Sorry for basic question, Maple newbie here and I could not find answer using google.

I understand in Maple one uses the back quote key (or rather the apostrophe, 0X27) to prevent one time evaluation of expression. Hence when writing

'sin(Pi)'; #this remain sin(Pi)
%; # now we get 0

But when I tried it on fraction, it did not hold it:

'16/4'; #maple replied with 4

This might indicate that the front end parser did this simplification before the main evaluator got hold of it, so it was too late?

Either way, how would one make Maple return 16/4 when the input is '16/4'?

I am also not sure how to program a check in the code to determine that dsolve solved the differential equation.

For example

ode:=diff(y(x),x)-a*(x^n-x)*y(x)^3-y(x)^2=0;
sol:=dsolve(ode,y(x));

In this case, sol is () as it could not solve it. When I tried odetest, I get an error

odetest(sol,ode);

Error, invalid input: odetest uses a 2nd argument, ODE, which is missing

What is the correct way to check dsolve was successful that will work for all cases? I am looking for programmable method, no GUI use.

In Maple help on int, it says

"If Maple cannot find a closed form expression for the integral, the function call itself is returned"

But then, what is the correct way to check for this in the code? How do I know that the result returned is the call I made? (it would have been easier if these functions throw an error, or set some status code that one can check for success or failure).

Why this simplifes:

z1:=n*(Int(cos(x)^(n-2), x))-(Int(cos(x)^(n-2), x));

simplify(z1);

But when adding an extra term to z1, it no longer simplfies the above any more:

z2 := cos(x)^(n-1)*sin(x)+n*(Int(cos(x)^(n-2), x))-(Int(cos(x)^(n-2), x));

simplify(z2);

You can see the second term, which is z1, was not simplfied any more.

Why? And how would one go about simplifying z2 such that the second term gets simplfies as with z1, but while using z2 expression. It seems simplify stopped at first term and did not look ahead any more?

Maple 18.02, windows.

Given expr:=sin(n*x); and say I want to find its value at n=1,x=Pi/2, I found I can do

subs({n=1,x=Pi/2},expr); eval(%);

or combine into one

eval(subs({n=1,x=Pi/2},expr); ----(1)

But also one can write

eval(expr,{n=1,x=Pi/2}); ----- (2)

So if someone wants to just find value of expression as above, it seems (2) is much simpler than (1) since no need to call subs, it is done automatically. Is there anything else I am overlooking?

Only problem I see in the above, is that subs() uses different order: subs({},expr), while eval is:  eval(expr,{}); The other way around.

It would have been better if the order was the same in the API, to reduce confusion.