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Math Help - Laplace DE with unit step function

  1. #1
    Junior Member
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    Laplace DE with unit step function

    Just a quick question on a laplace transform question,

    I need to find the transform for

    y" +2y' + y = u(t-1) , y(0)=1, y'(0)= -1

    After using the known properties i obtained,

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

    using partial fractions, t-shifting and s-shifting and other known properties i arrived at

    y(t) = (1+exp(t-1)((t-1)-1))u(t-1) + exp(t)(1+t)

    Thus for 0<t<1

    y(t) =exp(t)(1+t)

    for t>1

    y(t)= 1+exp(t-1)((t-1)-1) + exp(t)(1+t)


    Is my answer and methodology correct? There is no answer in the book.

    Thanks for your help.
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  2. #2
    A Plied Mathematician
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    Re: Laplace DE with unit step function

    I think you might have some sign errors in there somewhere. Taking the LT of the original DE yields

    s^{2}Y+Y+2(sY-y(0))-sy(0)-y'(0)=\frac{e^{-s}}{s}, which leads to

    Y(s^{2}+2s+1)=\frac{e^{-s}}{s}+s+1.

    I end up with the LT being

    Y=\frac{e^{-s}}{s(s+1)^{2}}+\frac{1}{s+1}.

    You can use partial fractions on the first to obtain

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

    Can you finish from here?
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