I can't figure out how Maple obtained this solution and looking for some ideas to try.

It is first order non-linear ode in y(x), which is separable.

ode:=diff(y(x),x)=x*ln(y(x));
dsolve([ode,y(1)=1],y(x))

But the general solution is

sol:=dsolve(ode)

Setting up manually an equation using the given condition in order to solve for _C1, produces no solution. 

eq:=subs([y(x)=1,x=1],sol);
solve(eq,_C1)

Warning, solutions may have been lost
 

Also 

coulditbe(exp(RootOf(1 + 2*Ei(1, -_Z) + 2*_C1))=1)

   FAIL

So how did Maple solve for the constant of integration which results in particular solution y(x)=1 that is supposed to satisfy the condition y(1)=1?  

It is clear that y(x)=1 satisfies the ODE itself. But I am asking about how it also satisfies y(1)=1

(odetst says it does satisfy the ODE and condition as well. So Maple must have done something very smart under the cover)

Next I tried

ode:=diff(y(x),x)=x*ln(y(x));
sol:=dsolve(ode,y(x));
sol:=DEtools:-remove_RootOf(sol);
sol:=subs([y(x)=1,x=1],sol)

And now

solve(sol,_C1)

Error, (in Ei) numeric exception: division by zero
 

Just wondering how did Maple decide that y(x)=1 satisfies y(1)=1? I do not see it.

Using Maple 2020.1. But same result on Maple 2019

From help, it says

coulditbe routine returns true if there is a possible value of x1 that satisfies prop1

my question is, how to find out this condition/possible values that Maple found?  This infomration is very useful, but now I do not see how to obtain it. All what coulditbe retuirn is true or false.

Context of why I am asking:  Sometimes odetest do not verify its own solutions. And coulditbe can help in finding under what conditions the solution can satisfy the ode. Here is an example

restart;
ode:=diff(y(x),x) = abs(y(x))+1;
solExplicit:=dsolve(ode);
offset := odetest~([solExplicit],ode)

gives

[exp(-x)/_C1 - abs((-exp(-x) + _C1)/_C1) - 1, exp(x)*_C1 - abs(exp(x)*_C1 - 1) - 1]

Both solution fail odetest. 

coulditbe~(offset,0)

gives true

So there are assumptions/conditions which makes the solution satisfy the ODE. In this case, by inspection one can see what these conditions are. They are, for one solution:

(-exp(-x) + _C1)/_C1  >0

and for the other, the condition is

exp(x)*_C1 - 1 >0

Under these assumptions, odetest would have given 0 for each odetest.

And it is this information I wanted to obtain automatically from coulditbe.

In Mathematica, Reduce is used for this. Reduce gives conditions under which something is satisfied. For example, 

Reduce[ C[1] Exp[x] - Abs[C[1] Exp[x] - 1] - 1 == 0, {x, C[1]}, Reals]

Gives

C[1] >= Exp[-x]

While the above in  Maple

coulditbe( C[1]*exp(x)- abs( C[1]*exp(x)-1)-1 = 0)

gives true  only, but without the important information, true under what conditions.

Is there a different command in Maple which could give this information?

restart;
sub:=x/C;
expr:=1/2*x*(p^2+a)/p;

And now

subs(p = sub, expr)

But

algsubs(p = sub,expr)

Notice one "p" is still not replaced. 

This is very annoying. I looked at help and did not see anything about this. I could have missed it. It looks like it does not replace "p" when it is in denominator:

algsubs(p = x,1/p)

Remains 1/p but 

subs(p = x,1/p)

gives 1/x as expected.

May be this is documented somewhere? But why it does this?

This was generated when running some code on Maple 2020.1.

Just wondering if this might indicate some problem internally, or is this something that can happen.

restart;
ZZ:=Int(-(a*_a^2+(_a^4*a^2-4*_a*b*y(x))^(1/2))/(a*_a^3+_a*(_a^4*a^2-4*_a*b*y(x))^(1/2)+6*y(x)),_a = _b .. x)+Intat(-2/(a*x^3+x*(a^2*x^4-4*_f*b*x)^(1/2)+6*_f)-Int(2/(_a^4*a^2-4*_a*_f*b)^(1/2)*b*_a/(a*_a^3+_a*(_a^4*a^2-4*_a*_f*b)^(1/2)+6*_f)+(a*_a^2+(_a^4*a^2-4*_a*_f*b)^(1/2))/(a*_a^3+_a*(_a^4*a^2-4*_a*_f*b)^(1/2)+6*_f)^2*(-2*_a^2/(_a^4*a^2-4*_a*_f*b)^(1/2)*b+6),_a = _b .. x),_f = y(x))+_C1 = 0;

timelimit(30,value(ZZ))

Error, (in discont/zero) too many levels of recursion

The problem is that I am not able to trap the error. This does not work

try
ZZ:=Int(-(a*_a^2+(_a^4*a^2-4*_a*b*y(x))^(1/2))/(a*_a^3+_a*(_a^4*a^2-4*_a*b*y(x))^(1/2)+6*y(x)),_a = _b .. x)+Intat(-2/(a*x^3+x*(a^2*x^4-4*_f*b*x)^(1/2)+6*_f)-Int(2/(_a^4*a^2-4*_a*_f*b)^(1/2)*b*_a/(a*_a^3+_a*(_a^4*a^2-4*_a*_f*b)^(1/2)+6*_f)+(a*_a^2+(_a^4*a^2-4*_a*_f*b)^(1/2))/(a*_a^3+_a*(_a^4*a^2-4*_a*_f*b)^(1/2)+6*_f)^2*(-2*_a^2/(_a^4*a^2-4*_a*_f*b)^(1/2)*b+6),_a = _b .. x),_f = y(x))+_C1 = 0;
timelimit(30,value(ZZ));
catch:
  print("ignore");
end try;

Error, (in discont/zero) too many levels of recursion

Why can't one catch this error inside try/catch? It means the whole program can not  continue.

Maple 2020.1

Update Jan 9, 2025

This bug is still in Maple 2024.2. Will check again in 5 years. May be someone in Maplesoft will fix it by then.

Download not_fixed_jan_9_2025.mw

This works

restart;
patmatch(3*g(x),A::nonunit(algebraic)*B::function,'la');
la

[A = 3, B = g(x)]

But this fails

restart;
patmatch(3*g(x),conditional(A::nonunit(algebraic)*B::function, true ),'la');

Error, (in PatternMatching:-AlgStruct:-Match) improper op or subscript selector
 

I am not able to see why. In the above I used true for the condition, just to keep things simple. Any boolean valued expression will work.  

changing the above to

restart;
patmatch(3*g(x),conditional(A::anything*B::function, true ),'la');

it works. No error. [A = 3, B = g(x)]

What did I do wrong in the above syntax which generated the error?

Maple 2020.1