= Solution
For $x>0$ write $f(x)=x^{-a}I(x)$, where $I(x)=\int_0^xe^{-t}t^{a-1}\,dt$. Because $a>0$, the lower endpoint is <integrable>. At each positive upper endpoint the <fundamental theorem of calculus> gives $I'(x)=e^{-x}x^{a-1}$. The <product rule> therefore yields
$$
\boxed{f'(x)=\frac{e^{-x}}x-\frac{a}{x}f(x).}
$$
Differentiating once more and substituting this expression for $f'$ gives
$$
\boxed{f''(x)=-\frac{e^{-x}}x-\frac{(a+1)e^{-x}}{x^2}+\frac{a(a+1)}{x^2}f(x).}
$$
Substitution now cancels the $e^{-x}$ terms and the terms proportional to $f$, proving $xf''+(1+a+x)f'+af=0$.
The <change of variables> $t=xu$ identifies $f$ as the <scaled lower incomplete gamma integral> $\int_0^1e^{-xu}u^{a-1}du$. It has a smooth extension at zero, with $f(0)=1/a$, $f'(0)=-1/(a+1)$ and $f''(0)=1/(a+2)$. Thus the apparent singularities in the derivative formulas cancel there; those formulas with denominators $x$ were derived for $x>0$.
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