The task is to find the upper and lower bounds of the distance between any two neighbouring non-trivial zeroes of
$$y^{\prime\prime} + 2xy = 0, \qquad 20 \leq x \leq 45$$
The given answer is that
$$\frac13< d < \frac12$$
I am trying to achieve this bound using the Sturm–Picone comparison theorem. First thing I did was to double the interval since the coefficient function of $y$ is $2x$, thus we're evaluating
$$y^{\prime\prime} + 2xy = 0, \qquad 40 \leq 2x \leq 90 \\ y^{\prime\prime} + p(x)y = 0, \qquad 40 \leq p(x) \leq 90 $$
Next, I divided the interval $[40;90]$ into four sub-intervals according to the perfect squares between. Therefore, we have the following sub-intervals and their lengths below.
$$[40;90] = [40;49) + [49; 64) + [64;81) + [81;90] \\ 50\,\,\,\,\,\,\, = \,\,\,\,\,\,\,\,9 \,\,\,\,\,\, + \,\,\,\,\,\,\,\, 15\,\,\,\,\,\,\, +\,\,\,\,\,\,\,\, 17\,\,\,\,\ +\,\,\,\,\,\,9\,\,\,\,\,\,\,\,\,\,$$
Next, I started with finding the upper bound: Starting with the fourth interval, consider $y_4 : y^{\prime} + 81y = 0$. Then the distance between the zeroes of the solutions to $y_4$ converges to $\dfrac{\pi}{\sqrt{81}} = \dfrac\pi9$. Thus, by the comparison theorem, this is an upper bound for the distance between solutions of the initially given equation. Therefore, to find the approximate number of zeroes between this sub-interval, we divide by the length of the interval
$$ d_4 < \dfrac{9}{\frac\pi9} = \frac{9\cdot9}{\pi} \approx \frac{3\cdot3\cdot9}{3} = 27$$
Similarly, for the third sub-interval, we look at $y_3 : y^{\prime} + 64y = 0$ to get that
$$d_3 < \dfrac{17}{\frac\pi8} = \frac{17\cdot8}{\pi} < \frac{18\cdot8}{\pi} \approx \frac{9\cdot3\cdot8}{3} = 72 $$
For $d_2$
$$d_2 < \dfrac{15}{\frac\pi7} = \frac{15\cdot7}{\pi} < \approx \frac{5\cdot3\cdot7}{3} = 35 $$
And lastly for $d_1$
$$d_1 < \dfrac{9}{\frac\pi{\sqrt{40}}} \approx \frac{9\cdot6}{\pi} \approx \frac{3\cdot3\cdot6}{3} =18 $$
Thus we have the following upper bound for the total number of roots: $$27 + 72 + 35 + 18 = 152 $$ We can subtract $3$ to account for the case of coincident roots at the boundary points. The length of the whole interval is $50$, thus, the upper bound over the whole interval is $$\frac{149}{50} < 3 $$
Now, before I decided to evaluate the lower bound, I noticed that this seems like an overapproximation in comparison to $0.5$. How can I reduce the bound? And is my application of the theorem correct?
I think in the task they where just using the general estimate $$ (2\pi)^2<40\le 2x\le 90 <(3.1\pi)^2 $$ so that the distance between roots is $$ \frac1{3.1}\le d\le\frac12. $$ However, $(3\pi)^2$ is only $88.826..$, so that one would have to carefully investigate to get a better bound on the remaining interval $[44.413,45]$