Question: Find the general solution of the given differential equation: $y'' - y' - 2y = cosh(2t)$ Hint: $cosh(t) = \frac{e^t + e^{-t}}{2}$
Write the differential equation into:
$$y'' - y' -2y = \frac{1}{2}e^{2t} + \frac{1}{2}e^{-2t}$$
homogeneous solution:
\begin{align}
r^2 - r - 2 &= 0 \notag\\
(r - 2)(r + 1) &= 0 \notag
\end{align}
$r_1 = 2$ and $r_2 = -1 \implies y_c = c_1e^{2t} + c_2e^{-t}$
$y'' - y' - 2y = \frac{1}{2}e^{2t} $:
\begin{align}
y_{p1} &= Ate^{2t}\notag\\
y' &= (A + 2At)e^{2t}\notag\\
y'' &= (4A + 4At)e^{2t}\notag
\end{align}
\begin{align}
(4A + 4At)e^{2t} - (A + 2At)e^{2t} - 2Ate^{2t} &= \frac{1}{2}e^{2t}\notag\\
4A + 4At - A - 2At - 2At &= \frac{1}{2}\notag\\
3A &= \frac{1}{2}\notag\\
A &= \frac{1}{6}\notag
\end{align}
$y'' - y' - 2y = \frac{1}{2}e^{-2t} $:
\begin{align}
y_{p2} &= Be^{-2t}\notag\\
y' &= (-2B)e^{2t}\notag\\
y'' &= (4B)e^{2t}\notag
\end{align}
\begin{align}
(4B)e^{2t} - (-2B)e^{2t} - 2Be^{-2t} &= \frac{1}{2}e^{-2t}\notag\\
4B + 2B - 2B &= \frac{1}{2}\notag\\
4B &= \frac{1}{2}\notag\\
B &= \frac{1}{8}\notag
\end{align}
from $y_c = c_1e^{2t} + c_2e^{-t}$, $y_{p1} = \frac{1}{6}te^{2t}$, and $y_{p2} = \frac{1}{8}e^{-2t}$ we get:
$$y = c_1e^{2t} + c_2e^{-t} + \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t}$$
