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Homework Help: Inverse Laplace Transform, can't use partial fractions

  1. Jun 15, 2011 #1
    Hi, this is my first post, and I'm not sure if I'm posting this in the right place.

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

    I have a 2nd order diff. equation:
    3y''(t) + 2y'(t) + 5y(t) = 3
    with initial values: y'(0) = 1, y(0) = 0

    2. Relevant equations
    After using Laplace transform I get:

    Y(s) = (3 + 3s) / (s*[3s2 +2s +5])

    I believe it is correct, but even if it's not, what I'm interested in is how to solve this particular inverse laplace transform where I can't use partial fractions due to 3s2 +2s +5 not having any roots.

    3. The attempt at a solution
    I'm kind of stuck here, but maybe I could split the fraction like:

    Y(s) = 3 / (s*[3s2 +2s +5]) + 3 / ([3s2 +2s +5])

    but the problem would remain.

  2. jcsd
  3. Jun 15, 2011 #2


    User Avatar

    Staff: Mentor

    You could split it as

    A/s + (Bs + C)/(3s2 + 2s + 5)
  4. Jun 15, 2011 #3
    I would us an online solver such as WolframAlpha type in:

    DSolve[{3y''[t]+2y'[t]+5y[t]==3,y'[0]==1,y[0]==0},y[t],t] and enter
  5. Jun 15, 2011 #4
    Re-reading my post I think I didn't explain my problem correctly. My problem is in calculating the inverse transform of such rational function.
    I could split it like that, but then I would have no idea on how to solve:
    L-1[ (Bs + C)/(3s2 + 2s + 5) ]

    Since I'm studying for a final exam, I really can't use a solver, but thanks.
  6. Jun 15, 2011 #5
    Thanks for the clarification. Looking at the WolframAlpha solution can still be helpful. Their solution looks very similar to a homogeneous second order diff. eqn with complex roots. Might be a place to start.
  7. Jun 15, 2011 #6

    Ray Vickson

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    Science Advisor
    Homework Helper

    Of course you can use partial fractions. The function 3s^2 + 2s + 5 has the form 3*(s-r1)*(s-r2), where r1 and r2 are the two roots of the quadratic; they are complex: r1 = -(1/3) + (1/3)*sqrt(14)*I and r2 = -(1/3) - (1/3)*sqrt(14)*I, where I = sqrt(-1). Then 1/[(s-r1)(s-r2)] = [1/(s-r1) - 1/(s-r2)]/(r1 - r2). The inverse Laplace transform of 1/(3s^2+2s+5) is [(1/3)/(r1-r2)]*[exp(r1*t) - exp(r2*t)] = (1/3)*exp(-t/3)*[(2i)sin(w*t)]/(2i*sqrt(14)/3) = (3/14)*exp(-t/3)*sin(wt), where w = sqrt(14)/3.

    R.G. Vickson
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