Difference between revisions of "HWK6MA527Fall2010" - Rhea

Revision as of 20:19, 6 October 2010

Homework 6 collaboration area

Here is something to get you started:

$ \mathcal{L}[f(t)]=\int_0^\infty e^{-st}f(t)\ dt $

$ \mathcal{L}[f'(t)]= sF(s)-f(0) $

p. 226: 1.

$ \mathcal{L}[t^2-2t]= \frac{2}{s^3}-2\frac{1}{s^2} $

Odd solutions in the back of the book.

p. 226: #2: who can validate this?

$ \mathcal(t^2 - 3)^2 $

$ \mathcal[F(s)] = {L}f(t) = \int_0^\infty e^{-st}f(t)\ dt $

$ \mathcal[f(t)] = (t^2 - 3)(t^2 - 3) = t^4 - 9t^2 + 9 $

$ \mathcal[F(s)] = \int_0^\infty e^{-st}(t^4 - 9t^2 + 9)\ dt $

$ \mathcal[F(s)] = \int_0^\infty [t^4 e^{-st} - 9t^2 e^{-st} + 9 e^{-st}]\ dt $

$ \mathcal[F(s)] = [\frac{24}{s^5} - \frac{18}{s^3} + \frac{9}{s}]e^{-st} $

$ \mathcal [e^{-st}] = 1 $ therefore $ \mathcal[F(s)] = [\frac{24}{s^5} - \frac{18}{s^3} + \frac{9}{s}] $ DONE?

p. 226: #4: who can validate this?

$ \mathcal [sin^2 4t] $

$ \mathcal[F(s)] = {L}f(t) = \int_0^\infty e^{-st}f(t)\ dt $

$ \omega=4 $

$ \mathcal[f(t)] = sin^2 4t $

-Does anyone have a hint on solving #23?

Even solutions (added by Adam M on Oct 5, please check results):

p. 226: 10.

$ \mathcal{L}[-8sin(0.2t)]=\frac{-1.6}{s^2+0.04} $

p. 226: 12.

$ \mathcal{L}[(t+1)^3]=\frac{6}{s^4}+\frac{6}{s^3}+\frac{3}{s^2}+\frac{1}{s} $

p. 226: 30.

$ inverse \mathcal{L}[\frac{2s+16}{s^2-16}]=2cosh(4t)+4sinh(4t) $

(AJ) I have the same solutions for p 226 #10 and #12, but on #30, I factored the denominator and used partial fraction decomposition to get

$ inverse \mathcal{L}=-e^{-4t}+3e^{4t} $

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