Answer
$y(x)=c_1e^{4x}+c_2e^{-4x}-\cos 4x$
Work Step by Step
$y''-16y=20\cos 4x$
The complementary function is
$y_c(x)=c_1e^{4x}+c_2e^{-4x}$
We consider the complex differential equation
$z''-16z=20e^{4xi}$
An appropriate complex-valued trial solution for this differential equation is:
$z_p(x)=A_0e^{4xi}$
where $A_0$ is a complex constant. The derivatives of $z_p$ is
$z''_p(x)=4(i)^2A_0e^{4xi}=A_0e^{4xi}$
so that $z _p$ is a solution if and only if
$-4A_0e^{4xi}-16A_0e^{4xi}=20e^{4xi}\\
-20A_0e^{4xi}=20A_0e^{4xi}\\
A_0=-1$
Substitute: $z_p(x)=-e^{4xi}=-\cos 4x-i\sin 4x$
A particular solution is:
$y_p(x)=Re_{\{z_p\}}=-\cos 4x$
so that the general solution is
$y(x)=c_1e^{4x}+c_2e^{-4x}-\cos 4x$
