UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
ASSIGNMENT 01
Year Module – 2026
⋄
Module Code: APM3706
Module Name: Ordinary Differential Equations
Assignment No.: 01
Due Date: 2026
Semester: Year Module 2026
Submitted in partial fulfilment of the requirements for
Ordinary Differential Equations (APM3706)
at the University of South Africa.
,UNISA | APM3706 Assignment 01 – ODEs
Question 1
Question 1.1 – Degeneracy Analysis
Question: Determine whether the system
ẋ + ẏ + y = et , ẍ + ÿ + ẏ = et
is degenerate. In the degenerate case, decide whether it has no solution or infinitely many
solutions. If it has no solution explain why, else find the general form of the solutions.
Solution:
Step 1: Write the system in operator notation.
d
Let D = dt . The system becomes
D[x] + D[y] + y = et (1)
D2 [x] + D2 [y] + D[y] = et (2)
Step 2: Differentiate equation (1).
Applying D to (1):
D2 [x] + D2 [y] + D[y] = et
This is exactly equation (2). Implying that equation (2) carries no new information; the two
equations are dependent.
Step 3: Conclude degeneracy.
Since the second equation is simply the derivative of the first, the system is degenerate. The
determinant of the operator matrix is zero, confirming linear dependence.
Step 4: Solve the single independent equation.
We work with equation (1):
D[x] + D[y] + y = et
which can be rewritten as
D[x] = et − D[y] − y
Page 1 of 17
, UNISA | APM3706 Assignment 01 – ODEs
Step 5: Introduce a free function.
Because only one constraint exists for two unknowns, we may choose y(t) freely. Let y(t) be
any differentiable function. Then
ẋ = et − ẏ − y
Integrating both sides with respect to t:
Z
et − ẏ − y dt = et − y(t) + c,
x(t) = c∈R
Step 6: State the general solution.
The system is degenerate and has infinitely many solutions. The general form is
x(t) = et − y(t) + c,
where y(t) is an arbitrary differentiable function and c ∈ R is an arbitrary constant.
Page 2 of 17
College of Science, Engineering and Technology
⋄
ASSIGNMENT 01
Year Module – 2026
⋄
Module Code: APM3706
Module Name: Ordinary Differential Equations
Assignment No.: 01
Due Date: 2026
Semester: Year Module 2026
Submitted in partial fulfilment of the requirements for
Ordinary Differential Equations (APM3706)
at the University of South Africa.
,UNISA | APM3706 Assignment 01 – ODEs
Question 1
Question 1.1 – Degeneracy Analysis
Question: Determine whether the system
ẋ + ẏ + y = et , ẍ + ÿ + ẏ = et
is degenerate. In the degenerate case, decide whether it has no solution or infinitely many
solutions. If it has no solution explain why, else find the general form of the solutions.
Solution:
Step 1: Write the system in operator notation.
d
Let D = dt . The system becomes
D[x] + D[y] + y = et (1)
D2 [x] + D2 [y] + D[y] = et (2)
Step 2: Differentiate equation (1).
Applying D to (1):
D2 [x] + D2 [y] + D[y] = et
This is exactly equation (2). Implying that equation (2) carries no new information; the two
equations are dependent.
Step 3: Conclude degeneracy.
Since the second equation is simply the derivative of the first, the system is degenerate. The
determinant of the operator matrix is zero, confirming linear dependence.
Step 4: Solve the single independent equation.
We work with equation (1):
D[x] + D[y] + y = et
which can be rewritten as
D[x] = et − D[y] − y
Page 1 of 17
, UNISA | APM3706 Assignment 01 – ODEs
Step 5: Introduce a free function.
Because only one constraint exists for two unknowns, we may choose y(t) freely. Let y(t) be
any differentiable function. Then
ẋ = et − ẏ − y
Integrating both sides with respect to t:
Z
et − ẏ − y dt = et − y(t) + c,
x(t) = c∈R
Step 6: State the general solution.
The system is degenerate and has infinitely many solutions. The general form is
x(t) = et − y(t) + c,
where y(t) is an arbitrary differentiable function and c ∈ R is an arbitrary constant.
Page 2 of 17
