Let's say we define the Fourier transform as (sorry for the abuse of notation)
$$\mathcal{F}[f(t); \omega] = f(\omega) = \frac{1}{2\pi}\int_{-\infty}^\infty dt f(t) e^{-i\omega t},$$ which then means that the inverse Fourier would be defined as $$\mathcal{F}^{-1}[f(\omega); t] = f(t) = \int_{-\infty}^\infty d\omega f(\omega) e^{i\omega t}. $$
Next, consider the following ODE $$M\frac{d}{dt}f(t) + \eta f(t) = g(t), $$ where $M, \eta > 0$ and $f, g$ are Fourier transformable. One can find the particular solution by Fourier transforming both sides (I want to use Fourier transforms for practice) which then yields (again with abuse of notation) $$f(t) = \mathcal{F}^{-1}[\frac{g(\omega)}{Mi\omega + \eta}; t], $$ which can be rewriten as $$f(t) = g(t)\ast \mathcal{F}^{-1}[\frac{1}{Mi\omega + \eta}; t].$$ Now my question is that it seems to me that if I chose the factor of $\frac{1}{2\pi}$ to be with the inverse Fourier then I would get a different result by a factor of $\frac{1}{2\pi}$ (based on this answer: Inverse Fourier transform of $ \frac1{a+\mathrm{j}w} $ which I agree with). This seems extremely strange to me since this prefactor of $\frac{1}{2\pi}$ is a convention. What is happening here?