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.
Response from Alfred Piggott APiggott
Thanks Mickey!
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
Question 13 on pg. 223: I think the answer of the book is not correct >> 2[1+u(t-�pi)]sin3t. Instead of "+" between 1 and u(t-pi) it should be "-". Am I right? Thanks!
response from Mickey Rhoades Mrhoade
I made the same mistake the fist time through, but I checked the answer in Maple so I reworked it and found my mistake. When you use the triginometric difference formula, you will have one term vanish from sin (3pi) and the other has a cos(3*pi). This is a (-1) and it multiplies times the -2 * u(t-pi) to make it positive. --Mick
Question from Tlouvar
On Lesson 19 #26, building on above: I understand factoring out the s^2, but I'm confused why you need to integrate twice. Maybe someone can explain that part a little better. Thanks, Tim.
Response from Mickey Rhoades Mrhoade
Tlouvar,
You need to "differentiate" twice. The method to solving these problems is to get a funtion on the right that looks like the original function so you can substitute in an f(t), but it is derivative of some order. With exponentials, this can usually be accomplished with one differentiation, however, with triginometric functions, it takes two differentiations to produce a function that looks like the original. You need to get a function on the right you can sub in f(t) for so that when you perform the Laplace transform you get F. The trick, or the big fact (as Dr. Bell likes to say) is that you can do away with all the t's and come up with right and left sde of all s's. Once you are dealing with all s's it becomes an algebra problem. I hope this explanation is sufficient... -Mick
