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Math Help - [SOLVED] prove using laplace transforms

  1. #1
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    [SOLVED] prove using laplace transforms

    cant figure out how to show that they are equal. can anyone help?
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  2. #2
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    Quote Originally Posted by mathprincess24 View Post

    cant figure out how to show that they are equal. can anyone help?

    \mathcal{L}^{-1} \left(\frac{1}{s^2 + k^2} \right) = \frac{\sin (kt) - kt \cos (kt))}{2k^3}
    \frac{2k^2}{(s^2 + k^2)^2}= \frac{s^2 + k^2 + k^2 - s^2}{(s^2 + k^2)^2}=\frac{1}{s^2 + k^2} + \frac{k^2 - s^2}{(s^2 + k^2)^2}. but we have that \frac{k^2 - s^2}{(s^2 + k^2)^2}=\frac{d}{ds}\left(\frac{s}{s^2 + k^2} \right)=\frac{d}{ds} \mathcal{L}(\cos (kt))=\mathcal{L}(-t\cos(kt)). \ \ \color{red} (*)

    thus: \frac{2k^2}{(s^2 + k^2)^2}= \frac{1}{k} \mathcal{L}( \sin(kt)) + \mathcal{L}(-t \cos(kt))=\mathcal{L} \left(\frac{\sin(kt) - kt \cos(kt)}{k} \right), and the claim follows.


    \color{red} (*): recall that in general: \frac{d}{ds} \mathcal{L}(f(t))=\mathcal{L}(-tf(t)).
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  3. #3
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    thanks so much. that makes it a lot more clear now.
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