@nm 

After the described substitution u = y^2, the original equation becomes a conventional linear differential equation. This is easily solvable using a textbook recipe (e.g., Polyanin/Zaitsev, Chapter 1). The general solution, with a constant C determined from the initial condition, is:
y^2(t) = C*1/t + 1/t*arctan(t).
How should C be determined from y(0) = 0?

@nm 

...that due to the continuity of the right-hand side of y' = f(t,y), solutions exist in the continuity domain of f(t,y). However, according to the rules of "classical" analysis, there exists no solution for the initial value (0;0) that is everywhere differentiable and satisfies the initial condition. I can therefore only confirm what Rouben Rostamian wrote about this.

However, if you are looking for a solution with "different" quality (so-called weak/generalized solutions), you must apply the rules of functional analysis. To do this, the differential equation y´ = f(t,y) must be transformed into the integral equation 

y = y(0) + Int(from 0 to t)(f(x,y)dx) using a more general integral (no longer a Riemann integral). Using a scalar product in a function space (also realizable by an integral), a functional equivalent to the original integral equation can then be constructed, the minimum solution of which yields weak/generalized solutions (e.g., see: Michlin, Lehrgang math. Physik, Variationsmethoden...). However, I don't know whether this is feasible in Maple—I'm just a Maple beginner.

My impression is that Maple successfully performed the substitution test in your first post without checking the initial condition.

The substitution u=y^2 simplifies the ode to the linear
ode u´ + (1/t)*u - 1/(t^3+t) = 0
(hopefully I didn't miscalculate).

@Alfred_F 

I forgot to mention the result: sqrt(pi)/2

@vv 

I had to spend a while working with the powerful "convert" command. It is obviously the key to the solution. With the exception of the attached task, I have been able to tidy up my collection of "ancient" tasks. Now I would like to ask for your help one last time with the "series" issue.

test.mw

@vv 

Thank you in advance for your advice.

1.) "Why don't you express the series in Maple, i.e.  sum( arctan(2/n^2), n=1 .. infinity) ?"

I'm not very good with computers and software, I'm just an old theorist. I have particular difficulties with uploading files. But I promise to improve ;-) .

2.) "Of course this is not a proof.  It is easy to obtain one in Maple, but only if the user knows it mathematically.
I omit it because you probably know it."

Your assumption is correct. Maple is obviously very result-oriented and the convenient commands omit solutions or only appear when specifically requested. The mathematical modeling of states and processes especially for safety-relevant calculations as well as theoretical backgrounds are not foreign to me. Unfortunately, experience has taught me to be thorough in the details.

SCNR

@Mariusz Iwaniuk 

This solution helps me a lot to understand the way Maple works and its logical structure a little better. As a newcomer to this field, I am trying to learn this using interesting example tasks. So it is not homework, solution paths according to theory (Lit.: e.g. "Knopp") and solutions are of course known to me in advance at my age.
Thanks also to @ dharr and @ vv.

It is gratifying to see how easy it is to find solutions using Maple. I still need to practice a lot. But the steps on the way to the solution are interesting. In your solution, it is important to know under what conditions series can be integrated term by term. That is OK here.

@vv 

...... of the theorems of Lebesgue, Levi, Fatou. ... (Lit.: Hewitt/Stromberg, Natanson, Shilov/Gurevich) You are right with the reference to "almost everywhere" in a more general type of convergence. Up to now, however, I have assumed that we are working in "classical" analysis. I will bear this in mind in future.

@vv 

...but it fails at the point (1;1), but it is initially a clever solution. Nevertheless, the limit of the double integral over the entire square can be calculated.

My questions are not about finding fault. Rather, I would like to be able to convince myself of the plausibility of the calculations using example calculations. I certainly won't want to see any source code.

@nm 

Thank you very much. Thanks to the logical working structure of Maple that you have described, the solution is testable and reliable. (It is probably an old occupational hazard that I attach importance to such things ;-). )

@nm 

How does Maple arrive at the correct result in a flash? Does Maple calculate or does it look in a database?

I'm asking because a solution using pen and paper is very tiring. On the other hand, it should be known that the integral to be calculated is related to a famous function that has been the subject of a lot of research and publishing.

@Kitonum @Scot Gould

The power of Maple commands is impressive.
But:
Is there a way to make the intermediate steps on the way to the solution visible and verifiable? What do Maple commands do "behind the scenes"?

@Kitonum 

My solution to this old (1878) problem is much more complicated. It was an unwieldy Diophantine equation to solve. I had hoped that Maple would be used for this. :-(

@Alfred_F 

The calculation result from (3) according to (6) assumes that the formation of the limit and the definite integration can be interchanged. Irrespective of the fact that the calculation using the antiderivative leads to the correct result in an elegant way, a formal addition is necessary. For every natural n, the integrand (2) or (3) in [0; 1] is continuous and bounded. According to Heine's theorem, it is then even uniformly continuous. And according to another theorem, the limit and the integral can then be interchanged. Only then does the limit move into the exponents of the integrand and, as proven, deliver the result.
A wonderful task worth remembering :-).