hw5sol - 5%. £7 W) # S’ 14. Note the transform pair. 3...

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Unformatted text preview: 5%. £7 W) # S’ 14. Note the transform pair. 3 52+4 cos 2: <-+ By Proposition 2.12, s——1 6t COS 216-) Hence, ~ _1 4(s—1) M‘fi {(s—1)2+4} _ —1 __§;1___ "M {(s—1)2+4} =4e’c052t. 18. Let £(y) = Y(s). Since y(0) = O, and y’(0) = —-1, £01" a y' — 2y) = szY(s)+1-« sY(s) — 2Y(s). In addition, 2 (s —2)3‘ [,(tzeZ‘) = Solving (sz—s—2)Y(S)+1= 2 (S~2)3 for Y, we get V/ \ *1 | 2 1 \S} —— . T 7 ' ' (5 +1)(s * 2) (S +1)(S ~ 2)4 Tho farmn A“ «Inn 1‘ n-lnf ka‘ 9 Annnmnnmhnng lllv [8.11.1113 U1] “1U llglll ll VV “UVUILIFUDJUUA —1 1/3 1/3 and 2 2/81 2/81 2/27 f : ——-“" " + 2 2 <s+n(s—2)4 s+1 3‘2 ‘3”) 2/9 + 2/3 ‘ (s~2)3 9—94 Therefore, 29/81 29/81 2/27 Y(s)=s+1—s_2 (3—2)2 9 2 3 _ .91.. + / 36. Find a partial fraction decomposition S (s + 2)2(s2 + 9) A B Cs + D + ._.___ s+2 (s+2)2 + s2+9 Equating numerators s = A(s + 2)(s2 + 9) + 3(32 + 9) + (Cs + D)(s + 2)2 = (A+C)s3+(2A+ B+4C+D)s2 + (9A +4C+4D)s + (18A +93 +40) Thus, A+C=0 2A+B+4C+D=0 9A+4C+4D=1 18A+93+4D=0 and A = 5/169, B = —2/13, C = ~5/169, and D 2 36/169. Thus, _ 5/169 ~2/13 y(t) =5 1{ + s + 2 (S + 2)2 ——(5/169)s 36/169} + M. s2 + 9 + 52 + 9 , Note the transform pair. [95. By Proposition 2.12, 1 —21 t . e ‘9 (s+2)2 Thus, 5 -2t 2 —2t 5 12 ' :_ _._ ___ 3t —— 3t. W) 1696 13“ 169COS +169sm 8. Let C(y) = Y. Then £0" — 2y) = £(e"’ cost). Note the transform pair. s cost <—> s2+1 Then by Proposition 2.12, “ cost 4+ S +1 e -+—————~. (5 +1)2 +1 Thus, 5 +1 L ’ — 2L? = (y) (y) (5+1)2+1 3 +1 ~ Y ~ 0 '— 2Y = s (s) y<> (s) Mm“ But y(0) = —2, so 3 +1 ~ 2 Y 2 = —-————~— (s )(s)+ (S+1)2+1 *2 s +1 Y(s) = s—2+ ((s+1)2+1)(s—2)' Find a partial fraction decomposition. s+1 _ A + Bs+C (3—2)((s+1)2+1)_s-—2 (s+1)2+1 Equating numerators, 5+1=A((s+1)2+1)+(Bs+C)(s—2) =(A+B)s2+(2A—-ZB+C)S +(2A—2C) Thus, A+B=0 2A—ZB+C=1 2A—2C=1, and A = 3/10, B = —3/10, and C 2 ~1/5. Thus, “2 + 3/10 w (3/10)s —1/5_ 5—2 5—2 (s+1)2+1 _ —17/10 —(_3/10)s — 1/5 — ,s—2 + (s+1)2+1 Y(s) = Note the transform pairs 5 s2+1 1 32+1 COSI 6 sint <—> $3: ,9? and 29 29 2 t : ——— ‘t ._.. ._ 2’ 21‘ y() 81c 81e +—-27te 1 2 2: 1 3 2 _ ._t _ t 9 e + 92‘ e ‘ Proposition 2.12 provides 1 e" cost <-> L (s + 1)2 + 1 1 e“ sint <—> ————~———. (s +1)2 +1 32. (a) Let y = e". Then, y//+4y : 0 rze" +4e" = 0 r2+4 :0. Thus, r : iZi leads to the independent solu- tions y = cos 2t and y :2 sin 2t. Therefore the general solution is 3’1: = C1 COSZt + C2 sin2t. (b) We examine z” + 4g 2 2ei3t. Let z = AeiBt- Then z, = 3Aiei3’ and Z” ._ _9Aei£ and _9Aei3t + 4Aei3t : 2ei3t —9A+4A=2 -—5A=2 A=_3 5 Thus, 2,, = (—2/5)ei3' and the real part of zp IS a solution of y” + 4y = 2 cos 31‘. Thus, 2 = —— cos 3t. )7}; 5 (c) The general solution and its derivative are 3’ = )7]: + y p = C1C082t +Czsin21—gcos3t y’ = —~2C1 sin 2t + 2C2 cos 2t + 2 sin 3!. Thus 7 J’(O)=1=>C1-—:-= y'(0)=—1=>2C2 = *1. Therefore C] = 7/5 and C2 = ~1/2 and the solution is 7 l . 2 y = ECOSZI — istt ~ 300831. ((1) Let My) 2 Y(s). Since y(0) = 1 and ya» = —1, £(y’l + 4y) 2 Y(s)(s2 + 4) — 3 +1. In addition 2.3 s2+9’ £(2 cos 3t) = SO Y(s)(s2 +4) —s +1 = 32 + 9. Solving for Y, we get s — 1 + 23 52 +4 (s2 ~l-4)(s2 + 9)' The fraction on the right has decomposition 2s _ (2/5)s _ (2/5)s (521%)“2 +9) “3 s2 +4 s2 +9' Y(s) = 515 Therefore, (7/5)s —1 a 2/SS n3)— 32+4 s2V+9 7 s 1 2 _ 3 ‘ 32 + 4 2 s2 + 4 5 s2 + 9 and y(t) = COS 2! w ésin 2t .— gcos 3t. 8. We have 3 2t 3” + 3” cos t.— cos 2 2 3t 37: OS 3:1 : -— *— C .- cos 2 2 ' 3t 3” sin 3” Sm 2 2 3n = ' 31‘ -—~ fl sm( 2 ) . II = sin 3 (t a a) . Therefore, H(t—%)cos3z=H(r—%)sin3(t—g). Because f (t) = sin 3t has transform 3 s2 + 9’ then g(t) = H(t —7r/2) sin 3(t ~7r/2) has transform G(s) = (gem/2%) Fm = 3 v ~7r3/2 e s2 + 9 3e—ns/2 : a2 ! n 0 T7 20. Note the transform pair . 2 f(t) = s1n2t <—> F(s) = 32 +4. 24. A partial fraction decomposition. Then, s—0=>‘A~1 " ‘4 s-2:>C-I ’2 s—1=>B- 1 — 4 and 8—5 e—s e—s 6—; Note the transform pairs: 1 f(t)=1<->F(S)=; g(r) 2 32‘ <—> 0(5) 2 1 s —2 __ 2: fl 1 k(t) _ te <—-> K(s) _ (S _2)2. Thus, *1 i 6‘3 ‘ ‘3 1mm —2>2 } ‘0 _ ~1 —-s _1__ 1 “I: [e [43 4(s—2) -u\ l .. + 2(s — 32””) = flme—Smnm — £“l{e"sG(s)}(t) + 2£—1{e“sK(s)}(t)] 1 = Z[H(t—1)f(t— 1)-— Ha — 1>g<r — 1) + 2H(t — 1)k(t — 1)] = flaw — 1) — H(t — Dem—1) + 2H(t—1)(t—— new—D] _l _ _§ _ 20—1) —4H(t 1) 4110 De 1 + 5H0 — 1)te2(’“1) g? 18. Note the expression 1 __ 1 _~ 1 1 S2—3S ~s(s—3) — s 's—3’ = F(s)G(s). We have the transform pairs F(s) = % <==> f(t) = 1, 0(5) = 1 <==> g(t) = e3’. s — 3 Thus, 5‘ l 2 1 }<0 = £‘1{F(s)G<s>}<t), 5 —3s = f * g0), 33 g 3% g x 35 g 28. We start by computing the impulse response function e(t). y" + 5y' + 4)’ = 30?). y(0) = y'(0) = 0 By Theorem 6.10, 1 1 13(3) = ___« = 1 = P(s) s2 +55 +4 (s +4)(s +1) —1/3 1/3 " 5 +4 + s +1' Thus, 1 1 e(t) = -—§e"4‘ + 3e”’, and the derivative is 1 III, :_ —4t____ ——l e ( ) 3e 33 From Theorem 7.16, the solution to y” + 5)” + 4y = g0), y(0)=1, y’(0) = 0 is W) = e * 80) + ayoe’(t) + (ayl + byo)e(t), = e * g0) + (1)(1)e'(t) + [(1)(0) + (5)(1)]e(t)9 = 6 >1: g(t) + e’(t) + 5e(t), ...
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This note was uploaded on 01/12/2010 for the course MATH 53 at Stanford.

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hw5sol - 5%. £7 W) # S’ 14. Note the transform pair. 3...

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