Number of zeroes of solution to $y''(x)+e^{x^2}y(x)=0$ in $[0,3π]$ The question is to investigate the number of zeroes of $y''(x)+e^{x^2}y(x)=0$ in $[0,3π]$.
Solving this ODE would not be an easy task as one has to use the power series solution and then investigating the zeroes of the solution will require more analysis. I thought it to compare this ODE with the standard $y''(x)+y(x)=0$ whose solution has three or four zeroes in the interval $[0,3π]$.
Since the coefficient of $y(x)$ is $e^{x^2}\ge 1$ for $x\in [0,3π]$ so the solution of given ODE must have atleast three zeroes in $[0,3π]$. However, what  I thought was in the lights of Sturm-Comparison theorem so I am not sure.
Am I correct to interpret this?

(Moved from a deleted duplicate question, answered Feb 18 '17 at 8:44, since it contains a more elementary approach)
See the Sturm-Picone comparison theorem which tells you that you have at least as many roots as $\cos x$ on $[0,3π]$. 
You could apply it to the segments $[0,π]$, $[π,2π]$ and $[2π,3π]$ separately to get a better lower bound for the root numbers as you then compare to $y''+e^{(k\pi)^2}y=0$, $k=0,1,2$ so that you get on the respective intervals at least as many roots as $\cos(e^{k^2\pi^2/2}x)$ where the frequencies have numerical values $1,\; 139.045636661,\;
373791533.224$.
With a finer subdivision one can drive this lower bound up to $6.5·10^{17}$ roots inside the interval.

Details on the application of the Sturm-Picone comparison theorem (2/21/17): On $[0,3\pi]$ use $q_1(x)=1$ and $q_2(x)=e^{x^2}$. Then $q_1\le q_2$ and $p_1=1=p_2$, so the theorem applies and any solution $v$ of $v''+q_2v=0$ has at least one root between any two consecutive roots $x_k=k\pi$, $k=0,1,2,3$ of the solution $u(x)=\sin x$ of $u'+q_1u=0$. The roots of $\cos x$ have this property, Which is why one can say that $v$ has at least as many roots as $\cos x$ in that interval.
