APM2616 Assignment 3 2026 Solutions Year Module 2026

UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology



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Applied Mathematics
Assignment 03

Initial Value Problems, Numerical Estimates & Series Solutions


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Module Code: APM2616

Module Name: Applied Mathematics

Assignment No.: 03

Due Date: August 2026

Semester: 2026




Submitted in partial fulfilment of the requirements for Applied Mathematics
at the University of South Africa.

,UNISA | APM2616 Assignment 03



Question 1

Determine the solution y(x) for each of the following initial value problems.


1.1


y ′ − yx cos x = 0, with y ′ (π) = 1.


Rearranging,
y ′ = yx cos x.


Separating the variables,
dy
= x cos x dx.
y

Integrating 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,
y = Cex sin x+cos x .


Differentiating,

d
y ′ = Cex sin x+cos x (x sin x + cos x) = Cex sin x+cos x (sin x + x cos x − sin x) = Cxex sin x+cos x cos x.
dx


Since y = Cex sin x+cos x , this confirms


y ′ = yx cos x.




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, UNISA | APM2616 Assignment 03


Applying the given condition y ′ (π) = 1,


1 = y(π)(π) cos π.



Since cos π = −1,
1
1 = −πy(π) ⇒ y(π) = − .
π

From y(π) = Ceπ sin π+cos π ,
1
− = Ce0−1 = Ce−1 .
π

Multiplying by e,
e
C=− .
π

Therefore,

e 1
y(x) = − ex sin x+cos x or equivalently y(x) = − ex sin x+cos x+1 .
π π


1.2

y
2y ′ + = 0, with y ′ (1) = π.
x

Rearranging,
y y
2y ′ = − ⇒ y′ = − .
x 2x

Separating the variables,
dy 1
= − dx.
y 2x

Integrating,
Z Z
1 1 1
dy = − dx,
y 2 x
1
ln |y| = − ln |x| + C.
2

Exponentiating,
y = Cx−1/2 .




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