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anonymous
 4 years ago
Differential Equations> Laplace> Piecewise Defined functions. Solve the initial value problem.
anonymous
 4 years ago
Differential Equations> Laplace> Piecewise Defined functions. Solve the initial value problem.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0L{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 \[f(x) * e^{3x} = \int\limits_{0}^{x}f(u) e^{3(x  u)}du.\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[= e^{3x}\int\limits\limits_{0}^{x}e^{3w} \sin w dw\]\[= (e^{3x} + 3 \sin x  \cos x)/10.\]The solution, then, is\[y(x) = (e^{3x} + 3 \sin x  \cos x)/10\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for \[x \in [0, \pi), \]\[(e^{3 \pi} + 1)/10, x \in [\pi, \infty).\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ewwww, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My solution is 3 pages long....somebody please take a look at it and give me some feedback. See attached.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Page 3 (sorry, it was giving me trouble when trying to upload all 3 at one time.)
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