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Nisha318

  • 3 years ago

Differential Equations-> Laplace-> Piecewise Defined functions. Solve the initial value problem.

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  1. Nisha318
    • 3 years ago
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  2. abtrehearn
    • 3 years ago
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    L{y'+3y} = (s - 3)Y(s) - 1.\[L \left\{ f(x) \right\} = \int\limits_{0}^{\pi}e ^{-s x}\sin x dx\]\[= (e^{-\pi x} - 1)/(s^{2} + 1),\]so \[(s - 3)Y(s) - 1 = (e^{-\pi s} - 1)/(s^{2} + 1),\]giving us\[Y(s) = 1/(s + 3) + \[(e^{-\pi s} - 1)/[(s^{2} + 1)(s + 3)]\].\]No problem finding \[L^{-1}\left\{ 1/(s + 3 \right\}.\]To find\[L^{-1}\left\{ (e^{-\pi s} - 1)/[(s^{2} + 1)(s + 3) \right\}, \]we can use the convolution theorem. We know that\[L^{-1}\left\{ (e^{-\pi s} - 1)/(s^{2} + 1)\right\} = f(x),\]so the inverse laplace transform works out to \[f(x) * e^{-3x} = \int\limits_{0}^{x}f(u) e^{-3(x - u)}du.\]

  3. abtrehearn
    • 3 years ago
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    \[= e^{-3x}\int\limits\limits_{0}^{x}e^{3w} \sin w dw\]\[= (e^{-3x} + 3 \sin x - \cos x)/10.\]The solution, then, is\[y(x) = (e^{-3x} + 3 \sin x - \cos x)/10\]

  4. abtrehearn
    • 3 years ago
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    for \[x \in [0, \pi), \]\[(e^{-3 \pi} + 1)/10, x \in [\pi, \infty).\]

  5. Nisha318
    • 3 years ago
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    ewwww, my solution was wayyyy off.....we haven't learned convulution theorem yet....think we're covering that next. I'll post my solution in a minute....I tried working it out along with my professor's video.

  6. Nisha318
    • 3 years ago
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    My solution is 3 pages long....somebody please take a look at it and give me some feedback. See attached.

  7. Nisha318
    • 3 years ago
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    Page 1

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  8. Nisha318
    • 3 years ago
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    Page 2

  9. Nisha318
    • 3 years ago
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  10. Nisha318
    • 3 years ago
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    Page 3 (sorry, it was giving me trouble when trying to upload all 3 at one time.)

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  11. Nisha318
    • 3 years ago
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    IN PDF

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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