= Solution
Separate the transported initial datum from the gain term. Define
$$
F(f_0)(t,x,v)=e^{-t}f_0(x-tv,v),
$$
and the <Boltzmann Volterra operator>
$$
(\tau f)(t,x,v)=\int_0^t e^{-(t-s)}\int_{\mathbb R^d}
k\bigl(s,x-(t-s)v,v,v_*\bigr)f\bigl(s,x-(t-s)v,v_*\bigr)\,dv_*\,ds.
$$
The <Duhamel principle> makes the <linear Boltzmann equation> equivalent, whenever the <integrals> and solution are legitimate, to
$$
\boxed{f=F(f_0)+\tau f}.
$$
The gain samples velocities $v_*$ at the backward spatial position of the outgoing velocity $v$; replacing that spatial shift by one depending on $v_*$ would give a different equation.
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