APM1514 Assignment 7 |COMPLETE ANSWERS & SOLUTIONS|- Due 15 August 2025

APM1514
Assignment 7

Due 15 August 2025

, ASSIGNMENT 07

Due date: Friday, 15 August 2025

Total Marks: 100

Question 1

We are given:

𝑑𝑥 𝑑𝑦
= 𝑥 + 9, = −𝑥𝑦
𝑑𝑡 𝑑𝑡

Step 1: Isoclines

• Vertical isoclines (dx/dt = 0): 𝑥 + 9 = 0 → 𝑥 = −9 Along this vertical line, there
is no horizontal movement.

• Horizontal isoclines (dy/dt = 0): −𝑥𝑦 = 0 → 𝑥 = 0 or 𝑦 = 0 These are the y-axis
(x = 0) and the x-axis (y = 0).

Step 2: Equilibrium points

Equilibrium points occur when both dx/dt = 0 and dy/dt = 0.

From 𝑥 + 9 = 0 → 𝑥 = −9 From −𝑥𝑦 = 0 → 𝑥 = 0 or 𝑦 = 0

The only common solution is: 𝑥 = −9 and 𝑦 = 0 → Equilibrium Point: (−9,0)

Step 3: Stability Analysis

We use the Jacobian matrix:

∂𝑓 ∂𝑓
∂𝑥 ∂𝑦
𝐽=
∂𝑔 ∂𝑔
[ ∂𝑥 ∂𝑦]

, Here:

• 𝑓(𝑥, 𝑦) = 𝑥 + 9 , 𝑔(𝑥, 𝑦) = −𝑥𝑦
∂𝑓 ∂𝑓
• = 1, =0
∂𝑥 ∂𝑦

∂𝑔 ∂𝑔
• = −𝑦, = −𝑥
∂𝑥 ∂𝑦


At (−9,0):

1 0
𝐽=[ ]
0 9

Eigenvalues: 𝜆1 = 1, 𝜆2 = 9 → both positive. This is an unstable node (source).

Step 4: Direction of motion in regions

We split the plane along x = -9, x = 0, and y = 0:

• If 𝑥 > −9 , dx/dt > 0 → motion to the right.

• If 𝑥 < −9 , dx/dt < 0 → motion to the left.

• If 𝑦 > 0 and 𝑥 > 0 , dy/dt < 0 → motion downward.

• If 𝑦 > 0 and 𝑥 < 0 , dy/dt > 0 → motion upward.

• If 𝑦 < 0 and 𝑥 > 0 , dy/dt > 0 → motion upward.

• If 𝑦 < 0 and 𝑥 < 0 , dy/dt < 0 → motion downward.