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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 = —2ao + 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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 Spring '07
 TERRELL,R
 Math

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