= Solution
Set $u=1/y$, so $u'=-y'/y^2$. The equation becomes $x(x^2-1)u'+u=-x^3$. For $x>1$ its <integrating factor> is $\sqrt{x^2-1}/x$, giving
$$
\frac{d}{dx}\left(\frac{\sqrt{x^2-1}}x u\right)=-\frac{x}{\sqrt{x^2-1}},\qquad
u=-x+\frac{Cx}{\sqrt{x^2-1}}.
$$
The initial value is $u(2)=-2$, forcing $C=0$. Hence
$$
\boxed{y=-\frac1x.}
$$
Substitution into the original equation verifies this answer throughout $x>0$, including $x=1$ where the divided linear equation was singular. Its maximal connected interval through the initial point is $(0,\infty)$.
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