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Differential equation with Heaviside step function

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the solution to the differential equation using Laplace transforms.


    2. Relevant equations

    [tex]y'' - y' - 2y = 12 \theta{(t-\pi)}\sin{t}[/tex]

    [tex]y(0) = 1[/tex]

    [tex]y'(0) = -1[/tex]

    3. The attempt at a solution

    Using Laplace transforms...

    [tex]\mathcal{L}(y'') = s^2F - s + 1[/tex]
    [tex]\mathcal{L}(-y') = -sF-1[/tex]
    [tex]\mathcal{L}(-2y) = -2F[/tex]

    [tex]\mathcal{L}(12\theta{t-pi}\sin{t}) = \mathcal{L}(-12\theta{(t-\pi)}\sin{(t-\pi)}) = \frac{-12e^{-\pi s}}{s^2+1}[/tex]

    [tex](s^2-s-2)F = \frac{-12e^{-\pi s}}{s^2+1}[/tex]

    I rearrange and end up with partial fractions...

    [tex]F = e^{-\pi s}[\frac{-12}{(s^2+1)(s-2)(s+1)}] = e^{-\pi s}(\frac{A}{s-2} \frac{B}{s+1} \frac{Cs+D}{s^2+1})[/tex]

    From this point, I can rearrange everything and end up with 4 equations, 4 unknowns. I am thinking, however, that there is an easier way, simply by using one of the common transforms found in a Laplace transform table. Is there a better way to go about this problem without having to resort to partial fractions?
     
    Last edited: Sep 19, 2010
  2. jcsd
  3. Sep 19, 2010 #2

    vela

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    You seem to have omitted the terms due to the initial conditions. Also, you should check the signs of those terms. I think you flipped one.

    Do you know how to find the inverse Laplace transform using complex analysis? If not, partial fractions is your best bet.
     
  4. Sep 19, 2010 #3
    where is the step function ? :)
     
  5. Sep 19, 2010 #4
    Crap, I didn't surround the initial step function with parentheses. Also, I forgot the initial conditions. I'll fix this soon.
     
  6. Sep 19, 2010 #5
    It's fixed. I don't see anywhere where I forgot a sign, and I do not know how to figure out the inverse using complex analysis. So, if partial fractions are the way to go, then that's what I'll do!
     
  7. Sep 19, 2010 #6

    vela

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    You have a sign error in your transform for -y'. You also left out the plus signs between the terms of the partial fraction expansion.
     
  8. Sep 19, 2010 #7
    For the second one, yikes, I wrote that on the paper, but left them out when transferring them over.


    [tex]F = e^{-\pi s}[\frac{-12}{(s^2+1)(s-2)(s+1)}] = e^{-\pi s}(\frac{A}{s-2}+\frac{B}{s+1}+\frac{Cs+D}{s^2+1})[/tex]

    As for the sign error in the transform...

    [tex]\mathcal{L}(y') = s\mathcal{L}(y) - y'(0)[/tex]

    So

    [tex]\mathcal{L}(-y') = -s\mathcal{L}(y) + y'(0)[/tex]

    So taking the initial condition that [tex]y'(0) = -1[/tex]...

    [tex]\mathcal{L}(-y') = -s\mathcal{L}(y) - 1[/tex]

    Three negatives = negative. Is this not right?

    EDIT: wait, I see where I messed up. I got the original formula wrong, so I switched the 1 with a -1. I am fixing this now.
     
  9. Sep 19, 2010 #8
    So, for the equations and unknowns, I am getting...

    [tex]A - 2B + 0C - 2D = -12[/tex]
    [tex]A + B - 2C - D = 0[/tex]
    [tex]A - 2B - C + D = 0[/tex]
    [tex]A + B + C + 0D = 0[/tex]

    For the [tex]-12e^{\pi s}[/tex] term, and

    [tex]A - 2B + 0C - 2D = -2[/tex]
    [tex]A + B - 2C - D = 0[/tex]
    [tex]A - 2B - C + D = 0[/tex]
    [tex]A + B + C + 0D = 0[/tex]

    For the [tex]2[/tex] term.
     
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