Given an expression, how to best find the constants _C1, _C2, etc.... in this expression?  Currently I do the following, but I think there should be a better way.

restart;
sol:=dsolve(diff(y(x),x$2)+y(x)=1);
indets(sol);
select(x->`if`(type(x,symbol) and convert(x,string)[1..2]="_C",true,false),indets(sol))

The above works, at least on the few examples I tried it on, but is there a better way to do it?  All constants will have the form _Cn where n is integer. 

Which one of these two versions, is the recommened one to use? For example, to check for type  integer*x, where x is literal x. i.e. identical(x)

restart;
TypeTools:-AddType('type_1', `&*`(integer,identical(x)));
type(3*x,type_1);

restart;
TypeTools:-AddType('type_1', '`*`'({integer,identical(x)}));
type(3*x,type_1)

Both work.  The difference is that `*` needs {} while `&*` does not.

I read the help page  and I do not understand what it says about the difference, and when is one supposed to use `*` vs. `&*`. it says 

              | `*`(type)  a product of factors of the given type
              | `&*`(type*)  a product of factors in which the nth factor is of the nth type specified in type*

Could possibly someone please explain in simple plain english what is the difference? If I use `&*`  vs. `*` with {}, will they work the same all the time or are there cases when to use one vs. the other?  Any rules of thumb to follow?

I need to simplify terms such as (cos(x)^2+sin(x)^2) to 1 if present in input, but I do not Maple to also do any other simplification rewriting polynomials that might be present in the expression.

And example will make it clear. Given this

expr:= (cos(x)^2+sin(x)^2)+5+(1+x+x^2+x^3)*(cos(x)^2+sin(x)^2)*exp(x);

I want expr to become  6+(1+x+x^2+x^3)*exp(x).   i.e. only simplify trig terms

But simplify(expr,trig); gives

                 6 + (x + 1)*(x^2 + 1)*exp(x)

Which is not what I want. Then I tried the trick of thaw and freeze to tell Maple to freeze polynomial type, like this

restart;
expr:= (cos(x)^2+sin(x)^2)+5+(1+x+x^2+x^3)*(cos(x)^2+sin(x)^2)*exp(x);
thaw(simplify(subsindets[flat](expr,satisfies(Z->type(Z,polynom(integer, x))),(freeze))));

And it actually worked, giving

           6 + (x^3 + x^2 + x + 1)*exp(x)

Question is: Why did simplify(expr,trig) not do what expected, which is to only simplify trig terms in expression and not mess around with the polynomial there? 

Is the above method of thaw/freeze to control which parts of expression gets simplify a recommended way to work around this, or is there a better way?

I can't figure how to make my own type, which is rational and greater than one.

There are buildin types for postive and posint, and so on. But what if I want to make for rational and greater than some value, say 1?

This is easy to do using patmatch, using the conditional. But do not know how to do it to make my own type.

Here is a simple example. I want to check for sin(x)^n, where n is rational and must be >1.  Using pathmatch

restart;
patmatch(sin(x)^(1/2),conditional(sin(x)^a::rational,a>1),'la');la;

does it. Using structured type, the best I could do is this

restart;

TypeTools:-AddType('my_sin',specfunc(sin)^And(rational,positive));
type(sin(x)^(1/2),'my_sin');

But this does not check for >1, only positive.

Any suggestions? I know I could do this using other means, by direct parsing, using op, and so on. But I'd like to learn how to do it using structured type, just to learn the syntax if there is one.

is it possible to use conditional with structured types? But need  name to do that, like with patmatch, but there is no such syntax in structured types. 

ps. I think conditional does not make much sense with structured type. But I need to figure how to make my own type, which is rational and say >1, or integer and say >2 and so on. I just do not know how to do that yet. But I am sure there is a way. Will try to figure it out.

I can't get pdsolve to solve this pde. Here are my tries below. One of them works, but the analytical solution Maple gives is wrong. So I am not sure if it needs some additional hints or some other help to make it give the correct solution.

This is Laplace pde but with non-zero on RHS. On disk centered at origin and with radius 1, with given BC on edge of disk.

restart;
the_rhs := r^2*cos(theta)*sin(theta);
pde := VectorCalculus:-Laplacian(u(r,theta),'polar'[r,theta])=the_rhs;
bc := u(1, theta) = cos(theta)*sin(theta);
sol:=pdsolve([pde, bc], u(r, theta));

Maple gives

But this does not look right. There is a complex exponential on its own there. The solution should be real. 

I tried to solve it using pdsolve numerically, but could not make it work. It kepts complaining about missing BC. I am not good with numerical solvers in Maple. May be someone could try to do it.

evalf(eval(rhs(sol),[r= sqrt(1/4 + 1/4),theta=Pi/4]))

gives 0.5951891582 which is wrong,. It should be  0.23958

It tried to give it the symmetry conditions on theta, but now it did not solve it

restart;
the_rhs := r^2*cos(theta)*sin(theta);
pde := VectorCalculus:-Laplacian(u(r,theta),'polar'[r,theta])=the_rhs;
bc := u(1, theta) = cos(theta)*sin(theta),u(r,0) = u(r,2*Pi),(D[2](u))(r, 0) = (D[2](u))(r, 2*Pi);
sol:=pdsolve([pde, bc], u(r, theta));

No solution. I tried adding HINT = boundedseries(r=0) but that also did not help. No solution.

Could someone solve this PDE numerically in Maple and check the solution at the above location? It should be 0.23958 which shows the analytical solution given is not correct. Can Maple solve this numerically?

Maple 2021.1

ps. Analytical solution was obtained by Mathematica. I did not solve this by hand.