APM2611 Assignment 2 solution 2026
UNISA
APM2611 ASSIGNMENT 02 FULL COMPLETE SOLUTIONS
Due date: Wednesday, 17 June 2026
ONLY FOR YEAR MODULE
HIGHER-ORDER DIFFERENTIAL EQUATIONS, INITIAL-VALUE PROBLEMS,
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
, QUESTION 1
We use substitution:
𝑑𝑢
𝑢 = 𝑦 ′ ⇒ 𝑦 ′′ =
𝑑𝑥
1.1
𝑦 ′′ + (𝑦 ′ )2 + 1 = 0
Substitute:
𝑑𝑢
+ 𝑢2 + 1 = 0
𝑑𝑥
Separate variables:
𝑑𝑢
= −𝑑𝑥
𝑢2 +1
Integrate:
arctan(𝑢) = −𝑥 + 𝐶
Substitute back 𝑢 = 𝑦 ′ :
arctan(𝑦 ′ ) = −𝑥 + 𝐶
𝑦 ′ = tan(−𝑥 + 𝐶)
Integrate:
𝑦 = ∫ tan(𝐶 − 𝑥) 𝑑𝑥
Let 𝑘 = 𝐶 − 𝑥, then:
𝑦 = −ln ∣ cos(𝐶 − 𝑥) ∣ +𝐶1
Final answer:
𝑦 = −ln ∣ cos(𝐶 − 𝑥) ∣ +𝐶1
UNISA
APM2611 ASSIGNMENT 02 FULL COMPLETE SOLUTIONS
Due date: Wednesday, 17 June 2026
ONLY FOR YEAR MODULE
HIGHER-ORDER DIFFERENTIAL EQUATIONS, INITIAL-VALUE PROBLEMS,
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
, QUESTION 1
We use substitution:
𝑑𝑢
𝑢 = 𝑦 ′ ⇒ 𝑦 ′′ =
𝑑𝑥
1.1
𝑦 ′′ + (𝑦 ′ )2 + 1 = 0
Substitute:
𝑑𝑢
+ 𝑢2 + 1 = 0
𝑑𝑥
Separate variables:
𝑑𝑢
= −𝑑𝑥
𝑢2 +1
Integrate:
arctan(𝑢) = −𝑥 + 𝐶
Substitute back 𝑢 = 𝑦 ′ :
arctan(𝑦 ′ ) = −𝑥 + 𝐶
𝑦 ′ = tan(−𝑥 + 𝐶)
Integrate:
𝑦 = ∫ tan(𝐶 − 𝑥) 𝑑𝑥
Let 𝑘 = 𝐶 − 𝑥, then:
𝑦 = −ln ∣ cos(𝐶 − 𝑥) ∣ +𝐶1
Final answer:
𝑦 = −ln ∣ cos(𝐶 − 𝑥) ∣ +𝐶1
