How can I solve this ordinary differential equation? What are the Initial/Boundary Conditions used?

Question

I am having trouble following the solution to this problem, I can understand how to obtain the general solution but cannot figure out the conditions used to obtain the particular solution with the constants given, can anyone explain please?
The evolution of H is described by a reaction-diffusion equations (which is in diffusive equilibrium):
0=$r_t$T-$r_v$VH+$D_h$$\frac{d^2H}{R^2}$
V=0 for R<$R_2$ and V=$K_v$ elsewhere.
T=$K_t$ for $R_1$$<$R$<$$R_2$
We have q=$\sqrt{\frac{r_vK_v}{D_h}}$ and $H_0$=$\frac{r_tK_t}{r_vK_v}$
We can non-dimensionalize the equation with r=qR and h=$\frac{H}{H_0}$.
We obtain 3 equations (primes denote derivative w.r.t r):
0$<$r$<$$r_1$: $r^2$h''+2rh'=0
$r_1$$<$r$<$$r_2$:$r^2$h''+2rh'=-$r^2$
$r_2$$<$r: $r^2$h''+2rh'=$r^2$h
We assume $lim_{r→∞}$h(r)=0, h and its derivate are continuous at $r_1$ and $r_2$ and $lim_{r→0}$h(r) is finite. The general solutions are:
0$<$r$<$$r_1$: h(r)=$c_1$
$r_1$$<$r$<$$r_2$:h(r)=$c_2$-$\frac{c_3}{r}$-$\frac{r^2}{6}$
$r_2$$<$r: h(r)=$\frac{c_4e^{-r}}{r}$
Constants are:
$c_1$=$\frac{2r^3_1+3r^2_2+r^3_2}{6(r_2+1)}$-$\frac{r_1^2}{2}$
$c_2$=$\frac{2r^3_1+3r^2_2+r^3_2}{6(r_2+1)}$
$c_3$=$\frac{r_1^3}{3}$
$c_4$=$\frac{e^{_2}(r^3_2-r^3_1)}{3(r_2+1)}$
The problem is from a journal article, "The role of acidity in solid tumour growth and invasion" by Smallbone et al. 2005.