but neither (deprecated) 'student[changevar]' nor Maxima 5.15 'changevar' are better. It seems to me that limits and integration paths in the (extended) complex plane were not given yet a serious enough thought.

What I observe tracing with:

nmax:=15:
Order:=nmax+1:

infolevel[all]:=5:
trace(dsolve):
trace(combine):
trace(`dsolve/IC/_C`):
trace(`dsolve/IC/_C/do`):
printlevel:=10:
dsolve(subs(k=nmax+1,{des[k],bcs(k-1)}));

is a long "solution" expression in terms of an inert quadruple integral of a sum of powers of eta and four arbitrary integration constants (ie before solving them by means of the boundary conditions).

PS: I have copy-pasted this inert integral, evaluated and solved for the integration constants. Ie, nothing seems basically wrong with the calculation of F[15].

What I observe tracing with:

nmax:=15:
Order:=nmax+1:

infolevel[all]:=5:
trace(dsolve):
trace(combine):
trace(`dsolve/IC/_C`):
trace(`dsolve/IC/_C/do`):
printlevel:=10:
dsolve(subs(k=nmax+1,{des[k],bcs(k-1)}));

is a long "solution" expression in terms of an inert quadruple integral of a sum of powers of eta and four arbitrary integration constants (ie before solving them by means of the boundary conditions).

PS: I have copy-pasted this inert integral, evaluated and solved for the integration constants. Ie, nothing seems basically wrong with the calculation of F[15].

at least in Physics parlance, is tipically the derivative with respect to time. May be that this is what is meant?

at least in Physics parlance, is tipically the derivative with respect to time. May be that this is what is meant?

of this blog?

As F[0] and F[1] do not match smoothly at x=0:

F[1]:= x -> x^3 - 2*x^2;
F[0]:= t -> t^3 - 2*t^2 - 3*t;
D(F[0])(0);
D(F[1])(0);

                                  -3

                                  0

df/dx is undefined at x=0. Then, what you get this way is a solution within the open intervals i< x < i+1, i integer, but not on the whole line.

As F[0] and F[1] do not match smoothly at x=0:

F[1]:= x -> x^3 - 2*x^2;
F[0]:= t -> t^3 - 2*t^2 - 3*t;
D(F[0])(0);
D(F[1])(0);

                                  -3

                                  0

df/dx is undefined at x=0. Then, what you get this way is a solution within the open intervals i< x < i+1, i integer, but not on the whole line.

F[1](x)=x^3 - 2*x^2 is not a solution for x in [0,1), but just for x=1, to start with.

F[1](x)=x^3 - 2*x^2 is not a solution for x in [0,1), but just for x=1, to start with.

But f(x) = x^3 - 2*x^2 is not a solution for x in [0,1] but only for x=1:

eq:=diff(f(x),x)=f(x)-f(x-1):
(expand@eval)(eq,f=(x->x^3 - 2*x^2));
solve(%);

                        2            2
                     3 x  - 4 x = 3 x  - 7 x + 3


                                  1

May be that you mean a solution just for integers? I mean a solution for all real x.

But f(x) = x^3 - 2*x^2 is not a solution for x in [0,1] but only for x=1:

eq:=diff(f(x),x)=f(x)-f(x-1):
(expand@eval)(eq,f=(x->x^3 - 2*x^2));
solve(%);

                        2            2
                     3 x  - 4 x = 3 x  - 7 x + 3


                                  1

May be that you mean a solution just for integers? I mean a solution for all real x.

This equation is first order as differential and as recurrence. And it is linear. So, it sounds to me that the general solution has two arbitrary constants in a linear combination. Hence it seems that there is no other. But really I would like to see if the theorem says what I guess.

PS: Geometrically, the slope equals the secant only for the stright lines.

This equation is first order as differential and as recurrence. And it is linear. So, it sounds to me that the general solution has two arbitrary constants in a linear combination. Hence it seems that there is no other. But really I would like to see if the theorem says what I guess.

PS: Geometrically, the slope equals the secant only for the stright lines.

Begin with simpler circuits, of one or two loops. You have two basic standard methods for writing the system of equations: by branch current and by mesh current. Besides your textbooks, you may find by googling for these methods plenty of tutorials. Just as a sample I have found these ones:

http://www.allaboutcircuits.com/vol_1/chpt_10/2.html

http://www.allaboutcircuits.com/vol_1/chpt_10/3.html