Fredholm Alternative for Singular ODE Consider the following inhomogeneous boundary value problem, 
$$t^2 u'' + tpu' +qu = f(t), \ t \in [-1,1], \ \ u(1) = \alpha, \ u(-1) = \beta,$$
where $p$ and $q$ are constants.
I would like to determine a condition for the existence of a solution to this problem using the Fredholm alternative. To use it, I need to express the ODE above in self-adjoint form, $$-(a(t)u'(t))' + b(t)u = \tilde{f}.$$ However, doing so involves steps which are irreversible, i.e multiplying both sides of the differential equation by a power of $t$, which may take the value $t = 0$. This means that the self-adjoint equation is not equivalent to the original. Does this render the Fredholm alternative unusable? 
How could I determine a condition for the existence of a solution to this BVP?
 A: As RHowe remarked, 
the homogeneous equation $t^2 u'' + t p u' + q u = 0$ is Cauchy-Euler.   Its indicial roots are $r_\pm = (1-p \pm \sqrt{(1-p)^2 - 4 q})/2$.  Thus if those are distinct, the general solution of the homogeneous equation for $t > 0$ is $c_+ t^{r_+} + c_- t^{r_-}$.  Now since you want a solution on an interval containing $0$, a lot depends on whether $r_+$ and $r_-$ have positive, negative or $0$ real parts.  A solution $t^a$ with $\text{Re}(a) > 0$ can be continuously continued to $t < 0$ as $c |t|^a$ (the fact that this may not be differentiable at $t=0$ is not significant because the coefficients of $u'$ and $u''$ are $0$ there).  Moreover, there's an extra degree of freedom because the coefficient $c$ is arbitrary.  On the other hand, a solution with $\text{Re}(a) \le 0$ can't be made continuous at $t=0$ (except in the case $a=0$, where we take $t^0=1$ for all $t$)).  
For example, consider the case $p=q=1$, where the indicial roots are $\pm i$.  The general solution of the homogeneous equation for $t > 0$ is $c_1 \cos(\ln(t)) + c_2 \sin(\ln(t))$, which has no limit as $t \to 0+$ unless $c_1 = c_2 = 0$, so there are no nontrivial solutions of the homogeneous equation.  If the non-homogeneous equation has a solution (and it does for every polynomial, as $u = t^n/(n^2+1)$ solves $t^2 u'' + t u' + u = t^n$), that solution is unique; it may or may not satisfy the boundary conditions.
