= Solution
With $\Re\alpha<0$, the causal exponential is integrable and its <Fourier transform> is
$$
\widetilde h(\omega)=\int_0^\infty e^{(\alpha-i\omega)t}\,dt
=\left[\frac{e^{(\alpha-i\omega)t}}{\alpha-i\omega}\right]_0^\infty
=\boxed{\frac1{i\omega-\alpha}}.
$$
For the pendulum system, a <transfer function> is the <Fourier transform> of its causal <impulse response>, equivalently the multiplier relating the transformed forcing to the transformed zero-initial-data response. Transforming <derivatives> gives
$$
\widetilde R(\omega)=\frac1{(i\omega)^2+2i\omega+5}
=\frac1{(i\omega+1-2i)(i\omega+1+2i)}.
$$
Partial fractions therefore give
$$
\boxed{\widetilde R(\omega)=\frac1{4i}\left[\frac1{i\omega+1-2i}-\frac1{i\omega+1+2i}\right].}
$$
The exponential transform identifies the causal response as $R(t)=\tfrac12e^{-t}\sin2t$ for $t>0$ and zero for $t<0$. It solves the homogeneous equation for $t>0$, with $R(0^+)=0$ and $R'(0^+)=1$, the jump required by a unit impulse. The forcing is understood as switched on at zero; its transform is $1/(i\omega+4)$, and <convolution> with $R$ gives, if the time response is wanted,
$$
\theta(t)=\frac1{13}e^{-4t}-\frac1{13}e^{-t}\cos2t+\frac3{26}e^{-t}\sin2t\quad(t\geq0).
$$
This has both stated initial values equal to zero.
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