Solution
= Solution
The real equation has domain $x>0$. Its <integrating factor> is
$$
\mu(x)=\exp\left(\int(1+\ln x)\,dx\right)=e^{x\ln x}=x^x.
$$
Thus the <linear ordinary differential equation> becomes $(x^xy)'=x^xx^{-x}=1$. After integration, $x^xy=x+C$. The <initial condition> gives $2=1+C$, hence
$$
\boxed{y(x)=(x+1)x^{-x},\qquad x>0.}
$$