Let $$g(t) =\begin{cases} t & \text{if $t \leq6π$} \\ 6\pi & \text{if $t>6\pi$} \end{cases} $$
Solve $y''+ 16y = g(t)$ where $y(0) = 9$ and $y'(0) = 4$ using Laplace transforms.
I got
$y(t) = \frac{4t+63\sin(4t)}{64} + 9\cos(4t)$ for $t\leq6\pi$
$y(t) = \frac{3π}{8} + \frac{416-3\pi}{32\sin(4t)}$ for $t>6\pi$
2) seems to be wrong
I got $$F(s) = \frac{3\pi}{8s} + \frac{104-3π}{8(s^2+4^2)}=\frac{3\pi}{8s} + \frac{26(4)}{8(s^2+4^2)} - \frac{3\pi}{8(s^2+4^2)}$$
Here I used the inverse Laplace Transforms
$$\frac{3\pi}{8s} \to \frac{3\pi}{8}$$
$$\frac{26(4)}{8(s^2+4^2)} \to \frac{13}{4}\sin(4t)$$
$$\frac{3\pi}{8(s^2+4^2)} \to \frac{3\pi}{32}\sin(4t)$$
Is this wrong?