= Solution
After division by $z$, the coefficients are $P=2/z-1$ and $Q=-1/z$. At $0$, $zP$ and $z^2Q$ are <analytic functions>, so $0$ is a <regular singular point>. Its <indicial equation> is $r(r-1)+2r=r(r+1)=0$. There are no other finite <singular points of a differential equation>. At infinity the <regular singular point criterion for a second-order equation> would require $P=O(z^{-1})$, but $P\to-1$; hence infinity is an <irregular singular point>.
Substituting $y=f/z$ reduces the equation to $f''-f'=0$, so
$$
\boxed{y(z)=\frac{A+B e^z}{z},\qquad z\ne0.}
$$
A <linearly independent> pair is $1/z$ and $(e^z-1)/z$. Their <Wronskian> is $e^z/z^2\ne0$. The second has a removable singularity at $0$ and value $1$, whereas the first has a simple pole, agreeing with the two <characteristic exponents at a regular singular point>, $0,-1$. The <integer> difference does not require a <logarithm>. At infinity $e^z/z$ has an <essential singularity>, consistent with the <irregular singular point>; the existence of the simpler solution $1/z$ does not make infinity regular.
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