Zeroes of ODE in normal form

I need to find the number of zeroes of an ODE: $$y"+q(x)y=0$$ when $q(x)>0$ but monotonic decreasing. For instance:

$$y"+ \frac{y}{x^2} =0$$ where $x\in [1, \infty)$.

How can I use the Strum-Comparision Theorem to estimate the number of zeroes of it's solution?

(P.S. Here the number of zeroes won't be infinitely many contrary to the general case of $q(x)\gt 0$ and monotonic increasing.)

The example you can solve exactly, as that is Euler-Cauchy, $$ x^2y''(x)+y(x)=0\implies y(x)=Ax^{\frac12}\cos\left(\frac{\sqrt3}2\ln(x)+B\right) $$ giving infinitely many zeros.

In the more general case, you can get a first impression by using the WKB approximation $$ y\approx Aq(x)^{-\frac14}\cos\left(B+\int_1^x \sqrt{q(s)}\,ds\right) $$ to get a general idea if the number of zeros will be finite or infinite.

The WKB approximation can also be worked backwards to get expressions that are compatible with an exact approximation of the comparison theorem, see https://math.stackexchange.com/a/2075892 for an example. Note that this works originally for $q$ becoming very large, opposite to the current case.