@Carl Love 

How can sin(k*x)*cos(x)^k be expanded into a series of sin and cos? Using convert?

@Carl Love 

I'm afraid ;-) you've found the same solution I achieved a long time ago, partly with the help of terrible formulas from Gradstein/Ryshik. Fortunately, transforming trigonometric terms and calculating improper integrals are now much easier to do on a computer. After all, it was my goal to find out whether symbolic calculations could make such problems easier to solve.

Would you please make your code available for practice purposes?

@nm 

Thanks, learned something new again. Then I want to calculate a table for a specific k-sequence, e.g., k = 10, 20, 30, ...100. How do I do that?

@nm 

MaplePrimes is finally working again. Thank you for the link to the book. I had a quick "browsing" look. I found the problem and its solution.
Summary:
The book contains a lot of interesting information, is aimed at interested beginners, and whets the curiosity for more in-depth content not covered in the book. Fortunately, there are classics such as Kamke, Walter, Pontrjagin, etc. Given the large number of example problems, it may happen that a typo goes unnoticed. So, I'll stick with the statement that your equation for the initial value y(0)=0 has no solution. In principle, you've already provided a proof by contradiction yourself. And theoretically, even for the initial value y(0)=1, the general solution only becomes the final solution when one considers the continuation of the solution from the open continuity domain of the "right-hand side" to the boundary. But that's splitting hairs and a different chapter of theory, and it's not worth putting a lot of effort into simple problems. The book authors should comment and the odetest would have to be examined on a case-by-case basis.

@nm 

In the general solution with still undetermined C, t is in the denominator of the explicit solution y^2 = .... Multiplying the solution by t is therefore only possible for t not equal to zero, and only then can it be reversed to obtain the explicit solution structure. Otherwise, nothing can be calculated from 0 = 0. The calculation of C = 0 from the equation t*y^2=C+arctan(t) for t=0 is therefore incorrect.
Does your book contain concrete solution hints for this problem? Should it perhaps cover the problem in the complex plane? Please forgive my meticulous precision. And if there are any errors in my arguments, please let me know and correct them.

@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.