UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
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APM2616: Applied Mathematics
Assignment 03 | 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM2616
Module Code:
Applied Mathematics
Module Name:
Initial Value Problems, Numerical Esti-
Assignment Topic:
mates, Series Solutions & Chebyshev
Polynomials
03
Assignment Number:
August 2026
Due Date:
100
Total Marks:
Submitted in partial fulfilment of the requirements for APM2616, UNISA 2026
,UNISA | APM2616 Assignment 03 – Initial Value & Numerical Estimates
Question 1: 20 Marks
Determine the solution y(x) for each of the following initial value problems.
(1.1)
y ′ − yx cos x = 0, with y ′ (π) = 1.
y ′ − yx cos x = 0
Rearrange:
y ′ = yx cos x
Separate the variables:
dy
= x cos x dx
y
Integrate both sides:
Z Z
1
dy = x cos x dx
y
Using integration by parts:
Z
x cos x dx = x sin x + cos x
Therefore,
ln |y| = x sin x + cos x + C
Exponentiating:
Page 2 of 27
,UNISA | APM2616 Assignment 03 – Initial Value & Numerical Estimates
y = Cex sin x+cos x
Differentiate:
d
y ′ = Cex sin x+cos x (x sin x + cos x)
dx
y ′ = Cex sin x+cos x (sin x + x cos x − sin x)
y ′ = Cxex sin x+cos x cos x
Since
y = Cex sin x+cos x
we have
y ′ = yx cos x
Apply the given condition y ′ (π) = 1:
1 = y(π)(π) cos π
Since
cos π = −1
1 = −πy(π)
Page 3 of 27
,UNISA | APM2616 Assignment 03 – Initial Value & Numerical Estimates
1
y(π) = −
π
From
y(π) = Ceπ sin π+cos π
1
− = Ce0−1
π
1
− = Ce−1
π
Multiply by e:
e
C=−
π
Therefore,
e
y(x) = − ex sin x+cos x
π
or
1
y(x) = − ex sin x+cos x+1
π
(1.2)
y
2y ′ + = 0, with y ′ (1) = π.
x
Rearrange:
y
2y ′ = −
x
Page 4 of 27
, UNISA | APM2616 Assignment 03 – Initial Value & Numerical Estimates
y
y′ = −
2x
Separate the variables:
dy 1
= − dx
y 2x
Integrate:
Z Z
1 1 1
dy = − dx
y 2 x
1
ln |y| = − ln |x| + C
2
Exponentiating:
y = Cx−1/2
Therefore,
y = Cx−1/2
Differentiate:
1
y ′ = − Cx−3/2
2
Apply y ′ (1) = π:
1
π = − C(1)−3/2
2
1
π=− C
2
Page 5 of 27
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM2616: Applied Mathematics
Assignment 03 | 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM2616
Module Code:
Applied Mathematics
Module Name:
Initial Value Problems, Numerical Esti-
Assignment Topic:
mates, Series Solutions & Chebyshev
Polynomials
03
Assignment Number:
August 2026
Due Date:
100
Total Marks:
Submitted in partial fulfilment of the requirements for APM2616, UNISA 2026
,UNISA | APM2616 Assignment 03 – Initial Value & Numerical Estimates
Question 1: 20 Marks
Determine the solution y(x) for each of the following initial value problems.
(1.1)
y ′ − yx cos x = 0, with y ′ (π) = 1.
y ′ − yx cos x = 0
Rearrange:
y ′ = yx cos x
Separate the variables:
dy
= x cos x dx
y
Integrate both sides:
Z Z
1
dy = x cos x dx
y
Using integration by parts:
Z
x cos x dx = x sin x + cos x
Therefore,
ln |y| = x sin x + cos x + C
Exponentiating:
Page 2 of 27
,UNISA | APM2616 Assignment 03 – Initial Value & Numerical Estimates
y = Cex sin x+cos x
Differentiate:
d
y ′ = Cex sin x+cos x (x sin x + cos x)
dx
y ′ = Cex sin x+cos x (sin x + x cos x − sin x)
y ′ = Cxex sin x+cos x cos x
Since
y = Cex sin x+cos x
we have
y ′ = yx cos x
Apply the given condition y ′ (π) = 1:
1 = y(π)(π) cos π
Since
cos π = −1
1 = −πy(π)
Page 3 of 27
,UNISA | APM2616 Assignment 03 – Initial Value & Numerical Estimates
1
y(π) = −
π
From
y(π) = Ceπ sin π+cos π
1
− = Ce0−1
π
1
− = Ce−1
π
Multiply by e:
e
C=−
π
Therefore,
e
y(x) = − ex sin x+cos x
π
or
1
y(x) = − ex sin x+cos x+1
π
(1.2)
y
2y ′ + = 0, with y ′ (1) = π.
x
Rearrange:
y
2y ′ = −
x
Page 4 of 27
, UNISA | APM2616 Assignment 03 – Initial Value & Numerical Estimates
y
y′ = −
2x
Separate the variables:
dy 1
= − dx
y 2x
Integrate:
Z Z
1 1 1
dy = − dx
y 2 x
1
ln |y| = − ln |x| + C
2
Exponentiating:
y = Cx−1/2
Therefore,
y = Cx−1/2
Differentiate:
1
y ′ = − Cx−3/2
2
Apply y ′ (1) = π:
1
π = − C(1)−3/2
2
1
π=− C
2
Page 5 of 27
