= Solution
Now $\dot m=\beta$ and $\dot M=\alpha-\beta$, so
$$
M(t)=M_0+(\alpha-\beta)t,\qquad
M\dot v+\alpha v=u\beta.
$$
For $\alpha>0$, set $w=u\beta-\alpha v$. Then $\dot w=-\alpha w/M$ and $w(0)=u\beta$. For $\alpha\ne\beta$, integration gives
$$
w=u\beta\exp\left(-\alpha\int_0^t\frac{ds}{M(s)}\right)
=u\beta\left(\frac{M}{M_0}\right)^{\alpha/(\beta-\alpha)}.
$$
Consequently
$$
\boxed{v=\frac{u\beta}{\alpha}
\left[1-\left(\frac{M}{M_0}\right)^{\alpha/(\beta-\alpha)}\right].}
$$
This solution of the <rocket equation with stationary mass accretion> covers increasing or decreasing rocket mass, as long as $M>0$. For <equal-rate rocket burning and accretion>, in the case $\alpha=\beta>0$, the mass is constant and the linear equation instead gives
$$
\boxed{v(t)=u\left(1-e^{-\alpha t/M_0}\right).}
$$
It is also the continuous equal-rate limit of the preceding expression when written in terms of $t$.
For the no-accretion limit, expand the power as an exponential:
$$
\left(\frac{M}{M_0}\right)^{\alpha/(\beta-\alpha)}
=1+\frac{\alpha}{\beta}\log\frac{M}{M_0}+O(\alpha^2).
$$
Substitution gives
$$
\boxed{\lim_{\alpha\to0}v=u\log\frac{M_0}{M},}
$$
recovering (a). At a fixed time with $M_0-\beta t>0$, the limiting mass is $M=M_0-\beta t$, so the same calculation recovers the time-dependent no-dust solution.
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