Solution
= Solution
Set $y=\sum_{j\geq0}a_jx^j$ and equate coefficients in the <power series>. The constant coefficient gives $a_2=0$. For $m\geq1$, the coefficient of $x^m$ gives
$$
(m+2)(m+1)a_{m+2}-[(m-1)(m-2)+6(m-1)+6]a_{m-1}=0.
$$
The square bracket is $(m+1)(m+2)$, so the <recurrence relation> is $a_{j+3}=a_j$ for $j\geq0$. The <initial conditions> give $a_0=1$, $a_1=0$, $a_2=0$. Hence
$$
\boxed{y(x)=1+x^3+x^6+x^9+\cdots,\qquad |x|<1.}
$$