# Laplace Transforms (Partial Fractions - Real Roots)

1. Dec 8, 2008

### Nubcakes

Recently I asked a question about Laplace Transforms and got several very informative answers. I had thought I was as deep as they went... I couldn't have been more wrong!

This time I got problem from the section labeled "Laplace Transforms - Partial Fractions (Real Roots)" It's as follows:

Note; assume f(t)=f(y),{t=y} and f(s)=f(x),{s=x}. I noticed my Ts look like addition symbols and my S(s) look like 5s, so to prevent confusion I changed the variables.

Since the section was called partial fractions I decided to break it apart as much as I could and then pursue to use the first part of the integration technique(Partial Fractions);

I thought I had did it correctly, but for some reason my answer does NOT match the given answer in the book. Sadly they don't show any work so I have no clue where I went wrong or if I am wrong at all:

If anyone who knows this topic well has a moment, I would really appreciate it if you could point out where I messed up.

2. Dec 8, 2008

### gabbagabbahey

Your solution is correct. So either (1) the answer in the book is wrong, (2) you looked at an answer to a different question by mistake, or (3) you read the question incorrectly.

Also, I'd like to point out that writing things like $$e^{-4y}=\frac{1}{x+4}-2\frac{2}{x(x+4)}$$ (or even just writing $$e^{-4y}=\frac{1}{x+4}$$) is incorrect and confusing.

What is true is that $$\mathcal{L}[e^{-4y}]=\frac{1}{x+4}$$

And so $$\frac{x-2}{x^2+4x}=\mathcal{L}[e^{-4y}]-2\frac{2}{x(x+4)}$$.

3. Dec 8, 2008

### Defennder

I believe it's supposed to be $$\frac{3/2}{x+4}$$.

4. Dec 8, 2008

### gabbagabbahey

Which is the same thing as $$\mathcal{L}[e^{-4y}]+\frac{1}{2}\mathcal{L}[e^{-4y}]$$

5. Dec 8, 2008

### Dick

You definitely have a problem with the partial fractions part. If (x-2)/(x^2+4x)=A/x+B/(x+4), that's the same as (x-2)=A(x+4)+Bx. I don't get the same A and B as you do.

6. Dec 8, 2008

### Nubcakes

Yea, I broke the fraction up...

(x-2)/(x^2+4X) becomes { x/(x^2+4x) - 2/(x^2+4x) }

I think this is legal...

7. Dec 8, 2008

### gabbagabbahey

He/she set it up so that $$\frac{x-2}{x^2+4x}=\frac{1}{x+4}-\frac{2}{x(x+4)}=\frac{1}{x+4}-\left(\frac{A}{x}+\frac{B}{(x+4)}\right)$$ which although a little longwinded gives the same final result as your method would; but with different A and B values of course.

8. Dec 8, 2008

### Nubcakes

Err, The answer provided is incorrect then. There is none even close to this problem on the page so it looks like the publishers messed up.

Yea, sorry about that. I gotta make my work more clear and less confusing. I didn't quite mean to imply that. Sorry for the confusion.

Thanks again! That makes me feel alittle safer doing these problems.

9. Dec 8, 2008

### Dick

Oh. So he/she did. Sorry.

10. Dec 8, 2008

### Dick

Oh. So he/she did. Sorry.