Solution
= Solution
{c}
With the <change of variables> $z=y^2$, multiplication of the <differential equation> by $2y$ gives $z'+2xz=2x$. The <integrating factor> is $e^{x^2}$, so
$$
(ze^{x^2})'=2xe^{x^2},\qquad z=1+Ce^{-x^2}.
$$
The <initial condition> gives $C=3$. Since $y$ starts negative and $z\geq1$, <continuity> selects the negative <square root> throughout:
$$
\boxed{y(x)=-\sqrt{1+3e^{-x^2}}\quad(x\in\mathbb R).}
$$
This never reaches the singular value $y=0$.