However, the $\Omega$ parameters are not all independent since they are related by the condition $$\label{constraint}
\Omega_{\rho}+\Omega_{k}+\Omega_{\Lambda}+\Omega_{\lambda} + \Omega_C =1\, .$$ Taking this constraint into account leaves us with a four-dimensional space and only four eigenvalues of the static solution to be found.

Introducing the primed time derivative $$' = \frac{1}{H} \frac{d}{dt} \, ,$$ one obtains the system of first-order differential equations [@CS] $$\begin{array}{l}
\Omega_{\rho}'=[2(1+q)-3(1+w)]\Omega_{\rho}\, , \smallskip \\
\Omega_{k}'=2q\Omega_{k}\, , \smallskip \\
\Omega_{\Lambda}'=2(1+q)\Omega_{\Lambda}\, , \smallskip \\
\Omega_{\lambda}'=2[(1+q)-3(1+w)]\Omega_{\lambda}\, , \smallskip \\
\Omega_C' = 2 (q - 1) \Omega_C \, ,
\end{array}$$ where $$q = \frac{1 + 3w}{2} \Omega_{\rho} - \Omega_{\Lambda} + (3w + 2)
\Omega_{\lambda} + \Omega_C\, .$$ The behavior of this system of equations in the neighborhood of its stationary point is determined by the corresponding matrix of its linearization. The real parts of its eigenvalues tell us whether the corresponding cosmological solution is stable or unstable with respect to the homogeneous perturbations.

As was said above, only four parameters $\Omega$ are independent because of constraint (\[constraint\]). In each of the cases listed below, we choose independent parameters conveniently, and our choice is reflected in the indices of the corresponding eigenvalues.

[**(1)**]{}  The Friedmann case, or $(\Omega_{\rho},\Omega_{k},\Omega_{\Lambda},\Omega_{\lambda},\Omega_C) =
(1,0,0,0,0)$. We 