# Inverse Laplace Tranform

1. Dec 6, 2008

### Ithryndil

1. The problem statement, all variables and given/known data
$$Y(s) = \frac{1}{(s-2)^{2}(s+1)^{3}}$$

2. Relevant equations
...

3. The attempt at a solution

Ok, so my goal is to find the Laplace transform of the above. I need to use partial fraction decomposition to break up the denominator. With that in mind, I have the following:

$$Y(s) = \frac{1}{(s-2)^{2}(s+1)^{3}} = \frac{A}{s-2}+ \frac{B}{(s-2)^{2}} + \frac{C}{s+1} + \frac{D}{(s+1)^{2}}+ \frac{E}{(s+1)^{3}}$$

After this I multiplied by the common denominator and came up with 5 equations. Those equations I came up with are:

A + C = 0
A + B - 2C + D = 0
-3A + 3B - 3C -3D + E = 0
-5A + 3B + 4C -4E = 0
-2A + B + 4C + 4D + 4E = 1

Now, once solving those I get:

A = 4/27
B = 8/27
C = -4/27
D = 7/27
E = 3/27

However, that's wrong, I am wondering if my setup is correct. I can post all my algebraic work if necessary.

2. Dec 6, 2008

### gabbagabbahey

so far so good...

This can't be right; using these values, $A+B-2C+D==27/27=1\neq0$ which clearly contradicts your second equation....

3. Dec 6, 2008

### Ithryndil

This post...unnecessary.

4. Dec 6, 2008

### Ithryndil

Hmm...well, let me double check my matrix in my calculator. Thanks for pointing that out to me.

I found my problem, I had swapped a 1 with a 0 in my matrix on my calculator. My new values are:

A = -1/27
B = 1/27
C = 1/27
D = 2/27
E = 1/9

Thank you, after looking at my work for awhile, and double checking the work tends to blur together. I needed an extra pair of eyes. Thanks again!

Last edited: Dec 6, 2008
5. Dec 6, 2008

### gabbagabbahey

That's better

6. Dec 6, 2008

### Ithryndil

Yeah, I have yet to do the inverse transform, but those values are definitely better and look like they will work.