= Solution
Set $v=y'$ to reduce the <second-order linear differential equation> to $xv'+v=0$. Equivalently, $(xv)'=0$, so $v=C_1/x$ on any <interval> excluding zero. The <derivative> condition at $x=1$ gives $C_1=2$.
Integrating on the positive <interval> containing the initial point yields $y=2\ln x+C_2$. The value $y(1)=-1$ then fixes $C_2=-1$, giving
$$
\boxed{y(x)=2\ln x-1,\qquad x>0.}
$$
Its <derivative> is $2/x$, and direct substitution verifies the equation. The selected solution diverges as $x\downarrow0$ and cannot extend through zero as a finite-valued <function>.
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