@Heeka 

(1) Y_vals produces a list of y(x), for x=0, 0.1, 0.2, ..., 1.0.

(2) Comparison exact and numeric solution:

restart;
DE := (x+b*y(x))*diff(y(x),x) + y(x) = 0;
BC := y(1)=1:
b := 1/10:
# exact solution:
sol1 := dsolve( {DE,BC}, y(x) ): Y1 := unapply( subs( sol1, y(x) ), x );
# numeric solution:
sol := dsolve( {DE,BC}, y(x), numeric, output=listprocedure ):
Y := subs( sol, y(x) ):
# comparison exact and numeric sulution
plot( Y1-Y, 0..1 );

Notice that in the display statement you must provide the plots as a sequence or list. If you use a set:

  plots:-display]( {A,B,C,E}, ...

the plots, and consequentlly the corresponding legends, are put in random order.

Notice that in the display statement you must provide the plots as a sequence or list. If you use a set:

  plots:-display]( {A,B,C,E}, ...

the plots, and consequentlly the corresponding legends, are put in random order.

... and if you want a plot of the derivative dpH/dVb:

plot( [VB,1/diff(VB,ph), ph=ph0..ph40], labels=["Vb",typeset(Diff(pH,Vb))] );

... and if you want a plot of the derivative dpH/dVb:

plot( [VB,1/diff(VB,ph), ph=ph0..ph40], labels=["Vb",typeset(Diff(pH,Vb))] );

@emma hassan If you want to use the assigned valies of the table B in the matrix, you have to remove the unassign command B := 'B':

@emma hassan If you want to use the assigned valies of the table B in the matrix, you have to remove the unassign command B := 'B':

@Carl Love I was trihggered by the words  "Existence and uniqueness" , which leads me automatically to IVP's!

@Carl Love I was trihggered by the words  "Existence and uniqueness" , which leads me automatically to IVP's!

@fias In this special case the for-loop is (perhaps the most) efficient.

But I supposed that your problem (because it is easily done by hand) was a simple example of more complicated recurrence equations.

@fias In this special case the for-loop is (perhaps the most) efficient.

But I supposed that your problem (because it is easily done by hand) was a simple example of more complicated recurrence equations.

Thank you, Carl, this helps.
I think it is not a bug, bacause when the function is not one-to-one there can be multiple solutions:

h := piecewise( x<1,x,x-1): solve(h=y,x): lprint(%);
piecewise(y < 0, [y], y < 1, [1+y, y], 1 <= y, [1+y])

Thank you, Carl, this helps.
I think it is not a bug, bacause when the function is not one-to-one there can be multiple solutions:

h := piecewise( x<1,x,x-1): solve(h=y,x): lprint(%);
piecewise(y < 0, [y], y < 1, [1+y, y], 1 <= y, [1+y])

@williamov 

(1) In the procedure Basisvector you initialize num6 als a number. This has to be a Vector:

BasisVector := proc(a, b, num)
  local result6, i;
  result6 := Vector(num);
  for i from 1 to num do
    result6[i] := 0;
  od;
  for i from 1 to num do
    result6[i] := Basis(convert(a(i, 1 .. -1),list), b);
  od;
  return result6;
end proc;

(2) Now we can see what Q is:

 BasisVector(<g1|g2|g3|g4|g5>, tdeg(x, y, h121, h122, h123, h124, h125), 2),
tdeg(x, y, h121, h122, h123, h124, h125),
'Q',2):
lprint(Q);

[-h124*y^3-h124*y-h125*x+h125*y-h121+1]

So Q is a list, not a rtable!

 nops(Q);
                               1

@williamov 

(1) In the procedure Basisvector you initialize num6 als a number. This has to be a Vector:

BasisVector := proc(a, b, num)
  local result6, i;
  result6 := Vector(num);
  for i from 1 to num do
    result6[i] := 0;
  od;
  for i from 1 to num do
    result6[i] := Basis(convert(a(i, 1 .. -1),list), b);
  od;
  return result6;
end proc;

(2) Now we can see what Q is:

 BasisVector(<g1|g2|g3|g4|g5>, tdeg(x, y, h121, h122, h123, h124, h125), 2),
tdeg(x, y, h121, h122, h123, h124, h125),
'Q',2):
lprint(Q);

[-h124*y^3-h124*y-h125*x+h125*y-h121+1]

So Q is a list, not a rtable!

 nops(Q);
                               1