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Exam (elaborations) APM2611 Assignment 4 (COMPLETE ANSWERS) 2024 - DUE 25 September 2024 • Course • Differential Equations - APM2611 (APM2611) • Institution • University Of South Africa • Book • Differential Equations APM2611 Assignment 4 (COMPLETE ANSWER

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Course
Differential Equations

Institution
University Of South Africa

Book
Differential Equations

Exam (elaborations) APM2611 Assignment 4 (COMPLETE ANSWERS) 2024 - DUE 25 September 2024 • Course • Differential Equations - APM2611 (APM2611) • Institution • University Of South Africa • Book • Differential Equations APM2611 Assignment 4 (COMPLETE ANSWERS) 2024 - DUE 25 Septemb...

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apm2611

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, Exam (elaborations)
APM2611 Assignment 4 (COMPLETE ANSWERS) 2024 -
DUE 25 September 2024
Course
 Differential Equations - APM2611 (APM2611)
 Institution
 University Of South Africa
 Book
 Differential Equations

APM2611 Assignment 4 (COMPLETE ANSWERS) 2024 - DUE 25 September
2024 ;100 % TRUSTED workings, explanations and solutions. ...........

Question 1 1. Find the radius and interval of convergence of the following
series: ��X n=1 (−1) n−1 x2n−1 (2n − 1)! 2. Rewrite the expression below
as a single power series: ∞X n=2 cn+1 x n−2 − ∞X n=1 4cn x n−1 . 3. 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. 4. Use the power series method to solve the initialvalue problem.In
particular, find c 0 , c1 , c2 , c3 and c4 in the equation y(x) = P ∞ n=0 cn x n .
y 00− x 2 y = 0; y(0) = 3, y 0 (0) = 7.

Question 1

Find the radius and interval of convergence of the series:

∑n=1∞(−1)n−1x2n−1(2n−1)!\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n-1}}{(2n -
1)!}n=1∑∞(2n−1)!(−1)n−1x2n−1

Solution:

1. Consider the general term an=(−1)n−1x2n−1(2n−1)!a_n = \frac{(-1)^{n-1} x^{2n-1}}
{(2n-1)!}an=(2n−1)!(−1)n−1x2n−1.
2. Apply the ratio test: ∣an+1an∣=∣(−1)nx2(n+1)−1(2(n+1)−1)!⋅(2n−1)!
(−1)n−1x2n−1∣=∣x2n+1(2n+1)!⋅(2n−1)!x2n−1∣=∣x2(2n+1)(2n)∣\left| \frac{a_{n+1}}
{a_n} \right| = \left| \frac{(-1)^n x^{2(n+1)-1}}{(2(n+1)-1)!} \cdot \frac{(2n-1)!}{(-
1)^{n-1} x^{2n-1}} \right| = \left| \frac{x^{2n+1}}{(2n+1)!} \cdot \frac{(2n-1)!}
{x^{2n-1}} \right| = \left| \frac{x^2}{(2n+1)(2n)} \right|anan+1=(2(n+1)−1)!
(−1)nx2(n+1)−1⋅(−1)n−1x2n−1(2n−1)!=(2n+1)!x2n+1⋅x2n−1(2n−1)!=(2n+1)(2n)x2

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