APM3706
ASSIGNMENT 1
Ordinary Differential Equations
FULL SOLUTIONS
COMPLETE SOLUTIONS
MEMORANDUM
UNISA 2026
Page 1 of 13
, QUESTION 1
1.1 Determine whether the system is degenerate
𝑥̇ + 𝑦̇ + 𝑦 = 𝑒 𝑡
{
𝑥̈ + 𝑦̈ + 𝑦̇ = 𝑒 𝑡
𝑑
Let 𝐷 = 𝑑𝑡. The system becomes:
𝐷𝑥 + (𝐷 + 1)𝑦 = 𝑒 𝑡
{
𝐷2 𝑥 + (𝐷2 + 𝐷)𝑦 = 𝑒 𝑡
𝐷 𝐷+1
The operator matrix is [ 2 ]. Its determinant is:
𝐷 𝐷2 + 𝐷
Δ = 𝐷(𝐷2 + 𝐷) − 𝐷2 (𝐷 + 1) = 𝐷3 + 𝐷2 − 𝐷3 − 𝐷2 = 0
Since Δ = 0, the system is degenerate.
From the first equation: 𝐷𝑥 = 𝑒 𝑡 − (𝐷 + 1)𝑦. Differentiating: 𝐷2 𝑥 = 𝑒 𝑡 − 𝐷(𝐷 + 1)𝑦 =
𝑒 𝑡 − (𝐷2 + 𝐷)𝑦. Substituting into the second equation:
[𝑒 𝑡 − (𝐷2 + 𝐷)𝑦] + (𝐷2 + 𝐷)𝑦 = 𝑒 𝑡 ⇒ 𝑒 𝑡 = 𝑒 𝑡
The system is consistent, therefore it has infinitely many solutions.
From the first equation: 𝐷𝑥 = 𝑒 𝑡 − 𝑦 ′ − 𝑦. Integrating with respect to 𝑡:
𝑥 = ∫ 𝑒 𝑡 𝑑𝑡 − ∫ 𝑦 ′ 𝑑𝑡 − ∫ 𝑦 𝑑𝑡 + 𝐶 = 𝑒 𝑡 − 𝑦 − ∫ 𝑦 𝑑𝑡 + 𝐶
Let 𝑌(𝑡) = ∫ 𝑦 𝑑𝑡, so 𝑌 ′ = 𝑦. Then 𝑥 = 𝑒 𝑡 − 𝑌 ′ − 𝑌 + 𝐶.
𝑦(𝑡) is arbitrary, 𝑥(𝑡) = 𝑒 𝑡 − 𝑦(𝑡) − ∫ 𝑦(𝑡) 𝑑𝑡 + 𝐶
where 𝐶 is an arbitrary constant.
Page 2 of 13
ASSIGNMENT 1
Ordinary Differential Equations
FULL SOLUTIONS
COMPLETE SOLUTIONS
MEMORANDUM
UNISA 2026
Page 1 of 13
, QUESTION 1
1.1 Determine whether the system is degenerate
𝑥̇ + 𝑦̇ + 𝑦 = 𝑒 𝑡
{
𝑥̈ + 𝑦̈ + 𝑦̇ = 𝑒 𝑡
𝑑
Let 𝐷 = 𝑑𝑡. The system becomes:
𝐷𝑥 + (𝐷 + 1)𝑦 = 𝑒 𝑡
{
𝐷2 𝑥 + (𝐷2 + 𝐷)𝑦 = 𝑒 𝑡
𝐷 𝐷+1
The operator matrix is [ 2 ]. Its determinant is:
𝐷 𝐷2 + 𝐷
Δ = 𝐷(𝐷2 + 𝐷) − 𝐷2 (𝐷 + 1) = 𝐷3 + 𝐷2 − 𝐷3 − 𝐷2 = 0
Since Δ = 0, the system is degenerate.
From the first equation: 𝐷𝑥 = 𝑒 𝑡 − (𝐷 + 1)𝑦. Differentiating: 𝐷2 𝑥 = 𝑒 𝑡 − 𝐷(𝐷 + 1)𝑦 =
𝑒 𝑡 − (𝐷2 + 𝐷)𝑦. Substituting into the second equation:
[𝑒 𝑡 − (𝐷2 + 𝐷)𝑦] + (𝐷2 + 𝐷)𝑦 = 𝑒 𝑡 ⇒ 𝑒 𝑡 = 𝑒 𝑡
The system is consistent, therefore it has infinitely many solutions.
From the first equation: 𝐷𝑥 = 𝑒 𝑡 − 𝑦 ′ − 𝑦. Integrating with respect to 𝑡:
𝑥 = ∫ 𝑒 𝑡 𝑑𝑡 − ∫ 𝑦 ′ 𝑑𝑡 − ∫ 𝑦 𝑑𝑡 + 𝐶 = 𝑒 𝑡 − 𝑦 − ∫ 𝑦 𝑑𝑡 + 𝐶
Let 𝑌(𝑡) = ∫ 𝑦 𝑑𝑡, so 𝑌 ′ = 𝑦. Then 𝑥 = 𝑒 𝑡 − 𝑌 ′ − 𝑌 + 𝐶.
𝑦(𝑡) is arbitrary, 𝑥(𝑡) = 𝑒 𝑡 − 𝑦(𝑡) − ∫ 𝑦(𝑡) 𝑑𝑡 + 𝐶
where 𝐶 is an arbitrary constant.
Page 2 of 13