APM1514 Assignment 7 2025 (Exceptionally Crafted) Due 15 August 2025

APM1514
Assignment 7
Due 15 August 2025

,Student Name: APM1514 Assignment 7


APM1514 Assignment 7
Due Date: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)


Direction of Motion
- Vertical direction: sign of dy/dt = −xy - If x > 0, y > 0: dy/dt < 0 (down) - If
x > 0, y < 0: dy/dt > 0 (up) - If x < 0, y > 0: dy/dt > 0 (up) - If x < 0, y < 0:
dy/dt < 0 (down) - Horizontal direction: sign of dx/dt = x + 9 - If x > −9: right - If
x < −9: left

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, Student Name: APM1514 Assignment 7


Phase Diagram
x = −9 y 0
x=




x
y=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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