how to find equlibrium point of 5 non linear system with numerical method

I have a system of 5 non linear ordinary differential equations with variable coefficients . I am trying to find the equilibrium points by hand but it seems like it is not possible without the help of a numerical method. What would be a good method to calculate equilibrium points of the system?

Another question (somehow related to the problem above): Would it be possible to check the stability of the equilibrium points and then draw a bifurcation diagram? If so, please suggest some way out!

 e1 = 13;
    g  = 0.0125;
    h  = 0.284253;
    f  = 0.05;
    q1 = 0;
    k1 = 10;
    d1 = 0.0412;
    e2 = 0.0188;
    j  = 0.0082;
    q2 = 0;
    k2 = 10;
    d2 = 0.0288;
    b  = 2;
    d4 = 0.1152;
    e3 = 0.166667;
    a  = 1.7;
    q3 = 0;
    k3 = 10;
    d3 = 0.1152;
    r  = 0.5;
    m  = 1.02;
    q4 = 0;
    k4 = 10;
    dydt(1) = e1 + (g*y(3)*y(1)/(h + y(3))) + (f*y(3)*y(1)) - (y(1)*(1 + ((q1/k1)*y(1)))) - (d1*y(1));
    dydt(2) = e2*y(2) + (f*y(1)*y(3)) - (j*y(2)) - (y(2)*(1 + ((q2/k2)*y(2)))) - (d2*y(2));
    dydt(3) = (b*e2*y(2)) - (d4*y(3));
    dydt(4) = e3*y(4) + (j*y(2)) - (a*y(4)) - (y(4)*(1 + ((q3/k3)*y(4)))) - (d3*y(4));
    dydt(5) = (r*y(5)*(1 - (m*y(5)))) + (a*y(4)) - y(5)*(1 + ((q4/k4)*y(5)));

Answer 1

[Torsten]

I don't know if the result is as expected.

format long
yequi = fsolve(@fun,rand(5,1))

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

yequi = 5×1

12.485593545908634 0.000000000000008 0.000000000000003 0.000000000000000 0.000000000000000

fun(yequi)

ans = 1×5

1.0e-13 * -0.657252030578093 -0.063585506384309 -0.000000000537866 0.000000000086870 -0.000243217417049

function dydt = fun(y)
e1 = 13;
g  = 0.0125;
h  = 0.284253;
f  = 0.05;
q1 = 0;
k1 = 10;
d1 = 0.0412;
e2 = 0.0188;
j  = 0.0082;
q2 = 0;
k2 = 10;
d2 = 0.0288;
b  = 2;
d4 = 0.1152;
e3 = 0.166667;
a  = 1.7;
q3 = 0;
k3 = 10;
d3 = 0.1152;
r  = 0.5;
m  = 1.02;
q4 = 0;
k4 = 10;
dydt(1) = e1 + (g*y(3)*y(1)/(h + y(3))) + (f*y(3)*y(1)) - (y(1)*(1 + ((q1/k1)*y(1)))) - (d1*y(1));
dydt(2) = e2*y(2) + (f*y(1)*y(3)) - (j*y(2)) - (y(2)*(1 + ((q2/k2)*y(2)))) - (d2*y(2));
dydt(3) = (b*e2*y(2)) - (d4*y(3));
dydt(4) = e3*y(4) + (j*y(2)) - (a*y(4)) - (y(4)*(1 + ((q3/k3)*y(4)))) - (d3*y(4));
dydt(5) = (r*y(5)*(1 - (m*y(5)))) + (a*y(4)) - y(5)*(1 + ((q4/k4)*y(5)));
end

[Latifah]

Is it possible for the equilibrium point to be negative?

[Torsten]

I left it to you to glue the solutions together.

syms e1 g h f q1 k1 d1 e2 f j q2 k2 d2 b e2 d4 e3 a q3 k3 d3 r m q4 k4
y = sym('y',[5 1]);
dydt(1) = e1 + (g*y(3)*y(1)/(h + y(3))) + (f*y(3)*y(1)) - (y(1)*(1 + ((q1/k1)*y(1)))) - (d1*y(1));
dydt(2) = e2*y(2) + (f*y(1)*y(3)) - (j*y(2)) - (y(2)*(1 + ((q2/k2)*y(2)))) - (d2*y(2));
dydt(3) = (b*e2*y(2)) - (d4*y(3));
dydt(4) = e3*y(4) + (j*y(2)) - (a*y(4)) - (y(4)*(1 + ((q3/k3)*y(4)))) - (d3*y(4));
dydt(5) = (r*y(5)*(1 - (m*y(5)))) + (a*y(4)) - y(5)*(1 + ((q4/k4)*y(5)));
y2 = solve(dydt(3)==0,y(2))
y1 = simplify(solve(subs(dydt(2),y(2),y2)==0,y(1)))
y3 = solve(subs(dydt(1),y(1),y1)==0,y(3),'MaxDegree',3)
y31 = y3(1);
y32 = y3(2);
y33 = y3(3);
y11 = subs(y1,y(3),y31);
y12 = subs(y1,y(3),y32);
y13 = subs(y1,y(3),y33);
y21 = subs(y2,y(3),y31);
y22 = subs(y2,y(3),y32);
y23 = subs(y2,y(3),y33);
y41 = solve(subs(dydt(4),y(2),y21)==0,y(4))
y411 = y41(1);
y412 = y41(2);
y42 = solve(subs(dydt(4),y(2),y22)==0,y(4))
y421 = y42(1);
y422 = y42(2);
y43 = solve(subs(dydt(4),y(2),y23)==0,y(4))
y431 = y43(1);
y432 = y43(2);
y511 = solve(subs(dydt(5),y(4),y411)==0,y(5))
y512 = solve(subs(dydt(5),y(4),y412)==0,y(5))
y521 = solve(subs(dydt(5),y(4),y421)==0,y(5))
y522 = solve(subs(dydt(5),y(4),y422)==0,y(5))
y531 = solve(subs(dydt(5),y(4),y431)==0,y(5))
y532 = solve(subs(dydt(5),y(4),y432)==0,y(5))

[Torsten]

Is it possible for the equilibrium point to be negative?

It's the numerical error in the equation fun(yequi) that is negative, not the solution itself (yequi) (if this is what you mean with your question).

[Latifah]

thank you , it helps me a lot