Given $y''' - 5y'' - y' + 5y = 3e^{-x}$, find the general solution.
I found the roots for the homogeneous solution to be 5, 1, and -1:
$$(r - 5)(r + 1)(r - 1)=0$$
$$y_h(x) = c_1e^x + c_2e^{-x} + c_3e^{5x}$$
Setting up the particular solution, I have:
$$g(x) = 3e^{-x}$$
$$y_p(x) = Axe^{-x}$$
$$y_p'(x) = Ae^{-x} - Axe^{-x} = A(1 - x)e^{-x}$$
I know I need to use the product rule to continue differentiating $y'$, but is there an easier method to do so?