Problems 18 through 25 deal with the predator–prey system

for which a bifurcation occurs at the value ∈ = 0 of the parameter ∈ . Problems 18 and 19 deal with the case ∈ = 0, in which case the system in (6) takes the form

and these problems suggest that the two critical points (0,0) and (5,2) of the system in (7) are as shown in Fig. 9.3.16- a saddle point at the origin and a center at (5,2). In each problem use a graphing calculator or computer system to construct a phase plane portrait for the linearization at the indicated critical point. Do your local portraits look consistent with Fig. 9.3.16?

Show that the coefficient matrix of the linearization x' = 2x, y' = -5y of (7) at (0,0) has the positive eigenvalue λ₁ = 2 and the negative eigenvalue λ₂ = -5. Hence (0,0) is a saddle point for the system in (7).

Transcribed Image Text:

dx dt = 2x - xy + ex(5-x), dy = −5y+xy, dt (6)