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Diff. Eq. La place transform

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Evaluate the Laplace of {tcos4t} using the derivative of a transform

    Ofcourse i know the shortcut way of doing this, but I need to do it the long way.

    2. Relevant equations

    shortcut way
    t cos bt = [itex]\frac{s^2-b^2}{(s^2+b^2)^2}[/itex]

    long way transform of a derivative
    (-1)^n [itex]\frac{d^n}{ds^n}[/itex] F(s)

    F(s)=L{f(t)}


    3. The attempt at a solution

    n=1 f(t)=cos4t
    f(s)= Laplace of cos4t

    L{cos4t}= [itex]\frac{s}{s^2+16}[/itex]

    -[itex]\frac{d}{ds}[/itex] [itex]\frac{s}{s^2+16}[/itex]

    Quotient rule

    -[itex]\frac{s^2+16-2s^2}{(s^2+16)^2}[/itex]

    This is as far as I get on my own.

    In my notes for class shes goes a step further and I'm not quite sure what happens to the
    -2s^2 in the next step

    my notes continue as follows:
    -[itex]\frac{-s^2+16}{(s^2+16)^2}[/itex]

    now we distribute the negative

    and get
    [itex]\frac{s^2-16}{(s^2+16)^2}[/itex]

    I checked the answer using a Laplace transform table(easy way) and did receive [itex]\frac{s^2-16}{(s^2+16)^2}[/itex]

    but when i do it the long way i dont know what happens to -2s^2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 29, 2014 #2
    wow im so stupid. i missed such an easy thing. combing the like terms s^2 and -2s^2. i get it now.
     
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