Answer
See below
Work Step by Step
We are given $x^2y''-2y'+5y=7e^{x}\cos x+\sin x$
We have: $F(x)=G(x)+H(x)=7e^{x}\cos x+\sin x$
Obtain: $7e^{x}\cos x(D^2-2D+2)=0$
and $\sin x(D^2+1)=0$
So, $D^2-2D+2$ is the annihilator of $7e^{x}\cos x$ and $D^2+1$ is the annihilator of $\sin x$
Therefore,
$(D^2+1)(D^2-2D+2)(D^2-2D+5)=0$
The trial solution is
$y_p(x)=A_0\sin x+A_1\cos x+A_2e^x\sin x+A_3e^x\cos x$
