I wanted to remove entry from a list that contain y=y or x=x in it. Here is an example

f:= (x-1)*y^4/(x^2*(2*y^2-1));
S:=[singular(f)]

Where I wanted to remove those entries highlighted above to obtain

This is below how I ended up doing it. I'd like to ask if there is a better or more elegent way. I had to use map, since could not get remove() to work on the original list in one shot. 

foo:= z->remove(has,z,{y = y,x = x});
map(foo,[singular(f)])

Which gives the output above.

Is there a better way to do this? I always learn when I find how to do something better.

Maple 2019.1

Are there commands in Maple to find the order and degree of an ODE?  Searching help I could not find anything so far.

For an example, given 

restart;
ode:=(1+diff(y(x),x)^2)^(3/2)=diff(y(x),x$2)

I want the command to return 2 for the order of the ODE and degree is also 2 in this case.

I looked at DEtools package and googled. I am sure Maple have build in commands to do this without me having to parse the ODE myself to find out.

fyi, there seems to be a problem here. Maple 2019, Physics version 395 on windows 10.

The solution given to this wave PDE by Maple as sum that starts from zero, has "n" in the denominator. When n=0, this gives division by zero.  Is this a bug?

Download bug_july_11_2019.mw

I solved this PDE by hand to verify Maple's solution. I think Maple solution is wrong. This PDE is the heat PDE on a bar (1D) with boundary coditions on both ends are function of time and zero initial conditions.

unassign('A,B,x,t,L,k,f');
pde := diff(u(x,t),t)= diff(u(x,t),x$2):
bc := u(0, t) = A(t), u(1, t) = B(t):
ic := u(x, 0) = 0:
sol1:=pdsolve([pde, ic, bc], u(x, t));

#now try when A(t)=sin(t),B(t)=t, use 20 terms for the sum
sol2:=simplify(subs([infinity=20,B(tau)=tau,A(tau)=sin(tau),A(0)=0,B(0)=0,A(t)=sin(t),B(t)=t],sol1)):
sol3:=simplify(value(subs(t=1,sol2))):
evalf(subs(x=0.5,sol3))

Also doing pdetest(sol1,pde); on the above solution does ot return zero as expected.

To verify more, I solved the same PDE again, but now using an explicit values for the boundary conditions A(t), B(t). Using A(t)=sin(t), B(t)=t. Then found the value again of the solution u at x=0.5 and t=1 like in the above, and it gives different value:

unassign('A,B,x,t,L,k,f');
pde := diff(u(x,t),t)= diff(u(x,t),x$2):
bc := u(0, t) = sin(t), u(1, t) = t:
ic := u(x, 0) = 0:
sol4:=pdsolve([pde, ic, bc], u(x, t));
sol5:=simplify(subs(infinity=20,sol4)):
sol6:=simplify(value(subs(t=1,sol5))):
evalf(subs(x=0.5,sol6))

Then I typed my hand solution into Maple and for the same values x=0.5, t=1 and same number of terms, I also get the same value 0.819. 

I do not see at all where the function sin integral should come into play in this solution. 

Could some Maple expert please check to see what is going on with this solution to Maple? 

Using Maple 2019.1 and Physics version 370

I do not understand why Maple can simplify this expression below when told that n is integer and also positive using a "," to separate the assumptions, but does simpify the same expression when using "and" to build the assumptions.

Here is an example

restart;
result:=int(x*cos(n*Pi/5*x),x=0..5)
simplify(result) assuming n::integer and n>0

But this works

simplify(result) assuming n::integer, n>0

What are the semantic differences between writing assuming "n::integer and n>0" and "n::integer,n>0" ? I thought these would be the same, but clearly they are not.

Maple 2019.1 on windows.