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Find the inverse Laplace transform of $F(s)=\dfrac{5e^{−6s}}{s^2+4}$

$f(t)=$ __________?

Here is my work: $L{(5/2) \sin(2t)} = 5/(s^2 + 4)$, we have by the shifting theorem $f(t) = (5/2) \sin(2(t - 6)) u(t - 6)$

Please help me with the correct solution. Thanks

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You have already gotten the correct solution if by $u(x)$ you mean Heaviside unit step function $\theta(x)$.
If you want to eliminate Heaviside function you could as well write $$f(t)=\left\{\begin{matrix}0,&\ \text{if}~x<6\\ \frac{5}{2}\sin(2(t-6)),&\ \text{if}~x>6\end{matrix}\right.$$

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  • $\begingroup$ what would the u equal in this case? $\endgroup$ – Michael Rametta Apr 3 '13 at 21:58
  • $\begingroup$ @Michael $u(x)=\theta(x)=\left\{\begin{matrix}1: x>0\\ 0: x<0\end{matrix}\right.$ $\endgroup$ – Ruslan Apr 3 '13 at 22:05
  • $\begingroup$ so how would i set up f(t) properlly $\endgroup$ – Michael Rametta Apr 3 '13 at 22:12
  • $\begingroup$ Almost the same as you've already written: $$f(t)=\frac{5}{2}sin(2(t-6))\theta(t-6)$$ $\endgroup$ – Ruslan Apr 3 '13 at 22:14
  • $\begingroup$ In fact, your solution was already correct as I now see that $u(x)$ is also a common notation for Heaviside step function. $\endgroup$ – Ruslan Apr 3 '13 at 22:21

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