APM1514 Assignment 5 (Comprehensive Solutions) Due 15 July 2025

APM1514
Assignment 05

Due 15 July 2025

, APM1514 Assignment 05: Due 15 July 2025



Question 1 (10 Marks)
Problem Statement:
Determine all the equilibrium point(s) of the following differential equation using an analytical approach:
dx 5 p
=√ − 2 x2 − 1 − 3
dt x2 − 1

Step 1: Define equilibrium points
An equilibrium point occurs when:
dx
=0
dt
Set the right-hand side equal to zero:
5 p
√ − 2 x2 − 1 − 3 = 0
x2 −1

Step 2: Substitution
Let: p
u= x2 − 1 (note that this is defined only for x2 > 1)
Then the equation becomes:
5
− 2u − 3 = 0
u

Step 3: Multiply through by u
Multiply both sides by u (where u > 0):

5 − 2u2 − 3u = 0
Rewriting:

−2u2 − 3u + 5 = 0
Multiply both sides by −1:

2u2 + 3u − 5 = 0




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, Step 4: Solve the quadratic equation
Using the quadratic formula:
√
−b ± b2 − 4ac
u=
2a
where a = 2, b = 3, c = −5:
p √ √
−3 ± 32 − 4(2)(−5) −3 ± 9 + 40 −3 ± 49
u= = =
2(2) 4 4
−3 ± 7
u=
4
So the solutions are:
−3 + 7 −3 − 7
u1 = = 1, u2 = = −2.5
4 4
√
Since u = x2 − 1 > 0, we discard u = −2.5. Thus, the valid solution is:

u=1


Step 5: Solve for x
Recall that:
p p √
u= x2 − 1 ⇒ 1 = x2 − 1 ⇒ x2 − 1 = 1 ⇒ x2 = 2 ⇒ x = ± 2


Final Answer:
√ √
x= 2 and x = − 2
These are the equilibrium points of the differential equation.


Question 2.1 (10 Marks)
Problem Statement:
Draw the phase line of the following differential equation. List all equilibrium points and classify them
as stable or unstable:
dx
= e2 ln(x) + 5eln(x) − 36
dt

Step 1: Simplify the expression
Use logarithmic and exponential identities:

e2 ln(x) = x2 , eln(x) = x
Thus, the equation becomes:
dx
= x2 + 5x − 36
dt




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