Question

Use a table of Laplace transforms to find the Inverse Laplace transform of

F(s)=
`(3s+9)/(s^2+9)`


f(t)=_____

Answer 1

The inverse Laplace transform of
`F(s) = (3\cdot s + 9)/(s^(2)+9)` is
`f(t) = 3\cdot \cos \omega t + 3\cdot \sin \omega t`.

Explanation:

In this case, we should use the following direct and inverse Laplace Transforms:

`F(s) = G(s) + H(s)` (1)

`\mathcal{L}^(-1) \{(s)/(s^(2)+\omega^(2)) \} = \cos \omega t` (2)

`\mathcal{L}^(-1)\{(\omega)/(s^(2)+\omega^(2)) \} = \sin \omega t` (3)

`\mathcal\{L\}^(-1)\{c\cdot F(s)\} = c\cdot \mathcal\{L\}^(-1)\{F(s)\}` (4)

Then, we apply all these trasforms:

`F(s) = (3\cdot s + 9)/(s^(2)+9)`

`F(s) = 3\cdot \left((s)/(s^(2)+9) \right)+3\cdot \left((3)/(s^(2)+9) \right)`

`f(t) = 3\cdot \cos \omega t + 3\cdot \sin \omega t`

The inverse Laplace transform of
`F(s) = (3\cdot s + 9)/(s^(2)+9)` is
`f(t) = 3\cdot \cos \omega t + 3\cdot \sin \omega t`.