Question

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


F(s)= $\frac{3s+9}{s^2+9}$


f(t)=_____

Answer 1

Answer:

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

Step-by-step explanation:

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

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

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

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

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

Then, we apply all these trasforms:

$F(s) = \frac{3\cdot s + 9}{s^{2}+9}$

$F(s) = 3\cdot \left(\frac{s}{s^{2}+9} \right)+3\cdot \left(\frac{3}{s^{2}+9} \right)$

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

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