform. to ﬁt} is continuous on [t:-, no}. Since Embargo—‘1‘ = o. for all 31 we conclude that ﬁt) has n Lap-lace transform. Use the linorn-ityI of ﬁle tapioce transfonn: L{t — e"): Mt} — Me") = :1;- — ﬁf LIsin[2t]}= 1—. Thus L[e' sin{2t)] = m Use the principle that when G[s)= L{g{t}} then LEok'glftﬂ = G{s - stannmatstti— 3t+5}— — ﬁ—fﬁi Hence Lonﬂni— at+5n= is+3Jl- EH 355! + E5 mm 2m: ﬁ-N owl Staluﬁm‘i in :5 31. Let yIt} denote the solution of the M. and put YES] = My}. Then ..L—- Lly’J = 31’ — yﬂl] = 51* and my") = 321* — 9wa - ﬁn} = 391’ — 1. The Laplace transform ufthe one is {321/ — 11+ 451’ + 41’ = u. Solving forY, Y = m = 17:15:35. The solutionofthuWPis thteilwurse Laplacu u'ausform of Y. 3.: = 16‘” 34. Let gilt) denote tho solution of the WP, and put st} = Lﬁy}. Thou Lit“) = sY—yiﬂ) = 31’ Liar”) = sEY-sylﬂl—y’iﬂl = EYE“! Lia“) = 33?“. Th: Laplace transfonn of the arm is 331’ — 3311* + 331* .. Y = —L 5—1' Solving for Y, Y = F—LIF The solution of the NP is thu'invorso Laplace mfot'm of Y. y = 11:33:. lo)! 2. The partial frantic-n5 rxpunsiou. with undotermiood poofﬁoients. is 23 + 9 A B s=+3s+2=s+1+a+2' Multiplying through by qIEs] = {a + 1H: + 2], we have 23+9=A(3+‘Z)+B{3+1} Setting 5 = —1 in this Bquaﬁtln. wt: have A = T. Putting s = —2 in equation loads to B = —~5. Thus 25+9 _ r 5 52+35+2—5+1 3+2' and hunmL'lfF] = 7a“ — 53‘3“. 3. Th: partial fractions upmioo. with undotemiined coefﬁcients, is 3+5 _A{r+3}+E 33+ﬁs+1u“ {5+3}=+1' Multiplying through by 9113] = {a + 312 + 1, we have 3+5=Afs+3l+3 . _ Equating the ooeﬁiciouts of powers of s we have A = 1, B = 2. Thus 3+5 _ [s+3}+2 52+ﬁa+1u _ {5+3}=+1' and hmwoL“[F] = cos{t}e4‘ + Esiuftjo'm. ...
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HW9soln - Mot-H1 2-5315 “N 44:51 Shim-item‘s E 4...

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