The Jacobian Form Of The ODE

ASSIGNMENT ANSWERS.
SOLUTION ONE
(a) X1(1)= 4x-y
Y1(1)= 2x+y
Initial conditions x (0)=6 and y(0)=7
Unstable equilibrium α1> α2>0 (node)
(b) As t→ ∞ any solution with nonzero initial data approaches the limit cycle.
(c ) Solving
J= d/dx(4x-y) d/dy(4x-y)
d/dx(2x+y) d/dy(2x+y)
J= 4 -1
2 1
The characteristic equation is given by
4- α -1 =(4- α)(1- α)+2=0
2 1- α 4-4 α- α+ α2=0
α 2-5 α+6=0
α1=2
α2=3 (Unstable node)
QUESTION FOUR

X1=2x-y-x(3-x2-y2)
Y1=x+2y-y(3-x2-y2)
Using Jacobian Method
J= d/dx(2x-y-3x+x3+xy2) d/dy(2x-y-3x+x3+xy2)
d/dx(x+2y-3y+x2y+y3) d/dy(x+2y-3y+x2y+y3)
= -1+3x2+y2 -1+2xy
1+2xy -1+x2+3y2
= -1 -1
1 -1
The characteristic equation is
-1- α -1
1 -1- α =0
(-1- α)(-1- α)+1=0
1+ α+ α+ α2+1=0
α 2+2 α+2=0
Using quadratic expression
α 1,2=-2± 4-4(1)(2)
2
α 1,2=-1±i
Given that -1 is below 0, then it is an Attractive Focus given that the
sequence starting near the point will approach it.
The stability given that -1±i is a complex Eigen values.
It is spiral sink.