if I run the below code in MATLAB 2024a, jacobian is zero. However, this is wrong.
If I use "clear" instead of "clear all", the answer is correct. However, in 2023b in both cases (clear or clear all) I get the correct answer?
is this a bug? or what is the point?
clear all
clc
syms a b c d m n s r x y k
dy1 = -x*( r*x.^2 – r*(a+b)*x + a*b +(c*y-r*d));
dy2 = s + m*x-n*y ;
dr = r-x;
F=[dy1;dy2;dr]
% equilibrium_points = solve(F == 0, [x, y, r]);
% equilibrium_points = …,
% [equilibrium_points.x, equilibrium_points.y, equilibrium_points.r];
J=jacobian(F,[x y r])if I run the below code in MATLAB 2024a, jacobian is zero. However, this is wrong.
If I use "clear" instead of "clear all", the answer is correct. However, in 2023b in both cases (clear or clear all) I get the correct answer?
is this a bug? or what is the point?
clear all
clc
syms a b c d m n s r x y k
dy1 = -x*( r*x.^2 – r*(a+b)*x + a*b +(c*y-r*d));
dy2 = s + m*x-n*y ;
dr = r-x;
F=[dy1;dy2;dr]
% equilibrium_points = solve(F == 0, [x, y, r]);
% equilibrium_points = …,
% [equilibrium_points.x, equilibrium_points.y, equilibrium_points.r];
J=jacobian(F,[x y r]) if I run the below code in MATLAB 2024a, jacobian is zero. However, this is wrong.
If I use "clear" instead of "clear all", the answer is correct. However, in 2023b in both cases (clear or clear all) I get the correct answer?
is this a bug? or what is the point?
clear all
clc
syms a b c d m n s r x y k
dy1 = -x*( r*x.^2 – r*(a+b)*x + a*b +(c*y-r*d));
dy2 = s + m*x-n*y ;
dr = r-x;
F=[dy1;dy2;dr]
% equilibrium_points = solve(F == 0, [x, y, r]);
% equilibrium_points = …,
% [equilibrium_points.x, equilibrium_points.y, equilibrium_points.r];
J=jacobian(F,[x y r]) jacobian, symbolic toolbox MATLAB Answers — New Questions

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