inverse transform

Problem 02 | Inverse Laplace Transform

Problem 02
Find the inverse transform of   $\dfrac{5}{s - 2} - \dfrac{4s}{s^2 + 9}$.
 

Solution 02
$\mathcal{L}^{-1} \left[ \dfrac{5}{s - 2} - \dfrac{4s}{s^2 + 9} \right]= 5\mathcal{L}^{-1}\left( \dfrac{1}{s - 2} \right) - 4\mathcal{L}^{-1}\left( \dfrac{s}{s^2 + 9} \right)$

$\mathcal{L}^{-1} \left[ \dfrac{5}{s - 2} - \dfrac{4s}{s^2 + 9} \right]= 5e^{2t}\mathcal{L}^{-1}\left( \dfrac{1}{s} \right) - 4\mathcal{L}^{-1}\left( \dfrac{s}{s^2 + 3^2} \right)$