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Math Help - Laplace transform to solve the initial value

  1. #1
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    Laplace transform to solve the initial value

    I'm not sure with my solution and I cannot get the final answer.

    y'+y=f(t), y(0)=0 where f(t) = 1 when 0≤t<1 and f(t) = -1 when t≥1

    f(t) = 1(u(t-0)-u(t-1))+(-1)u(t-1)

    f(t) = 1 - 2u(t-1)

    L{f(t)} = \frac{1}{s} - \frac{2e^{-s}}{s}

    Then,

    sY(s) - y(0) + Y(s) = \frac{1}{s} - \frac{2e^{-s}}{s}

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

    After that,

    y(t) = L^{-1}[\frac{1}{s(s+1)} - \frac{2e^{-s}}{s(s+1)}]

    I used partial fraction: \frac{1}{s(s+1)} = \frac{1}{s}-\frac{1}{s+1}

    Then,
    y(t) = L^{-1}[\frac{1}{s}]-L^{-1}[\frac{1}{s+1}]-2L^{-1}[\frac{e^{-s}}{s(s+1)}]

    y(t) = 1 - e^{-t}-2L^{-1}[\frac{e^{-s}}{s(s+1)}]

    Is that correct? And how can I solve for L^{-1}[\frac{e^{-s}}{s(s+1)}]?

    Thank you
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  2. #2
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    Quote Originally Posted by noppawit View Post
    I'm not sure with my solution and I cannot get the final answer.

    y'+y=f(t), y(0)=0 where f(t) = 1 when 0≤t<1 and f(t) = -1 when t≥1

    f(t) = 1(u(t-0)-u(t-1))+(-1)u(t-1)

    f(t) = 1 - 2u(t-1)

    L{f(t)} = \frac{1}{s} - \frac{2e^{-s}}{s}

    Then,

    sY(s) - y(0) + Y(s) = \frac{1}{s} - \frac{2e^{-s}}{s}

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

    After that,

    y(t) = L^{-1}[\frac{1}{s(s+1)} - \frac{2e^{-s}}{s(s+1)}]

    I used partial fraction: \frac{1}{s(s+1)} = \frac{1}{s}-\frac{1}{s+1}

    Then,
    y(t) = L^{-1}[\frac{1}{s}]-L^{-1}[\frac{1}{s+1}]-2L^{-1}[\frac{e^{-s}}{s(s+1)}]

    y(t) = 1 - e^{-t}-2L^{-1}[\frac{e^{-s}}{s(s+1)}]

    Is that correct? And how can I solve for L^{-1}[\frac{e^{-s}}{s(s+1)}]?

    Thank you
    I haven't checked your work but assuming the final result is correct you can use partial fractions and the shifting theorem to get the inverse Laplace transform.
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  3. #3
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    Hello everyone!

    Quote Originally Posted by mr fantastic View Post
    I haven't checked your work but assuming the final result is correct you can use partial fractions and the shifting theorem to get the inverse Laplace transform.
    I have checked his work, it seems to be correct.

    But the following lines have a typo in it:






    Of course it is f(t) = u(t) - 2 u (t-1)

    He also could use an approximation

    e^{-s} = 1-s+\frac{s^2}{2}-\frac{s^3}{6}+...

    Regards
    Rapha
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