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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?

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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 $\mathcal{L}_{t}\left[e^{at}\cos(\omega t)\right]_{(s)}=\frac{s-a}{(s-a)^2+\omega^2}$:

$$\mathcal{L}_{s}^{-1}\left[\frac{s-7}{s^2-14s+73}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{s-7}{(s-7)^2+24}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{s-7}{(s-7)^2+\sqrt{24}}\right]_{(t)}=$$ $$\mathcal{L}_{s}^{-1}\left[\frac{s-7}{(s-7)^2+\left(2\sqrt{6}\right)^2}\right]_{(t)}=e^{7t}\cos\left(2\sqrt{6}t\right)$$

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  • $\begingroup$ how can this formula explain the example with " s-7" $\endgroup$ – privetDruzia Nov 22 '15 at 12:35
  • $\begingroup$ @privetDruzia I'll edit my answer with your second problem! $\endgroup$ – Jan Nov 22 '15 at 12:38
  • $\begingroup$ thx for the update but, The expected answer for the second problem is e^(7t) cos(5t). No roots $\endgroup$ – privetDruzia Nov 22 '15 at 13:30

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