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?
Best Answer
How to differentiate $y'$ using the product rule? $y''=A(1-x)'e^{-x}+A(1-x)(e^{-x})'$, and so on.
It is easier (IMO) to use annihilator method [1] which says: