Cfrgtkky

$$ left(frac{d^2 u}{d x^2} right)+frac{2}{x+x_0} frac{d u}{d x}=frac{lambda}{k} u; quad x in [0,L], k,x_0>0 $$

$$u(0)=0$$
$$u(L)=0$$

Rewrite

$$ (x+x_0)^2frac{d^2u}{dx^2} + 2(x+x_0)frac{du}{dx} - frac{lambda}{k}(x+x_0)^2u = 0 $$

A non-trivial solution only exists for $lambda < 0$. The substitution $z = sqrt{frac{-lambda}{k}}(x+x_0)$ gives

$$ frac{d^2u}{dz^2} + 2zfrac{du}{dz} + z^2 = 0 $$

The solution is in the form of spherical Bessel functions for $n=0$ which happen to have a closed form

$$ u(z) = c_1j_0(z) + c_2y_0(z) = frac{c_1sin z - c_2cos z}{z} $$

The boundary conditions are $u(alpha x_0) = u(alpha (L+x_0)) = 0$ where $alpha = sqrt{frac{-lambda}{k}}$. We need to pick $alpha$ such that two of the zeroes of this function align with the boundary.

Case 1: If $c_2=0$, then the zeroes are the zeroes of $sin z$, which are $z_n = npi$. Since any 2 roots always differ by an integer multiple of $pi$, we have

$$ alpha(L+x_0) - alpha x_0 = alpha L = m pi implies alpha = frac{mpi}{L} $$

where $m$ is an integer. Note that $alpha x_0 = frac{mpi x_0}{L}$ must also be an integer multiple of $pi$, therefore $frac{x_0}{L}$ must be rational.

Case 2: If $c_1=0$, then the zeroes are the zeroes of $cos z$, which are $z_n = (n+frac12)pi$. Similarly, these roots also differ by an integer multiple of $pi$, so we have the same eigenvalue as Case 1, with the condition

$$ alpha x_0 = frac{mpi x_0}{L} = (2n+1)frac{pi}{2} $$

or $frac{2x_0}{L} = frac{2n+1}{m}$ must be rational.

Case 3: If $c_1,c_2 ne 0$, then we can write

$$ u(z) = frac{c_1(sin z - mu cos z)}{z} $$

where the zeroes occur at $tan z_n = mu$ for some $mu in (0,infty)$. These roots also differ by an integer multiple of $pi$, so yet again we have $alpha = frac{mpi}{L}$.

Combining the 3 cases, the eigenfunction can be simplified as

$$ u_m(z) = frac{cos z_m sin z - sin z_m cos z}{z} $$

where $z_m = alpha_m x_0 = frac{mpi x_0}{L}$