In this section we showed that one solution of Bessels equation of order zeroL[y] = x2y + xy + x2y = 0is J0, where J0(x) is given by Eq. (7) with a0 = 1. According to Theorem 5.6.1, a secondsolution has the form (x > 0)y2(x) = J0(x)ln x +n=1bnxn.(a) Show thatL[y2](x) = n=2n(n 1)bnxn +n=1nbnxn +n=1bnxn+2 + 2xJ0(x). (i)(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).

Math 1550 - 004 Spring 2017 Exam I Review We'll do our first midterm test on Friday, February 17 . Sections that will be tested on this exam: 0.1: Linear Equations a. Be able to solve Applied Linear Expression Problems (Mixture, Distance, Geometry) 0.2: Quadratic Equations a. Know the Quadratic Formula (I will...