Differential equations? solve using power series

I'm having problems with this one differential equation, can someone help me out?

y''-2xy'+6y=0

The equation has a solution which is a polynomial of degree 3 with the leading term 8x^3

Assume that y = Σ(n = 0 to ∞) a(n) x^n.

Substituting this into the DE yields
[Σ(n = 2 to ∞) n(n-1) a(n) x^(n-2)] - 2x[Σ(n = 1 to ∞) na(n) x^(n-1)] + 6 Σ(n = 0 to ∞) a(n) x^n = 0.

Simplifying and re-indexing the first series:
Σ(n = 2 to ∞) n(n-1) a(n) x^(n-2) + Σ(n = 1 to ∞) -2na(n) x^n + Σ(n = 0 to ∞) 6a(n) x^n = 0
Σ(n = 0 to ∞) (n+2)(n+1) a(n+2) x^n + Σ(n = 1 to ∞) -2na(n) x^n + Σ(n = 0 to ∞) 6a(n) x^n = 0

So, we have
[2 a(2) + 6 a(0)] + Σ(n = 1 to ∞) [(n+2)(n+1) a(n+2) - (2n - 6) a(n)] x^n = 0.

Assuming that a(0) and a(1) are arbitrary constants, equating like coefficients yields
2 a(2) + 6 a(0) = 0 ==> a(2) = -3 a(0).

For n > 0:
(n+2)(n+1) a(n+2) - (2n - 6) a(n) = 0
==> a(n+2) = 2(n-3) a(n) / [(n+2)(n+1)].
Note that this recurrence is in steps of 2.
---------------
(i) If n is even:
a(2) = -3 a(0), from above.
a(4) = 2(-1) a(2) / (4 * 3) = 2 * (-3)(-1) a(0) / (4 * 3) = 2^2 (-3)(-1) a(0) / 4!
a(6) = 2(1) a(4) / (6 * 5) = 2^3 * (-3)(-1)(1) a(0) / 6!
...
a(2k) = 2^k * (-3)(-1)...(2k-5) a(0) / (2k)! for k = 2, 3, ...

(ii) If n is odd:
a(3) = 2(-2) a(1) / (3 * 2) = (-2/3) a(1).
a(5) = 2 * 0 a(3) / (5 * 4) = 0.
==> a(2k+1) = 0 for all k = 2, 3, 4, ...

Therefore,
y = a(0) [1 - 3x^2 + Σ(k = 2 to ∞) 2^k * (-3)(-1)...(2k-5) x^(2k)/(2k)!
+ a(1) [x - (2/3)x^3].
-----------------
In particular, any polynomial solution is of the form y = C(x - (2/3)x^3).
Setting C = -12 yields y = 8x^3 - 12x.

Partial double check:
48x - 2x(24x^2 - 12) + 6(8x^3 - 12x) = 0, as required.

I hope this helps!