@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.

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Do the exercise now, before continuing to read this post

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[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.

Please try uploading  your file again.

Do not post the same thing in both Posts and Questions. This is obviously a Question. Posts are mostly expository.

@maple2015 Your equation can be easily solved. Maple is fully prepared to deal with complex-valued solutions, often with no additional effort required by the user. Indeed, it often works more naturally with complex-valued functions than with real-valued ones. Your complex solution can be easily split into real and imaginary parts using commands evalc, Re, and Im. Those parts will both be real-valued functions. This will also often cause the appearance of sin and cos in the solution. You can also use convert(..., sincos) to force re-expression in terms of sin and cos.

@maple2015 Yes, that's correct, there's no way to get a purely real-valued solution under those circumstances. This is a mathematical limitation, not a Maple limitation. You seem surprised by this. Why? The space of complex-valued functions is much larger than the space of real-valued functions, so it's no surprise to me.

@mmcdara Here's my "external eye": 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.

I don't see any Maple bugs revealed by your worksheet.

@elen

Are you trying to right-click on the input (the black text) or the output (the blue text)? It should be the output.

Disclaimer: I don't know if the following test will do anything in your Maple 13:
========= Test that works in modern Maple =================================
As a diagnostic for your situation, try this. Let expr be the equation or expression for which you are trying to get a context menu. Then type this command:

ConxtextMenu:-Test:-GetGeneratedMenuAndAction(expr);

and report here the results.

If (and only if) the expression was the most-recent thing that you typed in, you could also do

ConxtextMenu:-Test:-GetGeneratedMenuAndAction(%);
========== End of test that works in modern Maple ============================

Condidering the age of your Maple, you may need to learn to just type the commands rather than using context menus. We can help you with that here.

Try first using a left click to select the object for which you want a contact menu, then right click.

@radaar Nobody in this thread "confirmed that it won't converge"; there was merely skepticism expressed about it. For my part, I wanted a strong theoretical reason for its convergence before I spent considerable effort getting a numerical result, a result which would be worthless if it didn't converge.

The link to your uploaded worksheet doesn't work. (Probably not your fault.) Please try uploading it again.