I have questions regarding the solution to the following question.
For which $\alpha$ will the equation $x''+9x=0$ be disconjugate - that is, no non-trivial solution will have two or more zeros - on $J=[0,\alpha)$?
The general solution is $c_1\cos(3t)+c_2\sin(3t)=c\sin(3(t+a))$? Given any $0<\beta<\alpha$, $x''+9x=0$ is disconjugate on $[0,\beta]$ $\iff$ there exists a positive sol'n of $x''=9x=0$ on $[0,\beta]$.
For any $0<\beta<\pi/3$, let $x=\sin(3(t+a))$, where $a = \frac{1}{4}(\frac{\pi}{3}-\beta)$, then $x(t)>0$ in $[0,\beta]$ and x is a solution of $x''+9x=0$. Therefore, x is disconjugate in $[0,\beta]$ and x is a solution of $x''+9x=0$. Therefore, $x''+9x=0$ is disconjugate in $[0,\beta]$ for all $0<\beta<\frac{\pi}{3}$. On the other hand, if $\beta\geq\frac{\pi}{3}$, then $x(t)=sin(3t)$ has at least two zeros in $[0,\beta]$. Thus the eq $x''+9x=0$ is disconjugate on $J=[0,\alpha)$ only for any $0<\alpha<\frac{\pi}{3}$.
My question follows below:
Why is the inequality $0<\beta<\alpha$ a strict inequality?
Because you are considering the positivity of solutions on $[0,β]$, while the requirement for dis-conjugacy is that there is at most one solution on $[0,α)$.
You have the example solutions in your post, $\sin(3(t+\varepsilon))$ is positive on $[0,β]$ if $ε>0$ and $β+2ε\le α=\frac\pi3$, which requires $β<α$, and $\sin(3t)$ has roots at $0$ and $\frac\pi3$, which also prevents taking $β=α$.