In this problem you will solve the non-homogeneous differential equation $$y''+16 y =\sec^{2}\!\left(4x\right)$$ on the interval $-\pi/8 \lt x \lt \pi/8$.

(1) Let $C_1$ and $C_2$ be arbitrary constants. The general solution of the related homogeneous differential equation $y''+16 y =0$ is the function $y_h(x)=C_1\ y_1(x)+C_2\ y_2(x)=C_1$ $+C_2$.

(2) The particular solution $y_p(x)$ to the differential equation $y''+16 y =\sec^2(4 x)$ is of the form $y_p(x)=y_1(x)\ u_1(x)+y_2(x)\ u_2(x)$
where $u_1'(x)=$ and $u_2'(x)=$ .

(3) It follows that
$u_1(x)=$ and $u_2(x)=$ ;
thus $y_p(x)=$ .

(4) Therefore, on the interval $(-\pi/8,\pi/8)$, the most general solution of the non-homogeneous differential equation $y''+16 y =\sec^2(4 x)$
is $y=C_1$ $+C_2$$+$.