TAM310quiz02sol - OsloMOX Quiz; 2_ SOLUTIONS: j W 0 h”...

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Unformatted text preview: OsloMOX Quiz; 2_ SOLUTIONS: j W 0 h” —+ H7’+ 157:0 This )5 6W?” DE' M 7"?” t ‘1: Y 7‘1“) 7”: 7‘7”; S’M/lohg V? ‘W 3)V€J, (“IN/I) + 9” + if) {:0 :vle'f 1Y'7‘uu’o FIGURE 8.2.1. Radius of convergence as distance to nearest singularity. WPLE 5 Find the general solution in powers of x of (x2 — 4)y” + 3xy’ + y = 0. Then find the particular solution with y (0) = 4, y’(0) = 1. The only singular points of Eq. (7) are :|:2, so the series we get will have radi convergence at least 2. Substitution of OO 00 00 Eb y 2 chx", y’ = chnxn’l, and y” = an - 1)C,1x"_2 n20 71:1 11:2 in Eq. (7) yields 00 00 27101 — UCnX" — 4 flew —1)cnx"_2 + 3 incnx” + Z Cnx'l = O. n22 n=l 11:0 11:2 We can begin the first and third summations at n 2 O as well, because no nor terms are thereby introduced. We shift the index of summation in the second by +2, replacing n with n + 2 and using the initial value n = O. This gives 00 oo oo 00 Z 7101 — l)cnx” — 4 2m + 2)(n + l)cn+2x” + 3 chnx” + Z cnx” an n=0 n=0 n20 After collecting coefficients of cn and on”, we obtain 2 [(112 + 2n + 1)C,, — 4(n + 2)(n +1)Cn+2]x" : 0. ":0 The identity principle yields (n + 1>2cn — 4m + m +1)cn+z : 0, 530 Chapter 8 Power Series Methods which leads to the recurrence relation s (n + l)c,, W : __.._...__ . cn+2 401 + 2) for n g 0. With n : O, 2, and 4 in turn, we get C0 362 3C0 564 3 ' 5C0 :—--y =~v—‘-:-——--~—y d ‘:———=—————-"—_ Q 4.2 C“ 4.4 42-24 an ‘6 4-6 43-2-4‘6 Continuing in this fashion, we evidently would find that l-3‘S---(2n—l)‘ 4"-2-4---(2n) ‘0' CZrL '3 With the common notation (2n + l)! 2" r n! (2n+l)!l=l-3>5---(2n+l)z and the observation that 2 - 4 - 6- - - (2n) = 2" - 71!, we finally obtain 2 — l! ‘ 4L1,“ (9) , ‘} C2" 23" - n! . . ~. (We also used the fact that 4” - 2” = 23”.) With n = l, 3, and Sin Eq. (8), we get 201 C 463 2 - 401 and 6C5 2 - 4 > 661 C=__.—’ =~—--:——-—, C:——= ' 3 4.3 5 4-5 42-3-5 7 47 43-3.5.7 It is apparent that the pattern is 2-4~6~~2n n! ’ " C2n+1 == ( ) c1 = Cl. (10) 4”-1-3-5---(2n+l) 2"-(2n+1)!! . A The formula in (9) gives the coefficients of even subscript in terms of co; the formula in (10) gives the coefficients of odd subscript in terms of Cl. After We separately collect the terms of the series of even and odd degree, we get the general solution ( ) C 1+i(2n—1)!!x2n + +i’: n! 2"“ (11) x = c x x . ' y 0 "=1 23" -n! 1 n=l 2" - (2n +1)” Alternatively, l 1 +61 (x+—x3+_x5+—~—x7+--->. (11/) /\ 5- X27” + x I L ‘ 7 + x "-L} to ( 4 7 M 7; g 64 X’Hv/ wzzmoaww ‘1” Z mm) mm) Q XMM‘ hag Hajj}? 3M?) 01:. Y Y 2 (“+1>/NY’I) Q KM + 20‘“) G‘XMT 2 CKXMflL—l. é CAX “:0 me two A, hzo N \u/ E CIVIXMY h>+Z Far A10!) Only Ill 2.9 fummHaM loni’fibmfl- [Q‘rtfl (WW4) +{M+1)"Zrl Ca ‘1 D "f" 0" [fwflLai] a“ :0 4 Mao {Ll-2 i .3; ' ...
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TAM310quiz02sol - OsloMOX Quiz; 2_ SOLUTIONS: j W 0 h”...

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