I have a question about laplace transformation.
$\frac{8s+4}{s^2+23}$
I tried to split them. $\frac{8s}{s^2+23}$ is the image of a cosine and $\frac{4}{s^2+23}$ is the image of a sine.
Here is what I did :
$\frac{8s}{s^2+(\sqrt{23})^2}$ is the image of $8\cos(\sqrt{23}t)$ and $\frac{4}{s^2+(\sqrt{23})^2}$ is the image of $\frac4{\sqrt{23}}\sin(\sqrt{23}t)$
But according to homework, I am wrong. But I am sure I have the correct answer, there is no mistake.
If you don't mind some Residue theory, we can check use that to check your solution. \begin{align} \mathcal{L}^{-1}\biggl\{\frac{8s+4}{s^2+23}\biggr\}&=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{8s+4}{s^2+23}e^{st}ds\\ &=\sum\text{Res} \end{align} The poles in the $s$-plane occur at $s=\pm i\sqrt{23}$ both of order one. Then we have \begin{align} \mathcal{L}^{-1}\biggl\{\frac{8s+4}{s^2+23}\biggr\}&= \lim_{s\to i\sqrt{23}}(s-i\sqrt{23})\frac{8s+4}{s^2+23}e^{st}+\lim_{s\to -i\sqrt{23}}(s+i\sqrt{23})\frac{8s+4}{s^2+23}e^{st}\\ &= \frac{8i\sqrt{23}+4}{2i\sqrt{23}}e^{it\sqrt{23}}+\frac{-8i\sqrt{23}+4}{-2i\sqrt{23}}e^{-it\sqrt{23}}\\ &=4e^{it\sqrt{23}}+4e^{-it\sqrt{23}}+\frac{2}{i\sqrt{23}}e^{it\sqrt{23}}-\frac{2}{i\sqrt{23}}e^{-it\sqrt{23}}\\ &=8\cos(t\sqrt{23})+\frac{4}{\sqrt{23}}\sin(t\sqrt{23}) \end{align} Thus, the answer agrees with your by tables.