Solving the inhomogeneous wave equation in polar coordinates?

Imagine you have the following wave equation, with an excitation term $$\mu/r^2$$: $$\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2} - \Delta \psi = \frac{\mu}{r^2}$$ $$\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2} - \frac{\partial^2 \psi}{\partial r^2} - \frac{1}{r}\frac{\partial \psi}{\partial r} - \frac{1}{r^2} \frac{\partial^2 \psi}{\partial \theta^2} = \frac{\mu}{r^2}$$ Using separation of variables: $$\psi(r, \theta, t) = \phi(r, \theta) T(r)$$: $$\frac{d^2T}{dt^2} = - k^2 c^2 T\\ \; \\ \frac{\partial^2 \phi}{\partial r^2} + \frac{1}{r}\frac{\partial \phi}{\partial r} + \frac{1}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} - \frac{\mu}{r^2} =-k^2\phi$$ The first equation is simply harmonic motion: $$T(t) \propto e^{ikc t}$$ The second one is a little bit trickier. Because of the linearity of the equation I started solving the homogeneous equation: $$\frac{\partial^2 \phi}{\partial r^2} + \frac{1}{r}\frac{\partial \phi}{\partial r} + \frac{1}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} = -k^2\phi \\ \; \\ \phi(r, \theta) \propto J_m(kr)e^{im\theta}$$ Where $$J_m(x)$$, is the Bessel Function of first kind. Then I assumed a particular solution would look like: $$\phi(r, \theta) = u(r, \theta) J_m(kr)e^{im\theta}$$ And my problem would be reduced to find $$u(r, \theta)$$. But, I cannot continue. Could someone help? What I'm doing wrong? Any hint?

If you have in inhomogenous term only dependent on $$r$$, you should definitly do a seperation of variables for $$r$$ and not $$t$$. In fact, as @naturallyInconsistent has already pointed out, you can even do a seperation of variables for all three of them. I don't know if this is what you want, or if you want to keep it more general by using $$\phi(r,\vartheta)$$, but just in case I have written down the calculations for seperating all three variables:
With $$\psi(t,r,\vartheta)=T(t)R(r)\Theta(\vartheta)$$ you get: $$\frac{\mathrm{d}^2T}{\mathrm{d}t^2} =-k^2c^2T$$ $$\frac{\mathrm{d}^2R}{\mathrm{d}r^2} +\frac{1}{r}\frac{\mathrm{d}R}{\mathrm{d}r} =\frac{1}{r}\frac{\mathrm{d}}{\mathrm{d}r}\left(r\frac{\mathrm{d}R}{\mathrm{d}r}\right) =-\left(\frac{C}{r^2}+k^2\right)R$$ $$\frac{\mathrm{d}^2\Theta}{\mathrm{d}\vartheta^2}+\mu =C\Theta$$ with a constant $$C\in\mathbb{R}$$. (Keep in mind here, that from your second equation to your forth equation, the sign in front of $$\mu$$ has switched. I kept the upper sign.) If you now multiply the first equation with $$R\Theta$$ and divide by $$c^2$$, multiply the second equation with $$T\Theta$$ and multiply the third equation with $$TR$$ and divide by $$r^2$$ and then substract the second and the third equation from the first, you get exactly your wave equation.
• The equation for $$T$$ can be solved quite easily with the exponential function as you have already explained. The solution is $$T(t)=e^{ikct}$$.
• The equation for $$\Theta$$ can be simplified, when replacing $$\Theta$$ with $$\widetilde\Theta=\Theta-\frac{\mu}{C}$$ to get $$\frac{\mathrm{d}^2\widetilde\Theta}{\mathrm{d}\vartheta^2} =C\widetilde\Theta$$. Since $$\Theta$$ (and therefore $$\widetilde\Theta$$) is periodic with $$\Theta(\vartheta+2\pi)=\Theta(\vartheta)$$ (or $$\widetilde\Theta(\vartheta+2\pi)=\widetilde\Theta(\vartheta)$$), a suitable choice for $$C$$ is $$-n^2$$ with a natural $$n$$, so that $$\Theta(\vartheta)=a\sin(n\vartheta)+b\cos(n\vartheta)+\frac{\mu}{n^2}$$ with constants $$a,b\in\mathbb{R}$$.