= Solution
Substitute the <power series> and its termwise <derivatives> into the <ordinary differential equation>. The coefficient of $x^n$, for $n\geq0$, is
$$
(n+2)(n+1)a_{n+2}+[n(n-1)-6n+12]a_n=0.
$$
The bracket factors as $(n-3)(n-4)$, so the <power-series solution of a differential equation> satisfies
$$
\boxed{a_{n+2}=-\frac{(n-3)(n-4)}{(n+1)(n+2)}a_n.}
$$
The odd and even <recurrence relations> terminate: $a_3=-a_1$, $a_5=0$, and $a_2=-6a_0$, $a_4=a_0$, $a_6=0$. Thus the full solution is $y=a_0(1-6x^2+x^4)+a_1(x-x^3)$. The two <polynomials> have independent initial data at zero, and the leading coefficient $1+x^2$ never vanishes on the real axis, so every real <solution of a differential equation> for this equation is their <linear combination> on the entire real axis. The given initial data select $a_0=0$, $a_1=1$:
$$
\boxed{y(x)=x-x^3.}
$$
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