Number of zeroes of solution to $y''(x)+e^{x^2}y(x)=0$ in $[0,3π]$

For any function $\eta: [0,3\pi] \to \mathbb{R}$, consider functions of the form $$y_\eta(x) = \Im \left[ \exp\left( -\eta(x) + i\int_0^x e^{2\eta(t)} dt + K\right)\right]$$ where $K$ is an arbitrary constant. It is easy to check $y_\eta(x)$ satisfies an ODE

$$y_\eta''(x) + p_\eta(x)y_\eta(x) = 0 \quad\text{ where }\quad p_\eta(x) = e^{4\eta(x)} + \eta''(x) - \eta'(x)^2\tag{*1} $$

When $\eta(x) = \frac{x^2}{4}$, this reduces to $$y_\eta''(x) + \left(e^{x^2} + \frac{2-x^2}{4}\right)y_\eta(x) = 0\tag{*2}$$

Compare this to the ODE at hand

$$y''(x) + p(x)y(x) = 0 \quad\text{ where }\quad p(x) = e^{x^2}\tag{*3}$$

They look "approximately" the same. More precisely, the relative difference between the coefficients of the two ODE, $\frac{p_\eta(x) - p(x)}{p(x)} = \frac{2-x^2}{4}e^{-x^2}$, become tiny as $x$ increases. For example, at $x = 2$, this relative difference already falls below $1\%$. This suggests we can use solutions of $(*2)$ to bound the number of zeros for solutions of $(*3)$.

Notice $p(x) \ge p_\eta(x)$ for $x \in [\sqrt{2},3\pi]$. By Sturm-Picone comparison theorem, we have

$$\#\big\{ x \in [\sqrt{2},3\pi] : y(x) = 0\big\} \ge \#\big\{ x \in [\sqrt{2},3\pi] : y_\eta(x) = 0\big\} - 1$$

In general, if $x_1$ is a zero for $y_\eta(x)$ and $x_2 > x_1$ satisfies $\int_{x_1}^{x_2} e^{2\eta(t)} dt = \pi$, then $x_2$ is also a zero.
When we set $K$ to $-i\int_0^{\sqrt{2}}e^{t^2/2}dt$, $\sqrt{2}$ becomes a zero of $y_\eta(x)$. For this particular choice of $K$,

$$\begin{align} & \#\big\{ x \in [\sqrt{2},3\pi] : y_\eta(x) = 0\big\} = \left\lfloor \frac{1}{\pi}\int_{\sqrt{2}}^{3\pi} e^{t^2/2} dt\right\rfloor + 1\\ \implies & \#\big\{ x \in [0,3\pi] : y(x) = 0\big\} \ge \#\big\{ x \in [\sqrt{2},3\pi] : y(x) = 0\big\} \ge \left\lfloor \frac{1}{\pi}\int_{\sqrt{2}}^{3\pi} e^{t^2/2} dt\right\rfloor \end{align} $$ Together with $\displaystyle\;\frac{1}{\pi}\int_0^{\sqrt{2}}e^{t^2/2}dt \approx 0.658425\;$, we obtain a lower bound for number of zeros. $$\#\big\{ x \in [0,3\pi] : y(x) = 0\big\} - \frac{1}{\pi}\int_0^{3\pi}e^{t^2/2} dt > -1.66$$

To obtain an upper bound, we choose a new $\tilde{\eta} = \frac14\log\left(e^{x^2} + \frac{x^2}{4}\right)$. With help of an CAS, one can verify the corresponding $p_{\tilde{\eta}}(x) \ge p(x)$ for all $x \in [0,3\pi]$. By Sturm-Picone comparison theorem again, we obtain

$$\#\big\{ x \in [0,3\pi] : y_{\tilde{\eta}}(x) = 0\big\} \ge \#\big\{ x \in [0,3\pi] : y(x) = 0\big\} - 1 $$ By choosing a suitable $K$, we can make $y_{\tilde{\eta}}$ satisfy

$$ \#\big\{ x \in [0,3\pi] : y_{\tilde{\eta}}(x) = 0\big\} = \left\lfloor\frac{1}{\pi}\int_0^{3\pi} e^{2\tilde{\eta}(t)}dt\right\rfloor = \left\lfloor\frac{1}{\pi}\int_0^{3\pi} \sqrt{e^{t^2} + \frac{t^2}{4}} dt\right\rfloor $$ Since $\displaystyle\;\sqrt{e^{t^2} + \frac{t^2}{4}} \le e^{t^2/2} + \frac{t^2}{8}e^{-t^2/2}\;$, this leads to an upper bound for number of zeros.

$$\#\big\{ x \in [\sqrt{2},3\pi] : y = 0\big\} -\frac{1}{\pi}\int_0^{3\pi} e^{t^2/2} dt \le 1 + \frac{1}{8\pi}\int_0^{3\pi} t^2 e^{-t^2/2} dt \approx 1.0498678 $$ Combine these, we obtain an estimate of the number of zeros of $y(x)$ in terms of an integral.

$$-1.66 < \#\big\{ x \in [\sqrt{2},3\pi] : y(x) = 0\big\} -\frac{1}{\pi}\int_0^{3\pi} e^{t^2/2} dt < 1.05$$

According to WA, the integral evaluates to $$\frac{1}{\pi}\int_0^{3\pi} e^{t^2/2} dt \approx 663789642044913452.583222...$$ Since number of zeros is always an integer, we can conclude $$ 663789642044913451 \le \#\big\{ x \in [\sqrt{2},3\pi] : y(x) = 0\big\} \le 663789642044913453 $$