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Math Help - Laplace Transform of a 2nd order Differential Equation

  1. #1
    Super Member craig's Avatar
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    Laplace Transform of a 2nd order Differential Equation

    Another Laplace question, this time a second order DE.

    We have \ddot{x} + 3\dot{x} + 2x = 1, with zero initial conditions. The Laplace Transform gives:

    s^2\mathcal{L}\{x\} -sx_0 - \dot{x}_0 + 3(s\mathcal{L}\{x\} - x_0) + 2\mathcal{L}\{x\} = \frac{1}{s}

    (s^2 + 3s + 2)\mathcal{L}\{x\} = \frac{1}{s}

    \mathcal{L}\{x\} = \frac{1}{s(s+1)(s + 2)} = \frac{1}{2s} - \frac{1}{s+1} + \frac{1}{2s + 4} by partial fractions.

    So using the inverse Laplace:

    \mathcal{L}\{x\} = \frac{1}{2} \frac{1}{s} - \frac{1}{s-(-1)} + \frac{1}{2} \frac{1}{s -(-2)}

    x(t) = \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2}

    Does this seem about right?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Laplace Transform of a 2nd order Differential Equation

    Quote Originally Posted by craig View Post
    x(t) = \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2} Does this seem about right?
    Right.
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  3. #3
    Super Member craig's Avatar
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    Re: Laplace Transform of a 2nd order Differential Equation

    Thankyou!
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