Suppose now system of differential equations, namely, $$ \begin{equation} \ddot{y}(t) + \omega^{2}y(t) = \dot{z}(t) \\ \dot{z}(t) = (-A+\dot{y}(t))z(t)\end{equation} $$ I want to check, for which $y(t)$ the second derivative $\ddot{y}$ is zero. Combining the first and the second equations, I obtain following differential equation: $$ (-A+\dot{y}(t))z(t) - \omega^{2}y(t) = 0 $$ The question: may I divide on $z(t)\neq 0$, and after that differentiate obtained equation with respect to $t$, again setting $\ddot{y}$ to zero?
This question arises, because if I don't divide on $z(t)$, then the equation on $\dot{y}(t)$ will look differently after derivating with respect to $t$, and I don't understand, why.
If you are looking for approximate solution [according to your comment] in the case $\omega=0$ see the solution below :
$c_1$ , $c_2$ , $c_3$ are constants.