APM3706 Assignment 4 2026 Solutions Due August 2026

UNIVERSITY OF SOUTH AFRICA (UNISA)
Department of Mathematical Sciences



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Ordinary Differential Equations
Assignment 04 — Year Module


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Module Code: APM3706

Module Name: Ordinary Differential Equations

Assignment No.: Assignment 04

Semester: Year Module

Department: Department of Mathematical Sciences




Submitted in partial fulfilment of the requirements for APM3706
at the University of South Africa.

,UNISA | APM3706 Assignment 04



Question 1: Exercise 6.7

Q: Given X(t) = et e2t , determine the coefficients Xk such that

∞
X
X(t) = Xk tk .
k=0



A:

Given
X(t) = et e2t .


Use the law of exponents:
et e2t = et+2t = e3t .


So,
X(t) = e3t .


The Maclaurin series for ex is
∞
X xk
ex = .
k!
k=0


Substitute x = 3t:
∞
3t
X (3t)k
e = .
k!
k=0


Expand:
∞
X 3k tk
e3t = .
k!
k=0


Compare this with
∞
X
X(t) = Xk tk .
k=0


Match the coefficients of tk :
3k
Xk = , k = 0, 1, 2, . . .
k!

Calculate the first few coefficients:

30 1
X0 = = = 1,
0! 1


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, UNISA | APM3706 Assignment 04


31 3
X1 = = = 3,
1! 1
32 9
X2 = = ,
2! 2
33 27 9
X3 = = = ,
3! 6 2
34 81 27
X4 = = = .
4! 24 8

Hence,
9 9 27
X(t) = 1 + 3t + t2 + t3 + t4 + · · ·
2 2 8

Final Answer
3k
Xk = , k = 0, 1, 2, . . .
k!




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