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Inverse Laplace transforms

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data
    find the partial fractions and thus the inverse of the following
    6s^2-2s-11/(s-1)(s^2-1)

    and

    7s^2+8s+16/(s+2)(s^2+3)


    2. Relevant equations

    answer tutor gave for the fist one was 3e^2t + 3cosht + sinht
    and second was 4e^-2t+3cos sqrt3t+ 2/sqrt3 sin sqrt3

    3. The attempt at a solution


    skipping a few steps for the first inverse transform i factored (s^2-1) to (s+1)(s-1) and got the following partial fractions

    6s^2-2s-11=a/(s-2) + b/(s+1)(s-1) +c(s^2-1) and eventually was able to derive the 3e^-2t but struggled to get the other values

    i didn't really no where to star with the 2nd one

    please help
     
  2. jcsd
  3. Oct 18, 2012 #2

    LCKurtz

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    The first one you should break into$$
    \frac A {s+1} + \frac B {s-2} + \frac C {(s-1)^2}$$
    The second would be like this:$$
    \frac A {s+2} +\frac{Bs+C}{s^2+3}$$
     
  4. Oct 20, 2012 #3
    i put in the wrong transform for the first one it is 6s^2-2s-11/(s-2)(s^2-1)
     
  5. Oct 20, 2012 #4

    LCKurtz

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    So factor the rest of the way to 3 linear factors in the denominator.
     
  6. Oct 20, 2012 #5

    Ray Vickson

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    For the first one: what you have written--when read using standard priority rules--is
    [tex] 6s^2 - 2s - \frac{11}{(s-1)(s^2-1)}, \text{ or possibly }
    6s^2 - 2s - \frac{11}{s-1} (s^2-1),[/tex] depending on whether you interpret a/bc as a/(bc) or as (a/b)c. (Some computer algebra systems may use one interpretation, and some may use the other.) However, possibly you really meant
    [tex]\frac{6s^2 - 2s - 11}{(s-1)(s^2-1)}. [/tex] To make thing clear, please use brackets, like this: (6s^2 - 2s - 11)/[(s-1)(s^2-1)].

    RGV
     
  7. Oct 21, 2012 #6
    {6s^2 - 2s - 11}/(s-2)(s^2-1)}. the problem i have is breaking it down to its partial fractions
     
  8. Oct 21, 2012 #7

    LCKurtz

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    $$\frac {6s^2-2s-11}{(s-2)(s+1)(s-1)}=\frac A {s-2}+\frac B {s+1}+\frac C {s-1}$$
     
  9. Oct 21, 2012 #8
    yes i thought that was the way but i just don't seem to get the answer my tutor has given i can get one of the numbers the 3e^2t but i can't find 3cosht and 4sinht
     
  10. Oct 21, 2012 #9

    LCKurtz

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    Remember that if you are getting answers with ##e^t## and ##e^{-t}##, you answers may be equivalent to answers in terms of ##\cosh t## and ##\sinh t##.
     
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