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Solving an Initial value problem using Laplace transform

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve the Initial value problem using Laplace transform
    [itex]\ddot{y}[/itex] +2y = 0, y0 = C1, [itex]\dot{y}[/itex] = C2


    2. Relevant equations

    [s2 - sy(0) - y'(0)] + a[sY - y(0)] + bY

    3. The attempt at a solution
    s2Y - sy(0) - y'(0) + 2y = 0
    s2Y + 2Y = sy(0) + y'(0)
    (s2 + 2)Y = s(C1) + (C2)
    Y = (s(C1))/[s2 + 2] + (C2)/[s2 + 2]

    And that is as far as I can get as I am unsure what to do now?
     
  2. jcsd
  3. Aug 22, 2011 #2

    rock.freak667

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    Well the laplace transform of what function gives the form s/(s2+k2)?

    Similarly what function's transform give k/(s2+k2)?



    (Hint:Think trig functions)
     
  4. Aug 22, 2011 #3
    Oh so I use cos[itex]\omega[/itex]t
     
  5. Aug 22, 2011 #4
    Actually I am confused does that mean for the first bit my answer is cos([itex]\sqrt{2}[/itex] t) what happens to the C1?
     
  6. Aug 22, 2011 #5

    rock.freak667

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    Right so cosωt would give ω/(s22), so comparing this to C1* s/(s2+2) what is ω ?

    The C1 should be there, even if your book says it isn't there, it should be there still.
     
  7. Aug 22, 2011 #6
    so is my answer just C1cos([itex]\sqrt{2}[/itex] t) otherwise I am really confused..
     
  8. Aug 22, 2011 #7
     
  9. Aug 22, 2011 #8

    rock.freak667

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    No, remember you have two functions.

    Sorry, I wrote the other one.

    s/(s22) compared to s/(s2+2) gives ω as?


    If you are confused as to what I am trying to get you to see, look up the laplace transforms for cosine and sine.
     
  10. Aug 22, 2011 #9
    so [itex]\omega[/itex]2 = 2 which implies [itex]\omega[/itex] = [itex]\sqrt{2}[/itex]?
     
  11. Aug 23, 2011 #10

    vela

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    Good. So you have [tex]Y(s) = C_1 \frac{s}{s^2+\omega^2} + (C_2/\omega)\frac{\omega}{s^2+\omega^2}[/tex]where [itex]\omega^2 = 2[/itex]. Note I multiplied and divided the second term by ω to get the Laplace transform to look like one in the table. The first term corresponds to [itex]C_1 \cos \omega t[/itex], as you noted earlier. What do you get for the second term?
     
  12. Aug 24, 2011 #11
    Is it right to do the second term as C2(1/(s2 + 2) which corresponds to (1/[itex]\sqrt{2}[/itex]) sin([itex]\sqrt{2}[/itex]t)?
     
  13. Aug 24, 2011 #12

    vela

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    Sure. I just wrote it with ω on top because that's how it most likely appears in the table, rather than 1/(s22).
     
  14. Aug 24, 2011 #13
    Thank you so much for all your help :D I am finally beginning to understand!!
     
  15. Aug 25, 2011 #14

    rude man

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    Y = (s(C1))/[s2 + 2] + (C2)/[s2 + 2]

    And that is as far as I can get as I am unsure what to do now?


    Just use tables! Of course you have to jiggle your constants around a bit in order to accommodate the particular form your tables happen to be in.
     
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