Let $a>0$ be a constant positive number. I am stuck trying to solve the following regular Sturm-Liouville problem:
$$\frac{\mathrm d}{\mathrm d x}((a+x)f'(x)) = -\lambda f(x),\qquad f(0)=f(1)=0$$
where $\lambda$ is the eigenvalue. I need to find the smallest eigenvalue $\lambda$ and the corresponding $f(x)$.
My attempt: According to Mathematica, the general solution of the ODE is:
$$f(x) = c_1 I_0(2\sqrt{-(a+x)\lambda}) + c_2 K_0(2\sqrt{-(a+x)\lambda})$$
where $I_0(x),K_0(x)$ are Bessel's modified functions of the first and second kind, respectively. However I cannot find a way to pick $c_1,c_2,\lambda$ so as to satisfy $f(0)=f(1)=0$, mainly because $K_0$ becomes complex and I don't know how to make it real! I am probably missing something obvious, but I keep obtaining that the only way to satisfy $f(0) = f(1)=0$ is to $c_{1,2}=0$, which of course violates the basic theorem of Sturm-Liouville theory (that a non-trivial solution must exist for regular problems).