= Solution
{c}
Divide by $x>1$. The <product rule> turns the <linear ordinary differential equation> into
$$
(\ln x\,y)'=\frac{\ln x}{x}.
$$
Equivalently, $\ln x$ is an <integrating factor> after division by $x\ln x$. An <antiderivative> gives $\ln x\,y=\tfrac12(\ln x)^2+C$. The <initial condition> at $e$ gives $C=-1/2$, hence
$$
\boxed{y(x)=\frac12\left(\ln x-\frac1{\ln x}\right),\qquad x>1.}
$$
Its <derivative> is $y'=[1+(\ln x)^{-2}]/(2x)$, so substitution gives $x\ln x\,y'+y=\ln x$, and $y(e)=0$. Every <solution of a differential equation> has the integrated form above, and the <initial condition> fixes its only constant. This agrees with the <uniqueness theorem for ordinary differential equations> on the stated <interval>.
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