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Inverse laplace transform

  1. Mar 5, 2012 #1
    Using the laplace transform, find the solution to the differential equation:

    y'' + y' + y = 0 , y(0)=0, y'(0)=1

    Using the laplace transform and its properties I end up with:

    f(s) = 1/(s2+s+1)

    How can I find the inverse of this/ does anyone know the inverse of it?

    Setting y=eax I got a characteristic equation of a2+a+1=0, which has a complex solution, so I suspec that the inverse above should be a combination of both a sine, cosine and an exponential..
  2. jcsd
  3. Mar 5, 2012 #2

    Ray Vickson

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    Factor the denominator, then use partial fractions.

  4. Mar 5, 2012 #3
    hmm okay so factoring the denominator I get:

    f(s) = 1/((s+½+i√(3)/2)(s+½-i√(3)/2))

    What do I then do? Sorry, but I don't know the term partial fractions :(
  5. Mar 5, 2012 #4

    Ray Vickson

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    Google "partial fractions", or look in any calculus textbook.

  6. Mar 5, 2012 #5


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    You could also complete the square in the denominator and then use the tables, but you should know how to invert the transform using either method.
  7. Mar 5, 2012 #6
    hmm using partial fractions I get:

    1/(s2+s+1) = -i√(3)/(s+½+i√(3)/2) + i√(3)/(s+½-i√(3)/2)

    But I don't see how that gets me anywhere.

    Instead I figured you might want to write it as:


    But can't find a specific transform looking like that.

    The issue is that I have already obtained the solution using the characteristic polynomial, which got the solution:

    e^(½x)(cos(√(3)/2 *x) + 1/√(3) * sin(√(3)/2 * x))

    But I can't find any transform tables with e^ax(sin(bx) or with cos(bx)
    Last edited: Mar 5, 2012
  8. Mar 5, 2012 #7
    If you write your solution all in terms of complex exponentials, you'll probably find correspondence in tables.
  9. Mar 5, 2012 #8
    No because I'm not considering complex solutions (there's no i in front of sine). Hmm this is getting too tedious anyways, Ima skip this assignment. After all I know what to do, and who cares I can't get it to fit completely.
  10. Mar 5, 2012 #9

    Ray Vickson

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    Well, you could work these out on your own; they are not difficult (but are a bit lengthy before the final result).

    Anyway, you have something of the form [itex] \frac{c_1}{s+a_1} + \frac{c_2}{s + a_2},[/itex] for constants [itex] c_1, c_2, a_1, a_2 [/itex] (which happen to be complex numbers, but that does not matter). Just invert each term separately. In the end, you will be able to combine the results to get a purely real result.

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