Homogeneous differential equation with undetermined coefficient Given the problem:
$y’’+16y=\cos(4x)$
It’s particular $y$ is equal to zero. I know how to get the complementary $y$ but I had problem with the particular $y. $
\begin{align*}
y_p&=A\sin(4x)+B\sin(4x)\\
y_p’&=4A\cos(4x)-4B\sin(4x)\\
y_p''&=-16A\sin(4x)-16B\cos(4x)
\end{align*}
As we substitute the $y_p''$ and $y_p’$ to $y''+16y=\cos(4x),$ we get:
\begin{align*}
-16A\sin(4x)-16B\cos(4x)+16(A\sin(4x)+B\sin(4x))&=\cos(4x)\\
-16A\sin(4x)-16B\cos(4x)+16A\sin(4x)+16B\sin(4x)&=\cos(4x)
\end{align*}
Notice that $-16A\sin(4x)$ will cancel with $16A\sin(4x),$ the same with the $-16B\cos(4x),$ which will also cancel with $16B\cos(4x).$ The $A$ will equal $0$ and also the $B$ will equal $0.$ 
Is there anything wrong with the equation?
Thanks for those who could help.
 A: $y_1 = \sin(4x)$ and $y_2 = \cos(4x)$ are solutions to the homogeneous equation. So by definition they are expected to cancel out. You need to try for a particular $y$ in the form of $y_p = Ax\sin(4x) + Bx\cos(4x)$.
A: The central problem here is that the RHS of your original DE contains only terms that are also in the homogeneous solution
$$y_h(x)=A\cos(4x)+B\sin(4x). $$
Thus, the LHS is going to annihilate your choice of a particular solution. What to do? This is precisely the situation in which variation of parameters comes in, and allows you to find a particular solution that is NOT annihilated by the LHS. It can be shown that the correct ansatz for your particular solution is
\begin{align*}
y_p(x)&=Cx\cos(4x)+Ex\sin(4x)\\
y_p'(x)&=\sin (4 x) (-4 C x+E)+\cos (4 x) (C+4 E x)\\
y_p''(x)&=8 \cos (4 x) (-2 C x+E)-8 \sin (4 x) (C+2E x).
\end{align*}
So we plug in to get
\begin{align*}
&8 \cos (4 x) (-2 C x+E)-8 \sin (4 x) (C+2E x)\\
+&16\left[Cx\cos(4x)+Ex\sin(4x)\right]\\
=&\cos(4x).
\end{align*}
Set things equal and solve for the constants, and you've got your particular solution.
