APM1514
Assignment 05
Due 15 July 2025
, APM1513 Assignment 5 Due 15 July 2025
July 10, 2025
Question 1: 10 Marks
By using an analytical approach (do not use inspection as marks will be de-
ducted), determine all the equilibrium point(s) of the following differential equa-
tion:
dx p p
= 5( x2 − 1)−1 − 2( x2 − 1) − 3.
dt
Step 1: Define Equilibrium Points
Equilibrium points occur where the rate of change of x with respect to t is zero,
i.e., dx
dt = 0. To find these points, set the right-hand side of the differential
equation equal to zero and solve for x.
p p
5( x2 − 1)−1 − 2( x2 − 1) − 3 = 0
Step 2: Simplify the Equation
√
To make the equation easier to handle, substitute u = x2 − 1, where u ≥ 0
since it is a square root. Then the equation becomes:
5u−1 − 2u − 3 = 0
Multiply through by u (noting u ̸= 0) to clear the denominator:
5 − 2u2 − 3u = 0
Rearrange into a standard quadratic equation:
−2u2 − 3u + 5 = 0
Multiply by −1 to make the leading coefficient positive:
2u2 + 3u − 5 = 0
1
, Step 3: Solve the Quadratic Equation
√
−b± b2 −4ac
Use the quadratic formula u = 2a , where a = 2, b = 3, and c = −5:
Discriminant = b2 − 4ac = 32 − 4 · 2 · (−5) = 9 + 40 = 49
√
−3 ± 49 −3 ± 7
u= =
2·2 4
Calculate the two solutions:
• u= −3+7
4 = 4
4 =1
• u= = −10
−3−7
44 = −2.5
√
Since u = x2 − 1 ≥ 0, discard the negative solution u = −2.5. Thus,
u = 1.
Step 4: Back-Substitute to Find x
√
Set u = x2 − 1 = 1: p
x2 − 1 = 1
Square both sides:
x2 − 1 = 1
x2 = 2
√
x=± 2
√ √
So, the equilibrium points are x = 2 and x = − 2.
Step 5: Classify Stability (Analytical Approach)
To classify the stability, consider the behavior of dx
dt near the equilibrium points.
Compute the derivative of the right-hand side with respect
√ to x (i.e.,√the Jaco-
bian or rate of change of the function). Let f (x) = 5( x2 − 1)−1 −2( x2 − 1)−
3. √ df
First, let u = x2 − 1, so f (x) = 5/u − 2u − 3. The derivative dx requires
the chain rule:
df d 5 5
= − 2u − 3 = − 2 − 2
du du u u
du √ x
Now, dx = (using the chain rule on u = (x2 − 1)1/2 ):
x2 −1
df df du 5 x
= · = − 2 −2 · √
dx du dx u 2
x −1
√
Substitute u = 1 at the equilibrium points x = ± 2:
√ √
√
df 5 2 2
√
= − 2 −2 · √ = (−5 − 2) · = −7 · 2
dx x= 2 1 22 − 1 1
2
Assignment 05
Due 15 July 2025
, APM1513 Assignment 5 Due 15 July 2025
July 10, 2025
Question 1: 10 Marks
By using an analytical approach (do not use inspection as marks will be de-
ducted), determine all the equilibrium point(s) of the following differential equa-
tion:
dx p p
= 5( x2 − 1)−1 − 2( x2 − 1) − 3.
dt
Step 1: Define Equilibrium Points
Equilibrium points occur where the rate of change of x with respect to t is zero,
i.e., dx
dt = 0. To find these points, set the right-hand side of the differential
equation equal to zero and solve for x.
p p
5( x2 − 1)−1 − 2( x2 − 1) − 3 = 0
Step 2: Simplify the Equation
√
To make the equation easier to handle, substitute u = x2 − 1, where u ≥ 0
since it is a square root. Then the equation becomes:
5u−1 − 2u − 3 = 0
Multiply through by u (noting u ̸= 0) to clear the denominator:
5 − 2u2 − 3u = 0
Rearrange into a standard quadratic equation:
−2u2 − 3u + 5 = 0
Multiply by −1 to make the leading coefficient positive:
2u2 + 3u − 5 = 0
1
, Step 3: Solve the Quadratic Equation
√
−b± b2 −4ac
Use the quadratic formula u = 2a , where a = 2, b = 3, and c = −5:
Discriminant = b2 − 4ac = 32 − 4 · 2 · (−5) = 9 + 40 = 49
√
−3 ± 49 −3 ± 7
u= =
2·2 4
Calculate the two solutions:
• u= −3+7
4 = 4
4 =1
• u= = −10
−3−7
44 = −2.5
√
Since u = x2 − 1 ≥ 0, discard the negative solution u = −2.5. Thus,
u = 1.
Step 4: Back-Substitute to Find x
√
Set u = x2 − 1 = 1: p
x2 − 1 = 1
Square both sides:
x2 − 1 = 1
x2 = 2
√
x=± 2
√ √
So, the equilibrium points are x = 2 and x = − 2.
Step 5: Classify Stability (Analytical Approach)
To classify the stability, consider the behavior of dx
dt near the equilibrium points.
Compute the derivative of the right-hand side with respect
√ to x (i.e.,√the Jaco-
bian or rate of change of the function). Let f (x) = 5( x2 − 1)−1 −2( x2 − 1)−
3. √ df
First, let u = x2 − 1, so f (x) = 5/u − 2u − 3. The derivative dx requires
the chain rule:
df d 5 5
= − 2u − 3 = − 2 − 2
du du u u
du √ x
Now, dx = (using the chain rule on u = (x2 − 1)1/2 ):
x2 −1
df df du 5 x
= · = − 2 −2 · √
dx du dx u 2
x −1
√
Substitute u = 1 at the equilibrium points x = ± 2:
√ √
√
df 5 2 2
√
= − 2 −2 · √ = (−5 − 2) · = −7 · 2
dx x= 2 1 22 − 1 1
2
