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p. 226: 2.
 
p. 226: 2.
<math>\mathcal[(t^2 - 3)^2)
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<math>\mathcal[(t^2 - 3)^2)</math>
  
  

Revision as of 19:21, 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. $ \mathcal[(t^2 - 3)^2) $


-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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