Heat equation: http://math.mit.edu/daimp/HeatEquation.html
Q. Today in class we had a few discussions that are not in lecture notes. Can we get questions about it in Midterm?
You may get a question similar to the homework problem (3.9.18). I will
not quiz you about experiments with tuning forks or anything that is
not in lecture notes (bonus questions are exceptions).
Here are two more links for Friday's lecture:
Breaking a wine glass
Q. Today in class we had a presentation of a spring vibration. What was the webpage?
Q. How to compute the 4-th roots of -3?
The textbook explains how to compute the 4-th roots of -1. On the other
hand -3 is the product 3*(-1). The 4-th roots of -3 are the products of
the real positive 4-th root of 3 and the forth roots of -1 (computed in
the textbook, page 221).
Q. Say we have a characteristic equation
r^3 - 4r^2 + 3=0.How do we solve it?
In general there is a formula for the roots, but we do not use it
in this course. Instead we, first, guess a root (among factors of the
free term 3), which is 1 in our case. Then we factor out (r-1). Or equivalently, we divide r^3 - 4r^2 + 3 by (r-1). To do so you can use polynomial long division. Or you can factor it as we did in class:
r^3 - 4r^2 + 3 = [ (r-1) * r^2 + r^2 ] - 4r^2 + 3
= (r-1)* r^2 - 3r^2 +3 = (r-1)*r^2 - [ (r-1)*3r + 3r] +3
(r-1)*r^2 - (r-1)*3r - 3r + 3
(r-1)(r^2 - 3r - 1)
Or you can write it as
r^3 - 4r^2 + 3 = (r-1) (ar^2 + br +c)
find the coefficients a,b,c by expanding the polynomial on the right
hand side and then equating it to the polynomial on the left hand side.
You may use any method you like.
Finally we find the other roots by solving the remaining equation
r^2 - 3r -1 =0.
Q. Why is linear independence of solutions y1, y2 important for writing a general solution c1y1 + c2y2?
A. There is a theorem for equations of the form
ay'' + by' + cy = 0
asserting that if y1, y2 are linearly independent, then c1y1 + c2y2 is a general solution.
Linear independence is crucial here. Let us suppose that y1, y2
are linearly dependent, i.e., there are constants k1
and k1, not all zero, such that
k1y1 + k2y2 = 0.
Suppose k1 is not zero. Then by dividing each term by k1 we may assume that k1 is 1. In other words,
y1 = - k2y2.
Of course for such a choice of y1, the family c1y1 + c2y2 will be missing many solutions of the equation.
Q: For second order linear differential equations we say that solutions y1 and y2
form a fundamental set of solutions if the Wronskian W(y1, y2) is non zero. On the other hand, in general we say that functions y1, ..., yn form a fundamental set of solutions of a differential equation of order n if these functions are linearly independent. Do these definitions agree?
A: Yes, these two definitions agree. You may see a proof sketched in Problem 4.1.25.
Q: We had many Existence and
Uniqueness theorems. It is getting a bit confusing...
A: Well, Existence and Uniqueness theorems tell us
on which intervals a solution to a differential equation certainly
exists and unique. Basically at this moment we have two Existence and
Uniqueness theorems. A very strong one for linear equations and a
weaker one for non-linear first order equations.
If our equation is linear (of order one or two),
y' = f(x, y)
Q: Suppose I am given a
problem in the coming quiz. Suppose it is of the first order, but it is
not linear, it is not separable and it is not exact. So I try to find an
integrating factor mu(x) that would make the equation
such a factor does not exist. Then I try to find an integrating
Suppose that such
a factor does not exist either. What should I do next?
y'' = f(x, y, y') y(x0)=y0, y'(x0)=y'0,
then a solution exists and unique at
least on the interval containing x0 where the function f is continuous. For non-linear first order equations, the Existence and Uniqueness theorem states that a solution of
y' = f(x, y)
certainly exists only on a small interval about x0 if the function f is smooth enough in a box neighborhood about (x0, y0).
For precise statements, see the textbook or lecture notes.
A: Hmmm... All equations in the quiz will
be either linear (of the first or second order), or separable or exact.
So the only type of equations for which you may need to find an
integrating factor will be the linear one.
Q: How many questions will we have
in the quiz?
A: There will be 4 problems, but some problems may
have several questions.
Q: How long is the quiz
A: It will be 30 min long.
Q: When we solve a
equation and obtain an implicit solution, should we write a solution in
the explicit form, or it is OK to leave it in the implicit form?
A: If it is not explicitly stated in the problem, you may leave your
solution in the implicit form (you will not loose points).
It is however better to write a solution in the explicit form (if
possible) because of a few reasons. The main reason is that an explicit
really gives us a function that we a looking for, while an implicit
only gives an expression that the function satisfies. Of course in
practice explicit solutions are what we want to have.
Q: What does it mean
that a function f(x, y)
is linear in variable y?
A: It means that if we regard x to be a constant,
then each term in f
is either a constant or a multiple of a constant and y. For example, a
linear function does not have terms y^2.
Similarly a differential equation f(x,
y, y')=0 is linear if each term of f is linear in y and x, i.e., if we
regard x to
be a constant, then each term of f
is either a constant or a multiple of a constant and one of the two
Q: In L2 we have N'=n
where N and n are two functions. And then we get
d/dx (N) = N'dy/dx. Why?
Note that the function N depends on y, which in its turn is a function
of x. For example suppose n=3y^2. Then N=y^3. So N'=(y^3)'=3y^2.
d/dx (y^3)=3y^2* (dy/dx).
Say, suppose in this example
Q: The textbook recommends that I
use software to solve one of the assigned homework problems. Should I
use a computer?
In this course we will use little or no computers. I suggest
you do homework problems by hand; note that you may get a similar
problem in a quiz and you will be expected to be able to
the problem by hand. On the other hand, if you want to use a computer
when you do your homework, it is up to you.
Q: Will we always get
I would continue to distribute notes for lectures if it is helpful. At
the moment I plan to distribute notes for at least another
week and after that I will see if it is
Q: I will not be able to attend a
particular lecture. Would it affect my grade?
A: College regulation
class attendance is expected of all students. Though academic schedules
may sometimes conflict with College-sponsored or College-recognized
extracurricular events, there are no excused absences for participants
in such activities. Students who participate in athletics, debates,
concerts, or other activities should check their calendars to see that
these events do not conflict with their academic schedules. Should such
conflicts occur or be anticipated, each student is responsible for
discussing the matter with his or her instructor at the beginning of
the appropriate term. Instructors may be accommodating if approached
well in advance of the critical date.
accommodations can be made only when the conflict occurs because of a
scheduled College-sponsored or College-recognized event. No participant
should expect to be excused in order to attend a team meeting or
orientation session, practice session, meal, or other such activity.
other words, you should notify me about time conflicts now (at
beginning of the term). Otherwise there is no problem.
Please make sure that you do not have time conflicts with quizzes.
Q: I have to leave
Hanover before the Final. Is it possible for me to take the exam
A: This is not possible, you have
to take the exam on December 9th (see the College regulations).