Solutions of the Laplace equation How do I find solutions $u=f(r)$ of the two-dimensional Laplace equation $u_{xx}+u_{yy}=0$ that depend only on the radial coordinate $r= \sqrt{x^2+y^2}$

..and here's how you arrive at the polar coordinate version:
$$ 
\begin{align}
& u := f(r(x,y)), \\
& u_x = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}, \\
& u_{xx} = \left(\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}\right)_x = \frac{\partial^2 f}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 + \frac{\partial f}{\partial r}\frac{\partial^2 r}{\partial x^2}, \\
& u_{xx} + u_{yy} = \frac{\partial^2 f}{
\partial r^2}\left(\left(\frac{\partial r}{\partial x}\right)^2 + \left(\frac{\partial r}{\partial y}\right)^2\right) + \frac{\partial f}{
\partial r}\left(\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2}\right), \\
& u_{xx} + u_{yy} = \frac{\partial^2 f}{
\partial r^2}\left(\frac{x^2}{x^2 + y^2} + \frac{y^2}{x^2 + y^2}\right) + \frac{\partial f}{
\partial r}\left(\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2}\right), \\
& u_{xx} + u_{yy} = \frac{\partial^2 f}{\partial r^2} + \frac{\partial f}{
\partial r} \left( - \frac{x^2 + y^2}{(x^2 + y^2)^{3/2}} + \frac{2}{\sqrt{x^2 + y^2}} \right), \\
& u_{xx} + u_{yy} = \frac{\partial^2 f}{\partial r^2} + \frac{\partial f}{
\partial r} \frac{1}{r} = 0.
\end{align}
$$
