Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 8

Use reduction of order by setting $y=xv$. Then $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$, so the differential expression simplifies toIn the homogeneous case $x{v}^{\prime }$ is constant. Hence $v=c_1+c_2\ln x$, and a basis of the solution space isTheir Wronskian is $x\ne 0$ on $[1,e]$, proving linear independence. No nonzero homogeneous solution satisfies both endpoint conditions: $y(1)=0$ sets $c_1=0$, and $y(e)=0$ then sets $c_2=0$.

Choose the homogeneous solutions $u(x)=x\ln x$, which vanishes at $1$, and $v(x)=x(\ln x-1)$, which vanishes at $e$. Their Wronskian in the order $u{v}^{\prime }-u^{\prime }v$ is $x$. For a source point $1<\xi <e$, seek the Green's function asIt is continuous at $x=\xi$; otherwise its second distributional derivative would contain a derivative of the Dirac delta function, which the source does not contain. Integrating the differential equation across the source, using continuity and the leading coefficient one, gives the jump conditionContinuity and this derivative jump requireSolving with the Wronskian $\xi$ gives $A=v(\xi )/\xi =\ln \xi -1$ and $B=u(\xi )/\xi =\ln \xi$. ThereforeThe two values at $x=\xi$ are identical. The derivative jump is explicitlyBoth endpoint values vanish and the differential expression is zero off the diagonal, completing the verification of the Green's function.

By superposition, $y(x)={\int }_1^{e}G(x,\xi )f(\xi )d\xi$. Split the integral at $\xi =x$: the portion $\xi >x$ uses the first branch of the Green's function, and the portion $\xi <x$ uses the second. This yieldsFor $f(\xi )=1/\xi$, put $t=\ln x$. The two integrals areSubstitution and cancellation giveIndeed $y(1)=y(e)=0$. For a direct differential check,and thereforeThe homogeneous uniqueness shown in part (a) makes this the unique solution.