What is the serries solution of this second order differential equation problem

This was a question off of an old differential equations test.

Find the first 7 terms of the fundamental series (around 0) of the following differential equation.
xy'' + y'+xy = 0 , xo = 0

10 points to the most comprehensive and easy to follow reply!
:)

Let y = sum {an x^n} = a0 + a1*x + ...

y' = sum{ n*an*x^{n-1} }
= a1 + 2*a2*x + 3*a3*x^2 + ...

y'' = sum{ n*(n-1)*an*x^{n-2} }
= 2*a2 + 6*a3*x + 12*a4*x^2 + ...

Now, plugging into the DE,

sum{ n*(n-1)*an*x^{n-1} } + sum{ n*an*x^{n-1} } + sum{ an * x^{n+1} } = 0
a1 + (2*a2+2*a2 + a0)*x + (6*a3 + 3*a3 + a1)x^2 + ...( n*(n-1)*an + n*an + a{n-2})x^{n+1}... = 0

This series solution must be identically zero for all x; so after grouping like powers of x, each term will be set to zero:

O(1): a1 = 0.

O(x): 4*a2 + a0 = 0
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O(x^{n+1}): n^2*an + a{n-2} = 0

The choice of a0 is left arbitrary here, and a1=0 from the above condition. This will cause all odd coefficients (and thus all odd powers of x in the series) to vanish. The coefficients can be found in closed form:

an = (-1/n^2)*a{n-2}
=(-1/n^2)*(-1/(n-2)^2)*a{n-4}
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= (-1)^(n/2) * product ( 1/k^2 )

where the product is taken over all even numbers k from 2 to n. Thus,

a2 = -a0/2^2
a4 = a0/( 2^2 * 4^2 )
a6 = -a0/( 2^2 * 4^2 * 6^2 )
...

This should cover the first 7 terms:

y(x) = a0 + 0*x -a0/4 *x^2 + 0*x^3 +a0/64 * x^4 + 0*x^5 - a0/ 2304 * x^6 -...