homework 12 solutions

# homework 12 solutions - Section 9.3 10 a0 = if sintdt = —...

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Unformatted text preview: Section 9.3 10. a0 = if sintdt = — 7! o 4[1+cos§£] ~4/7z'(n2 -—4) for n odd an = —f: sintcos—dt= -—2—2‘“ = “8/”(nz—4) if n=4,8,12,--- ”(n —4) 0 if n=6,10,14,--- 1 . a2 —f s1ntcostdt = 0 7r . . l 4 1 t 1 3t 2 4t Cosme serles: f(t) = — —— —— ——cos— + —cos— + -——cos—— 7: 7r 3 2 5 2 12 2 1 5t 1 7t 2 8t +—cos——+——cos-——+———cos——+ 21 2 45 2 60 2 t —27r 271 471 Sn 8n 107T 43mg 0 for n>2 even b" = —J: sintsin—dt= -—(—2—255 = —4/7z'(n2—4) if n=1,5,9,-—- n'n — +4/7r(n2—4) if n=3,7,11,--- 1 1 b = — ‘ 2tdt = — 2 ”J: sm 2 2 b=— 'tht=1 1 ﬂfsm Sine series: f (t) = —-smt — -— 1.4[1.t1.3tlsi5t1.7t1 7r 5 221245 277 —27r 2n 7r 6n 7r 107T —sm— + s1n—-——— n—+—sm———sin—+--- 16. (a) Obviously xp (t) = t is a particular solution of x” + 4x = 4t, so a general solution is given by x (t) = Acos 21 + B sin 2! + t. We satisfy the endpoint conditions x(0) = x(2) = O by choosing A = 0 and B = —1/sin2. (b) The point is simply that the series in (31) is the Fourier sine series of the period 2 function deﬁned by ﬂt) = I ~ (sin 2t)/ (sin 2) for 0 < t < l: 2 gtsmmdt = W = _2(—1)" n 71' n71- I — ' 2 _ n - 2£sin2tsinn7rtdt = W : _2n_n(_9_(s4L2) n 71' - n ”- _ 2(-1)" 2mr(—1)" 3(_1)" 1’" = ‘ +'nz—2—‘ = -——-—2 2 ”77 -4 n7r(n 72' -4) _sin2t ___ _Z(—1)"sinn7rt for 0<t<1 sin2 "_ l "(”271- 19. The ﬁrst termwise integration yields 2 “° (—1)"cosnt — = 2 ———————+C , 2 "2:; n2 l and substitution of t = 0 gives C1 = 22 (—1)“ / n2 = 7:2 / 6, so "=1 t; (______— l)" cosnt+ 7r_2_ — = 22 n=l +6 A second termwise integration gives and substitution of t = 0 yields C3 = zz(—1)"/n4. n=l 20. 21. Substitution of t = 7: in the formula of Problem 19 above gives 71. (D (—1)", —— = —2 ————+——+2 24 "2:11;: in 71,4 an 1 no (_ ])Ix+l —— = 2 —+2 =4 24 En“ E "4,;«1’11 4 whichgives 1+i4+i4+—1;+--- = ”—. Then 3 5 7 96 F0) = 2b,, sinn—m which agrees with f(t) if 0 < t < L. Then 2 L mrt 2 ” nm b ———dt — 2L—t —-dt. " =2LJ foSin 2L IL+2LJ f( )szL The substitution u = 2L -t yields b n = %J: faking—ma? ———J: f(u)sinﬂ%I[—u)du 1 mrt (— 1)" mm —J: f(t)Sin3i_dt- —'—:J f(u)smEdu. =L L Now it is clear that 2 L mrt ——dt b,.= Ii f(t )cos u if n is odd, whereas b" = 0 if n is even. 22. We want to calculate the coefﬁcients in the period 4L Fourier cosine series °° t Gt “0+ ﬂ ()= 2 Zancos 2L "=1 which agrees with f(t) if 0 < t < L. Then 2 L rm! 2 2L mrt = ——dt —— 2L—t —dt 0" 27L foCOSzL tL+2LJ ﬂ )cos 2L The substitution u = 2L —t yields a": %J: f(t)cosn2—:£dt ——J: f(u)coswdu 1 rmt (—1)"LL mm —dt—- ~—d . =LJ: f(t )cos 2L L f(u)cos 2L u Now it is clear that 2 L mrt ———dt an= L] f ( )cos 2L if n is odd, whereas an = 0 if n is even (including n = 0). 25. t+2L 1% + 2L) — N) = / [f(s) — gums t+2L = / f(.3) ds — Lao t = Lao -— Lao :0 (b) By the fundamental theorem of calculus, F' (t) : f(t), so by integration by parts, An=— :/_: F< ><os[(:—”t)dt n M N: :3 \$11 31TH EL 3 lag: LL—L/LLdt F'(t) si11(£7r——t)dt:l 27rt ‘———/_:L m )sin<—>dt [I and L 3,, = l F(t t)sin("—’1t—)dt = l _: F(t)i[— icos(27T—t)]dt dt n17 =————[F(t)cos(E—E)LL —/_: F'(t)cos(1—zﬂ)dt:| —~/_: f( “04.213” : “ﬂan n7r 3H Notice that these computations involve division by n, so they are not valid for n = 0. (C) t t t 1 °° t A f(.s) d3 = —2-ao + F(t) = gao + 5A0 + ,2”: [an 31M?) " bn 604%] Taking t = 0, 1 °° L 0 = _A "' —‘bn> 2 0 ”:31 mt so 1 °° L ~Ao — 2: _b,, Section 9.4 9. The natural frequency is we = 2, and FU) = i(sint+sm3t+sm5t+sm7t+m - 72' 3‘ 5 7 Because the sin 2t term is missing from the Fourier series of F (t), resonance will not occur. l7. 3,, = 60/n7: for n odd, 3,, = 0 for n even xspa) z 0.5687 5mm — 0.0562) + 0.4271 sin(37n‘ — 0.3891) + 0.1396 sin(5m — 2.7899) + 0.0318 sin(7m‘ — 2.9874) +. . . xsp(5) z 0.248 ft z 2.98 in. ...
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