= Solution
The perturbed <Hamiltonian> is $H'=(Q^2+P^2)/2-g(t)Q$. <Hamilton's equations> give $\dot Q=P$, $\dot P=-Q+g(t)$, hence
$$
\boxed{\ddot Q+Q=g(t).}
$$
In the paper's time units the natural frequency is one. The <retarded Green function> is $G_R(t)=\Theta(t)\sin t$: it vanishes before the source, is continuous at zero and has a unit derivative jump there. With Fourier convention $e^{-i\omega t}$,
$$
\boxed{G_R(\omega)=\frac1{1-(\omega+i0^+)^2}.}
$$
Both real-frequency poles are displaced below the real axis. <Causality> therefore selects the convolution
$$
Q_R(t)=\int_{-\infty}^t\sin(t-t')\frac{\omega_0}{2}e^{-\omega_0|t'|}\,dt'.
$$
For $t<0$ the exponential rises as $e^{\omega_0t'}$ throughout the integral. For $t\ge0$, split the integral at $t'=0$. Evaluation yields
$$
\boxed{Q_R(t)=\frac{\omega_0}{2(1+\omega_0^2)}
\begin{cases}
e^{\omega_0 t},&t<0,\\
e^{-\omega_0t}+2\omega_0\sin t,&t\ge0.
\end{cases}}
$$
The response is nonzero before $t=0$ because the prescribed force is already nonzero then; this does not violate <causality>. Both $Q$ and $\dot Q$ are continuous at zero. The force eventually decays, leaving the free oscillation $\omega_0^2\sin t/(1+\omega_0^2)$. A general solution adds $A\cos t+B\sin t$, but the retarded condition of no incoming free oscillation as $t\to-\infty$ fixes $A=B=0$.
The oscillator equation is well-defined on the full $(Q,P)$ plane. Interpreted through the positive-$p$ chart of part (c), it describes the original real variables only while $P>|Q|$, with $P=\dot Q$. The requested retarded oscillator solution is a continuation in the new variables; it is not a claim that this canonical chart remains valid at all times for every $\omega_0$.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2015/paper-1-retarded-oscillator.png]
{title=Original retarded response to a two-sided exponential force; the pre-zero response follows earlier nonzero forcing}
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