UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3711: Numerical Methods II
Assignment 01 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3711
Module Code:
Numerical Methods II
Module Name:
Assignment 01
Assignment Number:
30 April 2026
Due Date:
36
Total Marks:
Submitted in partial fulfilment of the requirements for APM3711 — UNISA 2026
, UNISA | APM3711 Numerical Methods II — Assignment 01
Question 1: Modified Euler (Predictor–Corrector) Method [14 marks]
Question: Consider the initial value problem
y ′ = −2xy, y(0) = 1.
2
The exact solution is y(x) = e−x . Use the Modified Euler (predictor–corrector) method with
step size h = 0.25 to approximate the solution up to x = 0.5. At each step perform two
correction iterations.
1(a): Predictor and Corrector Formulas
Question (a): Write down the predictor formula and the corrector formula used in the Modi-
fied Euler method.
Answer:
For y ′ = f (x, y), the Predictor (Euler step) is:
(0)
yn+1 = yn + h f (xn , yn )
The Corrector (iterated trapezoidal rule) is:
(k+1) hh (k)
i
yn+1 = yn + f xn , yn + f xn+1 , yn+1 , k = 0, 1, . . .
2
where f (x, y) = −2xy for this problem.
1(b): Numerical Approximation at x = 0.25 and x = 0.5
Question (b): Compute the approximate numerical values y(0.25) and y(0.5) using the Mod-
ified Euler method.
Answer:
Given: f (x, y) = −2xy, h = 0.25, x0 = 0, y0 = 1.
Page 2 of 14
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3711: Numerical Methods II
Assignment 01 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3711
Module Code:
Numerical Methods II
Module Name:
Assignment 01
Assignment Number:
30 April 2026
Due Date:
36
Total Marks:
Submitted in partial fulfilment of the requirements for APM3711 — UNISA 2026
, UNISA | APM3711 Numerical Methods II — Assignment 01
Question 1: Modified Euler (Predictor–Corrector) Method [14 marks]
Question: Consider the initial value problem
y ′ = −2xy, y(0) = 1.
2
The exact solution is y(x) = e−x . Use the Modified Euler (predictor–corrector) method with
step size h = 0.25 to approximate the solution up to x = 0.5. At each step perform two
correction iterations.
1(a): Predictor and Corrector Formulas
Question (a): Write down the predictor formula and the corrector formula used in the Modi-
fied Euler method.
Answer:
For y ′ = f (x, y), the Predictor (Euler step) is:
(0)
yn+1 = yn + h f (xn , yn )
The Corrector (iterated trapezoidal rule) is:
(k+1) hh (k)
i
yn+1 = yn + f xn , yn + f xn+1 , yn+1 , k = 0, 1, . . .
2
where f (x, y) = −2xy for this problem.
1(b): Numerical Approximation at x = 0.25 and x = 0.5
Question (b): Compute the approximate numerical values y(0.25) and y(0.5) using the Mod-
ified Euler method.
Answer:
Given: f (x, y) = −2xy, h = 0.25, x0 = 0, y0 = 1.
Page 2 of 14
