@dharr 

"0 means an exact result" That's exactly what I was asking for - thank you very much. (In return, I'll reveal solutions 11 and 7 ;-) ).

A possible arithmetic relationship between the numbers could be that the sum of two neighboring numbers in a row equals the number below it in the middle. Therefore, x=15968, y=15725, and z=47350 would be one solution. But it might not be the only one. Let's see what the prime factors say...

@salim-barzani 

Below, I would like to use the notation from your original post. Since it has now been established that f is a positive real number not equal to 1, as already mentioned, we have an ordinary differential equation. @janhardo cleverly reduced this to a Riccati-type differential equation using substitution and solved it.
The function U in your second post is very likely (I don't know the physical background) a linear combination of basis functions, which should be represented methodically more conveniently as exponential functions. This linear combination spans a finite-dimensional function space and is supposed to provide an approximate solution to another equation. In this specific case, the coefficients a must be determined according to criteria unknown to me. It is somewhat reminiscent of the Ritz/Galerkin method.
In any case, thanks to @janhardo, the function h has now been calculated in the exponent of f.

You write that f is an exponential function. However, the notation in equation (5) suggests that f is a number (regardless of whether it is real or complex). Or is f supposed to be a function of the variable phi from some arithmetic or other combination of the exponential function exp(phi)? This could easily be reformulated according to the power law. If f is a number, then equation (5) is a first-order ordinary differential equation with separated variables. Integrating the reciprocal of the right-hand side might then be somewhat challenging. But if f is meant to be a function f(phi), then a symbolic solution becomes very problematic.

Points P, Q, and R lie on the lines through BC, AC, and AB, respectively. Then there's a solution ;-).

@ZnMnCr 

... Some tips:
A good and practical introduction can be found in: Istvan Szabo, Introduction to Engineering Mechanics, starting on page 204, and Repertory and Exercise Book of Engineering Mechanics, starting on page 34. These books demonstrate problems like yours using conventional simplified structural models.
In simplified models, cable elongation is neglected. Simple equations then quickly lead to the solution of corresponding equations for smaller systems. However, if, for example, the guying of a 300-meter-high radio mast is to be calculated as a spatial system under combinations of external loads, then, in my experience, only numerical FEM methods are used, taking into account the large cable mass, sag, and cable elongation. There is powerful specialized software for this. In Maple, in my opinion, it would be worthwhile for you to calculate stresses and support forces in planar cable structures under simplified conditions according to the above-mentioned literature.

@dharr 

... that's exactly what interested me. One of the "strengths" of Zeta/polylog becomes clearly visible. Unfortunately, I'm still too clumsy to plot properly. Instead, I used derive like I did in the old days.

@nm 

... I would like to request an evaluation/transformation of your results with the restriction 0=a<b<2.14. With my limited Maple knowledge, I can't do this, so I'm asking for help. So, I'm not interested in the individual numerical calculations in tabular form.

@nm chini_dgl_a.mw

Happy Easter!

@Alfred_F 

This "unsolvable" equation has a numerical solution for the initial value (1;-1) in the interval [1; 2.700592...]. After a long calculation, I was unable to find a symbolic solution.

@Ronan 

How can point coordinates of the curve be directly retrieved in the plot of the function curve without tabulating (getdata)?

@Ronan 

I didn't know the command "rhs" yet.

@nm 

...exists, but not for every arbitrary initial value. A special solution exists for (sqrt(3);1). There is also another initial value (an incredibly long term).

Quote:

"...For this Abel ode   y'=x+y^3, which is known not to be solvable, Maple hangs on ..."

So the existence and uniqueness theorems of Peano, Picard, and Lindelöf do not apply to this equation if the right-hand side is continuous? But, contrary to the statement in the quote, these very theorems ensure the existence of at least a local solution.

@nm 

Reply to quotes:

"But without using the limit, how else will one find the constants of integration? Are you saying the series must be assumed to have  uniform convergence to use the limit as above?"

Yes.

"I assume Maple also used the limit internally to solve for C2 in this example and give the answer it did. How else could it have found the solution it did otherwise?"

That's probably the case. In such cases, a hint from the software would be helpful.