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

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 $f(t) = (5/2) \sin(2(t - 6)) u(t - 6)$

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For some basic information about writing math at this site see e.g. here, here. –  Américo Tavares Apr 3 '13 at 21:26
See here. This problem is more related. –  Mhenni Benghorbal Apr 3 '13 at 22:57

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 $$f(t)=\left\{\begin{matrix}0,&\ \text{if}~x<6\\ \frac{5}{2}\sin(2(t-6)),&\ \text{if}~x>6\end{matrix}\right.$$
@Michael $u(x)=\theta(x)=\left\{\begin{matrix}1: x>0\\ 0: x<0\end{matrix}\right.$ –  Ruslan Apr 3 '13 at 22:05
Almost the same as you've already written: $$f(t)=\frac{5}{2}sin(2(t-6))\theta(t-6)$$ –  Ruslan Apr 3 '13 at 22:14
In fact, your solution was already correct as I now see that $u(x)$ is also a common notation for Heaviside step function. –  Ruslan Apr 3 '13 at 22:21