Answer
See below
Work Step by Step
The general solution for the given differential equation is: $y(x)=c_1\cos 4x+c_2 \sin 4x+A_0 \sin x+B_0\cos x$
The trial solution for $y_p= A_0\sin x+B_0\cos x$ can be computed as by plugging back into the given differential equation.
So, we have: $(D^2+16)y_p(x)=4\cos x\\ (D^2+16)(A_0\sin x+B_0\cos x)=4\cos x\\ 15A_0\cos x +15B_0\sin x=4\cos x$
On comparing co-efficients, we get: $A_0=\frac{4}{15},B_0=0$
Therefore, the general solution for the given differential equation is: $y(x)=c_1\cos 4x+c_2 \sin 4x+\frac{4}{15}\sin x$
