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Laplace initial value problem> HELP! PLEASE!

  1. Feb 3, 2009 #1
    Laplace initial value problem.... HELP! PLEASE!


    Hello all!
    I'm stuck on this question:

    y''(t)+ 4y'(t) = sin2t

    y(0) = 0

    solve it using laplace transform,... my final is tomorrow, and its 2 am, i would appreciate a quick respone
    thanks in advance!
    Last edited: Feb 3, 2009
  2. jcsd
  3. Feb 3, 2009 #2
    Show your attempt at it...
  4. Feb 3, 2009 #3
    I can not!!
  5. Feb 3, 2009 #4
  6. Feb 3, 2009 #5
    Thank you very much
    But I have a problem is all Constants equal to zero!!
    What reason?
  7. Feb 3, 2009 #6
    Maybe because your initial condition works out that way? I didn't work your problem.
  8. Feb 4, 2009 #7
    The problem lies in the
    y'(0) = 0
    As a result, all the constants equal to zero
  9. Feb 4, 2009 #8
    If you say so :) Like I said I didn't work your problem.
  10. Feb 4, 2009 #9
    Thank you
    Your participation in the subject to others if allowed
  11. Feb 12, 2009 #10
    this solution is very straightforward so I am not going to solve it for you but I will tell you the steps you should do to get to the final solution. THis is a DE, u need to find y(t).

    Step 1)
    Find the laplace transforms for y'', y', y... note the LT of y is just Y(s)
    Find the laplace transform for the right hand side of the equation.
    Apply the initial conditions.

    Step 2)
    Put it all back in the algebraic equation and solve for Y(s),
    Step 3)
    find the laplace inverse of Y(s) and that will give you y(t)

    good luck, and I hope this is helpful and not to late.
  12. Feb 12, 2009 #11
    Didn't the OP already say they solved it?
  13. Feb 12, 2009 #12


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    Staff Emeritus
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    Am I the only one to notice that this is a second order d.e. and there is only one initial condition?

    There exist an infinite number of solutions that have different values of y'(0).

    y(t)= C- (C+ 1/4)e^{-4t}+ 1/2 cos(t)+ (1/4)sin(2t) satisfies this equation and y'(0)= 0 for any value of C.
  14. Feb 12, 2009 #13
    In post 7, the OP says y'(0) = 0 which seems to be in addition to y(0) = 0?
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