= Solution
For the exponent $\sigma=m$, the <linear recurrence relation> reduces to $(m+n)a_n=-a_{n-1}$. The excluded nonpositive integers ensure that no denominator is zero. Normalize $a_0=1$ and iterate:
$$
a_n=\frac{(-1)^n}{(m+1)(m+2)\cdots(m+n)}.
$$
The <ratio test> gives an infinite radius of convergence for the <power series> factor. Hence a solution is
$$
\boxed{y_m(x)=x^m\left(1+\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{(m+1)(m+2)\cdots(m+n)}\right).}
$$
For nonintegral $m$, choose instead the exponent $\sigma=0$. Since $n-m\ne0$ for every positive integer $n$, its <linear recurrence relation> gives $na_n=-a_{n-1}$. Thus
$$
\boxed{y_0(x)=e^{-x}.}
$$
The distinct nonintegral leading exponents establish <linear independence>. More explicitly, $W(e^{-x},y_m)=m e^{-x}x^{m-1}$ by the <Abel identity> and the leading coefficient at zero, so the <Wronskian> is nonzero. The factor $x^m$ uses the real branch on $x>0$.
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