= Solution
For $x>0$, division by $x$ gives
$$
y'-(2x+x^{-1})y=x.
$$
An integrating factor is $e^{-x^2}/x$, so
$$
y=xe^{x^2}\left(C+\int_0^xe^{-t^2}dt\right).
$$
Every choice except
$$
C=-\int_0^\infty e^{-t^2}dt=-\frac{\sqrt\pi}{2}
$$
grows like $xe^{x^2}$. The <bounded solution selected by a terminal condition> is therefore
$$
y_b(x)=-xe^{x^2}\int_x^\infty e^{-t^2}dt.
$$
Integration by parts, or the Gaussian-tail asymptotic, gives
$$
\int_x^\infty e^{-t^2}dt\sim\frac{e^{-x^2}}{2x},
\qquad y_b(x)\to-\frac12.
$$
It starts from $0$ at $x=0$, remains negative, and decreases monotonically toward the horizontal asymptote $-1/2$.
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