Answer
See below
Work Step by Step
We are given $y''+y=4\cos 2x+3e^x$
We have: $F(x)=G(x)+H(x)=4\cos 2x+3e^x$
Obtain: $4\cos x(D^2+4)=0\\
3e^x(D-1)=0$
So, $D^2+4$ is the annihilator of $G(x)=4\cos2x$ and $D-1$ is the annihilator of $H(x)=3e^x$
Therefore, the general solution for the given differential equation is:
$y(x)=C_1\cos 2x+C_2e^x+A_0\cos 2x+A_1\sin 2x+A_2e^x$
The trial solution is
$y_p(x)=A_0\cos 2x+A_1\sin 2x+A_2e^x$
