Answer
$y(x) = C_1\sin(4x) + C_2\cos(4x)$
Work Step by Step
We assume a solution of the form $y=e^{mx}$, where $m$ Is a constant.
$y = e^{mx}$ $\implies$ $y' = m(e^{mx})$ and $y'' = m^2(e^{mx})$
$y'' + 16y = 0$
$\implies$ $m^2(e^{mx}) + 16(e^{mx}) = 0 $
$\implies$ $e^{mx}(m^2+16) = 0$
since $e^{mx}$ cant be 0
$\implies$ $m^2 + 16 = 0$
$\implies$ $m = 4i$ or $m = -4i $
since we have complex roots, we have that :
$y(x) = C_1\sin(4x) + C_2\cos(4x)$
