Homework Help: 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.