Am 13.09.2013 03:15, schrieb firstname.lastname@example.org: >> Just interchange x and y, then get Max y for the new equation. > > x^4 + y^4 + A(x^2) - A(y^2) + 2(x^2)(y^2) - Bxy + C = 0 > > By substituting x for y the above equation becomes :- > x^4 + y^4 - A(x^2) + A(y^2) + 2(x^2)(y^2) - Bxy + C = 0 > > So presumably the new dy/dx becomes:- > dy/dx = -(4x^3 - 2ax + 4xy^2 - by)/(4y^3 + 2ay + 4x^2 y - bx) > > and in turn giving > 4x^3 - 2ax + 4x(y^2) - by = 0
4x^3 - 2Ax + 4x(y^2) - By = 0
> as the equation to be merged with the xy-altered equation:- > x^4 + y^4 - A(x^2) + A(y^2) + 2(x^2)(y^2) - Bxy + C = 0 > > Even if i'm right so far, I am again lost in finding the equivalent of > quasi's solution to the original equation.
Look closer: The only difference in the set of equations is that A is replaced by -A. Now look at the coefficients d8, d6, d4, d2, d0 from quasi's equation for y. What happens when you replace A by -A:
d8 -> d8 (nothing changes because of only even powers in A) d6 -> -d6 (switch sign because of only odd powers in A) d4 -> d4 d2 -> -d2 d0 -> d0.