I have to prove the following statement:
Show that, if $x(t) = \frac{1}{1+t}$ is a solution of the equation $x'' = f(x)$, then $\frac{1}{1 - t}$ is also a solution.
In a previous exercise I have shown that, if $x(t): ]a,b[\to \Bbb{R}^n$ is a solution for $x' = f(x)$ then for all $c \in \Bbb{R}$, $y(t) = x(t-c)$ is also a solution in the interval $]a+c, b+c[$.
This makes me believe that the proof I want will make use of that one in some way, or will have a similar reasoning. For that matter, what I tried doing was:
$$x(t) = \frac{1}{1+t}, y(t) = \frac{1}{1-t} \implies\\ y(t) = x(-t) = -x(t-2)$$
trying to get to the pattern where my alternative solution was a translation of the one I already know. But without success.
After that I noticed that in the previous exercise we have a first derivative, and thus I thought of a sort of "change of variables" where I would consider $x'$ and $y'$ instead, even though I didn't know how to formalize that change of variables in the statement. If the change were $X = x'$, would I do
$$x\ \text{is a solution for}\ x'' = f(x) \to\\X\ \text{is a solution for}\ X' = f(\int X)\text{?}$$
But even without being able to formalize that, I derived the functions to get
$$\begin{cases}y'(t) = -x'(-t)\\y'(t) = -x'(t-2)\end{cases}$$
but failed to take it any further.
After that, I assumed $x$ was a solution in the interval $]a, b[$ and thought that it would be reasonable to think that $y$ is a solution in $]-b, -a[$ given that $y(-t) = x(t), t \in ]-b, -a[\implies -t\in]a, b[$ but that made me think I could have problems with the domain.
Thus I did not manage to conclude the proof. Any help/hint is appreciated; furthermore I think a major problem I am having is that with representing the differential equation as $x'' = f(x)$. The "$f(x)$" is really confusing me a lot. Thank you for your time.