APM1514 ASSIGNMENT 07 2025 Due 15 AUGUST 2025

APM1514
Assignment 7
Due 15 August 2025

, APM1514 Assignment 7 — Due: Friday, 15 August 2025


APM1514 Assignment 7
Due Date: Friday, 15 August 2025


Question 1
We are given the system:
dx dy
= x + 9, = − xy
dt dt

Equilibrium Points
Equilibria occur when:
x+9 = 0 and − xy = 0

From x + 9 = 0 we get x = −9. From − xy = 0 we have x = 0 or y = 0. Combining,
the only common solution is:
E = (−9, 0)


Stability Analysis
Let f ( x, y) = x + 9, g( x, y) = − xy. The Jacobian is:
" #
1 0
J ( x, y) =
−y − x

At (−9, 0): " #
1 0
J (−9, 0) =
0 9
Eigenvalues: λ1 = 1, λ2 = 9 (both positive) =⇒ unstable node.


Isoclines
• dx/dt = 0 =⇒ x = −9 (vertical line)

• dy/dt = 0 =⇒ x = 0 or y = 0 (y-axis and x-axis)




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, APM1514 Assignment 7 — Due: Friday, 15 August 2025


Direction of Motion
Vertical: dy/dt = − xy

• x > 0, y > 0: down

• x > 0, y < 0: up

• x < 0, y > 0: up

• x < 0, y < 0: down

Horizontal: dx/dt = x + 9

• x > −9: right

• x < −9: left


Phase Diagram
x = −9 y 0
x=




yx = 0
E(−9, 0)




Outcome from Various Initial Points
- All trajectories move away from (−9, 0) since it is an unstable node. - In the right
half-plane (x > −9), motion is to the right; in the left half-plane (x < −9), motion is to
the left. - Above the x-axis: - Left of x = 0: y increases. - Right of x = 0: y decreases. -
Below the x-axis: - Left of x = 0: y decreases. - Right of x = 0: y increases.


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