MIT Power Series MITRES_18_008_partII_sol06.pdf - Solutions Block 2 Ordinary D i f f e r e n t i a l Equations U n i t 8 The U s e o f Power S e r i e s

MIT Power Series MITRES_18_008_partII_sol06.pdf - Solutions...

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Solutions Block 2 : Ordinary D i f f e r e n t i a l Equations Unit 8: The U s e of Power Series a . Since two of the t h r e e summations involve xn and only one involves n-1 x , it s e e m s l o g i c a l t o "jack up" the index i n xn'l by 1. That is, w e replace n by n + 1 i n s i d e the 1-sign and lower the s t a r t i n g point of t h e summation by 1. W e obtain s o t h a t W e then " s p l i t o f f " t h e f i r s t term i n each of the summation i n (2) which begin with n = 0 (in t h i s way, each summation begins with n = 1) . W e therefore rewrite (2) as: Note: Especially f o r those who may still be "edgy" about extensive use of the C -notation, it may be worthwhile t o show the equi- valence of (1) and (3) by long hand techniques. In t h i s way, one does not review what we've done without recourse t o the G -notation, and one may a l s o learn t o appreciate b e t t e r the compactness of the C -notation. To begin with, (1) may be expanded as:
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Solutions Block 2: Ordinary Differential Equations Unit 8: The U s e of Power Series 2.8.1(L) continued and t h i s agrees with (3) since the expansion of (3) yields equals It should also be noted that i n e i t h e r approach, we used the idea of absolute convergence t o justify our taking the liberty of re-arranging and combining terms i n the way we did. b. Replacing (1) by (31, we see t h a t implies t h a t OD Now, since the only way that two convergent power series can be identical is for t h e i r coefficients t o be equal term-by-term and since 03 w e see from (5) t h a t and S.2.8.2
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Solutions Block 2: Ordinary D i f f e r e n t i a l Equations Unit 8: The U s e of Power Series 2.8.1 (L) continued f o r n - > 1; (n + l)an+l - (n - l ) a n = 0. That is, n - > 1 implies t h a t From (6) we see t h a t a. may be chosen a t random, whereupon and we may now use (7) t o compute a2 from al, a3 from a2, a4 from a3, e t c . For example with n = 1, (7) becomes b u t once a 2 = 0, (7) t e l l s us t h a t f o r n > 2, an = 0. That is, from (7) an+l is a multiple of an s o t h a t an+l must be 0 once an = 0. Thus, w e have t h a t a. is a r b i t r a r y , al = -a ' and an = 0 f o r 0 n - > 2. Since %.re have that rlas a solution of the form
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Solutions Block 2: Ordinary Differential Equations Unit 8: The U s e of Power Series 2.8.1 (L) continued then so t h a t (10) becomes Notice t h a t equation (11) is precisely equation ( 4 ) of part (b) [where - the equality i n (11) is understood t o be our identity. We used the identity symbol i n p a r t (b) f o r emphasis, but other than that, we revert t o the usual notation that L(y) = 0 means L(y) 01. Hence, from part (b) we deduce from (11) t h a t y = a. (1 - X) . Note #1: A s usual we have elected t o s t a r t with an example t h a t could be solved more easily by a more familiar technique. In equation (10) we could separate variables t o obtain that f o r x # 1 and y # 0 , dy/y = dx/x - 1; o r lnlyl= lnlx - 11 + cl; o r y = + - ecl(x - 1) = c ( x - l), where c = + - e # 0.
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