Past exam of the mathematical methods I course of the University of Oxford / 2016 / 9 / b

The initial point lies in $x>1$, where division by $\ln x$ is valid. The first-order linear differential equationhas integrating factor $(\ln x{)}^2$. HenceThe condition at $x=e$ sets $C=0$, soThe solution cannot be continued regularly through $x=1$: its numerator tends to $-1$ while the denominator tends to zero. The real logarithm also requires $x>0$.