Mine was just a cleaned-up version ...

But if I use just a function instead of exp, then the result is different:

diff(u(x,t),t)=diff(u(x,t),x,x)+ f(t);
pdsolve(%);

  `&where`(u(x,t) = _F1(x)+_F2(t), [{diff(_F1(x),`$`(x,2)) = _c[1], diff(_F2(t),t) = _c[1]+f(t)}])

I do not see the product, which occures for f = exp. Hm ?

Being a moron in (P)DE and ignoring the initial condition at least I get

S := u(x,t) = (-a*sin(x)-1-b*cos(x))*exp(-t)+_F1(x)*_F2(t)

and have not looked into http://eqworld.ipmnet.ru/en/solutions.htm

Does it help at least a bit or is it too trivial (as non-solution)?

Being a moron in (P)DE and ignoring the initial condition at least I get

S := u(x,t) = (-a*sin(x)-1-b*cos(x))*exp(-t)+_F1(x)*_F2(t)

and have not looked into http://eqworld.ipmnet.ru/en/solutions.htm

Does it help at least a bit or is it too trivial (as non-solution)?

input: "tell me about A"

answer: "it is a Matrix"

input: "dimension ?"

answer: ...

if - for example - it is a matrix, then that would not work
(though it is 'clear' for us, what is meant):

B:=convert(A, matrix);
LinearAlgebra[Dimension](B);

    Error, (in LinearAlgebra:-Dimension) invalid input ...

input: "tell me about A"

answer: "it is a Matrix"

input: "dimension ?"

answer: ...

if - for example - it is a matrix, then that would not work
(though it is 'clear' for us, what is meant):

B:=convert(A, matrix);
LinearAlgebra[Dimension](B);

    Error, (in LinearAlgebra:-Dimension) invalid input ...

@barry314159 

At least the following indicates problems:
http://pari.math.u-bordeaux.fr/archives/pari-users-1007/msg00008.html

I suggest you search in the archives of their message board (have not
looked at them for over 6 years).

If it is _not_ implemented but needs scripts or code or similar in Pari,
then it would not make sense to interface with Maple (it becomes quite
complicated), but to use the other system completely and write results
to files.

Where - by the way - Pari has a very, hm, strange language for a Maple
user.

PS: that board still does not allow to simply reply to the according post.

@barry314159 

I would prefer questions and answers here at the board, that's one reason why it lives.

http://www.mapleprimes.com/posts/42290-External-Numerics-Using-Pari-From-Maple
some links have been broken since that board uses a new appearance

http://axelvogt.de/axalom/pari/index.html
gives source code and runtime library for Windows as well as example sheets for Maple

SW = SoftWare

Being no used to analytic number theory I yet would guess,
that there is SW specialist for that.

Years ago I played with interfacing PARI and Maple, because
that 'oldish' SW can be very fast.

I am not aware, whether it covers your need, you have to check
the manuals at http://pari.math.u-bordeaux.fr/

In case: let me know (after checking conventions for functions).

Anyway: may be to consider first whether it is more convenient
to let the PC run over night (some times) - or to dive into some
'optimization'.

I looked again at the package, and yes, it seems they mean to compute the usual
Krull dimension of FullPolynomialRing/Ideal using Hilbert functions. And that
dimension theory is complicated stuff.

It really would be worth to update the help, with more systematics and references,
though there is a PolynomialIdeals Package Example Worksheet, 

And caveats about passing from there to solutions of equations, especially since
the field (over which the polynomials are considered) is not specified:

  A:=<x^2+y^2>; HilbertDimension(A);

    1

Yes. But over the Rationals or Reals there is no solution, the user should be
reminded that switching between algebras and solutions needs algebraic closure,
so usually it has to be done over the Complex.

Also there is some weakness in that case: taking the vanishing ideal in the
2-dim points (x,y) = (0,0) and (x,y) = (1,1) I get  `<,>`(y-x,x^2-x).

Now I would expect a way to decompose (to find, that it stems from a union of
isolated points), yet ZeroDimensionalDecomposition does it not: it does some
representation as intersection. A somewhat missleading wording (a component
in geometry/topology is something different ...).

I looked again at the package, and yes, it seems they mean to compute the usual
Krull dimension of FullPolynomialRing/Ideal using Hilbert functions. And that
dimension theory is complicated stuff.

It really would be worth to update the help, with more systematics and references,
though there is a PolynomialIdeals Package Example Worksheet, 

And caveats about passing from there to solutions of equations, especially since
the field (over which the polynomials are considered) is not specified:

  A:=<x^2+y^2>; HilbertDimension(A);

    1

Yes. But over the Rationals or Reals there is no solution, the user should be
reminded that switching between algebras and solutions needs algebraic closure,
so usually it has to be done over the Complex.

Also there is some weakness in that case: taking the vanishing ideal in the
2-dim points (x,y) = (0,0) and (x,y) = (1,1) I get  `<,>`(y-x,x^2-x).

Now I would expect a way to decompose (to find, that it stems from a union of
isolated points), yet ZeroDimensionalDecomposition does it not: it does some
representation as intersection. A somewhat missleading wording (a component
in geometry/topology is something different ...).

@Markiyan Hirnyk 

I never liked the help on that 'dimensions' within PolynomialIdeals or Groebner,
they just should be clear (and give a reference).

HilbertDimension( , {x,y,z,v,w});

                                  4

That means they give the dimension of C[variables]/Ideal.

The (generic) number of equation of a reduced variety is the codimension,
so it is 2 in the example, no?

For the question it means the dimension is 3 in C^5 (if I understood it correctly).

But if it is not an irreducible variety it may contain components of lower
dimensions (and besides that the dimension depends only on the radical ideal,
while I guess that the above solution means for the actual ideal)

Hope I got it right

PS: that messy notation in the standard sheets do not export as *.mpl in a way,
which can be used a reasonable sheet.

PPS: And that image info at that board does not allow to copy code ... sigh

PPPS: solve returns an output for me in M15 after correcting the input

@Markiyan Hirnyk 

I never liked the help on that 'dimensions' within PolynomialIdeals or Groebner,
they just should be clear (and give a reference).

HilbertDimension( , {x,y,z,v,w});

                                  4

That means they give the dimension of C[variables]/Ideal.

The (generic) number of equation of a reduced variety is the codimension,
so it is 2 in the example, no?

For the question it means the dimension is 3 in C^5 (if I understood it correctly).

But if it is not an irreducible variety it may contain components of lower
dimensions (and besides that the dimension depends only on the radical ideal,
while I guess that the above solution means for the actual ideal)

Hope I got it right

PS: that messy notation in the standard sheets do not export as *.mpl in a way,
which can be used a reasonable sheet.

PPS: And that image info at that board does not allow to copy code ... sigh

PPPS: solve returns an output for me in M15 after correcting the input

Excel has it as bulit-in functionality for downloading data into sheets.

May be you can search for VBA code, which works on HTML tags

I played with it, using Excel, setting all parameters to 1 as initial value.

After a while I decided to consider only the first dozend pairs and then quickly
got a result, which looked not that bad for the overall range.

Now considering all pairs that refines and after that I used Maple to do a
NonlinearFit with the following result:

sol := [[a = 2.61748239286572, b = 1.71949328454291, c = 2.30924398872986, 
         d = 1.50333104337044, e = 1.84597270402404], .932182823605393e-2];


Taking only the the first dozend (to have a guess) however does not work in
Maple, so it may be just good luck w.r.t. to Excel's way that it worked.

I played with it, using Excel, setting all parameters to 1 as initial value.

After a while I decided to consider only the first dozend pairs and then quickly
got a result, which looked not that bad for the overall range.

Now considering all pairs that refines and after that I used Maple to do a
NonlinearFit with the following result:

sol := [[a = 2.61748239286572, b = 1.71949328454291, c = 2.30924398872986, 
         d = 1.50333104337044, e = 1.84597270402404], .932182823605393e-2];


Taking only the the first dozend (to have a guess) however does not work in
Maple, so it may be just good luck w.r.t. to Excel's way that it worked.