Soln_Ass4_AMATH351_topost

# Soln_Ass4_AMATH351_topost

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Unformatted text preview: x)2 x. 1 2. a) We compute p0 and q0 , as deﬁned in lecture, 2x 2 = = 1, 2 1−x 2 p(p + 1) = 0. 1 − x2 p0 = lim (x − 1) − x→ 1 q0 = lim (x − 1)2 x→ 1 So the associated Euler equation is, x2 y + xy = 0. The characteristic equation is r(r − 1) + r = 0 that implies r = 0 is a double root. 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 + (−2t − 2)Y + p(p + 1)Y = 0. Therefore, we look for a regular power series solution of the form, ∞ an tn Y= n=0 We substitute this into the equation and rewrite each term separately. ∞ p(p + 1)an tn , p(p + 1)Y = n=0 ∞ (−2n)an tn , −2tY = n=0 ∞ ∞ n−1 (−2n)an t −2Y = (−2(n + 1))an+1 tn , = n=0 n=1 ∞ −n(n − 1)an tn , −t2 Y = n=0 ∞ ∞ n−1 −2tY = −2n(n − 1)an t n=0 −2(n + 1)(n)an+1 tn . = n=0 We can combine these and get ∞ −2(n + 1)2 an+1 + {p(p + 1) − n(n + 1)} an tn = 0. n=0 This yields the recurrence relation of an+1 = p(p + 1) − n(n + 1) an 2(n + 1)2 From this we get two solutions in terms of the original variable x, y 1 ( x ) = a0 1 + p(p + 1) (x − 1) + 2 · 12 2 p(p + 1)(p(p + 1) − 2) 2 · 22 (x − 1)2 + · · · 3. a) We compute p0 and q0 , as deﬁned in lecture, p0 = lim (x − 1) − x→ 1 x 1 =, 2 1−x 2 q0 = lim (x − 1)2 x→1 p2 = 0. 1 − x2 1 So the associated Euler equation is x2 y + 2 xy = 0. The characteristic equation is r(r − 1) + 1 r = 0 that implies r = 0 and r = 1...
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## This document was uploaded on 01/26/2014.

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