Find the first four nonzero terms of the series solution to y\"+xy=0 about x0=0

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1) Assume that y = S(n = 0 to 8) a(n) x^n. Then, y'' - xy = S(n = 2 to 8) n(n-1) a(n) x^(n-2) - x S(n = 0 to 8) a(n) x^n = S(n = 2 to 8) n(n-1) a(n) x^(n-2) - S(n = 0 to 8) a(n) x^(n+1) = S(n = 0 to 8) (n+3)(n+2) a(n+3) x^(n+1) - S(n = 0 to 8) a(n) x^(n+1) = 2a(2) + S(n = 0 to 8) [(n+3)(n+2) a(n+3) - a(n)] x^(n+1) = 0 With a(0) and a(1) arbitrary, we have (i) 2a(2) = 0 ==> a(2) = 0 (ii) For n = 0, we have (n+3)(n+2) a(n+3) - a(n) = 0 ==> a(n+3) = a(n) / [(n+3)(n+2)]. Since the recurrence is in steps of 3, we have three cases. a(3) = a(0)/(3 * 2) = 1a(0)/3! a(6) = a(3) / (6 * 5) = 1a(0) / [6 * 5 * 3!] = (1 * 4) a(0) / 6! ... a(3k) = (1 * 4 * ... * (3k - 2)) a(0) / (3k)! ------------- a(4) = a(1)/(4 * 3) = 2 a(1)/4! a(7) = a(4) / (7 * 6) = 2 a(0) / [7 * 6 * 4!] = (2 * 5) a(1) / 7! ... a(3k+1) = (2 * 5 * ... * (3k - 1)) a(1) / (3k+1)! ------------- a(2) = 0 ==> a(5) = a(2)/(5 * 4) = 0 ... a(3k+2) = 0. Therefore, the general solution is y = a(0) S(k = 0 to 8) (1 * 4 * ... * (3k - 2)) x^(3k)/(3k)! + a(1) S(k = 0 to 8)(2 * 5 * ... * (3k - 1)) x^(3k+1) / (3k+1)!

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