hw10s - 5.2 163 Hence 00 DO 00 (1 — $2) 2 n(n —...

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Unformatted text preview: 5.2 163 Hence 00 DO 00 (1 — $2) 2 n(n — Dana?“2 = Z (n + 2)(n + 1)an+2 as“ — Z 7101 — Dan :17”. n:2 n:0 n=2 Note that when n = O and n = 1 , the coefficients in the second series are zero. So 00 DO (1 ~ .722) Z n(n — 1)anm”’2 = Z + 2)(n + 1)an+2 — n(n - Dan] :10". n: n:0 26. Clearly, 00 00 00 [)0 Z nan mn'l + m 2 an 1'” = Z nan m”#1 + Z an 55”“. n:1 n:0 n21 n:0 Shifting the index in the first series, that is, setting k7 = n — 1 , w k i nan con—1 = Z (k + 1)ak+1m . 11:1 k):0 Shifting the index in the second series, that is, setting k: : n + 1 , 00 00 E an 30”“ = E ak-1:£k. 71:0 19:1 ing the series, and starting the summation at n : 1 , Combin 00 OO 00 nan m”#1 + a: 2 an x” = a1 + Z [(n + 1)an+1 + an-1] :13“. 71:1 n:0 n: 1 27. We note that CO 00 00 00 a: Z n(n # 1)an 2:7”2 + Z an as” = Z Mn — Dan xn-l + E an:1:”. n=2 n:0 n22 n:0 Shifting the index in the first series, that is, setting It : n — 1 , ' 00 00 00 Z 7101 — Dan mnfll : Z Mk + 1)ak+1a:k = Z Mk + 1)ak+1mk, n22 k=1 k:0 = 0 is zero. Combining the series, since the coefficient of the term associated with k ()0 00 00 m 2 n(n — Dan gun-2 + 2 an x” : Z + Dan“ + an] at”. n:0 n:2 n:0 .+anx"+.... Then 1.(a,b,d) Let y = a0 + ala: + (12$? + .. 00 y" = 2 Mn ~ Danxn'z : n:2 00 (n + 2)(n + 1)an+2 cc". nZO 164 Chapter 5. Series Solutions of Second Order Linear Equations Substitution into the ODE results in 00 Z (n + 2)(n + 1)an+2 :0” — Z anm" = 0 n:0 n=0 or Z [(n + 2>(n + Dam — an] as" = 0. n20 Equating all the coefficients to zero, (n+2)(n+1)an+2*an=0, n:0,1,2,.... We obtain the recurrence relation an an 2‘, ,=0,1,2,.... +2 (n+1)(n+2) " The subscripts differ by two, so for k = 1, 2, . . . a _ a2k—2 H Oak—4 _ 2 Go 2’“ (2k — 1)2k (2k — 3)(2k — 2)(2k —1)2k "' (2k)! and a : a2k~1 _ a2k—3 : Z (11 2k“ 214215 +1) (2%: — 2)(2k — 1mm + 1) "' (212 + 1):' Hence 00 $2k 00 $2k+1 y = a0 2 . + “1 Z . k:0(2k:). k:0(2k+1) The linearly independent solutions are $2 $4 $6 yl:a0(1+§+1+a+...):a0coshm _ $3 $5 $7 _ . h yg—a1(x+§+a+fi+n.)—alsm x. (c) The Wronskian is aoal. 4.(a,b,d) Let y=a0 : (11:0 { a2$2 : l and?“ l Then y" = Z w — 1)anm"‘2 = 201+ mm + mam n22 n:0 Substitution into the ODE results in 00 00 :01 + 2)(n + Dan” :3" + k2x2 Z anx” = 0. n:0 n:0 Rewriting the second summation, 00 00 :01 + 2)(n +1)an+2 :L’” + 2 Hana :15" = 0, n:0 n:2 166 Chapter 5. Series Solutions of Second Order Linear Equations and y” = Z n(n~ Manx” : 20% 2)(n+1)an+2x”- n22 71:0 Substitution into the ODE results in 00 00 OO (2 +m2) 201+ 2)(n + 1)an+2 a?" — :3 201+ 1)an+1m” +4 Z anx” : 0. n=0 n:0 n:0 Before proceeding, write 00 00 272 Z (n -l- 2)(n + 1)an+2 :L‘" 2 Z n(n ; Manx" n22 71:0 and 00 00 I Z (n + llan+1$n = Z n20 n21 It follows that 4a0 + 4% n (3a1 + 12a3)$+ + Z [2(n + 2)(n + 1)an+2 W n(n — Dan * nan + 4am] m“ = 0. 11:2 Equating the coefficients to zero7 we find that a2 = —a0 , a3 = fol/4, and n2 i 2n+4 an =#-——— +2 2(n + 2)(n + 1) The indices differ by two, so for k = 0, 17 2, . . . (2k)2 ; 4k + 4 an, n=0,1,2,.... 0%” : ‘W and a (2k+1)2—4k+2 : —— a A . 2k+3 Zak: + 3)(2k + 2) 2k+1 Hence the linearly independent solutions are 364 176 : 1 — 2 — e — . . . mm “T + 6 30 + m3 7:05 193C7 y2(:c)=:c#—+ +.... 4 TGO_ 1920 (c) The Wronskian at 0 is 1. 7.(a,b,d) Let y—ao } alzv ‘. (L20: l l and?” l Then ()0 00 y, 2 Z nananl : Z + 1)an+1xn n:1 n:0 5.2 167 and 00 00 y" z 2 mm — Maw = 2: <71 + 2W + 1)an+2 71:2 n20 Substitution into the ODE results in ()0 OO 00 Z(n+2)(n+1)an+2 xn+$ Z(n+1)an+1$”+2 aux” :0. 71:0 n:0 n=0 First write 00 00 a: 1)an+1mn : Z manic”. 71:0 n: We then obtain 00 2a2 + 2&0 + Z + 2)(n +1)an+2 + nan + 26%] m” = 0. n21 It follows that (12 = #010 and an” 2 ~an/(n + 1), n 2 0,1,2,.... Note that the indices differ by two, so for k = 1, 2, . . . a _ (1216—2 # a2k—4 _ _ (‘Ukao 2’“ 2k—1 (2k—3)(2k:#1) 1-3-5 ...(2k—1) and a z -% Z 4%; z : flL 2k“ 2k (2k—2)2k 2-4-6...(2k)' Hence the linearly independent solutions are .’L' {17 CI} _1 i : 1: 91m 1 1-3 13-5 M 1 3 5 ..(2n—1) m3 m5 CE7 0° (_1)n$2n+1 92m“: 2 ‘2.4 246' SH“2:246 (2n) (c) The Wronskian at 0 is 1. 9.(a,b,d) Let y : a0 + alm + LLng + . . . + anrt” + .... Then 00 00 y, : Z nanx'rL—1 : :01 + 1)(ln+1$n 11:1 71:0 and Substitution into the ODE results in 00 00 (1 + as?) Z (n + 2)(n + nan+2 a?” — 41: Z (n + 1)an+1a:" + 6 Z anx" = 0. 71:0 ...
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This note was uploaded on 01/10/2012 for the course MATH 2233 taught by Professor Binegar during the Fall '08 term at Oklahoma State.

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hw10s - 5.2 163 Hence 00 DO 00 (1 — $2) 2 n(n —...

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