Difference between revisions of "HWK6MA527Fall2013" - Rhea

Revision as of 12:43, 7 October 2013

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 e^3t and then a transform of e^t.  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

Back to MA527, Fall 2013