= Solution
In the shifted variable $t=y-1$, substitute the <Frobenius solution> $x=\sum_{j\geq0}a_jt^{j+\sigma}$ into $t(t+2)x''+x'+x=0$. Equating the coefficient of $t^{j+\sigma}$ gives
$$
(j+1+\sigma)[2(j+\sigma)+1]a_{j+1}+[(j+\sigma)(j+\sigma-1)+1]a_j=0.
$$
Thus the <linear recurrence relation>, valid for $j=0,1,\ldots$, is
$$
\boxed{a_{j+1}=-\frac{(j+\sigma)(j+\sigma-1)+1}{(j+1+\sigma)[2(j+\sigma)+1]}a_j.}
$$
For the <indicial root> $\sigma=0$, this becomes
$$
a_{j+1}=-\frac{j^2-j+1}{(j+1)(2j+1)}a_j,\qquad \boxed{a_1=-a_0}.
$$
For the <indicial root> $\sigma=1/2$, rename the coefficients $b_j$; then
$$
b_{j+1}=-\frac{j^2+3/4}{(j+3/2)(2j+2)}b_j,\qquad \boxed{b_1=-\frac14b_0}.
$$
With $a_0=b_0=1$, the first terms of the two <Frobenius solutions> are
$$
\boxed{x_1=1-t+\frac16t^2+O(t^3),\qquad x_2=t^{1/2}\left(1-\frac14t+\frac7{160}t^2+O(t^3)\right).}
$$
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