# If it is singular determine whether it is regular or

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Unformatted text preview: ial equation. If r = 0, the recursion formula becomes an+1 = − 1 an . (2n + 1)(n + 1) Given a0 = 1, we ﬁnd that a1 = −1, a3 = − a2 = − 1 1 a2 = − , 5·3 (5 · 3)(3 · 2) 1 1 a1 = , 3·2 3·2 a4 = − 1 1 a3 = , 7·4 (7 · 5 · 3)4! and so forth. In general, we would have an = (−1)n 1 . (2n − 1)(2n − 3) · · · 1 · n! One of the generalized power series solution to (1.15) is y1 (x) = x0 1 − x + =1−x+ 12 1 1 x− x3 + x4 − · · · 3·2 (5 · 3)(3!) (7 · 5 · 3)4! 12 1 1 x− x3 + x4 − · · · . 3·2 (5 · 3)(3!) (7 · 5 · 3)4! If r = 1/2, the recursion formula becomes an+1 = − 1 1 an = − an . (2n + 2)(n + (1/2) + 1) (n + 1)(2n + 3) Given a0 = 1, we ﬁnd that 1 a1 = − , 3 a3 = − a2 = − 1 1 a1 = , 2·5 2·5·3 1 1 a2 = − , 3·7 3! · (7 · 5 · 3) and in general, an = (−1)n 1 . n!(2n + 1)(2n − 1) · · · 1 · n! 20 We thus obtain a second generalized power series solution to (1.15): 1 1 1 y2 (x) = x1/2 1 − x + x2 − x3 + · · · . 3 2·5·3 3! · (7 · 5 · 3) The general solution to (1.15) is a superposition of y1 (x) and y2 (x): y = c1 1 − x + 12...
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## This document was uploaded on 01/12/2014.

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