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How to create Jacobian matrix with given equilibrium points

I have this system. I have found the equilibrium points and I want to find the Jacobian Matrix of each point. But I am stuck. Could you help me, please?

r = 1;

b = 1;

c = 0.01;

μ = 0.4;

deq1 = x'[t] ==  r*x[t] - b*x[t]^2 - c*x[t]*y[t] - 0.75*x[t]*y[t]/(a + x[t])

deq2 = y'[t] == -μ*y[t] + 0.75*x[t]*y[t]/(a + x[t])

equilibrio = NSolve[{r*x[t] - b*x[t]^2 - c*x[t]*y[t] - 0.75*x[t]*y[t]/(a + x[t]) ==0, -μ*y[t] + 0.75*x[t]*y[t]/(a + x[t]) == 0}, {x[t], y[t]}]

jacobian := D[{f, g}, {{x, y}}]

f = r*x[t] - b*x[t]^2 - c*x[t]*y[t] - 0.75*x[t]*y[t]/(a + x[t])

g = -μ*y[t] + 0.75*x[t]*y[t]/(a + x[t])

MatrixForm[jacobian]
POSTED BY: Athanasios Paraskevopoulos
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You have to take derivatives with respect to x[t],y[t], not x,y, because that is the way f,g are written:

r = 1; b = 1; c = 1/100; \[Mu] = 4/10;
f = r*x[t] - b*x[t]^2 - c*x[t]*y[t] - 3/4*x[t]*y[t]/(a + x[t]);
g = -\[Mu]*y[t] + 3/4*x[t]*y[t]/(a + x[t]);
deq1 = x'[t] == f
deq2 = y'[t] == g
equilibrio = Solve[{f == 0, g == 0}, {x[t], y[t]}]
MatrixForm[jacobian = D[{f, g}, {{x[t], y[t]}}] // Simplify]
Simplify[MatrixForm[jacobian] /. equilibrio]
POSTED BY: Gianluca Gorni
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