@imparter 

misunderstanding of viable boundary conditions. Thje code

with(DETools):
with(plots):
with(IntegrationTools):
Nb:=1:Nt:=1:h(x):=exp(x):
Eq1 := (diff(f(x),x,x))+(((1/x)*(diff(f(x),x))+Nb*((diff(f(x),x))*diff(g(x),x))+Nt(diff(f(x),x)^2))):   
Eq2 := (diff(g(x),x,x))+(((1/x)*(diff(g(x),x))+(Nb/Nt)*((diff(f(x),x,x))+(1/x)*diff(f(x),x)))):       

# Boundary condition f(x)=0, g(x)=0 at x=h(x)" where h(x) is some function of x", 

 #f '(x)=0, g'(x)=0 at x=0

Cd1 := f(h(x)) = 0, (D(f))(0) = 0 :
dsys := {Cd1, Eq1}:
dsol := dsolve(dsys, numeric, output = operator):
plots[odeplot](dsol, [x, diff(f(x), x$1)], 0 .. 5, color = green):
Cd2 := g(h(x)) = 0, (D(g))(0) = 0:
dsys := {Cd1, Cd2, Eq1, Eq2}:
dsol := dsolve(dsys, numeric, output = operator):
plots[odeplot](dsol, [x, f(x)], 0 .. 5, color = red);
plots[odeplot](dsol, [x,g(x)], 0 .. 5, color = black);

contains two *interesting* definitions, namely

h(x):=exp(x)
f(h(x))=0

So just consider these for a moment. For any (real) value of x from -infinity to +infinity, exp(x) varies from 0 to +infinity, so h(x) is any value from 0 to +infinity - and the you define f(h(x))=0 - so f() for any value from 0 to +infinity is zero. You think this is a valid "boundary" condition??

@falkner 

there is no "tutor" available for use with dsolve().

One of the stategies used by  the dsolve() command is to try to identify whether a supplied ODE is of a certain "type".

The code

#
# Define system and solve
#
  sys:=[ diff(y(x),x) = y(x)^4 * cos(x) + y(x)*tan(x),
         y(0) = 0.5
       ]:
  DETools:-odeadvisor(sys[1]);
                          [_Bernoulli]

shows that Maple can identify the ODE as a Bernoulli equation. If you change the odeadviser() command in the above to

DETools:-odeadvisor(sys[1], help);

then Maple will display  help page describing the general strategy used for solving equations of this type

@Anthrazit 

there is very often more than one way to do something

is to always upload a worksheet which illustrates your problem!

Use the big green up-arrow in the Mapleprimes toolbar

@imparter 

to address the basic problem.

The ode you define in your worksheets as  'de1' only contains the dependent variable 'f(x)'. The ode you define in your worksheets as  'de2' contains the dependent variables 'f(x)' and 'g(x)'.

When you process the ode 'de1', to produce 'DE1', the latter contains the dependent variables

{b[0](x), b[1](x), b[2](x), b[3](x), b[4](x), b[5](x), b[6](x), b[7](x), b[8](x), b[9](x), b[10](x)}

However when you process the ode 'de2', to produce 'DE2', the latter contains the dependent variables (check the one highlighted in red)

{f(x), c[0](x), c[1](x), c[2](x), c[3](x), c[4](x), c[5](x), c[6](x), c[7](x), c[8](x), c[9](x), c[10](x)}

The mere existence of the undetermined function 'f(x)' in the definition of 'DE2' is the source of your problem.

See the attached for proof - it makes no attempt to solve anything - it just lists the indeterminates. Until you fix the existence of the indeterminate 'f(x)' in the definition of 'DE2' you are going to get precisely nowhere

Download indets.mw

@imparter 

actually looking at the system you are trying to solve?

The final execution group in your latest worksheet - ie this piece of code

for k from 0 to n do
    IBVC1 := select(has, CT, c[k]);
    {coeff(DE2, p, k), op(IBVC1)};
    slv1 := dsolve({coeff(DE2, p, k), op(IBVC1)});
    c[k] := unapply(rhs(slv1), x);
end do

For k=0, your code generates the ODE system

{c[0](0) - 1, diff(c[0](x), x, x), c[0](5)}

a second order ODE with two boundary conditions - which is fine, and Maple generates the solution

slv1 := c[0](x) = -x/5 + 1

For k=1, your code generates the ODE system

{diff(c[1](x), x, x) - f(x)/10, c[1](0), c[1](5)}

a second-order ODE with two boundary conditions, and a completely unknown function of the same independent variable f(x). What solution do you expect?

Maple returns a "formal" solution containing double integrals of the unknown function f(x)

@imparter 

you are trying to make.

1. Your original code generates an invalid ODE system, which means that dsolve() will produce an error.
2. This update to your code produces a valid ODE system, which means that dsolve() has a reasobale chance of coming up with a solution.

These two cases, both for loop index k=0 are illustrated in the attached. One errors, for reasons I explained previously and one doesn't.

Download diffODEs.mw

@FDS 

as in the attached "toy" example

Download getMean.mw

@Muhammad Usman 

that you were trying to achieve what is shown in the attached? I don't think this approach can be (easily) extended beyond first differences.

Download printStuff.mw

@ check my original comments on the typos in your equation and fix them properly before doing anything else.

I'd be surprised if an analytic solution can be found for the resulting ODE, so you may have to be satisfied with a numeric one - which means that you will need values for all parameters and a couple of initial/bondary conditions:-(

@JAMET 

the attached will show which months in a specified year have a Friday 13th. Output is in the form

[year, [list of months with Friday 13th]]

From which is pretty obvious that in 2022, the only month with a Friday 13th is month 5 - ie May!

Download fri_2.mw

@Will_iii 

when you type

convert(360*Unit(degrees), units, radians);

in Maple Flow - it doesn't work?

Because if this is true then all you have demonstrated ist that Maple Flow is unable to execute basic Maple commands - which is somewhat scary!

@Carl Love 

If you want to restrict the range variable to integers, then not only do you need the 'sample' option, but ialso to set the adaptive=false option. If the intent is to restrict the range variable to integers, why have a 'floor(n)' command in the function definition? Isn't this a bit - well - superfluous?

@ProfG 

was that in a plot() command, the plotting variable is continuous - it does not assume integer values. Kitonum circumvented this problem by plotting only points. This has the drawback that if you draw lines between the points (say by using style=pointline), then you will not get "vertical" lines, since x-values will always differ by one

I gave a (quick+dirty) workaround for this - and you can see the effect of superimposing my plot and Kitonum's in the attached. This gives an apparent "right shift to the curve on my plot. If this is a problem it is relatively easy to fix, see the final figure in the attached

Download sqPlot2.mw

@Tamour_Zubair 

at

https://en.wikipedia.org/wiki/Rouch%C3%A9%E2%80%93Capelli_theorem

which states (emphasis added)

A system of linear equations with n variables has a solution if and only if the rank of its coefficient matrix A is equal to the rank of its augmented matrix [A|b].^[1] If there are solutions, they form an affine subspace of ${\displaystyle \mathbb {R} ^{n}}$ of dimension n − rank(A). In particular:

1. if n = rank(A), the solution is unique,
2. otherwise there are infinitely many solutions.

It is trivial to produce the coefficient matrix, the augmented matrix and their ranks, using the code

sys:=[subs(x = xmin, uhat1) = rhs(bcf1),
      subs(x = xmax, uhat1) = rhs(bcf2),
      subs(p = pmin, uhat1) = rhs(bcf3),
      subs(p = pmax, uhat1) = rhs(bcf4)]:
  vars:=[Af[0, 1](t), Af[0, 2](t), Af[0, 0](t), Af[0, 3](t)]:
  coeffMat,b:=GenerateMatrix(sys, vars):
  augMat:=GenerateMatrix(sys, vars, augmented=true):
  Rank(coeffMat);
  Rank(augMat);

which shows that the coefficient matrix has rank 3, and the augmented matrix has rank 4. Thus the sytem has no solution