Consider a simple classical system that may execute simple harmonic motion, with equation of motion \begin{equation*} \frac{ d^2q(t) } { dt^2 }= - \omega^2q(t) \tag{1} \end{equation*}
I wish to prove, that if $x(t)$ is a solution to (1), then so is $x(-t)$, i.e. given \begin{equation*} \frac{ d^2x(t) } { dt^2 }= - \omega^2x(t) \tag{2} \end{equation*}
then we must also have
\begin{equation*} \frac{ d^2x(-t) } { dt^2 }= - \omega^2x(-t) \tag{3} \end{equation*}
I guess I could use the general solution of (1) to show that $x(-t)$ is a solution, if $x(t)$ is, but I would prefer not to think that way.
If we look at trying to get somewhere and do something with (3), we might start by considering it’s LHS.
So, start by considering $ \frac{ d^2x(-t) } { dt^2 }$ hence $ \frac{ dx(-t) } { dt } $
By the chain rule \begin{equation*} \frac{dx(-t)}{dt}= \frac {dx(-t)} {d(-t)}\frac {d(-t)} {d(t)} =-\frac {dx(-t)} {d(-t)} \end{equation*}
My question is: How should I understand $\frac {dx(-t)} {d(-t)}$ and relate it to $\frac {dx(t)}{dt}$?
I have a suspicion, that the understanding I am looking for, may involve the use of the idea of odd and even functions, but this suspicion may be based upon nonsense.
Other Information
I have an understanding of $\frac { dx} { dt }$. If we want a formula for it we may proceed as follows.
There is a function value $x(t)$. Let ‘$t$’ be changed by $\Delta t$, the function value changes to $x(t+\Delta t)$.
The change in the value of the function $\Delta x$, is given by
\begin{equation*} \Delta x = x(t+\Delta t) - x(t) \end{equation*}
Form the quotient $\frac{\Delta x } { \Delta t } $ \begin{equation*} \frac{\Delta x } { \Delta t }= \frac { x(t+\Delta t) - x(t) } { \Delta t} \end{equation*}
Take the limit of the quotient as $\Delta t \to 0$ and call this limit $\frac { dx} { dt }$. \begin{equation*} \frac { dx} { dt } =\lim_{\Delta \to 0} \frac { x(t+\Delta t) - x(t) } { \Delta t} \end{equation*}