Doraemonpaul's answer above is perfectly valid given the ODE originally suggested. However, a simpler method presents itself given the reformulation I have provided in the answers. For anyone interested, this is what I find.
We know that
$\left( \frac{1}{r}\frac{d}{dr} \left\{ r \frac{d}{dr} \right\} \right) R(r) = -k^2 R(r)$
has Bessel function solutions of the first and second kind, i.e. a linear combo of $J_0(kr)$ and $Y_0(kr)$ satisfies the above ODE. Because $n$ is an integer in the Bessel ODE we require both $J_0$ and $Y_0$ for a linearly independent solution. Next, note that
$\left( \frac{1}{r}\frac{d}{dr} \left\{ r \frac{d}{dr} \right\} \right) R(r) = k^2 R(r)$
is solved by a linear combo of $I_0(kr)$ and $K_0(kr)$, modified Bessel functions of the first and second kind. So, to solve
$\left( \frac{1}{r}\frac{d}{dr} \left\{ r \frac{d}{dr} \right\} \right) R(r) = m^4 R(r)$
we simply take a linear combination of all four. The general solution is given by
$R(r) = a_1 J_0(kr) + a_2 Y_0(kr) + a_3 I_0(kr) + a_4K_0(kr)$.
Then, we can apply boundary conditions. Noting that $Y_0(0)\rightarrow -\infty$ at the same rate as $K_0(0)\rightarrow \infty$ and that $J_0(0)=I_0(0)$, we require $a_2 = a_4$. The derivative yields
$R'(r) = -ka_1 J_1(kr) - ka_2 Y_1(kr) + ka_3 I_1(kr) - ka_2 K_1(kr)$.
Noting that $Y_1(0)$ and $K_1(0) \rightarrow \infty$ at the same rate, we find that $a_2 = a_4 \equiv 0$.
I will stop here, the other boundary conditions are easy to apply. You then get a second order system
$\left[\begin{array}{c c}
J_0(kR) - J_2(kR) & I_0(kR) + I_2(kR)\\
3J_1(kr) - J_3(kR) & 3I_1(kR) + I_3(kR) \\ \end{array} \right]$
$\left[\begin{array}{c}
a_1\\
a_3\end{array}\right]$
$=\left[\begin{array}{c}
0\\
0\end{array}\right]$
which can be solved numerically (unless anyone knows how to solve a coupled system of Bessel functions analytically? Or more specifically just the determinant. I would love to hear about it) to find allowable $kR$ values.
Best Answer
You can re-write your equation as $\dot{x}=-b|x|$, which has no turning points, i.e., $\dot{x}$ will never change sign. Then, split up your solution into two parts, one for $x>0$ and one for $x<0$:
$$x(t)=\begin{cases}Ce^{-bt}&\mbox{if }\,x>0 \\ Ce^{bt}&\mbox{if }\,x<0\end{cases}$$
Since in either solution $x(t)$ will be positive, the second case is a false solution, and so you get $x(t)=Ce^{-bt}$.