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<math>\mathcal[F(s)] = {L}f(t) = \int_0^\infty e^{-st}f(t)\ dt</math> | <math>\mathcal[F(s)] = {L}f(t) = \int_0^\infty e^{-st}f(t)\ dt</math> | ||
| − | <math>\mathcal[f(t)] = (t^2 - 3)(t^2 - 3) </math> | + | <math>\mathcal[f(t)] = (t^2 - 3)(t^2 - 3) = t^4 - 9t^2 + 9 </math> |
| − | <math>\mathcal[F(s)] = \int_0^\infty e^{-st}(t^ | + | <math>\mathcal[F(s)] = \int_0^\infty e^{-st}(t^4 - 9t^2 + 9)\ dt </math> |
| − | <math>\mathcal[F(s)] = \int_0^\infty [t^ | + | <math>\mathcal[F(s)] = \int_0^\infty [t^4 e^{-st} - 9t^2 e^{-st} + 9 e^{-st}\ dt </math> |
| − | <math>\mathcal[F(s)] = \frac{ | + | <math>\mathcal[F(s)] = \frac{24}{s^5(s-a)} - \frac{18}{s^3(s-a)} + \frac{9}{s(s-a)} </math> |
Revision as of 19:55, 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(s-a)} - \frac{18}{s^3(s-a)} + \frac{9}{s(s-a)} $
-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} $
