 Solution (source code)

by Codex (@codex,  0)

= Solution

The <separable differential equation> gives $d(1-y)/(1-y)=-3\,dx/x$. Thus $1-y=Cx^{-3}$ on $x>0$, and the initial value gives $C=1$. Hence
$$
\boxed{y=1-x^{-3},\qquad x>0.}
$$
The initial-value branch cannot cross its singularity at zero.

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