solve in series the equation d2y / dx2 + x2y = 04

Let y = S[n=0 ... infty] a_n.x^n, y'' = S[n=2 ... infty] n(n-1)a_n.x^(n-2)

inserting the above two expressions into the DE,

S[n=2 ... infty] n(n-1)a_n.x^(n-2) +x^2.S[n=0 ... infty] a_n.x^n - 4 = 0

S[n=2 ... infty] n(n-1)a_n.x^(n-2) +S[n=0 ... infty] a_n.x^(n+2) - 4 = 0

Let n = k+4,

S[k=-2 ... infty] (k+4)(k+3)a_(k+4).x^(k+2) +S[k=0 ... infty] a_k.x^(k+2) - 4 = 0

2a_2.x^0 +6a_3.x +S[k=0 ... infty] {(k+4)(k+3)a_(k+4) + a_k}.x^(k+2) - 4 = 0

Equating the coeffts of x to zero,

2a_2 - 4 = 0, a_3 = 0, {(k+4)(k+3)a_(k+4) + a_k} = 0, k = 0 ... infty

a_2 = 2, a_3 = 0

(k+4)(k+3)a_(k+4) = -a_k, k = 0 ... infty

k = 0: 12a_4 = -a_0 -> a_4 = -1/12.a_0

k = 1: 20a_5 = -a_1 -> a_5 = -1/20.a_1

k = 2: 30a_6 = -a_2 -> a_6 = -1/15

k = 3: 42a_7 = -a_3 -> a_7 = 0

The final solution will have two constants of integration, a_0 and a_1.

**Series solution is: y(x) = a_0 + a_1.x + 2.x^2 - (1/12).a_0.x^4 - (1/20).a_1.x^5 - (1/15).x^6 + ...**