Example 2.3.3.
We discovered in Example 2.3.2 that the complex solution of
\begin{equation*}
x'' + 6 x' + 5x = e^{2it}
\end{equation*}
to be \(x_c = A e^{2it}\text{,}\) where \(A = (1 - 12i)/145\text{.}\) Let us rewrite \(A\) in polar form. Since
\begin{equation*}
|A| = \frac{1}{\sqrt{145}},
\end{equation*}
we know that
\begin{equation*}
A = \frac{1}{\sqrt{145}} e^{i \theta},
\end{equation*}
where \(\theta = \arctan(-12) \approx -1.4877\text{.}\) Therefore,
\begin{equation*}
x_c = A e^{2it} = \frac{1}{\sqrt{145}} e^{i \theta} e^{2it} = \frac{1}{\sqrt{145}} e^{i (2t + \theta)}.
\end{equation*}
Our particular solution is the imaginary part of \(x_c\text{,}\)
\begin{equation*}
x_p(t) = \frac{1}{\sqrt{145}} \sin(2t + \theta) = \frac{1}{\sqrt{145}} \cos\left(2t + \theta - \frac{\pi}{2} \right) = \frac{1}{\sqrt{145} }\cos\left(2t - \phi \right),
\end{equation*}
where \(\phi \approx 3.058451\text{.}\) We say that \(\phi\) is the phase angle of our solution. The amplitude of our solution is \(1/\sqrt{145}\) and the period is \(\pi\) (Figure 2.3.4).
an oscillating steady state curve plotted against time