Past exam of the mathematical methods I course of the University of Oxford / 2019 / 4 / Solution

On the interval $x>0$ containing the initial point, divide the linear ordinary differential equation by $x$ to obtain $y^{\prime }-2y/x=x$. Its integrating factor is $e^{\int -2dx/x}=x^{-2}$, givingThe initial condition gives $1=C$. HenceDirect differentiation gives $y^{\prime }=x(3+2\ln x)$, so $x{y}^{\prime }-2y=x^2$ and $y(1)=1$. The coefficient of $y^{\prime }$ vanishes at zero; the boxed expression is the solution on the regular initial-value interval, rather than an assertion of unique continuation through zero.