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

Question

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

I dont really understand what we need to do for this question. Please help

@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

Amzoti · Accepted Answer · 2013-03-25T02:49:52.507

3

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?

edited Mar 25 '13 at 02:49

answered Mar 25 '13 at 02:12

Amzoti

+1. Have you seen this one? http://math.stackexchange.com/q/341507/8581 – Mikasa 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
so y(t)=e^(4t)? – Michael Rametta Mar 27 '13 at 20:10
I am not sure where $y(t)$ is coming from, but it would be $y(t) = e^{-4t} + e^{4t}$ (linear combination). If this is for an ODE you are solving, take this $y(t)$ and verify that it solves your equation. Clear? – Amzoti Mar 27 '13 at 20:18
im trying to find the inverse laplace transform i dont think that is correct for y(t) – Michael Rametta Mar 27 '13 at 20:23
has to be a simple mathematical error – Michael Rametta Mar 27 '13 at 20:27
let us continue this discussion in chat – Michael Rametta Mar 27 '13 at 20:52
1 Amzoti! ;-)

1 Answers1