I was indeed of the opinion that this should be a Post, and I made it so. And voted up.

@HS Our two most-recent Replies crossed each other. So, it seems that you understand what I mean by "nontrivial example".

@vv The reason that I brought up the issue of what the OP expects as output is that it seems as if they expect something to happen with respect to the extended finite field, as in the oft-repeated phrase "mod out by q", where q is the RootOf of the irreducible extension polynomial. It's not possible for the extension to have any effect on the output if it doesn't appear in the input; the computations within the degree-1 subfield suffice in that case. So, I was hoping for the OP to provide a nontrivial example. That being said, Normal is likely what will provide the OP's expected output for that nontrivial example. By nontrivial, I simply mean that the extension should come into play somehow.

@vv I meant that they did nothing special in the case at hand. Of course in the general case they do something special with respect to finite-field arithmetic, lest they wouldn't have been written.

@ianmccr No offense taken, not in the slightest! You have a genuine and serious issue here. And now that you've defined URI, I understand your problem. Other than that, I don't know a solution, other than to workaround by maintaining separate files in a folder, as we've done for decades. But you probably already realize that workaround.

@vv You're likely right about the OP's desired output, and this is not meant as a criticism of your Reply, but the commands Normal and Rem aren't doing anything special here with respect to finite-field arithmetic. The results are the same as would be obtained by doing the symbolic algebra over the rationals and then applying mod 13 to the coefficients alone. To wit:

(x-1)/(x-3) mod 13;
             (x + 12)/(x + 10)

r:= rem(x-1, x-3, x, 'q'):  q + r/(x-3) mod 13;
              1 + 2/(x + 10)

Note to the OP: Both VV and I have used q here for "quotient". This is not meant to be confused with q the finite-field element used elsewhere in this thread.

Thanks for this app: It's great to have a calculator form of Maple on my phone.

Is there a document or help page that tells what mathematical functions are implemented? While Answering a Question here last night, I discovered that erf is implemented if entered from the alphabetic keyboard, but BesselJ is not.

I hope that you'll add a feature to "lock on" the alternate keyboards. Otherwise, I need to select the alphabetic keyboard for each letter of B-e-s-s-e-l-J.

Also, it'd be great if it remembered previously entered expressions (simply in a most-recents list, as other Android calculators do).

@HS Okay, Expand is not the right command for this; it expects a polynomial argument. Let's suppose that there was a command that did what you wanted. I'll call it ModOut. What would be your expected output of

ModOut((x-1)/(x-3)) mod 13;

?

Do you have a factorization algorithm that you want to implement in Maple? Or were you planning to use Maple's ifactor?

@ I solved it both in WolframAlpha and Maple, then compared the results numerically using Maple, and I decided they were equal.

Do you have Maple? If so, you can look up the help on StruveH and LommelS1. I don't have time at the moment, but it may be possible to convert these to a more-familiar form using Maple's FunctionAdvisor and related tools. It might help if x can be restricted (to, say, x > 0, for example).

@Aisyah In situations where there is an extra equation, we would hope that there is a redundant equation---one that can be derived from some subset of the others. If we find and remove that equation, it doesn't change the system.

Unfortunately, that's not the case here: Eq2 and Eq3 are directly contradictory; at most one of them can be correct. To see this, do diff(Eq3, eta) and compare the result with Eq2. You'll see that 3, but not all 4, of the signs are different. So, you'll need to carefully review your source material to see where the mistake was made. It's probably just a change-of-sign error.

If you decide to remove Eq2, there's no longer any 2nd derivative of G in the system, so you need to remove the boundary condition on D(G) (which is the first of bcs2).

Then solve and plot solutions only for phi = 0. There's some more subtlety involved in solving for the other values of phi (that "Newton iteration is not converging" stuff), which we'll look at after having the phi = 0 solution. I think we may need to use that solution as a "bootstrap" to get the others.

@tomleslie You're right: I grossly underestimated the true size of the OP's plot. The options add a lot that I didn't account for. In addition to CodeTools:-Usage, it would be useful comparative information to apply length(...) to each plot.

Doing that:

• my 300x300 plot3d: 90,230 (roughly 300²)
• the OP's 500x70 matrixplot: 18,828,337
• your 500x70 matrixplot: 25,055,345 (!!!)
• exact same matrix in matrixplot with no options at all: 35,088 (roughly 500*70).

So, the OP's matrixplot is about 536 (!) times larger (datawise) than the exact same matrix plotted in matrixplot with no options.

You do have a valid Question about dsolve here, but it has nothing to do with the rkf45 algorithm. This is a boundary-value problem (BVP). The rkf45 algorithm is for initial-value problems (IVPs). At this point, however, you don't need to worry about this; Maple can already figure out that this is a BVP.

@Carl Love WolframAlpha (which is available free online for calculator-type queries, such as your integral) gives the result as 

(-(x*HypergeometricPFQ[{1/2, 1/2}, {3/2, 3/2, 2}, -x^2/4]) + x*HypergeometricPFQ[{1/2}, {3/2, 2}, -x^2/4]*Log[x])/2

That translated into Maple is

x/2*(ln(x)*hypergeom([1/2], [3/2,2], -x^2/4) - hypergeom([1/2,1/2], [3/2,3/2,2], -x^2/4))

Maple 2019 gives the result as (I also applied simplify(..., size))

1/2*Pi*x*(BesselJ(0,x)*StruveH(1,x)-StruveH(0,x)*BesselJ(1,x))*ln(1/2*x)-x*
BesselJ(0,x)*D[1](LommelS1)(1,1,x)+x*BesselJ(1,x)*D[1](LommelS1)(0,0,x)+1/2*(-
Pi*x*(ln(2)-ln(x)-1)*StruveH(0,x)-2*ln(x)-2)*BesselJ(1,x)+1/2*(Pi*(ln(2)-ln(x))
*StruveH(1,x)+2*ln(x))*BesselJ(0,x)*x

A quick numerical check over real ranges confirmed (although it's not iron-clad proof) that these two expressions are equal.
 

Would you please define "URI"?