Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 1 / c / Solution

For $y=x$, the first and second derivatives are one and zero, giving $-2x+2x=0$. To find an independent solution, use reduction of order and write $y=xv(x)$. Cancellation of the terms proportional to $v$ leavesWriting $w=v^{\prime }$ and using separation of variables givesSince $1/[x^2(1-x^2)]=x^{-2}+(1-x^2{)}^{-1}$, integration yields the general solutionThe second factor divided by $x$ has nonzero derivative $1/[x^2(1-x^2)]$, so the two solutions are linearly independent.