Question

I need help with this question of Differential Equation.

Thanks

Find and classify the equilibrium solutions and sketch the phase plane of the following system of ordinary differential equations. fx' = y(x - 1) y = x(y + 1)

Answer 1

x'= y(x-1) = +1849) y = x(4+1) - 91897) (say) given by , The equilibrium solutions are f=0 = y(x-1) = 0 → Y=0, x=1 → X(+1) = 0 if then, a(0+1)=0 x=0 If x=1 then 1.(+1)=0 →Y--1 . The equilibrium solutions are 10,0), (1,-1) af DY ox The jacobian is, I(Y) = 뾿 일 OX OY Y X-1 (y) = Y+1 x at 10,0), Ilona) 1:3] The eigen values are given by, -1 FO 0-

=+=0 X=Ii As eigen values are purely imagimary so so (0,0) is stable centre. O at, 11,-1), (1,-1) = 0 The eigen values are =-1, 1 as eigen values are (1,-1) is of opposite sign, saddle point . so a phase plane is given below: The

4.5 4 3.5 1 1 1 3 1 2.5) 1 1 2 1 ~ 1.5) 1 1 → / 1 0.5) → → 1 다 0 -0.5 -1 15 -21 -2.5 -3 -3.5 4 4.5 1 V - 5 1-4.5 -4 -3.5 -3 -2.5 -2 -1.5 1 -0.5 0 105 1.5 2 2.5 3 35 4 4.5