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?
