Let $\tau=|t-t'|$. Integrating the equation for $x$ and using independence of the two noises gives
$$
R(\tau)
=C^2\tau+
\int_0^\tau\int_0^\tau
f_0^2e^{-\alpha|s-s'|}\,ds\,ds'.
$$
With the supplied integral,
$$
\boxed{
R(\tau)
=C^2\tau+
\frac{2f_0^2}{\alpha^2}
\left(\alpha\tau-1+e^{-\alpha\tau}\right)
}.
$$
For $\alpha\tau\ll1$, the active contribution is ballistic, $f_0^2\tau^2+O(\tau^3)$, in addition to the Brownian term. For $\alpha\tau\gg1$,
$$
R(\tau)
=\left(C^2+\frac{2f_0^2}{\alpha}\right)\tau
-\frac{2f_0^2}{\alpha^2}+o(1),
$$
so the long-time motion is diffusive with an enhanced <diffusion coefficient>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 / 4 / c / Solution

{title2=$D$}

The diffusion coefficient sets the linear growth of mean-square displacement. In one dimension the convention $\mathbb E[(X_t-X_0)^2]=2Dt$ is common.


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 Incoming links: Diffusion coefficient