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Nisha318
Group Title
Differential Equations> Laplace> Piecewise Defined functions. Solve the initial value problem.
 3 years ago
 3 years ago
Nisha318 Group Title
Differential Equations> Laplace> Piecewise Defined functions. Solve the initial value problem.
 3 years ago
 3 years ago

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abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
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 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
\[= e^{3x}\int\limits\limits_{0}^{x}e^{3w} \sin w dw\]\[= (e^{3x} + 3 \sin x  \cos x)/10.\]The solution, then, is\[y(x) = (e^{3x} + 3 \sin x  \cos x)/10\]
 3 years ago

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

Nisha318 Group TitleBest ResponseYou've already chosen the best response.1
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.
 3 years ago

Nisha318 Group TitleBest ResponseYou've already chosen the best response.1
My solution is 3 pages long....somebody please take a look at it and give me some feedback. See attached.
 3 years ago

Nisha318 Group TitleBest ResponseYou've already chosen the best response.1
Page 3 (sorry, it was giving me trouble when trying to upload all 3 at one time.)
 3 years ago
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