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Laplace Transforms (Partial Fractions - Real Roots)

  1. Dec 8, 2008 #1
    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.

    [​IMG]


    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);

    [​IMG]


    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:

    [​IMG]


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

    Thank you for your time!~
     
  2. jcsd
  3. Dec 8, 2008 #2

    gabbagabbahey

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    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 [tex]e^{-4y}=\frac{1}{x+4}-2\frac{2}{x(x+4)}[/tex] (or even just writing [tex]e^{-4y}=\frac{1}{x+4}[/tex]) is incorrect and confusing.

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

    And so [tex]\frac{x-2}{x^2+4x}=\mathcal{L}[e^{-4y}]-2\frac{2}{x(x+4)}[/tex].
     
  4. Dec 8, 2008 #3

    Defennder

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    I believe it's supposed to be [tex]\frac{3/2}{x+4}[/tex].
     
  5. Dec 8, 2008 #4

    gabbagabbahey

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    Which is the same thing as [tex]\mathcal{L}[e^{-4y}]+\frac{1}{2}\mathcal{L}[e^{-4y}][/tex] :wink:
     
  6. Dec 8, 2008 #5

    Dick

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    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.
     
  7. Dec 8, 2008 #6
    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...
     
  8. Dec 8, 2008 #7

    gabbagabbahey

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    He/she set it up so that [tex]\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)[/tex] which although a little longwinded gives the same final result as your method would; but with different A and B values of course.
     
  9. Dec 8, 2008 #8
    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.
     
  10. Dec 8, 2008 #9

    Dick

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    Oh. So he/she did. Sorry.
     
  11. Dec 8, 2008 #10

    Dick

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    Oh. So he/she did. Sorry.
     
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