Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 1

After division by $x^2$, the linear ordinary differential equation becomesThe integrating factor is $e^{1/x}$. ThusThe initial condition $y(1)=1$ gives $C=e+1$, so

For $y=x$, the derivatives are $y^{\prime }=1$ and $y^{\prime \prime }=0$, and substitution gives $2(x-1)-2(x-1)=0$.

Use reduction of order: set $y=xv$, so $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$. Substitution and cancellation giveSince $x>0$, this reduces to $v^{\prime \prime }+2v^{\prime }=0$, hence $v^{\prime }=A{e}^{-2x}$ and $v=C_1+C_2{e}^{-2x}$. The general solution isThe two solutions are linearly independent, since their ratio $e^{-2x}$ is nonconstant; equivalently, their Wronskian is $-2x^2{e}^{-2x}\ne 0$.

The homogeneous linear differential equation $y^{\prime \prime }+y=0$ has general solution $C_1\cos x+C_2\sin x$. For a particular solution, use variation of parameters with $y_1=\cos x$, $y_2=\sin x$ and Wronskian one:The double-angle formulas and integration by parts giveSubstitution simplifies to $-x^2\cos x/4+x\sin x/4+\cos x/8$. Absorb the last term into the homogeneous linear differential equation solution. ThereforeIndeed, applying $d^2/d{x}^2+1$ to $x\sin x$ gives $2\cos x$, and applying it to $x^2\cos x$ gives $2\cos x-4x\sin x$; their difference divided by four is the required $x\sin x$.