I can't make sense of your Question. A vector field in the Euclidean plane is a function R^2 -> R^2 (i.e., 2 real inputs and 2 real outputs). For example:

VF1:= (x,y)-> < x^2 + y^2, x^2 - y^2 >;

I don't see anything like that in your Question.

@elen When you say "when I type enter, nothing will be done", do you mean that nothing at all happens---that there's no output at all? Or do you mean that you get output (blue text) that is equivalent to your input? If you get no output at all, then there is some serious low-level problem with your installation. But I'll guess that you have the other situation: Your output is equivalent to your input. Well, you didn't ask for anything to be done other than to name your equation eqn, which Maple did do, right? What do you want done with your equation? If you enter some simple arithmetic like 2+3, it's reasonable for Maple to guess that you want that simplified to 5. This is called automatic simplification. But with your eqn, there's no obvious thing to do with it. You need to ask for something. Do you want to solve the equation for a? Then enter

solve(eqn, a);

@max125 

Yes, if V is a list or set, then V[] returns the arguments (or entries). This only works for lists and sets.

Regarding the s: It does appear in eval([H,HD], s), which is nearly equivalent to subs(s, [H,HD]). Note that every value of s is a set of the form {x= ..., y= ...}, and H and HD do in general contain x and y. To evaluate an expression at particular values of its variables, it's preferred to use eval rather than subs. See ?eval, and note that there are two completely different commands named eval that are described there. The eval that you used in your Question is one of them; the eval that I'm using is the other.

@Britzel There are many ways to do that, and copy-and-pasting code is never needed or recommended. Most of the good ways of doing it involve having an initialization file. Do you have one? If not, see help page ?worksheet,reference,initialization. Create an initialization file, and get back to me. Or get back to me if you have trouble creating the file.

@raskr Indeed, your worksheet has convinced me that Maple 2017 is the lone exception. Thus, it is the only version of Maple for which displayprecision does something reasonable. Its behavior in Maple versions both before and after Maple 2017 is ridiculous.

I think that this is an intermediate representation that's internal to the GUI and external to the kernel. If I enter

_EXPSEQ(1, -1);

at a command prompt, and I have interface(prettyprint) set higher than 1, then I get just

      1, -1

So, the _EXPSEQ part seems to be some instruction for the GUI to process.

Please issue the commands

interface(version); kernelopts(version);

and check that the "Build ID" (the final number) of each matches.

@mmcdara I fully understand the situation that you're describing and have illustrated in your two most-recent worksheets "Still_not_clear" and "reply". Yes, I agree that those illustrate a weakness of Maple. I'm not sure that I'd call it a bug. I will get to addressing those issues. However, this is not the same issue as occurs in the worksheet "NeedHelp". My answers so far have applied to that worksheet alone (because it was the only worksheet at the time that I wrote my first two answers). In that worksheet, you used op to extract the polynomial from a RootOf and proceeded to get solutions for that polynomial (using subs, then solve, then allvalues). After you've explicitly extracted it, that polynomial is (and rightly ought to be) independent of the original radical equation from which it was derived. That polynomial has some roots that don't satisfy the original equation. As I've tried to illustrate, that's a mathematical necessity. So you can't blame Maple or its solve command for that; you can't say "Mr Maple has done only a part of the job" because solve (correctly) can only see the polynomial, not the original equation.

So please look over NeedHelp again very carefully and then tell me if you now understand and agree with my replies

• In the first case, you've used solve on an equation that contains radicals. In the second case, you've used solve on a non-radical polynomial equation derived from the radical equation. As is very well known, that process has the risk of introducing extraneous solutions.  Those solutions need to be verified in the original equation.
• It [The introduction of extraneous solutions] is a fundamental mathematical principle....

in light of the fact that that worksheet was the only thing that I had to work with at the time that I wrote those replies. When I said "Those solutions need to be verified," I did not mean that all solutions produced by solve need to be verified; I was only referring to those solutions produced by the process that I described in that paragraph.

After you reply, I'll address the other issue. Please only address the issues in the worksheet NeedHelp without adding any more context or worksheets.

@mmcdara If you've done the exercise as instructed, then you've converted the radical equation into a polynomial equation without radicals. Doing that involved two steps of squaring both sides of the equation (thus applying a non-injective function to both sides of the equation). The polynomial is a very simple quadratic with two roots. Only one of those two satisfies the original equation.

Now look at every RootOf expression in your two worksheets NeedHelp and Still_not_clear. Notice that the first argument of every one is a polynomial with integer coefficients and no radicals. The process of deriving those polynomials from the original radical equation involves steps equivalent to raising both sides of an equation to an integer power. So a non-injective function has been applied.

You have done different things in the two worksheets. The differences may seem insignificant to you, but they change the precise reason why you got a different number of results from solving in two different ways. My first two answers only apply to the worksheet NeedHelp because you hadn't yet posted the other.

(to be continued.)

@mmcdara The issue has nothing to with instantiation of parameters. The issue is just obfuscated by the presence of parameters M and lambda and the high-degree polynomial. The issue also has nothing to do with substitutions, such as _Z= U^(1/5). Please do by hand the exercise that I provided. I carefully constructed it to be the simplest possible exercise that completely illustrates the issue.

=============================================================

Do the exercise now, before continuing to read this post

=============================================================

[Continued in next Reply.]

@umangbedi Do you think that I'm going to type your formulas into Maple for you? You seem capable of typing a decent dsolve command into Maple, so why not type in these formulas? Type in all that you can, and then repost. We'll need numeric values for E and R before getting an actual solution.

[Three hours having elapsed with no response] In other words, if you post the values of eqs and ICs, then I will happily Answer your Question.

@mmcdara It is a fundamental mathematical principle that can be explained with simple set theory. I am not aware that it has a formal or proper name. In the USA, it is usually taught in a secondary-school course entitled "Algebra II" (so, for students aged 14 - 17 years).

Here's a brief explanation using more set-theoretic formality than is usually used for those students: Suppose that we have sets A and B and functions (f,g): A --> B that can be described algebraically. We want to solve f(x) = g(x) for x. Call this Eq1. Suppose that to facilitate its algebraic simplification, we apply to both sides of the equation a function h that is non-injective on (f(A) intersection g(A)), obtaining the equation h(f(x)) = h(g(x)). Call this Eq2. By the definition of function, any solution of Eq1 is also a solution of Eq2. But the converse is not necessarily true! For example, if h = y -> y^2 and there is a c such that f(c) = -1 and g(c) = 1, then x = c is a solution of Eq2 but not of Eq1. We call it an extraneous or spurious solution (or root).

And here's how it might be taught to the Algebra II students: If in the course of solving an equation you raise both sides to an integer power greater than 1, then any solutions need to be verified in the original equation.

To facilitate your understanding, you should obtain all complex solutions of 

sqrt(x) - sqrt(1-x) = 1

by hand using the systematic algebraic technique (i.e., no guessing and no ad hoc techniques) that you learned in secondary school: Convert the equation to a polynomial first.

How can it not be obvious to you that we need to see eqs and ICs in order to answer your question?

@Kitonum Although you have achieved the same final result as I did, I think that it's just a coincidence and that your method is invalid. Don't you need to split into two integrals with the split at t=1?

@tomleslie It's not a "shooting method" if it relies on an existing numerical BVP solver. Also, the smoothness of the error curves is a red flag that something is awry.