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Thread: Using Laplace

  1. #1

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    Using Laplace

    I am stuck on this one
    y''(t)-2y'(t)+10y(t)=0
    initial values y(0)=1; y'(0)=8

  2. #2
    Only Slightly Obsessive jemidiah's Avatar
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    Re: Using Laplace

    This is pretty basic as far as Laplace transform problems go. There's really not much to get stuck on. Still, here's my working.

    Recall these properties of the Laplace transform L:
    * My notation: L{x(t)} = X(s) by definition
    * Linearity, so L{ax(t) + by(t)} = a X(s) + b Y(s) for any constants a and b
    * L{0} = 0 from linearity
    * L{x'(t)} = s X(s) + x(0)
    * L{x''(t)} = s^2 X(s) - s x(0) - x'(0)
    * L{e^(at)} = 1/(s-a)

    Take the Laplace transform of your equation to get
    L{y''(t) - 2y'(t) + 10y(t)}
    = L{y''(t)} - 2 L{y'(t)} + 10 L{y(t)}
    = [s^2 Y(s) - s y(0) - y'(0)] - 2 [s Y(s) - y(0)] + 10 Y(s)
    = [s^2 - 2s + 10] Y(s) + [-s y(0) - y'(0) + 2y(0)]
    = [s^2 - 2s + 10] Y(s) - [s + 6]
    = L{0}
    = 0

    => Y(s) = (s+6) / (s^2 - 2s + 10)

    There are a few ways to take the inverse Laplace transform from here. You can factor the quadratic in the denominator (which has roots at s = 1 +/- 3i) and use partial fractions, which will give you something of the form 1/(s - a), which has inverse Laplace transform e^(at). You can also use the general form of the exponentially decaying sinusoids nearly directly. I'll pick the first method.

    I won't write out the partial fractions steps, but after doing the computation one finds

    Y(s) = (1/2 - 7/6 i) / (s - (1 + 3i)) + (1/2 + 7/6 i) / (s - (1 - 3i))

    so taking the inverse Laplace transform gives

    y(t) = (1/2 - 7/6 i) e^(1 + 3i)t + (1/2 + 7/6 i) e^(1 - 3i)t

    If you're more comfortable with sines and cosines, this expands out to

    y(t) = 7/3 e^t sin(3t) + e^t cos(3t)


    I've verified the result with a computer algebra system, and it's of course easy to check by hand.
    The time you enjoy wasting is not wasted time.
    Bertrand Russell

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