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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?

A. 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
Swing

Q. Today in class we had a presentation of a spring vibration. What was the webpage?

A:

Q. How to compute the 4-th roots of -3?

A. 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?

A.  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)
and 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, yimportant for writing a general solution c1y1 + c2y2?

A.  There is a theorem for equations of the form

a
y'' + 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)              y(x0)=y0,
or
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)              y(x0)=y0

certainly exists only on a small interval about x0 if the function is smooth enough in a box neighborhood about (x0, y0).

For precise statements, see the textbook or lecture notes.

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 exactSuppose such a factor does not exist. Then I try to find an integrating factor mu(y). Suppose that such a factor does not exist either. What should I do next?

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 on Wednesday?

A: It will be 30 min long.

Q: When we solve a differential 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 solution y=f(x) really gives us a function that we a looking for, while an implicit solution f(x,y)=0 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 variables y or y'.

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?

A: 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. But
d/dx (y^3)=3y^2* (dy/dx).
Say, suppose in this example y=x^2. Then
d/dx(y^3)=3y^2*(dy/dx)=3x^4*2x=6x^5.
Alternatively, d/dx(y^3)=d/dx(x^6)=6x^5.

Q:  The textbook recommends that I use software to solve one of the assigned homework problems. Should I use a computer?

A: In this course we will use little or no computers. I suggest that 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 solve 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 handouts?

A: 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 really necessary.

Q: I will not be able to attend a particular lecture. Would it affect my grade?

A: College regulation
(

Regular 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.

Such 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.

In other words, you should notify me about time conflicts now (at the 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 earlier.

A: This is not possible, you have to take the exam on December 9th (see the College regulations).