Transformasi Laplace

Carilah Nilai $f(t)$ dari $\mathcal{L}\left({\displaystyle \frac{6}{s^2-4s-5}}\right)$ 

Jawab: 


$$\begin{array}{rcl}\frac{6}{s^2-4s-5}& =& \frac{6}{(s-2{)}^2-9}\\ & =& \frac{3}{(s-2{)}^2-9}+\frac{3}{(s-2{)}^2-9}\end{array}$$

Karena $\mathcal{L}\sinh (at)={\displaystyle \frac{a}{s^2-a^2}}$ dan $\mathcal{L}{e}^{2t}={\displaystyle \frac{1}{s-2}}$ Maka :

 $$\begin{array}{rcl}\mathcal{L}\left({\displaystyle \frac{6}{s^2-4s-5}}\right)& =& \mathcal{L}(\frac{3}{(s-2{)}^2-9}+\frac{3}{(s-2{)}^2-9})\\ & =& e^{2t}\sinh (3t)+e^{2t}\sinh (3t)\\ \mathcal{L}\left({\displaystyle \frac{6}{s^2-4s-5}}\right)& =& 2e^{2t}\sinh (3t)\end{array}$$

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