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Math Help - 2nd Order Inhomogeneous DE

  1. #1
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    2nd Order Inhomogeneous DE

    x''-x = sin(t) x'(0)=x(0)=0
    I'm supposed to solve this using both Laplace and non-Laplace methods
    Using Laplace, I got that x(t) = .5cos(t)-.5sin(t)-.5e^(-t), which works with the given conditions

    However, for some reason when I try to find a solution by undetermined coefficients...I can't. I got a completely different answer (I think it was something like x(t)= -.5sin(t), or something like that) which obviously doesn't work.
    Solving for the homogeneous, I find that xh(t) = e^(-t) + e^t
    Then I guessed that xp(t) = acos(t) + bsin(t), plugged it into the equation, but then found that a = 0 and b= -.5
    Can anyone help me figure this one out? It seems silly that I can't do it...
    Thanks!
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  2. #2
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    Jan 2009
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    Quote Originally Posted by mistykz View Post
    x''-x = sin(t) x'(0)=x(0)=0
    I'm supposed to solve this using both Laplace and non-Laplace methods
    Using Laplace, I got that x(t) = .5cos(t)-.5sin(t)-.5e^(-t), which works with the given conditions

    However, for some reason when I try to find a solution by undetermined coefficients...I can't. I got a completely different answer (I think it was something like x(t)= -.5sin(t), or something like that) which obviously doesn't work.
    Solving for the homogeneous, I find that xh(t) = e^(-t) + e^t
    Then I guessed that xp(t) = acos(t) + bsin(t), plugged it into the equation, but then found that a = 0 and b= -.5
    Can anyone help me figure this one out? It seems silly that I can't do it...
    Thanks!
    Using the 'non-Laplace method' I get y = \frac{1}{4} e^t - \frac{1}{4} e^{-t} - \frac{1}{2} \sin t. So you've made a mistake in using the Laplace transforms. Did you check that solution by substitution? It didn't work when I checked it.
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  3. #3
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    Aug 2007
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    Figured them both out, thanks!
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