 Solution

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 $c_2=0$.