To perform a stability analysis, we first need to find the derivative of the differential equation with respect to n.

d(- n ln(n/K))/dn = - ln(n/K) - n * 1/n

[Remember that d(ln(n))/dn = 1/n and d(ln(n/K))/dn = 1/(n/K) * d(n/K)/dn = 1/(n/K) * 1/K = 1/n.]

We then plug in the equilibrium value for n when the species is present (=K):

d(- n ln(n/K))/dn |_n=K = - ln(K/K) - = - , using the fact that ln(1)=0.

Since this is a model in continuous time, stability of the equilibrium is guaranteed if this derivative is negative (ie when is positive). In this case, a perturbation away from the equilibrium decreases in size over time.

Conversely, the equilibrium will be unstable if the derivative is positive (ie when is negative). In this case, a perturbation away from the equilibrium increases in size over time.

To find the global solution, we perform a change of variables from n to y=ln(n/K) in the Gompertz equation. That is, we set n=K e^y in the equation to get:

d(K e^y)/dt = - K e^y ln(K e^y/K) = - K e^y y

But d(K e^y)/dt = K e^y dy/dt so

K e^y dy/dt = - K e^y y

Dividing both sides by K e^y, leads to:

dy/dt = - y

This is the same equation as in the exponential growth model. It can be solved in the exact same manner, by separation of variables. Moving dt to the left hand side and y to the right hand side, we get:

dy/y = - dt

Integrating both sides:

ln(y) = - t + c

where c is some constant of integration. Therefore,

y = e ^{- t +c} = c e ^{- t}

(where c is now some other constant).

This gives the general solution in terms of y, in terms of n, we have:

ln(n/K) = c e^{- t}

At t=0, we get ln(n₀/K) = c, which gives us a solution for c:

ln(n/K) = ln(n₀/K) e^{- t}

Solving this equation for n gives:

n/K = Exp[ln(n₀/K) e^{- t}] = Exp[ln(n₀/K)]^x

where x = e^{- t}. Therefore:

n = K (n₀/K)^x.