Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 6 / iii / Solution

Substituting $a_n=2^n/n!$ into the $r=0$ recurrence, and $a_n=1/n!$ into the $r=1/2$ recurrence, satisfies every coefficient equation. They give $e^{2x}$ and $\sqrt{x}{e}^x$. Thus on a positive interval not crossing a singular point,Their Wronskian is $e^{3x}(2x-1)/(2\sqrt{x})$, nonzero for $x>0$, $x\ne 1/2$, proving independence on each ordinary interval. Both functions are analytic near $1/2$, so this is an apparent regular singular point: a pole in normalized coefficients need not force a singular solution. The Wronskian can vanish at that singular point without making the two functions linearly dependent on an ordinary interval.