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Homework Help: Laplace problem (new) :|

  1. Nov 26, 2005 #1
    I have the following function with the following I.C. y(0)=1 & y'(0)=1
    y''-2y'+2y=cost
    I apply laplace.
    i get the following
    s^2Y(s)-Sy(0)-y'(0)-2[sY(s)-y(0)]+2Y(s)= s/(s^2+1)
    -------
    i apply the initial condtions
    and i obtain
    Y(S)[s^2-2s+2]=s-2+s/(s^2+1)
    therefore, Y(S)=(s-2)/((s-1)^2+1)+S/[(s-1)^2(s^2+1)]
    i am not sure how to procede forward.. any help would be appreciated.
    sorry for posting and asking so many questions in such a short period.
     
  2. jcsd
  3. Nov 27, 2005 #2

    benorin

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  4. Nov 27, 2005 #3

    benorin

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    Here you go

    From

    [tex]Y(s)=\frac{s-2}{\left( s-1\right)^{2}+1} + \frac{s}{\left( s-1\right)^{2} \left(s^{2}+1\right)}[/tex]

    Deal with the terms seperately: first put

    [tex]\frac{s-2}{\left( s-1\right)^{2}+1} = \frac{s-1}{\left( s-1\right)^{2}+1}-\frac{1}{\left( s-1\right)^{2}+1}[/tex]

    so that the right-hand side consists of terms on the table;

    next, use partial fraction decomposition on the other term, that is write

    [tex]\frac{s}{\left( s-1\right)^{2} \left(s^{2}+1\right)} = \frac{A}{s-1} + \frac{B}{\left( s-1\right)^{2}} + \frac{Cs+D}{s^{2}+1}[/tex]

    now cross-multiply to get

    [tex]s = A\left( s-1\right) \left( s^{2}+1\right) + B\left( s^{2}+1\right) + \left( Cs+D\right) \left( s-1\right)^{2}[/tex]

    Solve for A,B,C, and D as follows:

    If s=1, then [itex]1=2B[/itex];

    If s=0, then [itex]0=-A+B+D[/itex];

    If s=-1, then [itex]-1=-4A+2B-4C+4D[/itex];

    If s=2, then [itex]2=5A+5B+2C+D[/itex];

    Solving the above system of equations gives
    [tex]A=0,B=B=\frac{1}{2},C=0,D=-\frac{1}{2},[/tex]

    and hence

    [tex]\frac{s}{\left( s-1\right)^{2} \left(s^{2}+1\right)} = \frac{1}{2\left( s-1\right)^{2}} - \frac{1}{2\left( s^{2}+1\right) }[/tex]

    All togeather we have

    [tex]Y(s) = \frac{s-1}{\left( s-1\right)^{2}+1}-\frac{1}{\left( s-1\right)^{2}+1}+ \frac{1}{2\left( s-1\right)^{2}} - \frac{1}{2\left( s^{2}+1\right) }\right] [/tex]

    hence

    [tex]y(t) = L^{-1}\left\{ Y(s)\right\} = e^{t}\cos(t) - e^{t}\sin(t) + \frac{1}{2}te^{t} - \sin(t) [/tex]
     
    Last edited: Nov 27, 2005
  5. Nov 27, 2005 #4
    I made a mistake in copying the the laplace transformation

    Y(s)= (s-2)/((s-1)^2+1)+ s/(s^2+1)[(s-1)^2+1]

    I understood how you the first part of the laplace transform is a form i can look up, but the next part, the (s-1)^2+1, not sure how to deal with that one with PFDC..
     
  6. Nov 28, 2005 #5

    HallsofIvy

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    Your table certainly has a formula for the inverse Laplace transform of
    [itex]\frac{1}{s^2+ 1}[/itex] and also should tell you what to do with "s-1" rather than s.
     
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