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Math Help - Laplace Transforms

  1. #1
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    Laplace Transforms

    see the attachment for the questions i am stuck on.
    includes first and second order differential equations which need to be solved using Laplace Transforms.
    your help would be much appreciated
    thanks
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  2. #2
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    Quote Originally Posted by louboutinlover View Post
    see the attachment for the questions i am stuck on.
    includes first and second order differential equations which need to be solved using Laplace Transforms.
    your help would be much appreciated
    thanks
    (s-3)/ [(s+3)(s-1)] = A/(s+3) + B/(s-1)

    There is a quick trick to solving for the coefficients of this partial fraction. To find A for example I multiply both sides of the equation by (s+3), now evaluate the left hand side at s = -3 since B is gone.

    For A i get (s-3)/(s-1) evaluated at -3 gives 6/4 or 3/2. For B i do the same but multiply both sides by (s-1) and get -1/2

    So now take the inverse laplace transform by using a look up table if you wish and I get 1.5e^(-3t) - 0.5e^(-t)
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  3. #3
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    Actually I assumed your previous workings were corrrect but I spotted your denominator is incorrect. Your function in the Laplace domain should be F(s) = (5-s)/[(s-1)(s+3)]. By employing the same procedure as above I seperated the function into its partial fraction components, namely (5-s)/[(s-1)(s+3)] = A/(s-1) + B/(s+3) and found A = 1 and B = -2. The answer I finally got was
    F(x) = e^x - 2e^(-3x).
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  4. #4
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    HI I am extremely sorry but I made another small error. The question is very simple its just a small algabraic mistake both me and you have been making.

    Notice sX(s) - 1 + 3X(s) = 4/(s-1)

    X(s)(s+3) = 4/(s-1) +1

    now when you solve for X(s) you should obtain 1/(s-1) becausethe (s+3)'s cancel.

    Finally the inverse laplace transfrom of 1/(s-1) is indeed e^t.

    Sorry about the previous post, I have been running low on sleep the pastfew days.
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