A 2b and 1 1 a a 2 d in this case ax x4 bx

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Unformatted text preview: 1 ζ ζ ζ 4d 4d 2 Y dY + 2ζ 3 2 dζ dζ 2 Y +a dζ 2 1 ζ −b 1 ζ ζ2 2ζ 3 − b 1 ζ dY +c dζ ζ2 1 ζ dY +c dζ Y = 0. 1 ζ Y = 0. b) For an ordinary point we need that the following two functions are analytic (finite) about ζ = 0, 1 cζ 1 1 1 1 2 a − 2b and . 1 1 ζ ζ ζ ζ aζ a ζ ζ4 c) For a regular singular point we require the following two functions are analytic (finite) about ζ = 0, 1 cζ 2 1 1 1 ζ . a − 2b and 1 1 ζ ζ ζ ζ aζ a ζ ζ2 d) In this case a(x) = x4 , b(x) = 0, c(x) = α2 so the transformed equation is, ζ2 d2 Y dY + 2ζ + α2 ζ 2 Y = 0. 2 dζ dζ e) ζ = 0 is a regular singular point. f) Try a Frobenius expansion, ∞ an ζ n+r , Y= n=0 ∞ ζ 2Y = ζ dY = dζ 2 ζ2 dY = dζ 2 ∞ an ζ n+r+2 = n=0 ∞ an−2 ζ n+r , n=2 (n + r)an ζ n+r , n=0 ∞ (n + r)(n + r − 1)an ζ n+r . n=0 We sub this into our equation, ∞ r(r + 1)a0 xr + (r + 1)(r + 2)a1 xr+1 + (n + r)(n + r + 1)an + α2 an−2 ζ n+r = 0 n=2 4 For the first term to be nonzero we need r = 0 or r = −1. These are the roots to the indicial equation. For r = 0 we then need that a1 = 0 whereas for r = −1 we do not require that condition. The recurrence relation is, an = − α2 an−2 . (n + r)(n + r + 1) To solve the recurrence relation in general we get α2 a0 , (r + 2)(r + 3) α2 α4 a4 = − a2 = a0 . (r + 4)(r + 5) (r + 2)(r + 3)(r + 4)(r + 5) a2 = − For r = 0 we have that a1 and all the odd coefficients are zero and therefore y 1 = a0 1 − α2 2 α4 4 x + x + ··· . 3! 5! For r =...
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