# Finding the inverse laplace transform of $\frac{3s+4}{s^2-16}$

How would you find the inverse Laplace transformation of $\displaystyle \frac{3s+4}{s^2-16}$ when $s>4$? Thanks!!

• See here for different techniques. – Mhenni Benghorbal Mar 25 '13 at 2:47
• You really do not want to restrict $s$ to avoid the pole. – Ron Gordon Mar 25 '13 at 2:49
• @MichaelRametta: if you have an issue with what you posted above and the DEQ, that sounds like a new question. You can post the ODE and your solution and we can figure out where it went wrong. Regards – Amzoti Mar 27 '13 at 21:49

Hint:

Write the partial fraction fraction expansion as:

$$\displaystyle \tag 1 \frac{3s+4}{s^2-16} = \frac{1}{s+4} + \frac{2}{s-4}$$

Now, take the inverse Laplace of each of the terms on the right-hand-side (RHS) of $(1)$.

We have for $s \gt a$:

$$\mathcal{L}^{-1}\left(\frac{1}{s-a}\right) = e^{at}$$

Clear?

• +1. Have you seen this one? math.stackexchange.com/q/341507/8581 – mrs Mar 26 '13 at 14:49
• I hadn't, but now it has my curiosity up! Thanks for the reference! – Amzoti Mar 26 '13 at 15:19
• so what would this answer be – Michael Rametta Mar 27 '13 at 19:46
• $e^{-4t}$ for the positive term and $e^{4t}$ for the negative term. – Amzoti Mar 27 '13 at 19:48
• Great i got i got the same answers thanks – Michael Rametta Mar 27 '13 at 20:09