= Solution
Set $s=r_0+\mu t$. The proposed <velocity> and its <derivative> are
$$
v=\frac{g}{4\mu}\left(s-\frac{r_0^4}{s^3}\right),\qquad\dot v=\frac g4\left(1+\frac{3r_0^4}{s^4}\right).
$$
The remaining term in the <linear ordinary differential equation> is
$$
\frac{3\mu}{s}v=\frac{3g}{4}\left(1-\frac{r_0^4}{s^4}\right),
$$
so the sum is exactly $g$. Also $v(0)=g(r_0-r_0)/(4\mu)=0$, establishing both the equation and its <initial condition>:
$$
\boxed{v(t)=\frac g{4\mu}\left[(r_0+\mu t)-\frac{r_0^4}{(r_0+\mu t)^3}\right].}
$$
The <integrating factor> $(r_0+\mu t)^3$ explains where the expression comes from:
$$
\frac d{dt}(s^3v)=gs^3,\qquad s^3v=\frac{g}{4\mu}(s^4-r_0^4).
$$
The result has units of <velocity>. Its continuous $\mu\to0$ limit is $gt$, despite the apparent $1/\mu$ in the written expression.
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