How to find a PDE satisfying $\rho_{tt}=c^2(\rho_{rr}+2r^{-1}\rho_r)$ with $R=r\rho$?

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I'm working on a course problem,

In a compressible and uniform fluid the equilibrium density and pressure are $\rho_0$ and $p_0$, respectively. Due to the passage of a compressible perturbation, there are linear changes in density $\rho$ and pressure $p$, and the resulting fluid velocity is $u(x, t) = u(x, t)\textbf{i} + v(x, t)\textbf{j} + w (x, t)\textbf{k}$, where $u, v$ and $w$ are small.

[... parts (i) and (ii) ...]

(iii) In a particular case $\rho = \rho(r, t)$, where (as usual) $r^2 = x^2 + y^2 + z^2$. By changing the variables from Cartesians to spherical polars show that $$\frac{\partial^2\rho}{\partial t^2} = c^2\left(\frac{\partial^2\rho}{\partial r^2}+\frac{2}{r}\frac{\partial\rho}{\partial r}\right). \tag{$\ast$}$$ Find the partial differential equation satisfied by $R = r \rho$, and hence write down the general solution of $(\ast)$. (You may quote d’Alembert’s general solution of the one-dimensional wave equation.)

I've done the "show that" but I'm not sure how to go about solving $(\ast)$. I sub in $\rho=Rr^{-1}$ and get \begin{align} (Rr^{-1})_{tt} & = c^2R(r^{-1}_{rr}+2r^{-1}r^{-1}_r) \\ & = c^2R(2r^{-3}-2r^{-3}) \\ & = 0, \end{align} which doesn't seem useful. Could someone give me a pointer? I realise it'll be something to do with the d'Alembert solution.