Separation of variables for heat equation in cylindrical shell I already know how to apply the separation of variables method to solve transient radial heat equation inside a cylinder. But, when it comes to cylindrical shells, both Bessel J and Y functions appear in the solution and I don't know how to find the coefficients by taking advantage of orthogonality.
Here's the PDE:
$$u_t = \frac{k}{r}(ru_r)_r$$
with the following boundary/initial conditions:
\begin{align}
&r=r_3 &u_r = 0 \\
&r=r_2 &-ku_r + pu = 0 \\
&t=0   &u = u_0
\end{align}
After applying the separation of variables method, here's what I have:
\begin{align}
u(r,t) &= R(r)T(t) \\
T(t) &= A e^{-k\lambda^2t} \\
R(r) &= B J_0(\lambda r) + C Y_0(\lambda r)
\end{align}
I appreciate it if you could shed some light on how to proceed.

There are unique constants $A,B$ so that $R(r)=AJ_0(\lambda r)+BY_0(\lambda r)$ satisfies the endpoint conditions
$$
              R'(r_3) = 0,\;\; R(r_3)=1.
$$
This is because $R(r_3)=0=R'(r_3)$ implies $R\equiv 0$, and this is true for any constants $A,B$. Such a solution is
$$
          R(r) = \frac{J_0(\lambda r)Y_0'(\lambda r_3)-J_0'(\lambda r_3)Y_0(\lambda r)}{J_0(\lambda r_3)Y_0'(\lambda r_3)-J_0'(\lambda r_3)Y_0(\lambda r_3)}
$$
Indeed, the numerator $N$ is a solution of the Bessel equation with eigenvalue $\lambda$ such that $N'(r_3)=0$ and $N(r_3)\ne 0$. So $R$ is a solution of the Bessel equation with $R'(r_3)=0$ and $R(r_3)=1$. The eigenvalue equation for $\lambda$ is then determined by the condition
$$
              -kR'(r_2)+pR(r_2) = 0.
$$
