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Laplace transform, fraction expansion

  1. Oct 11, 2005 #1
    I have y"+y=t , y(0)=1, y'=0

    After Laplace transformation a got:

    After I made a partial fraction expansion

    (S^3+1)/(S^2(S^2+1))=a1/S^2+a2/S+a3/(S^2+1) (1)

    It comes to a system where

    a1+a3=0 (2)

    Here I am getting confused because according to (1) and (2)
    a2 is equal to 0 and 1. What I should use for inverse transformation: zero, one, or both?
  2. jcsd
  3. Oct 11, 2005 #2


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    First of all check out the LaTex tutorial thread here.

    Your problem is to exapand

    [tex] \frac {1 + s^3} {s^2 (s^2+1)} [/tex]

    Try seperating the polynomial in the numerator to get this

    [tex] \frac 1 {s^2 (s^2+1)} + \frac s {(s^2+1)} [/tex]

    IIRC the second fraction has a well known trasform, now use partial fractions on the first piece, it is pretty straight forward.
  4. Oct 11, 2005 #3
    I tried that and it looks like it works for me.

    Thank you!
  5. Oct 11, 2005 #4


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    Your basic formula for the "partial fractions" was wrong. If you have a quadratic, that cannot be factored into real factors, in the denominator, such as [tex]\frac{1}{s^2+1}[/tex], you will need a linear term in the numerator, not a constant: [tex]\frac{As+ B}{s^2+1}[/tex].
  6. Oct 11, 2005 #5


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    May I be so bold as to ask someone to be so kind to explain to me why/where/when/ Laplace Transforms are teached in K-12 grade? Really, that's so impressive.:smile: I expect great things from someone, well just some of them I suppose, alright, maybe one in a . . . several . . . thousand I guess, who study LT at such an early age.:smile:
  7. Oct 11, 2005 #6
    may be it is time for me to move up:smile:

    I'm not really familiar with this forum. Where should I post with stuff like that?
  8. Oct 11, 2005 #7
    I tried that. I must did something wrong because I was getting too many unknowns to solve equetions.
  9. Oct 11, 2005 #8

    Tom Mattson

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    In the College Level section.

    I found out not too long ago that to people outside the US, college=high school. But here college=university, which is why the other Homework Forum is named the way it is.
  10. Oct 11, 2005 #9


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    That's it! I was posting form work last night and did not have my book shelf to draw on. Looks like my solution gets the job done. There is always more then one way to approach problems.

    I was lucky, in general you need to learn the correct (Hall's ) approach. It will, and does in this case, always get you to the correct solution. What I did worked and in someways is actually a simpler method, but it is not general and may not be simpler or even work in every case.
  11. Oct 11, 2005 #10
    Your method worked out, but I can't figure a way HallsofIvy suggested.

    I'm getting

    (S-1)/(S^2+1) + 1/S^2.

    The second part of it is known combination. The first is not.
    Am I doing something wrong?
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