Exam (elaborations)
APM2611 Assignment 3 2024 - DUE 14 August 2024
- Institution
- University Of South Africa
APM2611 Assignment 3 2024 - DUE 14 August 2024 QUESTIONS AND ANSWERS
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APM2611Assignment 3 2024
- DUE 14 August
2024
QUESTIONS WITH COMPLETE ANSWERS
[DATE]
[COMPANY NAME]
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, APM2611 Assignment 3 2024 - DUE 14 August 2024
Question 1
1. Find the radius and interval of convergence of the following series: (
i) ∞X n=1 100n n! (x + 7) n
(ii) ∞X k=1 (−1) k 10k (x − 5) k
2. Rewrite the expression below as a single power series: ∞X n=2 n(n −
1)cn x n + 2 ∞X n=2 n(n − 1)cn x n−2 + ∞X n=1 ncn x n .
Question 2
1. Verify by direct substitution that the given power series is a particular
solution of the DE (x + 1)y 00+ y 0 = 0 ; y = ∞X n=1 (−1) n+1 n x n .
2. Use the power series method to solve the initialvalue problem (x + 1)y 00
− (2 − x)y 0 + y = 0, y(0) = 2, y 0 (0) = −1; where c0 and c1 are given by the
initial conditions. 16 APM2611/101/0/2024
Question 3
Calculate the Laplace transform of the following function from first
principles: 1. f (t) = sin t if 0 ≤ t < π 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 1
1. Find the radius and interval of convergence of the following series:
(i) ∑n=1∞100nn!(x+7)n\sum_{n=1}^{\infty} \frac{100^n}{n!} (x + 7)^n∑n=1∞
n!100n(x+7)n
To find the radius of convergence, we use the ratio test:
an=100nn!(x+7)na_n = \frac{100^n}{n!} (x + 7)^nan=n!100n(x+7)n
an+1an=100n+1(n+1)!(x+7)n+1100nn!(x+7)n=100⋅100n(n+1)n!⋅(x+7)(x+7)n(x+7)
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