Let us examine the transfer function more closely to see if we can use it to write our particular solution in a more useable form. First, let us write
\begin{equation*}
P(i \omega ) = (i\omega)^2 + 2c(i \omega) + \omega_0^2 = (\omega_0^2 - \omega^2) +
2ic \omega.
\end{equation*}
in polar form,
\begin{equation*}
P(i \omega) = R e^{i \phi} = R(\cos \phi + i \sin \phi),
\end{equation*}
where
\begin{equation*}
R = \sqrt{(\omega_0^2 - \omega^2)^2 + 4 c^2 \omega^2}
\end{equation*}
and \(\phi\) is the angle defined by the equations
\begin{align*}
\cos \phi & = \frac{\omega_0^2 - \omega^2}{\sqrt{(\omega_0^2 - \omega^2)^2
- 4 c^2 \omega^2}},\\
\sin \phi & = \frac{2c\omega}{\sqrt{(\omega_0^2 - \omega^2)^2 - 4 c^2
\omega^2}}.
\end{align*}
Since \(2 c \omega \gt 0\text{,}\) we know that \(\sin
\phi \gt 0\text{.}\) Equivalently, \(0 \lt \phi \lt \pi\text{.}\) Thus,
\begin{equation*}
\phi = \phi(\omega) = \cot^{-1} \left( \frac{\omega_0^2 - \omega^2}{2c \omega}
\right).
\end{equation*}
Therefore, we can write the transfer function as
\begin{equation*}
H(i \omega) = \frac{1}{P(i \omega)} = \frac{1}{R} e^{-i \phi}.
\end{equation*}
We define the gain to be
\begin{equation*}
G(\omega) = \frac{1}{R} = \frac{1}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4 c^2
\omega^2}},
\end{equation*}
and we will rewrite the transfer function as
\begin{equation*}
H(i \omega) = G(\omega) e^{-i \phi(\omega)}.
\end{equation*}
Thus, the solution to
\begin{equation*}
x_c'' + 2c x_c' + \omega_0^2 x_c = Ae^{i \omega t}
\end{equation*}
is
\begin{equation}
x_c(t) = H(i \omega) A e^{i
\omega t} = G(\omega) A e^{i(\omega t - \phi)}.\tag{2.4.17}
\end{equation}
Taking the real part of
(2.4.17), our particular solution is
\begin{equation*}
x_p(t) = \real(x_c(t)) = G(\omega) A \cos(\omega t - \phi ).
\end{equation*}