Are we using the same deq and ci?  What version of Maple are you using?

 deq := diff(x(t), t, t)+2*(diff(x(t), t))-5*x(t) = 0;
 ci := x(0) = 1, (D(x))(0) = -2;
 p:= dsolve({deq,ci},x(t),numeric, range=0..5, output= Array([0.1*j $ j=0..50]));
 

I tried several releases of Maple, and never got a warning.  In Maple 8 to Maple 10, something strange happened to the last row of the output:

> p[2,1][50..51, 1..3];

 [4.90000000000000034    359.488895369929650    521.075466325794764]
         [                                                                 ]
         [         ?                      ?                      ?         ]
 

In Maple 11 and 12, the result of this is

         [4.90000000000000034    359.488895369930104    521.075466325795332]
         [                                                                 ]
         [        5.             415.562142095951117    602.353062345626086]
 

Ah, I think I see the problem.  In addition to the procedure itself that dsolve returns, it seems you need to save `dsolve/numeric/data/modules` (or at least the part of it that your procedure references).

> P:= dsolve(....);
  save P, `dsolve/numeric/data/modules`, "dsolveoutput.txt";

Then...

> restart;
   read "dsolveoutput.txt";
   plots[odeplot](P, ...);

Ah, I think I see the problem.  In addition to the procedure itself that dsolve returns, it seems you need to save `dsolve/numeric/data/modules` (or at least the part of it that your procedure references).

> P:= dsolve(....);
  save P, `dsolve/numeric/data/modules`, "dsolveoutput.txt";

Then...

> restart;
   read "dsolveoutput.txt";
   plots[odeplot](P, ...);

I'm not sure what exactly goes in the "so on", but try

> M := Matrix(4,5):
   M(1.., 1):= <a,b,c,d>:
   for j from 2 to 5 do
      M(1.., j) := <0,0,0,1>
  end do:

I'm not sure what exactly goes in the "so on", but try

> M := Matrix(4,5):
   M(1.., 1):= <a,b,c,d>:
   for j from 2 to 5 do
      M(1.., j) := <0,0,0,1>
  end do:

See the help page ?LinearAlgebra,General,MVshortcut

Yes, ^%T is transpose.  You could avoid it, though, by using

> <M,<0|0|0|1>>;

"|" separates columns, "," separates rows, so <0,0,0,1> is a column vector and <0|0|0|1> is a row vector.

See the help page ?LinearAlgebra,General,MVshortcut

Yes, ^%T is transpose.  You could avoid it, though, by using

> <M,<0|0|0|1>>;

"|" separates columns, "," separates rows, so <0,0,0,1> is a column vector and <0|0|0|1> is a row vector.

It looks to me like this was a bug that only affected Maple 9.  Your Maple 11 should do this correctly.

It looks to me like this was a bug that only affected Maple 9.  Your Maple 11 should do this correctly.

Indeed, in this example the first derivative doesn't exist at x=0.  Going to the next interval, at x=1 the first derivative will be continuous but the second derivative will not exist, and in general at x=n you'll have n continuous derivatives.  But my next example, starting with a smooth function that is 0 on [0,1/3] union [2/3, 1], is C_infinity on the whole real line.

Indeed, in this example the first derivative doesn't exist at x=0.  Going to the next interval, at x=1 the first derivative will be continuous but the second derivative will not exist, and in general at x=n you'll have n continuous derivatives.  But my next example, starting with a smooth function that is 0 on [0,1/3] union [2/3, 1], is C_infinity on the whole real line.

Don't trust graphs too much: the picture would also seem to indicate a limit of 1 for my counterexample

f:= abs(x^2 - y)^(x^2 + y^2)

Don't trust graphs too much: the picture would also seem to indicate a limit of 1 for my counterexample

f:= abs(x^2 - y)^(x^2 + y^2)

It could be a solution for x in (0,1), if f(t) = t^3 - 2* t^2 - 3* t for t in [-1,0).

> F[1]:= x -> x^3 - 2*x^2;
   F[0]:= t -> t^3 - 2*t^2 - 3*t;

> D(F[1])(x) = normal(F[1](x) - F[0](x-1));

3*x^2-4*x = 3*x^2-4*x

It could be a solution for x in (0,1), if f(t) = t^3 - 2* t^2 - 3* t for t in [-1,0).

> F[1]:= x -> x^3 - 2*x^2;
   F[0]:= t -> t^3 - 2*t^2 - 3*t;

> D(F[1])(x) = normal(F[1](x) - F[0](x-1));

3*x^2-4*x = 3*x^2-4*x