= Solution
{c}
For $c_1=2$, $c_2=4$, $c_3=5$, the friction <memory kernel> and its <Laplace transform> are
$$
\kappa(t)=2(1+4t)e^{-5t},
\qquad
\widetilde\kappa(s)=\frac2{s+5}+\frac8{(s+5)^2}
=\frac{2(s+9)}{(s+5)^2}.
$$
Using the <generalized fluctuation-dissipation relation> with equilibrium <velocity> <variance> $5$, the <velocity> <autocorrelation function> has <Laplace transform>
$$
\begin{aligned}
\widetilde\chi(s)
&=\frac5{s+\widetilde\kappa(s)}\\
&=\frac{5(s+5)^2}{s(s+5)^2+2(s+9)}\\
&=\frac{5(s+5)^2}{s^3+10s^2+27s+18}\\
&=\frac{5(s+5)^2}{(s+1)(s+3)(s+6)}.
\end{aligned}
$$
All three poles are negative, so the stationary relaxation is stable. A <partial fraction decomposition> gives
$$
\widetilde\chi(s)=\frac8{s+1}-\frac{10/3}{s+3}+\frac{1/3}{s+6}.
$$
Taking the inverse <Laplace transform> gives
$$
\boxed{\chi(\tau)=8e^{-\tau}-\frac{10}3e^{-3\tau}+\frac13e^{-6\tau},\qquad \tau\geq0.}
$$
The negative middle coefficient is compatible with an <autocorrelation function>; individual terms in a <partial fraction decomposition> need not themselves be separate physical modes with positive weights. Several useful checks follow directly:
$$
\chi(0)=8-\frac{10}3+\frac13=5,
\qquad
\chi'(0)=-8+10-2=0,
\qquad
\chi''(0)=8-30+12=-10.
$$
The <Volterra integral equation> for $\chi$ predicts exactly $\chi'(0)=0$ and $\chi''(0)=-\kappa(0)\chi(0)=-10$. Its <integral> is also $8-10/9+1/18=125/18=5/\widetilde\kappa(0)$, agreeing with the <Green-Kubo formula for diffusion>.
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