@sursumCorda  Is convert(..., simplifier=F)  documented?

@Axel Vogt  Only under some assumptions, e.g. real.

@aroche  If was chosen the branch  diff(y(x), x) = (x-2)^(2/3)  instead of diff(y(x), x) = ((x-2)^2)^(1/3).
But then y(x) is complex for x<2, which is probably not wanted.

@nm It is 2 only because you want to be so. For the exponent 2, Maple says 4 (as you have complained).

But note that y(x) + x = 0 ==> y'(x)=-1, so you have:

(x + y(x)) * (1+diff(y(x),x)) = 0  <==> (1+diff(y(x),x)) = 0 <==>  y(x) = -x+c (for some constant c). 

@nm Mathematica is not the supreme authority.

P.S. With your statement,  (x + y(x))^7 * (1+diff(y(x),x)) = 0  has 8 solutions.

@nm It's not in the standard form (or reducible to)  d/dx y(x) = F(x, y(x)).

The reason is obvious. Anyway, I see ODESteps as designed for standard simple ODEs. 

@nm No bug without Physics updates.

@minhthien2016 Insert f:=simplify(f) assuming real; before calling extrema. 

@Carl Love Very nice, vote up! Unfortunately the rose is not quite 3d: rotating the plot, several gaps are visible and the stem is just "separately added".

@Hullzie16  Adding the option numeric, dsolve produces a nrmerical solution,
Your system has a lot of floats, so, approximations. In such situations, an exact (analytic) solution is usually useless (even when it exists). 

@Ali Hassani 

seriestorec(s, A(n)) 
   ==> for a given series s, finds a "suspected" recurrence relation of its coefficients A(n)
rectodiffeq(rec, A(n), f(z))
   ==> for a recurrence rec, finds the ode satisfied by the "generating function"
           f(z) = Sum(A(n)*z^n, n = 0 .. infinity).
 

@dharr  When checking the minimal polynomial we must use a much higher precision.

Digits:=200:
evalf(expr) - fsolve(poly)[2]: evalf[15](%);
#                        1.37954446875327 e-38

(it should be near 1e-200).
So, the first poly is incorrect.

The second one (of degree 20) is ok.

@sursumCorda  Why not use e.g.

Matrix( [ [x || (1 .. 2)] , [y || (1 .. 2)] , [1 $ 2] ] );

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