Here we can use Abel's Theorem to determine the value of the Wronskian
$$ \mbox{Therefore, } W = c e^{2t}, since p(t) = -2 $$
for part b, we must find the characteristic equation
$$ r^3 - 2r^2 - r + 2 = 0 $$
$$ (r-1)(r+1)(r-2) = 0 $$
$$ \mbox{Therefore, } r = 1, -1, 2 $$
$$ \mbox{Therefore, } y_1(t) = e^{t} y_2(t) = e^{-t} y_3(t) = e^{2t} $$
$$ \mbox{Therefore, the Wronskian} W(y_1, y_2, y_3)(t) = \begin{bmatrix} e^{t}&e^{-t}&e^{2t}\\ e^{t}&-e^{-t}&2e^{2t}\\e^{t}&e^{-t}&4e^{2t}\\ \end{bmatrix} $$
$$ det(\begin{bmatrix} e^{t}&e^{-t}&e^{2t}\\ e^{t}&-e^{-t}&2e^{2t}\\e^{t}&e^{-t}&4e^{2t}\\ \end{bmatrix}) = -6 e^{2t} $$
This is similiar to the solution found in part a, except that c = -6
for part c,
$$ \mbox{Assume} y(t) = Ate^{t} $$
$$ \mbox{Therefore, } y'(t) = Ae^{t}(1+t) $$
$$ \mbox{Therefore, } y''(t) = Ae^{t}(2+t) $$
$$ \mbox{Therefore, } y'''(t) = Ae^{t}(3+t) $$
Plugging this in the original equation
$$ Ae^{t}(t + 3 - 4- 2t - 1 - t - 2t) = 8Ae^{t} $$
$$ \mbox{Therefore, A = -4} $$
$$ \mbox{Therefore, the general equation is } y(t) = c_1e^{t} + c_2e^{-t} + c_3e^{2t} -4te^{t} $$
