To get the expression for speed(t) (assuming your previous code worked):

> spd := subs(sol3, speed(t));
   

To integrate it:

> pos:= int(spd, t);

And to plot:

> plot(pos, t = 0 .. 20);

You want diff(u(x,t),x,x) on the right side of he, not diff(u(x,t),t,t).  Also
u(x,0) = x, not u(0,t) = x. 

Your initial condition u(x,0) = x and your boundary condition u(1,t) = 0 contradict each other.  In the solution, u(x,t) is discontinuous at (1,0), with different limits as (x,t) approaches (1,0) on different paths.  Maple happens to take the value at (1,0) to be 1,
but that is not really significant.

You could use arrow in the plots package.  For example:

> with(plots):
   display(arrow([Re,Im](z1),colour=red), 
              arrow([Re,Im](z1),[Re,Im](z2),colour=blue),
              arrow([Re,Im](z1+z2),colour=yellow));

It's really hard to tell what went wrong: you just included a picture showing some of the last part of what must be a big worksheet.   It might help if you would upload the worksheet itself.  Is the problem unique to this worksheet, or do you encounter this sort of kernel error in other situations?

The yellow colour COLOR(RGB,1,1,0) is hard-coded in simplex[convexhull], but you can substitute some other COLOR specification for it in the output.  For example:

> subs(COLOR(RGB,1,1,0)=`plot/color`("Aquamarine"), convexhull(P,output=plot));

Suppose the Poisson process N(t) has rate lambda, and you want to find the distribution of the number R[a,b] of packet requests in time interval [a,b] (where s/r < a < a + s/r < b).  I'm assuming for simplicity that the first packet request is simultaneous with the file request - this is not important.  R[a,b] is (for first packets) the number of file requests in [a,b], plus (for second packets) the number of file requests in [a-1/r, b-1/r], plus ... (for s'th packets) the number of file requests in [a-(s-1)/r, b - (s-1)/r].  This can be written sum( k*N(B[k]), k=1..s) where B[1], ..., B[s] are disjoint sets of reals, N(B) is the number of file requests at times in B, and B[1],...,B[s-1] each have measure 2/r while B[s] has measure b - a - (s-1)/r.
Since the B[k] are disjoint, N(B[k]) are independent Poisson random variables.  The moment generating function of k times a Poisson random variable of parameter p is exp(p*(exp(k*t)-1)).  So the moment generating function M(t) of R[a,b] is

which Maple evaluates as

asympt(LambertW(0,x),x) didn't work in any version of Maple that I tried.  However, you could do this:

> asympt(LambertW(0,exp(t)),t);

> eval(%, t = ln(x));

Depending on what your algorithm is doing, you may want to insert code to have Maple tell you what progress it is making.  See this discussion.

Maple isn't "forgetting" anything, you are.  There's a difference between functions and expressions, and you're mixing the two up.  You could say

> Yt:= D[1](y); # note that y and Yt are both functions of two variables 
   A:= unapply( int(Yt(t,s)^2, s=0 .. zeta(t)), t);  # a function of only one variable
   D(A)(t);

   zeta(t)
  /
 |
 |         2 D[1](y)(t, s) D[1, 1](y)(t, s) ds
 |
/
  0

                                     2
     + D(zeta)(t) D[1](y)(t, zeta(t))

(I didn't use the Maple Math button because that insists on a bad conversion of the D's to diff's) 

Do the column vectors represent the points to be plotted (i.e. is the list [p[1], p[2], p[3], p[4], ..., p[2*n]] where you want to plot the segments from p[1] to p[2], p[3] to p[4], ..., p[2*n-1] to p[2*n], each p[i] being a column vector of two components?  Then try

> plot([seq(<p[2*i-1] | p[2*i]>^%T, i=1..n)], colour=blue);

Do you mean something like this?

> zip(assign, [seq(w[i],i=1..349)], L);

(where L is the list of 349 values to be assigned).

This sounds like a font problem on your system.  It works OK for me (under Windows 7).  You might contact Technical Support.

After substituting the other parameter values, A1 is an equation in the two variables S and k.  If you fix k=4 there will be just a discrete set of solutions for S, not a curve.

More can be said.  Consider

which will satisfy the differential equation

where

If we could get y(t), then x[1](t) or x[2](t) could be obtained by integration: in fact

Now with v = dy/dt, a constant of the motion is

For your initial conditions, the value of this constant is 2/5*k + 1/100.  We then have the first-order differential equation

for which dsolve yields the implicit solution

Using the initial condition y(0)=5, we get the value of the constant _C1:

We can't go much farther with closed-form solutions: clearly we shouldn't expect to be able to solve the equation for y(t) explicitly.

The main problem is that the string "C:\j.xls" is just a literal string; the value of j is not substituted for the j in the string.  You might do something like this using the sprintf function (and, by the way, you should use / rather than \ for directory separators)

Export(M[j], sprintf("C:/MyFolder/%d.xls",j));