UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
ASSIGNMENT 02
Year Module — 2026
⋄
Module Code: APM3706
Module Name: Ordinary Differential Equations
Assignment No.: 02
Semester: Year Module 2026
Submitted in partial fulfilment of the requirements for APM3706
at the University of South Africa.
,UNISA | APM3706 Assignment 02 — Ordinary Differential Equations
Question 1
1(i) — Exercise 2.4(b): Solve the ODE
Question: Solve the ordinary differential equation
dy y−x
= .
dx y+x
Solution:
The right-hand side depends only on the ratio y/x, so this is a homogeneous equation. Substi-
tute
y dy dv
v= =⇒ y = vx =⇒ =v+x .
x dx dx
The equation becomes:
dv vx − x v−1
v+x = = .
dx vx + x v+1
Isolating x dv/dx:
dv v−1 v − 1 − v(v + 1) v − 1 − v2 − v −1 − v 2
x = −v = = = .
dx v+1 v+1 v+1 v+1
Separating variables:
v+1 dx
dv = − .
v2 + 1 x
Integrating both sides:
Z Z
v+1 dx
dv = − .
v2 + 1 x
Split the left-hand integral:
Z Z
v 1
2
dv + dv = − ln |x| + C.
v +1 v2 +1
1
ln(v 2 + 1) + arctan(v) = − ln |x| + C.
2
Page 1 of 18
, UNISA | APM3706 Assignment 02 — Ordinary Differential Equations
Re-substitute v = y/x:
2
1 y y
ln 2 + 1 + arctan = − ln |x| + C.
2 x x
2
y + x2
1 y
ln + arctan = − ln |x| + C.
2 x2 x
1 y
ln(x2 + y 2 ) − ln(x2 ) + arctan
= − ln |x| + C.
2 x
1 y
ln(x2 + y 2 ) − ln |x| + arctan = − ln |x| + C.
2 x
The − ln |x| terms cancel:
1 y
ln(x2 + y 2 ) + arctan = C.
2 x
1(ii) — Exercise 2.32(c): Solve the ODE
Question: Solve the differential equation
(2xy − 3y 2 ) dx − (2xy − x2 ) dy = 0.
Solution:
Write M = 2xy − 3y 2 and N = −(2xy − x2 ) = x2 − 2xy.
Check exactness:
∂M ∂N
= 2x − 6y, = 2x − 2y.
∂y ∂x
Since ∂M/∂y ̸= ∂N/∂x, the equation is not exact. Look for an integrating factor.
Compute:
∂M/∂y − ∂N/∂x (2x − 6y) − (2x − 2y) −4y
= = .
N x2 − 2xy x(x − 2y)
This is not a function of x alone. Try:
∂N/∂x − ∂M/∂y (2x − 2y) − (2x − 6y) 4y 4
= 2
= = .
M 2xy − 3y y(2x − 3y) 2x − 3y
Page 2 of 18
College of Science, Engineering and Technology
⋄
ASSIGNMENT 02
Year Module — 2026
⋄
Module Code: APM3706
Module Name: Ordinary Differential Equations
Assignment No.: 02
Semester: Year Module 2026
Submitted in partial fulfilment of the requirements for APM3706
at the University of South Africa.
,UNISA | APM3706 Assignment 02 — Ordinary Differential Equations
Question 1
1(i) — Exercise 2.4(b): Solve the ODE
Question: Solve the ordinary differential equation
dy y−x
= .
dx y+x
Solution:
The right-hand side depends only on the ratio y/x, so this is a homogeneous equation. Substi-
tute
y dy dv
v= =⇒ y = vx =⇒ =v+x .
x dx dx
The equation becomes:
dv vx − x v−1
v+x = = .
dx vx + x v+1
Isolating x dv/dx:
dv v−1 v − 1 − v(v + 1) v − 1 − v2 − v −1 − v 2
x = −v = = = .
dx v+1 v+1 v+1 v+1
Separating variables:
v+1 dx
dv = − .
v2 + 1 x
Integrating both sides:
Z Z
v+1 dx
dv = − .
v2 + 1 x
Split the left-hand integral:
Z Z
v 1
2
dv + dv = − ln |x| + C.
v +1 v2 +1
1
ln(v 2 + 1) + arctan(v) = − ln |x| + C.
2
Page 1 of 18
, UNISA | APM3706 Assignment 02 — Ordinary Differential Equations
Re-substitute v = y/x:
2
1 y y
ln 2 + 1 + arctan = − ln |x| + C.
2 x x
2
y + x2
1 y
ln + arctan = − ln |x| + C.
2 x2 x
1 y
ln(x2 + y 2 ) − ln(x2 ) + arctan
= − ln |x| + C.
2 x
1 y
ln(x2 + y 2 ) − ln |x| + arctan = − ln |x| + C.
2 x
The − ln |x| terms cancel:
1 y
ln(x2 + y 2 ) + arctan = C.
2 x
1(ii) — Exercise 2.32(c): Solve the ODE
Question: Solve the differential equation
(2xy − 3y 2 ) dx − (2xy − x2 ) dy = 0.
Solution:
Write M = 2xy − 3y 2 and N = −(2xy − x2 ) = x2 − 2xy.
Check exactness:
∂M ∂N
= 2x − 6y, = 2x − 2y.
∂y ∂x
Since ∂M/∂y ̸= ∂N/∂x, the equation is not exact. Look for an integrating factor.
Compute:
∂M/∂y − ∂N/∂x (2x − 6y) − (2x − 2y) −4y
= = .
N x2 − 2xy x(x − 2y)
This is not a function of x alone. Try:
∂N/∂x − ∂M/∂y (2x − 2y) − (2x − 6y) 4y 4
= 2
= = .
M 2xy − 3y y(2x − 3y) 2x − 3y
Page 2 of 18
