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.

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].

Observe (in your head!) these features of the graph:

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!

Another student asked about problem #7

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