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Thanks - Mac | Thanks - Mac | ||
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| + | 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) | ||
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| + | -- Kunal | ||
Revision as of 10:57, 5 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
