# Inverse Laplace transforms

1. Oct 18, 2012

### wozzers

1. The problem statement, all variables and given/known data
find the partial fractions and thus the inverse of the following
6s^2-2s-11/(s-1)(s^2-1)

and

7s^2+8s+16/(s+2)(s^2+3)

2. Relevant equations

answer tutor gave for the fist one was 3e^2t + 3cosht + sinht
and second was 4e^-2t+3cos sqrt3t+ 2/sqrt3 sin sqrt3

3. The attempt at a solution

skipping a few steps for the first inverse transform i factored (s^2-1) to (s+1)(s-1) and got the following partial fractions

6s^2-2s-11=a/(s-2) + b/(s+1)(s-1) +c(s^2-1) and eventually was able to derive the 3e^-2t but struggled to get the other values

i didn't really no where to star with the 2nd one

2. Oct 18, 2012

### LCKurtz

The first one you should break into$$\frac A {s+1} + \frac B {s-2} + \frac C {(s-1)^2}$$
The second would be like this:$$\frac A {s+2} +\frac{Bs+C}{s^2+3}$$

3. Oct 20, 2012

### wozzers

i put in the wrong transform for the first one it is 6s^2-2s-11/(s-2)(s^2-1)

4. Oct 20, 2012

### LCKurtz

So factor the rest of the way to 3 linear factors in the denominator.

5. Oct 20, 2012

### Ray Vickson

For the first one: what you have written--when read using standard priority rules--is
$$6s^2 - 2s - \frac{11}{(s-1)(s^2-1)}, \text{ or possibly } 6s^2 - 2s - \frac{11}{s-1} (s^2-1),$$ depending on whether you interpret a/bc as a/(bc) or as (a/b)c. (Some computer algebra systems may use one interpretation, and some may use the other.) However, possibly you really meant
$$\frac{6s^2 - 2s - 11}{(s-1)(s^2-1)}.$$ To make thing clear, please use brackets, like this: (6s^2 - 2s - 11)/[(s-1)(s^2-1)].

RGV

6. Oct 21, 2012

### wozzers

{6s^2 - 2s - 11}/(s-2)(s^2-1)}. the problem i have is breaking it down to its partial fractions

7. Oct 21, 2012

### LCKurtz

$$\frac {6s^2-2s-11}{(s-2)(s+1)(s-1)}=\frac A {s-2}+\frac B {s+1}+\frac C {s-1}$$

8. Oct 21, 2012

### wozzers

yes i thought that was the way but i just don't seem to get the answer my tutor has given i can get one of the numbers the 3e^2t but i can't find 3cosht and 4sinht

9. Oct 21, 2012

### LCKurtz

Remember that if you are getting answers with $e^t$ and $e^{-t}$, you answers may be equivalent to answers in terms of $\cosh t$ and $\sinh t$.