On HW3, I couldn't find the eqm. points besides (0,0) and (77,95) in the predator prey problem - what are they and how do we find them?

Shreesh P. Mysore, Oct 28 02

Denote the right hand sides of the equation by f₁(x₁,x₂) and f₂(x₁,x₂) respectively. (x₁,x₂) in R² is an eqm point if f₁(x₁,x₂)=0 and f₂(x₁,x₂)=0.
(1) f₁(x₁,x₂)=0 => x₁=0 or ax₂+bx₁=br.
(2) f₂(x₁,x₂)=0 => x₂=0 or ax₁-bx₂=d_f.
Since we want both (1) and (2) to be satisfied, this can happen in any one of 4 ways (each condition from (1) can be paired with any one of 2 conditions from (2)). Therefore the four possible solutions are
(a) x₁=0 and x₂=0 (or)
(b) x₁=0 and ax₁-bx₂=d_f (or)
(c) ax₂+bx₁=br and x₂=0 (or)
(d) ax₂+bx₁=br and ax₁-bx₂=d_f
You found solutions (a) and (d).