It is possible to find non trivial solutions $f(t) \in C_c^\infty$ to $\dot{f}(t)=2f(2t+1)-2f(2t-1),\,f(0)=1$ for the whole $\mathbb{R}$ domain?

Question

It is possible to find non trivial solutions $f(t) \in C_c^\infty$ to $\dot{f}(t)=2f(2t+1)-2f(2t-1),\,f(0)=1$ for the whole $\mathbb{R}$ domain?

I am trying to find examples of solutions of finite duration to differential equations (meaning here, they achieve zero in finite time and stays there forever after), and some functions that shows a similar behavior are smooth bump functions $\in C_c^\infty$ (I hope that from an initial condition and their differential equation I could made a finite duration solution), and in MathStackExchange I found this question with a lot of examples of smooth bump functions, but so far, the only differential equation I found describing them without definition issues is the one given in this answer, which is the equation I am asking for.

There, the equation is taken from this paper by Juan Arias de Reyna (2017), where the author found the Fourier Transform of $f(t)$ (highly nonlinear), but the explicit/exact formula for $f(t)$ is not given.

I tried to research about this differential equation since I never saw before a differential equation with arguments different from a simple variable $t$, so I am not sure if this equation belongs to a Functional Equation or instead to a Delayed Differential Equation, or other as which are studied in the book Generalized Solutions Of Functional Differential Equations by Joseph Wiener (I tried to read it "diagonally" looking for something similar unsuccessfully - it works with things with explicit discontinuities).

I have no more related experience than reading the related Wikipedia pages, but so far, it looks like that the differential equation stand the zero trivial solution $f(t) = 0$, but also it will solved by every constant $f(t)= c\equiv f(0)$ (since $\dot{c}=0=c-c$), so it looks like Uniqueness of solutions is not hold (actually Wolfram-Alpha gives this solution on its numerical plots).

I don't even sure if the initial condition given is enough information to solve it, so if required, as is done I think in the paper, include the following additional requirement: $$f(t) = 0,\,|t|\geq 1$$

I have also played in Wolfram-Alpha trying to fit things like the other examples given in the mentioned question, or trying things like $y(x) = e^{-\log(1-x^2)^2}(1-x^4+|1-x^4|)$ which looks like the draw of the paper, but it didn't work so far (is really hard to make the two displaced and contracted versions to match their derivative).

So I hope you can answer this:

1. What kind of differential equation is this? (if it match any - this to look for references)
2. It is possible to find the exact solution $f(t) \in C_c^\infty$? (maybe someone already prove is not possible)
3. Obviously, which is the exact solution $f(t) \in C_c^\infty$?

---

Added later

My best attempt after trying "too many things" was this: $$y(x) = e^{-\log(\sqrt{1-x^4})^2}\left(\frac{1-x^4+|1-x^4|}{2}\right)^{14}$$ From what I am seeing, the derivative need to be more flat near $\pm 1/2$. The exponential function is so flat top that actually is the polynomial who is giving the form of the function, only been the exponential who made the function goes smoothly to zero at the edges $[-1,\,1]$. Maybe someone more experienced could made a polynomial fit.

Improved attempt: $$y(x) = e^{-\tanh^{-1}(x^4)^2}\left(\frac{1-|x|^\frac{5}{2}+|1-|x|^\frac{5}{2}|}{2}\right)^{4}$$ and as is seen in Desmos.

---

Added later 2

It looks the solution to the main differential equation behaves like a translated and extended version (with its own reflection to form a symmetric/odd function) of the also known Fabius function, on Wikipedia is shown some of their properties.

Answer 1

After some time I have learned that these kind of equation are known as Delayed Differential Equations, and for the question example the solution is a delayed version of the Fabius function: $$f(x)=\begin{cases} \text{FabiusF}(x+1),\quad |x|<1\\ 0,\quad |x|\geq 1\end{cases}$$ where some implementatations could be found here, as is used in this answer.