I am given that $r^7 - 6r^6 + 20r^5 - 56r^4 + 112r^3 - 160r^2 + 192r - 128 = (r-2)^3(r^2+4)^2$.
Then I know that the characteristic equation is then $r*(r-2)^3(r^2+4)^2 = 0$
This gives me $r = 0,2,2,2,+2i,-2i,+2i,-2i$
So I know the complimentary solution will be $y(t) = c_1 + c_2e^{2t} + c_3te^{2t} + c_4t^2e^{2t} + c_5\cos(2t) + c_6 \sin(2t) + c_7t\cos(2t) + c_8t\sin(2t)$.
However, what can I assume for my Y(t) for the particular solution?
Is $Ate^{2t} + Bt + C\sin(4t) + D\cos(4t)$ a good guess for $Y(t)$?
Are there more efficient ways of computing $Y^{(8)}(t)$ without a brute force approach of having to compute eight consecutive derivatives?
Thank You
The R.H.S. consists of a polynomial, a trigonometric function and a product of an exponential with a cube. Respectively
$$t^3\cdot e^{2t} + t + \sin(4t)$$
Therefore the right attempt would be
$$Y_p(t)~=~At + B + e^{2t}(Ct^3 + Dt^2 +Et +F) + G\sin(4t) + H\cos(4t) $$
But I do not know a way to compute the constants without derivate eight times.