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Math Help - Completing Laplace Transform problem.

  1. #1
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    Completing Laplace Transform problem.

    Could someone help me here?

    L[y] = \frac{2(s+5)}{(s+5)^2+1/2} + {10}{(s+5)^2+1/2}

    Solution to that gives

    y = 2e^-5t cos(\frac{t}{\sqrt{2}}) + 10\sqrt{2}e^-5t sin(\frac{t}{\sqrt{2}}).

    I understand all but the e^-5t in this solution.


    Any assistance greatly appreciated.
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  2. #2
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    Re: Completing Laplace Transform problem.

    Apologies. Darn Latex. Cannot get it to work for the life of me!


    L[y] = \frac{2(s+5)}{(s+5)^2+1/2} + {10}{(s+5)^2+1/2}

    Solution to that gives

    y = 2e^-5t cos(\frac{t}{\sqrt{2}}) + 10\sqrt{2}e^-5t sin(\frac{t}{\sqrt{2}}).
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  3. #3
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    Re: Completing Laplace Transform problem.

    Quote Originally Posted by paulbk108 View Post
    Apologies. Darn Latex. Cannot get it to work for the life of me!


    L[y] = \frac{2(s+5)}{(s+5)^2+1/2} + {10}{(s+5)^2+1/2}

    Solution to that gives

    y = 2e^-5t cos(\frac{t}{\sqrt{2}}) + 10\sqrt{2}e^-5t sin(\frac{t}{\sqrt{2}}).
    It's because of the horizontal shift theorem.

    \displaystyle \begin{align*} \mathcal{L}\left\{ e^{at}f(t) \right\} = \mathcal{L}\left\{ f(t - a) \right\} = F(t-a) \end{align*}, or the other way \displaystyle \begin{align*} \mathcal{L}^{-1}\left\{ F(t - a) \right\} = e^{at}\mathcal{L}^{-1}\left\{ F(t) \right\} = e^{at}f(t) \end{align*}.

    So in your case

    \displaystyle \begin{align*} \mathcal{L}^{-1}\left\{ \frac{2(s + 5)}{(s + 5)^2 + \frac{1}{2}} + \frac{10}{(s + 5)^2 + \frac{1}{2}} \right\} &= \mathcal{L}^{-1}\left\{ \frac{2[s - (-5)]}{[s - (-5)]^2 + \frac{1}{2}} + \frac{10}{[s - (-5)]^2 + \frac{1}{2}} \right\} \\ &= e^{-5t}\mathcal{L}^{-1}\left\{ \frac{2s}{s^2 + \frac{1}{2}} + \frac{10}{s^2 + \frac{1}{2}} \right\}\end{align*}
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