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I am having trouble finding the inverse Laplace transform of: $$\frac{1}{s^2-9s+20}$$

I tried writing it in a different way:

$L^{-1}\{\frac{1}{(s-\frac{9}{2})^2-(\frac{1}{2})^2}\}=2L^{-1}\{\frac{\frac{1}{2}}{(s-\frac{9}{2})^2-(\frac{1}{2})^2}\}$

Can someone help ?

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I'll give you a hint, because I think it might be useful for you to get practice. If you need more tell in comment: try rewriting this using partial fractions and then use the linearity of the inverse laplace.

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  • $\begingroup$ Ah you're right it's easier $\endgroup$ – user43758 Apr 13 '13 at 19:16
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I have done it: $$\begin{array}{l} X(s) = \frac{1}{{{s^2} - 9s + 20}} = \frac{1}{{(s - 4)(s - 5)}} = \frac{{ - 1}}{{s - 4}} + \frac{1}{{s - 5}} \\ = > x(t) = - {e^{4t}} + {e^{5t}} \\ \end{array}$$

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