Answer
See below
Work Step by Step
We are given $y''-18y'+16y=7e^{4x}$
We have: $F(x)=7e^{4x}$
Obtain: $7e^{4x}(D-4)=0$
So, $D-4$ is the annihilator of $7e^{4x}$
Therefore, the general solution for the given differential equation is:
$y(x)=C_1e^{4x}+C_2xe^{4x}+C_3x^2e^{4x}+A_0x^2e^{4x}$
The trial solution is
$y_p(x)=A_0x^2e^{4x}$
