Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 1 / a

Divide by $x>1$. The product rule turns the linear ordinary differential equation intoEquivalently, $\ln x$ is an integrating factor after division by $x\ln x$. An antiderivative gives $\ln xy=\frac{1}{2}(\ln x{)}^2+C$. The initial condition at $e$ gives $C=-1/2$, henceIts derivative is $y^{\prime }=[1+(\ln x{)}^{-2}]/(2x)$, so substitution gives $x\ln x{y}^{\prime }+y=\ln x$, and $y(e)=0$. Every solution of a differential equation has the integrated form above, and the initial condition fixes its only constant. This agrees with the uniqueness theorem for ordinary differential equations on the stated interval.