Answer
See below
Work Step by Step
Given: $y''+16y=f(t)$
Since $y(0)=\alpha\\
y'(0)=\beta$
obtain: $L[y]=\frac{F(s)}{s^2+16}+\frac{s\alpha}{s^2+16}+\frac{\beta}{s^2+16}$
Using the Convolution Theorem
$y(t)=L^{-1}[\frac{F(s)}{s^2+16}]+L^{-1}[\frac{\alpha}{s^2+16}]+L^{-1}[\frac{\beta}{s^2+16}]\\
=\frac{1}{4}\int^t_0 \sin(4\omega)f(t-\omega)d\omega+\frac{1}{4}\beta \sin 4t+\alpha \cos 4t$
