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Inverse Laplace (stuck @ Partial Fraction)

  1. Oct 1, 2015 #1
    1. The problem statement, all variables and given/known data

    Find the Inverse laplace transform of:

    http://www4c.wolframalpha.com/Calculate/MSP/MSP14541hg721e74730d4fb00004644i96f59549h1d?MSPStoreType=image/gif&s=30&w=201.&h=40. [Broken]

    Result http://www4c.wolframalpha.com/Calculate/MSP/MSP14591hg721e74730d4fb000042gcebh89c38eib7?MSPStoreType=image/gif&s=30&w=505.&h=36. [Broken]

    2. Relevant equations


    eq0011MP.gif = eq0012MP.gif

    empty.gif

    maybe e^(at) forms on the table too since there´s s-a form in a part of the bottom


    3. The attempt at a solution

    when getting to the final step of the partial fraction i get:

    http://www4b.wolframalpha.com/Calculate/MSP/MSP14771df8fbfi46b7gihh000059eh9936e24diiii?MSPStoreType=image/gif&s=24&w=267.&h=22. [Broken] (sorry about minus, idk what happen to the program ...)

    now i could asume that http://www5a.wolframalpha.com/Calculate/MSP/MSP172120i2249gd4beb17900003d9dfc9b6ef7i9e8?MSPStoreType=image/gif&s=54&w=63.&h=20. [Broken] that being 2i but it gets more complicated than i would expect, see, teacher sayd it is only 1 page long exercise, i would imagine there must be a much simpler way than going by imaginary roots.

    I would appreciate if possible a step by step explanation since as a fun fact we were given this exercise without any class about inverse transform and i myself did some research. Thanks in advance
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 1, 2015 #2

    Mark44

    Staff: Mentor

    You've omitted a lot of work, but it looks like you started with this:
    $$\frac{s}{(s^2 + 6s + 18)((s^2 + 4)} = \frac{As + B}{s^2 + 6s + 18} + \frac{Cs + D}{s^2 + 4}$$
    If you multiply both sides by ##(s^2 + 6s + 18)((s^2 + 4)##, you get something different from what you show above; namely,
    ##s = (As + B)(s^2 + 4) + (Cs + D)(s^2 + 6s + 18)##
    Note the extra parentheses I show that you don't show. These parentheses make a significant difference.
    Expand the right side, and group the terms by powers of s to solve for A, B, C, and D.
    Yes, there is. Don't break up the quadratics.

    The formulas you want are these:
    $$\mathcal{L}^{-1}[\frac{s}{s^2 + a^2}] = \cos(at)$$
    and
    $$\mathcal{L}^{-1}[\frac{1}{s^2 + a^2}] = \sin(at)$$

    You'll need to complete the square in the one with ##s^2 + 6s + 18## to make it fit one of these formulas.
     
    Last edited by a moderator: May 7, 2017
  4. Oct 1, 2015 #3
    Would you mind writing out all of your steps? It would make it easier to check.
     
  5. Oct 1, 2015 #4
    yes you're quite right thats were i started, i see now, with those parentheses there are less things to group up, ill give it a try thanks for that observation

    sorry about that, on mark post above there is the begining of the exercise, altho ill redo it from his point of view an post any doubt i have later on :P thanks both of you
     
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