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Homework Help: Fraction decomposition for inverse Laplace

  1. May 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the solution of the givien initial value problem and draw its graph
    y''+2y'+2y = δ(t-π) y(0) = 1, y'(0) = 0

    2. Relevant equations

    A Laplace transform chart would be very useful

    3. The attempt at a solution

    I chose to solve the equation with Laplace transforms. I took the laplace of the whole equation and then solved for the L(y):

    L(y) = (e^(-πs)+s+2)/(s^2+2s+2)
    I double checked to make sure I didn't make any errors in my Laplace calculation and I did the algerbra correctly. So I have the right equation. However, my issue is now decomposing this equation - how do you do that?

    You can't factor out the denominator? and I can't just set the numerator to As+B can I? The e^s term would make that useless...

    Any ideas?
  2. jcsd
  3. May 7, 2013 #2
    and just to explain, π is pi and δ is delta for dirac delta function
  4. May 7, 2013 #3
    I tried to use complete the square on the denominator to see if I could break it up more. I got s = -1 + i and s = -1 - i... so I don't know what to do with that at all. I'll keep trying though....
  5. May 7, 2013 #4


    Staff: Mentor

    If you complete the square in the denominator, you'll get s2 + 2s + 1 + 1 = (s + 1)2 + 1.
  6. May 7, 2013 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    L(y)(s) = exp(-πs)/(s^2+2s+2) + (s+2)/(s^2+2s+2). Can you find the inverse LT of 1/(s^2+2s+2)? If so, getting the inverse LT of exp(-πs)/(s^2+2s+2) is easy, just using some standard results about Laplace transforms. You can look them up for yourself.
  7. May 8, 2013 #6
    I got this. I also noticed my mistake in completing the square and solved that. I got (s+1)^2 +1 for the denominator now.

    So I should get the inverse laplace as uπ(t)*t*e^(-t) + ???

    I am getting (s+2)/((s+1)^2 + 1) which does not have an inverse laplace in the chart. The nearest thing is laplace of e^(at)*cos(bt) which is not exactly equivalent because the 'a' term doesn't match.
  8. May 8, 2013 #7

    Laplace inverse of s/((s+1)^2+1) + 2 * 1/((s+1)^2+1) does not work out either.

    The first term is close to the laplace of e^(at)*cos(bt); however, it is not exact again. The second term will be 2*e^(-t)*sin(t)

    I can't see any way to tinker with the algerbra without adding on new terms and creating more inverse laplaces...

    Would adding (1-1)/((s+1)^2+1) to y give you y = uπ(t)*t*e^(-t) + e^(-t)cos(t) + e^(-t)*sin(t)

    I'm pretty sure adding 0 is legal. Does that check out with you guys?
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