= Solution
Let $z$ measure height above the release point, and use $\mathcal V=4\pi a^3/3$ for the bubble volume to avoid confusing it with a <speed>. For constant radius, <Newton's second law>, the specified <buoyancy> and the given <Stokes law> give
$$
\alpha\rho\mathcal V\dot u=\rho g\mathcal V-6\pi\mu au.
$$
With positive $\alpha,\rho,\mu,a$, define
$$
\tau=\frac{\alpha\rho\mathcal V}{6\pi\mu a}
=\frac{2\alpha\rho a^2}{9\mu},\qquad
U=\frac{\rho g\mathcal V}{6\pi\mu a}
=\frac{2\rho g a^2}{9\mu}.
$$
The <differential equation> becomes $\dot u+(u/\tau)=U/\tau$. Its solution with $u(0)=0$ is $u(t)=U(1-e^{-t/\tau})$. Integration using $z(0)=0$ yields
$$
\boxed{z(t)=U\left[t-\tau\left(1-e^{-t/\tau}\right)\right],\qquad
\lim_{t\to\infty}u(t)=U=\frac{2\rho g a^2}{9\mu}.}
$$
An arbitrary initial height is added to this expression. The effective inertia changes the relaxation time $\tau$, but not the <terminal velocity> obtained by balancing <buoyancy> and drag.
For the dissolving bubble assume $\beta>0$ and use the specified instantaneous <terminal velocity>, rather than the constant-radius transient formula. The bubble disappears at $t_*=a_0^2/\beta$, and
$$
z(t)=\frac{2\rho g}{9\mu}\int_0^t(a_0^2-\beta s)\,ds
=\frac{2\rho g}{9\mu}\left(a_0^2t-\frac{\beta t^2}{2}\right),\qquad0\leq t\leq t_*.
$$
Its total rise is therefore $\boxed{z(t_*)=\rho g a_0^4/(9\mu\beta)}$. This second model neglects the initial <acceleration> and uses exactly the terminal-speed approximation stipulated for the shrinking bubble.
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