Students send me questions about the homework assignment. If I think the questions and answers are generally interesting, I will anonymize the questions, and post the questions and responses here.
Assignment due October 19
I have a question about the homework due tomorrow that I was hoping you could help me with please. For problem 51 of the homework, I got the inverse of the matrix to be
The (correct) answer has been removed.and then I multiplied this by the B values to get x1 and x2, but my answers don't come out to the same as the book's answers ... I'm sure that the inverse is correct and can't figure out why my answer differs ... For instance, I get x1 = 1/4 and x2 = -13/8 for the first B values of 5 and 4, but the book gets x1 = 9/10 and x2 = 13/20?!
I seem to get answers which are different from the textbook's also.
One of the reasons I leaned towards this book rather than another is that the other was the first edition and this was a second edition, so one could hope that most of the more gross (grosser?) errors had been found and corrected.
It certainly seems like the textbook's answers are incorrect here. Sorry.
Assignment due September 26
I have a quick question on the hw problem 4.5.3. I solved the problem and according to my work I get y(t)= sin(t-2pi)U(t-2pi) + sin(t). The answer in the back of the book is y(t)= sin(t)U(t-2pi) + sin(t). Are the two expressions equivalent or ...
Just remember that sine is a periodic function with period 2Pi. So sin(x+300Pi)=sin(x-96Pi)=... etc. And certainly sin(t-2Pi) is the same as sin(t).
This is a different sort of problem, called a boundary value problem. The Laplace transform works wonderfully well for initial value problems. In this case, you need to figure out a connection, if any, between those initial values (y''(0) and y'''(0)) and the desired values of t (or x) at L. One can try to be very clumsy (it will work, I think!) and just plug in A and B for the initial values for y''(0) and y'''(0) and see what happens at L, compare this to what is desired, and then go BACKWARDS. This is clumsy, but I think it will work. The two different functional descriptions for each half of the loaded beam occur because the load is "concentrated", modeled by a delta function.
Well, there are several things you can do.
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