Therefore we look for a regular power series solution

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Unformatted text preview: . If you consider x0 = −1 you will find 2 2 the same roots. b) Before we look for a power series solution I choose to substitute t = x − 1. If we rewrite Y (t) = y (x) then the equation is, (−t2 − 2t)Y + (−t − 1)Y + p2 Y = 0. Therefore, we look for a regular power series solution of the form Y = We substitute this into the equation and rewrite each term separately. p2 an tn+r , −(n + r)an tn+r , −tY = n=0 n=0 ∞ ∞ −(n + r)an tn+r−1 = −Y = an tn+r . ∞ ∞ p2 Y = ∞ n=0 −(n + r + 1)an+1 tn+r , n=−1 n=0 ∞ −(n + r)(n + r − 1)an tn+r , −t2 Y = n=0 ∞ ∞ n+r−1 −2(n + r)(n + r − 1)an t −2tY = −2(n + r + 1)(n + r)an+1 tn+r . = n=−1 n=0 We can combine these and get ∞ r−1 [−r(2r − 1)an+1 ] t −(n + r + 1)(2n + 2r + 1)an+1 + p2 − (n + r)2 an tn+r = 0. + n=0 The first term always vanishes because either r = 0 or r = −1/2. This yields the recurrence relation or an+1 = p2 − (n + r)2 an (n + r + 1)(2n + 2r + 1) For r = 0 we have the solution, y1 (x) = 1 + p2 (x − 1) − p(1 p)p(1 + p) (x − 1)2 + · · · . 3! For r = 1/2 we have 1 y2 (x) = |x − 1| 2 1 + 4p2 − 1 3·4 (x − 1) + 3 (4p2 − 1)(4p2 − 9) 3·4·5·8 (x − 1)2 + · · · . 4. a) We substitute ζ = 1/x and find, d dζ d d = = −ζ 2 dx dx dζ dζ d2 1 d2 d d2 2d + 4 2 = 2ζ 3 + ζ 4 2 =3 dx2 x dζ x dζ dζ dζ and Therefore, the equation is transformed into if we define Y (ζ ) = y (x), 1 ζ a or a...
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This document was uploaded on 01/26/2014.

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