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

With the change of variables $z=y^2$, multiplication of the differential equation by $2y$ gives $z^{\prime }+2xz=2x$. The integrating factor is $e^{x^2}$, soThe initial condition gives $C=3$. Since $y$ starts negative and $z\ge 1$, continuity selects the negative square root throughout:This never reaches the singular value $y=0$.

Separation of variables givesHere $x>1$ makes the argument of the second logarithm positive. Substituting the initial condition yields $3y^2+1=4(x^2-1{)}^3$. The positive branch is thereforeThis is the maximal connected interval containing $\sqrt{2}$ on which this branch solves the original differential equation. At its left endpoint $y=0$, the original right-hand side has a zero denominator, so one cannot extend the solution through that point merely by continuing the squared equation.

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.