= Solution
Take $x\geq0$ and a source position $\xi>0$. Before the source, the zero initial data require $G=0$. After it, set $G=e^{-2(x-\xi)}h(x-\xi)$; the homogeneous equation becomes $h''+a^2h=0$. Continuity of $G$ and a unit jump in $G_x$ at $x=\xi$ give $h(0)=0$, $h'(0)=1$. Thus the <causal Green function of a damped oscillator> is
$$
\boxed{G(x,\xi)=H(x-\xi)e^{-2(x-\xi)}\frac{\sin[a(x-\xi)]}{a}.}
$$
Here $H$ is the <Heaviside step function>. For $a=0$, take the continuous limit $\sin(as)/a\to s$, giving $G=H(x-\xi)(x-\xi)e^{-2(x-\xi)}$. The formula is also unchanged when a nonzero $a$ changes sign. Its continuity removes a delta derivative, and its unit first-derivative jump supplies the required <Dirac delta function>.
For the forced equation use $a=2$. The zero-initial-data <Green function> representation is
$$
y(x)=\frac12\int_0^xe^{-2(x-\xi)}\sin[2(x-\xi)]\sin2\xi\,d\xi.
$$
The sine-product identity reduces this to elementary exponential-trigonometric integrals. Evaluating gives
$$
\boxed{y(x)=\frac{\sin2x-2\cos2x+e^{-2x}(2\cos2x+\sin2x)}{20}.}
$$
For an independent check, the steady particular part is $\sin2x/20-\cos2x/10$, while the decaying homogeneous part has coefficients $1/10$ and $1/20$ that impose $y(0)=y'(0)=0$.
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