2/9/2004: Sent in response to a request for help doing problem #15 in section 3.5

O.k.: that's a nasty and rather realistic problem. Very often "real" problems aren't just initial value problems, they are "mixed" boundary value problems. A flexing beam (as mech. engineers should know or will learn) was analyzed by Euler several hundred years ago, and the analysis constructed makes a model which is a FOURTH ORDER o.d.e., and, depending how the beam is fixed or can move at both ends, leads to problems similar to this one.

So I just solved it. Here is what I did. To begin with, I ignored the condition y(L)=0. I don't know a simple Laplace transform technique which would enable me to build in a boundary condition at x=L.

I considered the problem

y(4)(x)=(Const)delta(x-a) where (Const) was the quotient M/EI.

I also used the initial conditions

y(0)=0 and y''(0)=0 and y(3)(0)=F_0 (given by the textbook statement) AND the initial condition y'(0)=B where B is a constant I will figure out later. I guessed that if I leave one parameter free in the problem statement, that, with luck, I will be able to deduce a value of B uniquely which will correspond to y(L)=0. This is very SNEAKY.

Then I took the Laplace transform of both sides of the equation. Then I imposed the initial conditions (including the "fictitious" one). I also assumed that 0 I solved for Y(s), and transformed back. The expressions involving s were not difficult to inverse Laplace transform. I then plugged in x=L and tried to get 0 by specifying B correctly, and I was able to do it. (Sometimes this is NOT possible -- I will try to mention this in class tomorrow.) I think I got something which looks very much like what is in the back of the book. All the darn letters make it much harder to write.

2/9/2004: Sent in response to a request for help doing problem #19 in section 3.5

There is some "physical" information in the problem statement. The first sentence ("8 pounds ... 6 inches") gives information about the spring constant.

Then the problem "kicks" the spring with a delta function. if you model it with delta(t-1) you would look at the velocity of the solution at t=1. If you try delta(t) (which I think I would do) then you need the velocity at t=0. Of course, the two numbers should be the same. Consideration of the period and magnitude corresponds.

1/22/2004: Sent in response to a request for help doing problem #3 on the Entrance Exam

Look at the function exp(-Lx^3). I will use L for lambda. Remember L is positive and x is in the interval [2,14].

1. The values of the function are values of exp, and therefore the values are all XXXXtive.
2. The input to exp is -Lx^3. L is positive and x is in [2,14] and there is a - sign. Therefore the input to exp is XXXXcreasing, and therefore exp(-Lx^3) is also XXXXcreasing. The largest value of the function in the interval [2,14] must occur at x=XXXX.
3. The length of the interval is 12 and the largest value of the function is XXXX (with L in it!) and therefore the integral is at most XXXX, the length multiplied by the max height.
4. This overestimate contains L and when L gets large, the overestimate gets small. Done!
I hope that this helps. If you would like further information (the identities of some of the XXXX's) let me know, please.

This is what Maple tells me is the antiderivative of x(cos(5x))7:

```        /            6                         4
1/5 x |1/7 cos(5 x)  sin(5 x) + 6/35 cos(5 x)  sin(5 x)
\

2            16         \                  7
+ 8/35 cos(5 x)  sin(5 x) + -- sin(5 x)| + 1/1225 cos(5 x)
35         /

5                  3   16
+ 6/4375 cos(5 x)  + 8/2625 cos(5 x)  + --- cos(5 x)
875
```
You certainly don't want to verify or use this. Please instead use symmetry and just try to answer the question (are the integrals 0 or positive or negative). You don't need to evaluate the integrals!