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Math Help - Stuck on a Laplace Transform Question

  1. #1
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    Stuck on a Laplace Transform Question

    Here's what I'm working on:

    y'' + 6y' + 5y = t; y(0) = a, y'(0) = b

    So far, I've done this. Taking the Laplace transform of both sides of the equation gives me:

    \mathfrak{L}\{y''\} + 6\mathfrak{L}\{y'\} + 5\mathfrak{L}\{y\} = \mathfrak{L}\{t\}

    Simplifying the above expression gets me to:

    s^2 Y(s) - sy(0) - y'(0) + 6sY(s) - 6y(0) + 5Y(s) = \frac{1}{s^2}

    Factoring out the Y(s) terms on the left and applying the initial conditions leads me to:

    Y(s)(s^2 + 6s + 5) - as - b - 6a = \frac{1}{s^2}

    Y(s)(s^2 + 6s + 5) = \frac{1}{s^2} + as + b + 6a

    Multiplying both sides by \frac{1}{s^2 + 6s + 5} gives me:

    Y(s) = (\frac{1}{s^2} + as + b + 6a)(\frac{1}{s^2 + 6s + 5})

    And this is where I'm stuck and need major help.
    Last edited by BlazingFire; April 30th 2009 at 06:59 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by BlazingFire View Post
    Here's what I'm working on:

    y'' + 6y' + 5y = t; y(0) = a, y'(0) = b

    So far, I've done this. Taking the Laplace transform of both sides of the equation gives me:

    \mathfrak{L}\{y''\} + 6\mathfrak{L}\{y'\} + 5\mathfrak{L}\{y\} = \mathfrak{L}\{t\}

    Simplifying the above expression gets me to:

    s^2 Y(s) - sy(0) - y'(0) + 6sY(s) - 6y(0) = \frac{1}{s^2}
    where is your 5Y??
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    where is your 5Y??
    I edited my original post to put it back. Sorry about that. Still doesn't change my original question, though.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by BlazingFire View Post

    Y(s) = (\frac{1}{s^2} + as + b + 6a)(\frac{1}{s^2 + 6s + 5})

    And this is where I'm stuck and need major help.
    the next step is to combine the fractions. note that you have

    Y = \frac {as^3 + (6a + b)s^2 + 1}{s^2(s + 5)(s + 1)}

    now, do partial fractions decomposition
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