= Solution
Write $u=t-t'$ and $\Omega=\sqrt{\omega_0^2-\gamma^2/4}$. With the printed <Fourier transform> convention, a time <derivative> transforms to $i\omega$. The <retarded Green function> is
$$
G(u)=\int_{-\infty}^\infty\frac{d\omega}{2\pi}
\frac{e^{i\omega u}}{\omega_0^2-\omega^2+i\gamma\omega}.
$$
Its <poles> are $\omega=\pm\Omega+i\gamma/2$, both in the upper half-plane. For $u>0$ close the contour above; for $u<0$ close below, where there are no <poles>. <Residues> give
$$
\boxed{G(t-t')=\Theta(t-t')e^{-\gamma(t-t')/2}
\frac{\sin[\Omega(t-t')]}{\Omega}.}
$$
It is continuous at $u=0$ and has <derivative> jump $G'(0^+)-G'(0^-)=1$, reproducing the <Dirac delta function> source. Thus the solution initially at rest is
$$
\eta(t)=A\int_0^t G(t-s)e^{-\alpha s}\,ds\quad(t\geq0),
$$
and vanishes for $t<0$. Set $D=\alpha^2-\gamma\alpha+\omega_0^2=(\alpha-\gamma/2)^2+\Omega^2>0$. Evaluating the integral gives
$$
\boxed{\eta(t)=\frac A D\left[e^{-\alpha t}
-e^{-\gamma t/2}\cos(\Omega t)
+\frac{\alpha-\gamma/2}{\Omega}e^{-\gamma t/2}\sin(\Omega t)\right],\quad t\geq0.}
$$
The <homogeneous solution> constants are $C_1=-A/D$ and $C_2=A(\alpha-\gamma/2)/(D\Omega)$. Substitution at zero gives $\eta(0)=\dot\eta(0)=0$. <Causality> selects this response; an arbitrarily excited prehistory would add a <homogeneous solution>.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2013/paper-1-tp1-causal-response.png]
{title=Illustrative retarded impulse response and initially resting response to a decaying drive}
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