Zeros of a differential equation 
Is the part highlighted in green correct? Could there not be infinitely many zeros in the region $x<N$?
 A: If there are infinitely many zeros in a finite interval, then there is a limit point of zeros, and at this limit point the solution and its derivative are zero. Which implies by uniqueness that the entire solution is zero.

Details: There is a sequence $t_k$ of zeros of $y(t)$ converging to the limit point $t^*$. Since $y$ is continuous, $y(t^*)=0$. Since $y$ is continuously differentiable ($q$ should be supposed to be continuous), by Rolle's theorem there is a zero $s_k$ of $y'(t)$ in between $t_k$ and $t_{k+1}$. The sequence of the $s_k$ also converges to $t^*$ leading to $y'(t^*)=0$. The IVP with $y(t^*)=0$, $y'(t^*)=0$ has the unique solution $y\equiv 0$.

More details: Assuming the implicit but surely implied condition that $q$ is continuous, the linear ODE satisfies, trivially because of linearity, the Lipschitz condition of the Picard-Lindelöf theorem. More specifically, this is satisfied for the equivalent first order system
$$
\frac{d}{dt}\begin{bmatrix}y\\v\end{bmatrix}
=
\begin{bmatrix}0&1\\q(t)&0\end{bmatrix}
\begin{bmatrix}y\\v\end{bmatrix}
$$
Thus when the initial conditions are $y(t^*)=0$ and $v(t^*)=y'(t^*)=0$, then the unique solution is $y\equiv 0$ (and $v\equiv 0$).
