Question

Anyone know diff eq? Laplace transform of sin^2(2t). it's the answer is 8/(s^3+16 s). But my math is all messed up and i don't know how it comes out that.

Answer 1

\[\sin^22t=\frac{1}{2}(1-\cos4t)\]
\[\begin{align*}\mathcal{L}\left\{\frac{1}{2}-\frac{1}{2}\cos4t\right\}&=\frac{1}{2}\mathcal{L}\left\{1\right\}-\frac{1}{2}\mathcal{L}\left\{\cos4t\right\}\\
&=\frac{1}{2s}-\frac{s}{2(s^2+16)}\\
&=\frac{s^2+16}{2s(s^2+16)}-\frac{s^2}{2s(s^2+16)}\\
&=\frac{16}{2s(s^2+16)}\\
&=\frac{8}{s^3+16s}\end{align*}\]