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I'm trying to calculate

$$L_s^{-1}\left({\cfrac{2s+12}{s^2+9}}\right)=2L_s^{-1}\left({\cfrac{s+6}{s^2+9}}\right)$$

But I do not know how to go from here. I have noticed that the bottom does look like it is in the form of $L({cos(wt))}$ where w=3 but I do not know where to go from here.

Can anyone point me in the right direction?

Thanks.

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$$\cfrac{2s+12}{s^2+9}=2\cfrac{s}{s^2+3^2}+\cfrac{12}{3}\cfrac{3}{s^2+3^2}$$ Here a good table for Laplace transform (note the inverse transform is linear)

Solution

$2\cos(3t)+4\sin(3t)$

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  • $\begingroup$ @Silynn, Thanks for your comment $\endgroup$ – Fabien Jul 1 '14 at 3:36

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