APM1514 Assignment 5 2025 - DUE 15 July 2025

APM1514
ASSIGNMENT 5 2025

UNIQUE NO.
DUE DATE: 15 JULY 2025

, APM1514 Assignment 5 2025
Due Date: 15 July 2025

Question 1 (10 Marks)

1.1 Determine all the equilibrium points of the differential equation:
dxdt=5x2−1−2x2−1−3\frac{dx}{dt} = \frac{5}{\sqrt{x^2 - 1}} - \frac{2}{\sqrt{x^2 - 1}} - 3

Simplify:
dxdt=3x2−1−3\frac{dx}{dt} = \frac{3}{\sqrt{x^2 - 1}} - 3

Set dxdt=0\frac{dx}{dt} = 0:
3x2−1−3=0⇒3x2−1=3⇒x2−1=1⇒x2−1=1⇒x2=2⇒x=±2\frac{3}{\sqrt{x^2 - 1}} - 3 = 0
\Rightarrow \frac{3}{\sqrt{x^2 - 1}} = 3 \Rightarrow \sqrt{x^2 - 1} = 1 \Rightarrow x^2 - 1 =
1 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt{2}

Equilibrium points: x=±2x = \pm\sqrt{2}




Question 2 (30 Marks)

2.1 dxdt=e2ln⁡ x+5eln⁡ x−36\frac{dx}{dt} = e^{2 \ln x} + 5e^{\ln x} - 36

Use identities:
e2ln⁡ x=x2,eln⁡ x=x⇒x2+5x−36=0e^{2 \ln x} = x^2, \quad e^{\ln x} = x \Rightarrow x^2 +
5x - 36 = 0

Factor:
(x+9)(x−4)=0⇒x=−9,x=4(x + 9)(x - 4) = 0 \Rightarrow x = -9, x = 4

x=4x = 4 is valid (domain x>0x > 0), x=−9x = -9 is invalid.