Answer
See below
Work Step by Step
$y''-y=10e^{2x}\cos x$
The complementary function is
$y_c(x)=c_1e^{x}+c_2e^{-x}$
We consider the complex differential equation
$z''-z=10e^{(2+i)x}$
An appropriate complex-valued trial solution for this differential equation is:
$z_p(x)=A_0e^{(2+i)x}$
where $A_0$ is a complex constant. The derivatives of $z_p$ is
$z''_p(x)=(2+i)^2A_0e^{(2+i)x}\\
z_p(x)=A_0e^{(2+i)x}$
so that $z _p$ is a solution if and only if
$(2+i)^2A_0e^{(2+i)x}-A_0e^{(2+i)x}\\
A_0=\frac{50}{6i-8}$
A particular solution is:
$y_p(x)=Re_{\{z_p\}}=2e^{2x}\sin x +e^{2x}\cos x$
so that the general solution is
$y(x)=c_1e^{x}+c_2e^{-x}+2e^{2x}\sin x +e^{2x}\cos x$
