Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 7E / Solution

After division by $z$, the coefficients are $P=2/z-1$ and $Q=-1/z$. At $0$, $zP$ and $z^{2}Q$ 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 $f^{\prime \prime }-f^{\prime }=0$, soA linearly independent pair is $1/z$ and $(e^z-1)/z$. Their Wronskian is $e^z/z^2\ne 0$. 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.