UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3706: Ordinary Differential Equations
Assignment 01 — 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3706
Module Code:
Ordinary Differential Equations
Module Name:
Assignment 01 (Study Guide: Chapter 1)
Assignment:
2026
Due Date:
Submitted in partial fulfilment of the requirements for APM3706 — UNISA 2026
,UNISA | APM3706 Assignment 01 – ODE Systems
Question 1
Question 1.1
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 so-
lutions. If it has no solution, explain why; else find the general form of the solutions.
Solution:
Step 1: Write the system in operator notation.
Using D = d/dt, the system becomes:
(D) x + (D + 1) y = et , (1)
(D2 ) x + (D2 + D) y = et . (2)
Step 2: Differentiate equation (1) to relate it to equation (2).
Apply D to equation (1):
D D x + (D + 1) y = D et
D2 x + (D2 + D) y = et .
This is exactly equation (2). Therefore equation (2) is not independent of equation (1); it fol-
lows directly from differentiating the first equation.
Step 3: Determine whether there are no solutions or infinitely many.
Since the second equation carries no new information, the entire system reduces to a single
equation:
ẋ + ẏ + y = et .
This is one equation in two unknown functions x(t) and y(t). One equation cannot uniquely
determine two unknown functions, so there are infinitely many solutions.
Page 2 of 19
, UNISA | APM3706 Assignment 01 – ODE Systems
Step 4: Find the general form of the solutions.
We can choose x(t) to be any differentiable function, say x(t) = ϕ(t) (an arbitrary function).
Then, from the single equation:
ẏ + y = et − ϕ̇(t).
This is a first-order linear ODE in y. The integrating factor is µ = et :
d t
e y = et et − ϕ̇(t) = e2t − et ϕ̇(t).
dt
Integrating both sides:
e2t
Z
et y = − et ϕ̇(t) dt + C.
2
Therefore:
et
Z
y(t) = − e−t et ϕ̇(t) dt + Ce−t .
2
Result
The general solution is the family of pairs
et
Z
− e−t et ϕ̇(t) dt + Ce−t ,
x(t), y(t) = ϕ(t),
2
where ϕ(t) is any differentiable function and C ∈ R is an arbitrary constant. The sys-
tem has infinitely many solutions.
Page 3 of 19
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3706: Ordinary Differential Equations
Assignment 01 — 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3706
Module Code:
Ordinary Differential Equations
Module Name:
Assignment 01 (Study Guide: Chapter 1)
Assignment:
2026
Due Date:
Submitted in partial fulfilment of the requirements for APM3706 — UNISA 2026
,UNISA | APM3706 Assignment 01 – ODE Systems
Question 1
Question 1.1
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 so-
lutions. If it has no solution, explain why; else find the general form of the solutions.
Solution:
Step 1: Write the system in operator notation.
Using D = d/dt, the system becomes:
(D) x + (D + 1) y = et , (1)
(D2 ) x + (D2 + D) y = et . (2)
Step 2: Differentiate equation (1) to relate it to equation (2).
Apply D to equation (1):
D D x + (D + 1) y = D et
D2 x + (D2 + D) y = et .
This is exactly equation (2). Therefore equation (2) is not independent of equation (1); it fol-
lows directly from differentiating the first equation.
Step 3: Determine whether there are no solutions or infinitely many.
Since the second equation carries no new information, the entire system reduces to a single
equation:
ẋ + ẏ + y = et .
This is one equation in two unknown functions x(t) and y(t). One equation cannot uniquely
determine two unknown functions, so there are infinitely many solutions.
Page 2 of 19
, UNISA | APM3706 Assignment 01 – ODE Systems
Step 4: Find the general form of the solutions.
We can choose x(t) to be any differentiable function, say x(t) = ϕ(t) (an arbitrary function).
Then, from the single equation:
ẏ + y = et − ϕ̇(t).
This is a first-order linear ODE in y. The integrating factor is µ = et :
d t
e y = et et − ϕ̇(t) = e2t − et ϕ̇(t).
dt
Integrating both sides:
e2t
Z
et y = − et ϕ̇(t) dt + C.
2
Therefore:
et
Z
y(t) = − e−t et ϕ̇(t) dt + Ce−t .
2
Result
The general solution is the family of pairs
et
Z
− e−t et ϕ̇(t) dt + Ce−t ,
x(t), y(t) = ϕ(t),
2
where ϕ(t) is any differentiable function and C ∈ R is an arbitrary constant. The sys-
tem has infinitely many solutions.
Page 3 of 19
