= Solution
After division by $x$, the equation is
$$
y''+\left(\frac1x-1\right)y'-\frac1xy=0.
$$
The <removal of the first derivative from a second-order differential equation> uses
$$
y=e^{-\frac12\int(1/x-1)\,dx}v
=e^{x/2}x^{-1/2}v.
$$
With
$$
P=\frac1x-1,\qquad Q=-\frac1x,
$$
the normal-form coefficient is
$$
Q-\frac12P'-\frac14P^2
=-\frac14-\frac1{2x}+\frac1{4x^2}.
$$
Consequently
$$
\boxed{
v''-\frac14\left(1+\frac2x-\frac1{x^2}\right)v=0},
$$
so the constants in the PDF's expression
$1+Ax^{-1}+Bx^{-2}$ are
$$
\boxed{A=2,\qquad B=-1}.
$$
To classify infinity, put $z=1/x$. Since
$$
\frac{d^2v}{dx^2}=z^4v_{zz}+2z^3v_z,
$$
the coefficient of $v$ in the transformed equation has a pole of order four at $z=0$. This exceeds the order two allowed at a regular singular point. Hence $x=+\infty$ is an <irregular singular point at infinity>.
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