Let me retype it in MathJax
First solve homogenous complement equation
\begin{align*}
y'' + 2y' + 5y &= 0\\
r^2 + 2r + 5 &= 0\\
r_1 = 1+2i,\ &r_2 = 1-2i
\end{align*}
And
\begin{align}
y_c = c_1e^{-t}\cos(2t) + c_2e^{-t}\sin(2t)
\end{align}
Now solve non homogenous part
\begin{align*}
y'' + 2y' + 5y &= 8e^{-t}\\
\end{align*}
Let $Y_1 = Ae^{-t}$, then $Y_1' = -Ae^{-t},\ Y_1'' = Ae^{-t}$
\begin{align*}
A - 2A + 5A &= 8\\
A &= 2\\
Y_1 = 2e^{-t}
\end{align*}
And
\begin{align*}
y'' + 2y' + 5y &= 34\sin(2t)\\
\end{align*}
Let
\begin{align*}
Y_2 &= B\cos(2t) + C\sin(2t)\\
Y_2' &= 2C\cos(2t) - 2B\sin(2t)\\
Y_2'' &= -4B\cos(2t) - 4C\sin(2t)\\
\end{align*}
and
\begin{align*}
-4B\cos(2t) - 4C\sin(2t) + 4C\cos(2t) - 4B\sin(2t) + 5B\cos(2t) + 5C\sin(2t) &= 34\sin(2t)\\
B\cos(2t) + C\sin(2t) + 4C\cos(2t) - 4B\sin(2t) &= 34\sin(2t)
\end{align*}
Solve the linear equation
\begin{align*}
B+4C&=0\\
C-4B&=34\\
B = -8 &,\ C=2\\
Y_2 = 2\sin(2t)&-8\cos(2t)
\end{align*}
Combining all we have
\begin{align*}
y = c_1e^{-t}\cos(2t) + c_2e^{-t}\sin(2t) + 2e^{-t} + 2\sin(2t)-8\cos(2t)
\end{align*}
