I am looking through an example and need some help to understand it.
For the problem $-u'' + q(x)u = 0$, where $0<x<1$, $q \in \mathcal{C}(0,1]$ and $x^2q(x) \rightarrow q_0$ as $x \rightarrow 0^{+}$, the goal is to show that any non-trivial solution only has finitely many zeros if $q_0 > -1/4$.
If $q_0 > -1/4$, then we can use Sturm's comparison theorem, and compare our equation with $-u'' - \frac{1}{4x^2}u = 0$, and the solution $\sqrt{x}$. We see that $\sqrt{x}$ has no zero in $(0,1]$, and since $q(x) > -1/4$ for sufficiently small $x$, we have that any non-trivial solution $u$ to $-u'' + q(x)u = 0$ can't have an accumulation of zeros at $x=0$. This part I understand.
Next, it says that it can also only have finitely many zeros in $(0,1]$, because otherwise the zeros would accumulate at some point $x_0$, and this would imply that $u(x_0) = u'(x_0)$ (which in turn implies that $u(x) \equiv 0$).
The part that I don't quite understand is why an accumulation of zeros at $x_0$ implies that $u(x_0) = u'(x_0)$.
I had a thought that if we didn't have $u(x) \equiv 0$, we would have a solution that kept oscillating for all future times around $x_0$ (if there were infinitely many zeros there) but I can't quite piece it together. Can such a solution not exist? Or does it have to do with something else entirely?