I had bigger (6 equations) system of ODE and reduced it to the follwing system to check if we cand find equilibruim points .
The code:
syms x y1 y2 real;assume (x>=0,y1>=0,y2>=0);syms m1 m2 epsilon1 epsilon2 positive;syms omega11 omega12 omega 21 omefa22 positive;syms b12 p12 gamma12 beta1 beta2 positive;F2=x*(1-x-y1/(m1*y1+x)-y2/(m2*y2+x))==0;F6=y1*(-epsilon1*(1+omega11*y1+omega12*y2)-(b12*y2)/(p12*y2+y1)+(beta1*x)/(m1*y1+x))==0; F7=y2*(-epsilon2*(1+omega21*y1+omega22*y2)+(gamma12*y1)/(p12*y2+y1)+(beta2*x)/(m2*y2+x))==0;eqns=[F2,F6,F7];diary('SolutionOfEquation_Subsystem_NonDim_RDFR_Condition.txt')S = solve(eqns,[x,y1,y2], 'returnconditions', true);x=latex(S.x)y1=latex(S.y1)y2=latex(S.y2)S.xS.y1S.y2
However Matlab gave one solution in term of z and others in terms of parameters ,what does that mean and how I can use the results? Is (z,z1,z2) a free solution?
Is it possible to get z's in terms of parameters ?
The solution is in the attached file.
I also tried to check if the solution is satisfies the equations but find out that just the first and second do and others no.
Best Answer