Answer
The solution to the boundary-value problem is
$$
y(x)=2 \sin(4x)-3 \cos(4x),
$$
Work Step by Step
$$
y^{\prime \prime}+16 y=0, \quad y(0)=-3, \quad y(\frac{\pi}{8})=2
$$
The given equation is a homogeneous linear equation. So the characteristic equation of the differential equation is
$$
r^{2}+16 =0
$$
whose roots are
$$
\:r_{1,\:2}=\frac{\pm \sqrt{-4\cdot \:1\left(16\right)}}{2\cdot \:1}= \pm 4i
$$
Therefore, the general solution of the given differential equation is
$$
y(x)=c_{1} \sin(4x)+c_{2}\cos(4x),
$$
Then
$$
-3=y(0)=c_{1} \sin(0)+c_{2}\cos( 0) \Rightarrow c_{2}=-3
$$
and
$$
2=y(\frac{\pi}{8})=c_{1} \sin(\frac{\pi}{2})+c_{2}\cos( \frac{\pi}{2}) \Rightarrow c_{1}=2
$$
Thus the solution to the boundary-value problem is
$$
y(x)=2 \sin(4x)-3 \cos(4x),
$$
