I am trying to understand the basics of easy inverse laplace transformations. On the first line is the "correct" answer. On the second line what i expected.
https://www.dropbox.com/s/j08inqvjw3v620i/IMAG1153.jpg?dl=0
Exactly the same scenario here:
https://www.dropbox.com/s/7chpoyjpgaqffvc/IMAG1154.jpg?dl=0
So it seems to me as if i am missing something. Could someone explain me why my expected solution is incorrect?
Using $\mathcal{L}_{t}\left[e^{at}\sin(\omega t)\right]_{(s)}=\frac{\omega}{(s-a)^2+\omega^2}$:
$$\mathcal{L}_{s}^{-1}\left[\frac{1}{s^2-6s+10}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{1}{(s-3)^2+1}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{1}{(s-3)^2+1^2}\right]_{(t)}=e^{3t}\sin(t)$$
Using $\mathcal{L}_{t}\left[e^{at}\cos(\omega t)\right]_{(s)}=\frac{s-a}{(s-a)^2+\omega^2}$:
$$\mathcal{L}_{s}^{-1}\left[\frac{s-7}{s^2-14s+73}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{s-7}{(s-7)^2+24}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{s-7}{(s-7)^2+\sqrt{24}}\right]_{(t)}=$$ $$\mathcal{L}_{s}^{-1}\left[\frac{s-7}{(s-7)^2+\left(2\sqrt{6}\right)^2}\right]_{(t)}=e^{7t}\cos\left(2\sqrt{6}t\right)$$