Transformasi Laplace

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

Jawab: 


\begin{eqnarray*} \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{eqnarray*}

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

 \begin{eqnarray*} \mathcal{L}\left(\dfrac{6}{s^2-4s-5}\right)&=&\mathcal{L}\left(\frac{3}{(s-2)^2-9}+\frac{3}{(s-2)^2-9}\right)\\ &=&e^{2t}\sinh(3t)+e^{2t}\sinh(3t)\\ \mathcal{L}\left(\dfrac{6}{s^2-4s-5}\right)&=&2e^{2t}\sinh (3t) \end{eqnarray*}

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