= Solution
Choose downward <velocity> $v$ positive, and assume the collected material is initially at rest in the cloud's <inertial frame>. This assumption is required for the equation in the question. During $dt$, <momentum conservation> for the droplet plus the material it collects gives
$$
(m+dm)(v+dv)=mv+mg\,dt,
$$
up to second-order increments. Thus the <variable-mass system> satisfies
$$
m\dot v+v\dot m=mg,\qquad\dot v+\frac{\dot m}{m}v=g.
$$
Since $m\propto r^3$, $\dot m/m=3\dot r/r=3\mu/(r_0+\mu t)$, giving
$$
\boxed{\frac{dv}{dt}+\frac{3\mu}{r_0+\mu t}v=g.}
$$
The notation $dv/dt$ is appropriate because $v$ here depends only on <time>. The extra term accounts for the <momentum> needed to accelerate newly collected <mass>; using $m\dot v=mg$ would omit it. For incoming material with downward <velocity> $u$, the more general equation is $m\dot v=mg+(u-v)\dot m$.
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