Difference between revisions of "HWK6MA527Fall2013" - Rhea

Revision as of 17:28, 6 October 2013

Homework 6 collaboration area

I am sort of stuck on Lesson 19 #26:

First of all, are the instructions correct when they say "...if L(F) equals:" --- Is it F or f(t)?

Secondly, I am not sure how to proceed with this. Thm 3 says the inverse Laplace is 1/s times the F(s) function. Therefore I can factor out a 1/s^3 to get : (1/s^3)*(1/(s^2-1)) ...Not sure where to go from here, or if my approach is wrong...

Thanks - Mac

I think you need to factor out s^2 to get (1/(s^2)*(s^2-1)), also question says inverse transform by integral which is done in example 3. If you will integrate inverse transform of (1/(s^2-1)) i.e., sinh t twice you will get inverse transform of (1/(s^4-s^2)) i.e., sinh t - t. (I am not sure, if I am right)

-- Kunal

Response from Mickey Rhoades (mrhoade~~)

I did the same thing and got the same answer. After the first integration of sinh(t) you get cosh(t) -1 and integrate that to find the solution of sinh(t )- t.

Question from T. Roe 21:28, 6 October 2013 (UTC):

I have a question on Lesson 19 #19. I have set up the equations as follows:

$ \mathcal{L}(f')=s\mathcal{L}(f)-f(0) $

$ f(0)=0 $

so

$ \mathcal{L}(f)=\mathcal{L}(f')/s. $

Using the information from class I get:

$ \mathcal{L}(f')=\frac{2\omega}{(s^2+4\omega^2)} $

and

$ \mathcal{L}(f)=\frac{2\omega}{s(s^2+4\omega^2)}, $

but the book as well as mathematica output a solution of

$ \mathcal{L}(f)=\frac{2\omega^2}{s(s^2+4\omega^2)}. $

Can someone explain to me where the extra $ \omega $ is coming from?

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