Example 3.10. In Example 3.8, c(t) and A(t) are determined by A=rA (r-p) The General Solution Consider a linear system i=a(x-)+b(y-0) (x-)+B(y-0) where a, b, a, BER and(I, g)is the steady state. The equation system can be written in a matrix form x-元 A (317) A The eigenvalues of A are u=5 d1+d1+ v(dn+d1)2-414l We have A< u. Let s and s be eigenvectors of A and u, respectively. The general solution of (3.17) is +c15e+c2 y(t) where Ci and C2 are arbitrary constants Stability Analysis Saddle-Point Stable Path with A<0. If [A<0, then A<0 and u>0, implying that the system is saddle-point stable. Convergent sequences must have C2=0Example 3.10. In Example 3.8, c(t) and A(t) are determined by A˙ = rA − c c˙ = (r − ρ) c α(c) .  The General Solution Consider a linear system x˙ = a(x − x¯) + b(y − y¯), y˙ = α(x − x¯) + β(y − y¯), where a, b, α, β ∈ R and (¯x, y¯) is the steady state. The equation system can be written in a matrix form: ⎛ ⎜⎝ x˙ y˙ ⎞ ⎟⎠ = A ⎛ ⎜⎝ x − x¯ y − y¯ ⎞ ⎟⎠ , (3.17) where A = ⎛ ⎜⎝ d1 a1 a2 d2 ⎞ ⎟⎠ . The eigenvalues of A are λ = 1 2 k d1 + d1 − s(d1 + d1)2 − 4 |A| l , μ = 1 2 k d1 + d1 + s(d1 + d1)2 − 4 |A| l . We have λ ≤ μ. Let ξ and ζ be eigenvectors of λ and μ, respectively. The general solution of (3.17) is ⎛ ⎜⎝ x(t) y(t) ⎞ ⎟⎠ = ⎛ ⎜⎝ x¯ y¯ ⎞ ⎟⎠ + c1ξeλt + c2ζeμt . (3.18) where c1 and c2 are arbitrary constants. Stability Analysis Saddle-Point Stable Path with |A| < 0. If |A| < 0, then λ < 0 and μ > 0, implying that the system is saddle-point stable. Convergent sequences must have c2 = 0. 3—8