= Solution
Use <reduction of order> by setting $y=xv$. Then $y'=v+xv'$ and $y''=2v'+xv''$, so the differential expression simplifies to
$$
y''-\frac{y'}x+\frac y{x^2}=xv''+v'=(xv')'.
$$
In the homogeneous case $xv'$ is constant. Hence $v=c_1+c_2\ln x$, and a <basis> of the solution space is
$$
\boxed{y_1=x,\qquad y_2=x\ln x.}
$$
Their <Wronskian> is $x\ne0$ 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 $c_2=0$.
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