@acer 

...

I noticed this and couldn't resist mentioning the source.

The problem isn't so much about the Kovacic algorithm. Rather, it's about how to generate a simplified equation through differentiation. I remember an old trick that might help here:

It's assumed that a continuous second derivative exists for the function in question. After calculating y'', the term structure of y' is found/generated within the term structure of y''. The term structure of y' in y'' is then replaced by the symbol y'. This results in a simpler equation with a solution.

In

https://www.12000.org/my_notes/my_paper_on_kovacic/paper.htm

is given a Maple-example of the algorithm in 5. 

Unfortunately, I can only partially understand the theory in the description of points 1 to 4. The algorithm was developed long after my time as a student, and I haven't had any experience with it. However, its structure is very interesting. Often necessary tricks for solving tricky equations are systematically avoided with the help of this algorithm.

... from:

"PDEtools:-Solve(my_sol_1,y(x)); map(X->odetest(X,[ode,IC]),[%]) # [[0, 1], [0, undefined]]"

Maple points out that there is a problem with satisfying IC for the two explicit solutions. The theory of ordinary differential equations that I know only deals with equations in explicit form and, since the 1960s, also with so-called semi-implicit equations and ensures the existence and uniqueness of the solution. For general implicit equations, individual case investigations are therefore always necessary. This is obvious in the current case.

@dharr 

...for me, as a Maple beginner. In particular, solving the nonlinear system will keep me busy in the future.

Translation correction:

The initially confusing/complex Diophantine equation dio intrigued me because, after transforming the lhs into a sum of three squares, which equals zero, each term must be zero. This makes the solution obvious, or rather, results in three simple equations. This trick is well-known from competitions and always worth a try.

@dharr 

...for pointing out the * for commutative multiplication instead of . for non-commutative multiplication. I hadn't noticed that before and just typed it out of habit. Now expand works.

...sorry, I made a mistake when transferring the formula from the paper to Maple. It should be -1440*x*y^2 (the zero was missing).

@vv 

... "A sufficient condition for F(x,y)  to be Lipschitz wrt  y near (x0,y0)  is that D[2](F) be without singularities near (x0,y0). " in the current case, the classic proof: abs(f/x,u)-f(x,v))<=L*abs(u-v) is locally valid replaced. With which constant L should the proof be carried out locally and globally? And that is what my original question refers to.

@dharr 

Please excuse my mistake. I promise to do better.

...for both advices (dharr 8722, janhardo 750). Now I have a lot to practice, understanding how to apply the theory I know "from paper" to the maple world.

@dharr 

...I found my mistake.

@dharr @janhardo 

Several tests have shown that a,...,f = -17, -3, -7, 371, 113, 131 is causing problems. The solution should be (21; -5). I would appreciate some advice.

@dharr @janhardo 

I got identical results for both problems :-) . Their solution methods are clear thanks to good structuring and the presentation of relationships. This should be particularly valuable for students.

In contrast, the solution algorithm I know from other software works "hard" in number theory. Naturally, the discriminant is the starting point. A subsequent transformation (Legendre) yields the constant term, and then its factorization on the right-hand side of the equation takes place. And now, in most cases, the Chinese remainder theorem (Maple recognizes it) is applied modulo the prime factors, followed by an inverse transformation.

@janhardo 

...correct the input error, it was c = -17.

@janhardo 

Let the coefficients a, b, ..., f = 14, -5, -17, 2, 11, -12 and 3, -5, 7, 2, -11, 19. What does your algorithm say? Will the result include not only natural numbers but also positive and negative integers? Problems involving Diophantine quadratics in two variables are very interesting :-) . I have already performed calculations using external software and would like to compare them.

@janhardo 

...Countably many further solution pairs. These are to be calculated starting from Your basic solution (1,0) and (-1,0) using recursion. In this case, this is based solely on number theory and can be calculated using a third-party program (not Mathematica) based on it.

(BTW: I don't know if and which "names" can be mentioned here for legal reasons.)