University Calculus: Differential Equations and Phase Plane Analysis

This assignment solution provides a detailed analysis of several differential equation problems. The first part explores the stability of equilibrium points using the Jacobian method, identifying an unstable node. The second part analyzes a non-linear system, determining the nature of its critical points and their stability, including the identification of an attractive focus and a spiral sink. The third part examines another system, computing the Jacobian at a critical point to determine it's an unstable saddle point and then analyzing the stability of the origin and a circle in polar form. The solutions involve finding eigenvalues, characteristic equations, and interpreting phase portraits to determine the behavior of solutions over time.

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.
SOLUTION TWO
X1(t)=x-xy2
Y1(t)=-x+2y
Let F= x-xy2 and G=-x+2y
Critical point are given by F=0 and G=0
x(1-y2)=0
-x+2y=0
X=0,0
Y=±1
The equilibrium point are (0,1) and (0,-1)
J=Δ f(x)= d/dx(x-xy2) d/dy(x-xy2) = 1-y2 -2xy
d/dx(-x+2y) d/dy(-x+2y) -1 2
At equilibrium point Δf(0,1)= 0 0
-1 2
The characteristic equation
- α 0
-1 2- α
(-α)(2- α)+1=0
-2 α+ α2+1=0
α2-2 α+1=0 Factors (-1,-1)

α2- α- α+1=0
α(α-1)-( α-1)=0
(α-1)( α-1)=0
(α-1)2=0 Type=Equal and Stability=Node
At equilibrium point Δf(0,-1)= 0 0
-1 2
The characteristic equation
- α 0
-1 2- α
(-α)(2- α)+1=0
-2 α+ α2+1=0
α 2-2 α+1=0 Factors (-1,-1)
α2- α- α+1=0
α(α-1)-( α-1)=0
(α-1)( α-1)=0
(α-1)2=0 Type=Equal and Stability=Node
By inspection, the type and stability at both (0, 1) and (0,-1) are equal.
SOLUTION (3)
We Compute the Jacobian at (0, 0) and determine the stability at this point
as follows;

(a)
J=
| ∂
∂ x ( x ' ) ∂
∂ y ( x ' )
∂
∂ x ( y ' ) ∂
∂ y ( y ' )|
=
| ∂
∂ x (− y + x2 + x y2 ) ∂
∂ y (− y+ x3 +x xy2 )
∂
∂ x (x + y x2 + y2 ) ∂
∂ y ( x + y x2 + x3) |
=|3 x2+ y2 −1+2 xy
1+2 yx x2 +3 y2 | (x, y) = (0,0)
=|0 −1
1 0 | where λ1=1 , λ2=-1
Unstable Saddle Point.
(b) Given the polar form of the ODE after solving its derivative is
R1=r3
Φ1=1
The origin r=0 is unstable spiral point and the circle r=1 is a stable since we
already know the phase plot analysis that it is.
(c) The Jacobian form of the ODE and the polar form of the ODE differ along
the stability points as analysed before.