Some questions and answers about homework assignments

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).

In section 4.5, #13, the initial conditions given are y(0), y'(0), y''(L), and y'''(L). How are the last two related to y''(0) and y'''(0)? I don't see how to insert those conditions into the equation I get after taking the Laplace transform of the original equation. Thanks.

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.

I have a question about one of the problems that you assigned, section 4.5 #11. When I took the Laplace transform of the function I got Y(s)=(something in numerator)/(s^2+4s+13). When I factored the denominator, I got [complex roots]. I was unsure about what to do from there ...

Well, there are several things you can do.

  1. Run it on Maple. Comment: fine for the real world, not for a course where you might be expected to do similar problems "by hand".
  2. The roots of s^2+4s+13 are P=-2+3i and Q=-2-3i. Then you could split up things using partial fractions with A/(s-P)+B/(s-Q): solve for A and B. Then take inverse Laplace transforms. The results will be a linear combination of e^Pt and e^Qt. Now by Euler's formula each of these exponentials can be rewritten. They look like e^(-2t)(cos(3t)+/-i*sin(3t). It will turn out that the linear combination can be written only in terms of real multiples of e^(-2t)cos(3t) and e^(-2t)sin(3t) (these "damped oscillation" terms should be familiar to you from previous work with diff'l equations).
  3. (best for last) I think the "simplest" way is to write STUFF/(s^2+4s+13) us to complete the square in the denominator, or bottom as I call it in class. So that turns out to be quite easy, since this is a textbook problem: s^2+4s+13=s^2+4s+4+9=(s+2)^2+3^2. Now you can use one of the shifting theorems to find the inverse Laplace transform, which will involve (not too surprisingly) cos(3t) and sin(3t) and also multiplication by e^(-2t) due to the shift. It actually all does work out, without terrible trouble.

Maintained by and last modified 9/26/2004.