Answer: [tex](d - 1)^3(d + 1)[/tex]
Explanation:
Let
[tex]f(x) = e^{-x} + 8xe^x - x^2 e^x[/tex].
[tex]d^n (f(x))[/tex] = nth derivative of f
Note that
[tex](d + 1) (e^{-x}) = d (e^{-x}) + e^{-x} = -e^{-x} + e^{-x} = 0
\\ \boxed{\Rightarrow (d + 1)e^{-x} = 0}
[/tex]
So, (d + 1) annihilates [tex]e^{-x}[/tex].
Note that
[tex](d + 1) (8x e^x - x^2 e^x) = d(8x e^x - x^2 e^x) + (8x e^x - x^2 e^x) \\ = (8e^x + 8x e^x - 2x e^x - x^2 e^x) + (8x e^x - x^2 e^x) \\ \boxed{(d + 1) (8x e^x - x^2 e^x) = 8e^x + 14x e^x - x^2 e^x}[/tex]
Let [tex]g(x) = 8e^x + 14x e^x - x^2 e^x[/tex]
For any real number [tex]\alpha[/tex] and positive integer n,
[tex](d - \alpha)^n (c_1 e^{\alpha x} + c_2 xe^{\alpha x} + c_3 x^2 e^{\alpha x} + ... + c_n x^{n - 1} e^{\alpha x}) = 0 [/tex] (1)
So, for g(x), [tex]\alpha = 1, n = 3[/tex]. Thus, using equation (1),
[tex](d - 1)^3 (8e^x + 14x e^x - x^2 e^x) = 0[/tex] (2)
Since, [tex]8e^x + 14x e^x - x^2 e^x = (d + 1) (8x e^x - x^2 e^x)[/tex] ,
[tex](d - 1)^3 ((d + 1) (8x e^x - x^2 e^x)) = (d - 1)^3 (8e^x + 14x e^x - x^2 e^x) = 0[/tex]
Moreover, since [tex](d + 1) (e^{-x}) = 0[/tex]
[tex](d - 1)^3(d + 1) (e^{-x}) = (d - 1)^3 (0) [/tex]
[tex]\\ \boxed{(d - 1)^3(d + 1) (e^{-x}) = 0 }[/tex] (3)
Hence based on equations (2) and (3),
[tex](d - 1)^3(d + 1) (e^{-x} + 8xe^x - x^2e^x)
\\ = (d - 1)^3(d + 1)(e^{-x}) + (d - 1)^3(d + 1)(8xe^x - x^2e^x)
\\ = 0 + 0
\\ \boxed{(d - 1)^3(d + 1) (e^{-x} + 8xe^x - x^2e^x) = 0}[/tex]
Therefore, the linear operator [tex](d - 1)^3(d + 1)[/tex] annihilates [tex]e^{-x} + 8xe^x - x^2e^x[/tex].
