Can someone check my work

I have to find the solution to the following second order differential equation using power series. The initial values are:

y(0) = 0 & y'(0) = 1.

Here is what I did:

y '' + x^2 y = 0

y = x^2 * Σ (c_n) * x^n from n = 0 to oo = Σ (c_n) * x^(n+2) (same bounds).
y' = Σ (c_n)n * x^(n-1) from n = 1 to oo.
y '' = Σ (c_n)(n-1)n * x^(n-2) from n = 2 to oo.

Solving for the initial values, I got:

...............oo
y(0) = 0 = Σ (c_n) * x^n = (c_0) * 0^0 + ... = 1; ==> c_0 = 1 since 0^0 = 1.
...............n=0

Next I solved for the other value.

................oo
y'(0) = 0 = Σ (c_n)n * x^(n-1) = (c_1)*1 * 0^0 + ... = 0; => (c-1)*1 = 0, => c_1 = 0.
................n=1.

Now I substituted back in for my functions of x.

y'' + x^2 y = 0

=> Σ(from n=2 to oo) (c_n)*n(n-1)*x^(n-2) + Σ(from n=0 to oo) (c_n)*x^(n+2) = 0

After shifting the index on both series, pulling out a couple of terms, and getting both powers of x to be the same, I got:

.............................oo
2(c_2) + 6(c_3)x + Σ{ (c_(n+2))*(n+1)(n+2) + c_(n-2)}x^n = 0
.............................n=2

==> c_2 = 0 & c_3 = 0.

==> c_(n+2) = - c_(n-2) / [(n+2)(n+1)], n = 2, 3, 4 ...

Plugging in values for n, gave me:

c_4 = -c_0 / (3*4) = -1 / (3*4), since c_0 = 1.
c_5 = -c_1 / (4*5) = 0, since c_1 = 0
c_6 = -c_2 / (5*6) = 0, since c_2 = 0
c_7 = -c_3 / (6*7) = 0, since c_3 = 0
c_8 = -c_4 / (7*8) = -(-1 / (3*4)) / (7*8) = 1 / (3*4*7*8).

Therefore,

c_(2n+4) = 0 for n = 1, 2, 3, ...
c_(2n+1) = 0 for n = 0, 1, 2, ...
c_(4n) = ((-1)^n * c_1) / (3*4*7*8 * ... *(4n)(4n-1)) for n = 1, 2, 3, ..

oo
Σ (-1)^n * x^(4n) / [(4n)(4n - 1)] = -x^4/12 + x^8/56 - x^12/132 + .. - ..
n=1

Is my answer right?

Thanks for any help

Assume solution is y = ∑ [n=0,∞] c(n)x^n