Homework 6 collaboration area
Question from A. Piggott
Section 6.1, problem 1 and 2, I solved with Integration by Parts.
Is this the least time consuming way to solve them? What is the best way?
Also, 6.1, #5, when I solve with integration by parts, I get a solution that always requires another integration by Parts so I am not getting anywhere. Can anyone help on this?
Response from Mickey Rhoades Mrhoade
I used integration by parts for 1,2, and 5. They weren't too terrible. Obviously, these transforms are covered in the table but going through the motions to transform them was the point. You can check your answers using the table. For #5, I broke sinh(t) into its exponential components and then you get two seperate integrals. you should have one transform of e3t and then a transform of et. You should get F(s) = .5/(s-3) - .5/(s-1). You get common denominators and this becomes:
1/(s-3)(s-1) = 1/(S^2-4S+3) you then add 1 and subtract 1 on the denominator to complete the square.
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 ω is coming from?
Response from Mickey Rhoades Mrhoade
I started by changing sin2(wt) into 1/2-cos(2wt)/2 I took f'(t) to be wsin(2wt) with f'=0. f''(t) = 2w2 * cos(2wt)
Then, I took L[f(t)] to get s2F=2w2(-2F + L[1]). The trick was to recognize that if f(t) = 1/2-cos(2wt)/2 you can rearrange and find cos(2wt) = -2*f(t) + 1. Then when you set the right side of the equation of the Laplace transform of the second derivative to be 2w2 * cos(2wt) you can substitute and make the right side 2w2(-2*f(t) + 1) which transforms to 2w2(-2F + 1/s). Set is equal to your earlier obtained s2F
(s2 + 4w2)F = 2w2/s
