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Y'' - 9y = 4t - δ_2(t); y() = -2, y'(0) = -1

  1. Mar 17, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Solve the differential equation y'' - 9y = 4t - δ_2(t); y() = -2, y'(0) = -1 for y(t) when 0 <= t < 2 as well as when t > 2.


    2. Relevant equations
    Dirac delta function. Heaviside function. Laplace and Inverse Laplace Transforms.


    3. The attempt at a solution
    I'll type the relevant information here since my handwriting for my work is ugly (but I am uploading it anyways just in case you need it). I get to Y(s) = 4/[s^2 * (s^2 - 9)] - e^(-2s)/(s^2 - 9) - (2s+1)/(s^2 - 9) and I'm stuck at getting the inverse Laplace Transform of the bolded/italicized term. Could someone please show me how to do that part? I know it involves the unit step/Heaviside function but I don't know what to do specifically.

    From Wolfram Alpha,
    The inverse Laplace Transform of that bolded/italicized term is:
    http://www.wolframalpha.com/input/?i=inverse+laplace+transform+of+{-exp(-2s)/(s^2+-+9)}
    The inverse Laplace Transform of the whole (s) = 4/[s^2 * (s^2 - 9)] - e^(-2s)/(s^2 - 9) - (2s+1)/(s^2 - 9) equality is:
    http://www.wolframalpha.com/input/?i=inverse+laplace+transform+of+{4/[s^2+*+(s^2+-+9)]+-+exp(-2s)/(s^2+-+9)+-+(2s%2B1)/(s^2+-+9)}

    (Just to note, the Wolfram Alpha answers would be wrong if my work leading up to the equation I retyped in this forum post from my work is incorrect.)

    Thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Mar 18, 2012 #2

    Mark44

    Staff: Mentor

    Have you tried splitting up 1/(s2 - 9) into two terms using partial fraction decomposition?
     
  4. Mar 18, 2012 #3

    s3a

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    I haven't because I didn't foresee any upcoming successful step.

    Here is my work for what you mentioned.
     

    Attached Files:

  5. Mar 18, 2012 #4

    Mark44

    Staff: Mentor

    That work looks OK.

    What you need to deal with the expressions you're asking about is this formula:

    [tex]L^{-1}(e^{-as}~G(s)) = u_a g(t - a)[/tex]

    Here G(s) = [itex]L {g(t)}[/itex]

    Edit: For some reason, the LaTeX isn't rendering correctly. Also, I neglected to indicate that I was talking about the inverse Laplace transform in the first equation above.
     
    Last edited: Mar 19, 2012
  6. Mar 19, 2012 #5

    s3a

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    Did you mean:

    L{u_c(t) f(t-c)} = e^(-cs) L{f(t)} ?
     
  7. Mar 19, 2012 #6

    Mark44

    Staff: Mentor

    The first formula should have been for the inverse Laplace transform. It's now fixed.
     
  8. Mar 24, 2012 #7

    s3a

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    Wrong thread sorry.
     
    Last edited: Mar 24, 2012
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