APM2611 Assignment 3 (COMPLETE ANSWERS) 2024 - DUE 14 August 2024

APM2611
ASSIGNMENT 3 2024
UNIQUE NO.
DUE DATE: 14 AUGUST 2024

, ASSIGNMENT 03
Due date: Wednesday, 14 August 2024
-

ONLY FOR YEAR MODULE

First order separable, linear, Bernoulli, exact and homogeneous equations. Higher order
homogeneous DE’s. Solving non-homogeneous DE’s using the undetermined
coefficients, variation of parameters and operator methods.

Answer all the questions. Show all your own and personalized workings, you get ZERO
to a question if we see that you have copied someone’s else solution word by word.

If you choose to submit via myUnisa, note that only PDF files will be accepted.

Note that all the questions will be marked therefore, it is highly recommended to attempt all of them.

Question 1

1. Find the radius and interval of convergence of the following series:
(i)
X∞ 100n n
(x + 7)
n=1
n!

(ii)
X∞ (−1) k k
k (x − 5)
k=1
10

2. Rewrite the expression below as a single power series:
X∞ n
X∞ X∞
n(n − 1)c n x + 2 n(n − 1)c n x n−2 + ncn x n .
n=2 n=2 n=1



Question 2

1. Verify by direct substitution that the given power series is a particular solution of the DE
00 0
X∞ (−1) n+1 n
(x + 1)y + y = 0 ; y = x .
n
n=1

2. Use the power series method to solve the initialvalue problem
00 0 0
(x + 1)y − (2 − x)y + y = 0, y(0) = 2, y (0) = −1;
where c0 and c1 are given by the initial conditions.



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, APM2611/101/0/2024


Question 3

Calculate the Laplace transform of the following function from first principles:

1.
sin t if 0≤t<π
f (t) =
0 if t≥π

2. f (t) = e −t sin t

3. Use Theorem 7.1 to find L{f (t)}

(i) f (t) = −4t 2
+ 16t + 9
(ii) f (t) = 4t 2
− 5 sin 3t
(iii) f (t) = (e t − e −t ) 2



Question 4


1. Use Theorem 7.3 to find the inverse transform:

(i)
2s − 4
L −1
(s2 + s)(s 2 + 4)
(ii)
s
L −1
(s + 2)(s 2 + 4)
2. Use the Laplace transform to solve the initialvalue problem

y00+ 5y 0 + 4y = 0, y(0) = 1, y0(0) = 0.



Question 5

1,
1. When g(t) = 1 and L{g(t)} = G(s) = s
the convolution theorem implies that the Laplace
transform of the integral of f is
Z t
F (s)
L f (τ ) dτ = .
0
s
The inverse form is
Z t
F (s)
f (τ ) dτ = L −1 .
0
s
Find


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