Consider the differential equation: $$x''(t) + q(t) x(t) = 0 \qquad , a\leq t\leq b$$ , where $q$ is a continuous function. If $\min_{t\in [a,b]} q(t) > k^2 \pi^2 (b-a)^{-2}$ for some positive integer $k$, I'm asked to show that $x(t)$ (real solution) has at least $k$ zeros in the interval $[a,b]$.
My work so far: WLOG set $[a,b]=[0,1]$ (any other closed interval can be approached by a linear transformation, and the study will apply). The real solutions of: $$y''(t) + \pi^2 k^2 y(t) = 0 \quad , 0\leq t\leq 1$$ , are written as: $$y(t) = C_1 \cos(k\pi t) + C_2 \sin (k\pi t) \qquad , 0\leq t\leq 1$$ Having in mind the Sturm comparison theorem, I would have to show that there is some solution in this family which has at least $k+1$ zeros in the interval $[0,1]$, since that would imply that for each pair of consecutive zeros of $y$ in $[0,1]$, there's one zero of $x$; resulting in $k$ zeros of $x$. Again, writting $w=k\pi t \in [0,k\pi]$ the situation is equivalent to describing the zeros of $C_1\cos(w)+C_2\sin(w)$ in the interval $[0,k\pi]$. The thing is, no matter how I choose $C_1,C_2\in \mathbb{R}$, the solutions displayed seem to just have $k$ zeros in the studied interval. Also, the result should be true and the book that I took the problem from suggest using the exact same argument as I did.
Any suggestion will be appreciated :)
You are nearly done. $y(t)=\sin(k\pi t)$ has $k+1$ roots $t=\frac{m}{k}$, $m=0,1,...,k$ in the interval $[0,1]$.