This pde used to be solved in 2018 as far as I know. Now it gives a strange new error

restart;
interface(showassumed=0);
pde :=  diff(u(x,t),t)+k*diff(u(x,t),x$2)+sin(2*Pi*x/L);
ic  :=  u(x,0)=f(x);
bc  :=  D[1](u)(0,t)=0, D[1](u)(L,t)=0;
sol :=  pdsolve({pde,ic,bc},u(x,t)) assuming L>0,t>0,k>0;

Error, (in assuming) when calling 'dsolve'. Received: 'found differentiated functions with same name but depending on different arguments in the given DE system: {F0_0(L), F0_0(x)}'

I am using  Physics:-Version();     MapleCloud version: 72

Do others get this erorr? Why does it show up now when it worked OK before?

update

Iam running on Linux. Here is screen shot

I want to match   a*x^b only if "a" and "b" are both not the symbol "y"

Therefore 5*x^y should fail to match, but this below matches.

restart;
expr:=5*x^y;
if patmatch(expr,conditional(a::anything*x^(b::anything),(a<>y and b<>y)),'la') then
   print("it matched, why??");
   map(assign,la):
   print("a = ",a, " b=",b);
fi;

it gives

                      "it matched, why??"
                      "a = ", 5, " b=", y

What did I do wrong? is my conditional wrong above?

In Maple this works

my_list :=[1,2,3];
my_list[1]:=0;

But this does not

my_set:={1,2,3};
my_set[1]:=0;

gives error "Error, cannot reassign the entries in a set". I suppose it has to do with the fact that sets, mathematically speaking, do not have an "oder" per say? But the elements must be ordered internally as they show, right? So the first element in the set is "1" and the second element in the set is "2", etc...

Had to use the ugly and unatural notation of

my_set := subsop(1=0, my_set)

To change the element of a set. The above implifies an order as well. So why the first did not work, but the above works?

The question is why is this inconsistency of notation?   It is much more natural to write "A[i]:=something". 

If using "subsop(1=0, my_set)" is a must for some reason, then why Can't Maple make a short cut, where if a user types "A[i]:=something" where A is a "set", then it will internally get modified/replaced by "A:=subsop(i=something, A)", so that one can still use the more natural syntax instead?

But it would be better just removing this restriction of using A[i]:=something on a set.

in this ODE, with initial conditions, I am trying to verify my solution, but I can't figure how Maple obtained the final answer.  I can obtain same solution, but before using the initial conditions to find the constant of integrations. The problem comes in solving for the constant of integration from initial conditions. I do not know how Maple did it, becuause when I do the normal steps, I get division by zero.

restart;
ode:=diff(y(x),x)=x*ln(y(x)):
ic:=y(1)=1:
sol:=dsolve({ode,ic},y(x));

gives "sol := y(x) = 1"

Now I solve without IC, then try to find _C1 by hand

sol:=dsolve(ode,y(x));

       sol := y(x) = exp(RootOf(x^2+2*Ei(1, -_Z)+2*_C1))

Now used remove_RootOf

sol2:=DEtools:-remove_RootOf(sol);

            sol2 := x^2+2*Ei(1, -ln(y(x)))+2*_C1 = 0

Now I followed the steps I learned at school, which is to plugin y=1 and x=1 in the solution to get an equation to solve for _C1

eq := subs({y(x)=1,x=1},sol2);

            eq := 1+2*Ei(1, -ln(1))+2*_C1 = 0

But Ei(1, -ln(1)) is a division by zero. So can't solve for _C1

solve(eq,_C1);
       Error, (in Ei) numeric exception: division by zero

So how would you solve for _C1 in the above? Or how did the smart Maple do it?

I tried using limits, but that did not help.  I tried using allvalues instead of remove_RootOf, and that did not get rid of RootOf.  Next I tried not to remove RootOf and keep it there to see what happens

sol:=dsolve(ode,y(x));
eq := subs({y(x)=1,x=1},sol);
             eq := 1 = exp(RootOf(1+2*Ei(1, -_Z)+2*_C1))

c1:=solve(eq,_C1, AllSolutions);
              c1 := -1/2-Ei(1, -(2*I)*Pi*_Z1)

subs(_C1=c1,sol);
             y(x) = exp(RootOf(x^2+2*Ei(1, -_Z)-1-2*Ei(1, -(2*I)*Pi*_Z1)))

simplify(%);
            x^2+2*Ei(1, -ln(y(x)))-1-2*Ei(1, -(2*I)*Pi*_Z1) = 0

DEtools:-remove_RootOf(%);
           x^2+2*Ei(1, -ln(y(x)))-1-2*Ei(1, -(2*I)*Pi*_Z1) = 0

solve(%,y(x));
          exp(RootOf(x^2+2*Ei(1, -_Z)-1-2*Ei(1, -(2*I)*Pi*_Z1)))

So I can't figure how Maple obtained y(x)=1. Maple must have used different method to solve for _C1.

What would be the Maple commands to use to obtain y(x)=1, starting from

            sol:=dsolve(ode,y(x));

and given that y(1)=1 ?

Can someone please explain in simple terms, why when I do

 r:=foo(r), where "r" is a Record, then the returned  "r" is not what is returned from foo()?  I want to write a proc foo() which takes in a Record variable, update some of its fields, and then return the updated Record to the caller

r:=Record('a','b');
foo:=proc(r)
    r:-b:=5;
    return(r);
end proc;

But now when I call the above as follows

r:-b:=99;
print(r);
r:=foo(r):
print(r);

The last print above just prints "r" and not the Record. It seems to have erased the Record.

But it works, if I change the name of the variable to return the result into, as 

r:-b:=99;
print(r);
r0:=foo(r):
print(r0);

When I do the same on say a Matrix, there is no problem

restart;
r:=<<1,1>>:
foo:=proc(r)
    r[1]:=5;
    return(r);
end proc;

And now

r;
r:=foo(r):
print(r);

Why it worked with a Matrix but not with Record? Why can't one overwrite the Record on the call return?

What would be the correct way to pass in a Record to a function, and have the function update some of its fields, and then return back the updated copy of the Record without having to make a new variable "r0" as above?