= Solution
On the interval $x>-1$ containing the <initial condition>, division by $1+x$ gives a <linear ordinary differential equation> with <integrating factor> $(1+x)^{-3}$. Consequently
$$
\frac{d}{dx}\bigl[(1+x)^{-3}y\bigr]=1+x,\qquad (1+x)^{-3}y=\frac{(1+x)^2}{2}+C.
$$
The initial value gives $C=1$. Thus
$$
\boxed{y=(1+x)^3\left[1+\frac{(1+x)^2}{2}\right].}
$$
This <polynomial> also satisfies the original equation at $x=-1$ and gives an extension to all real $x$. The <initial value problem> is uniquely determined on $x>-1$; the vanishing leading coefficient at $-1$ does not ensure unique continuation to its left.
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