Imagine, that I can prove $$\epsilon^\prime(x) \le -x\epsilon(x)$$ for a function $\epsilon:\mathbb R \rightarrow \mathbb R_0^{+}$. Does this imply $$\epsilon(x) \le c e^{-\frac{x^2}{2}}$$ whereby $c \in \mathbb R$ is a constant? (The solution of the DGL $f^\prime(x) = -xf(x)$ is $f(x) = c e^{-\frac{x^2}{2}}$).
When I also can prove $\epsilon(0) \le \alpha$, does this mean $\epsilon(x) \le \alpha e^{-\frac{x^2}{2}}$?
Yes, this is both true for $x \ge 0$. It even has a name and a Wikipedia page: Gronwall's inequality