Solved: In this section we showed that one solution of Bessels equation of order zero
Chapter 5, Problem 10(choose chapter or problem)
In this section we showed that one solution of Bessels equation of order zero L[y] = x2 y + xy + x2 y = 0 is J0, where J0(x) is given by Eq. (7) with a0 = 1. According to Theorem 5.6.1, a second solution has the form (x > 0) y2(x) = J0(x)ln x + n=1 bnxn. (a) Show that L[y2](x) = n=2 n(n 1)bnxn + n=1 nbnxn + n=1 bnxn+2 + 2xJ 0(x). (b) Substituting the series representation for J0(x) in Eq. (i), show thatb1x + 22b2x2 +n=3(n2bn + bn2)xn = 2n=1(1)n2nx2n22n(n!)2 . (ii)(c) Note that only even powers of x appear on the right side of Eq. (ii). Show thatb1 = b3 = b5 == 0, b2 = 1/22(1!)2, and that(2n)2b2n + b2n2 = 2(1)n(2n)/22n(n!)2, n = 2, 3, 4, ....Deduce thatb4 = 122 421 +12and b6 = 122 42 621 +12 +13.The general solution of the recurrence relation is b2n = (1)n+1Hn/22n(n!)2. Substitutingfor bn in the expression for y2(x), we obtain the solution given in Eq. (10).
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