Here is the questions： $$\frac{\mathrm{d^2x} }{\mathrm{d} t^2}+p(t)x=0$$ and $$ n^2<p(t)<(n+1)^2,$$ where $p(t)$ is a continuous function of $t$，and $n$ is an integer.
How to prove that the solution of this ordinary differential equation is not $2\pi$-periodic?
Here are some of my thoughts:
According to the Sturm comparison theorem，if $p(t)>m>0$ and $x=\varphi (t)$ is a solution of the ODE,then we can get that for any two adjacent roots of $x(t)$ the distance between them is less than $\frac{\pi }{\sqrt{m}}$, because we can use $x''+mx=0$ to compare to this ODE.
The same if $p(t)<m$ and $x=\varphi (t)$ is a solution of the ODE, then we can get that for any two adjacent roots of $x(t)$, the distance between them is larger than $\frac{\pi }{\sqrt{m}}$, because we can use $x''+mx=0$ compare to this ode.
I would be grateful if you could help me
If you shift the origin to the first root $x_0$, show that $$x_{2n}<2\pi<x_{2n+2}.$$ Note that for a periodic function, you would also need that not only the values, but also the derivatives repeat. However the odd-numbered roots have an opposite sign to $x_0$ in the derivatives.
You already enumerated all the necessary facts.