First we make the equation into the characteristic equation to find the roots:
$$ y''(t) + 4y'(t) + 5y(t) = e^{-2t} + 8sin(t) $$
$$ \mbox{Therefore, } r^{2} + 4r + 5 = 0$$
$$ \mbox{Therefore, } r = -2 \pm i $$
$$ \mbox{Therefore, } y(t) = c_1e^{-2t}cos(t)+ c_2e^{-2t}sin(t) $$
Now we change to non homogeneous equation and try to find the value of A:
$$y''(t) + 4y'(t) + 5y(t) = e^{-2t}$$
Let $$ y(t) = Ae^{-2t}$$
$$ \mbox{Therefore, } y'(t) = -2Ae^{-2t} \mbox{ and } y''(t) = 4Ae^{-2t}$$
$$ \mbox{Therefore, } 4Ae^{-2t} + 4(-2Ae^{-2t}) + 5Ae^{-2t} = e^{-2t}$$
$$ Ae^{-2t} = e^{-2t}$$
$$A = 1$$
Now change to non homogenous equation and find the value of B and C:
$$ y''(t) + 4y'(t) + 5y(t) = 8sin(t)$$
Let $$y(t) = Bcos(t) + Csin(t) $$
$$ \mbox{Therefore, } y'(t) = -Bsin(t) + Ccos(t) \mbox{ and } y''(t) = -Bcos(t) -Csin(t)$$
$$ \mbox{Therefore, } -Bcos(t) -Csin(t) + 4(-Bsin(t) + Ccos(t)) + 5(Bcos(t) + Csin(t)) = 8sin(t)$$
$$ \mbox{Therefore, } -Bcos(t) -Csin(t) + -4Bsin(t) + 4Ccos(t) + 5Bcos(t) + 5Csin(t) = 8sin(t)$$
$$ \mbox{Therefore, } 4Bcos(t) + 4Csin(t) - 4Bsin(t) + 4Ccos(t) = 8sin(t) $$
$$ sin(t)(4C - 4B) + cos(t)(4B + 4C) = 8sin(t) $$
$$ \mbox{Therefore, } 4C - 4B = 8 \mbox{ and } 4B + 4C = 0$$
$$ \mbox{Therefore, } B = -1 \mbox{ and } C = 1$$
$$ \mbox{Therefore, } y(t) = c_1e^{-2t}cos(t)+c_2e^{-2t}sin(t) + e^{-2t} + sin(t) - cos(t) $$
