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Given the following differential equation y" - x?y = 0 After plugging in the appropriate power series for "y" and "y'", we get the following power series. 00 E (n - 1) A, x* - 2) - EA, * + 2) = 0 n = 2 n = 0 1. Force both power series to have a simplified exponent. 2. Make sure that both power series start at the same value for "n". 3. Combine both power series together into a single power series.

Given the following differential equation y" - x?y = 0 After plugging in the appropriate power series for "y" and "y'", we get the following power series. 00 E (n - 1) A, x* - 2) - EA, * + 2) = 0 n = 2 n = 0 1. Force both power series to have a simplified exponent. 2. Make sure that both power series start at the same value for "n". 3. Combine both power series together into a single power series.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Given the following differential equation y" - x?y = 0 After plugging in the appropriate power series for "y" and "y'", we get the following power series. Σ 2 n (n - 1) A x (n - 2) n = 2 Σ. A x(m + 2) = 0 n = 0 Force both power series to have a simplified exponent. 2. Make sure that both power series start at the same value for "n". 3. Combine both power series together into a single power series. E [R (R – 1) A, (A A, + A,* R = 0 x + R = 2 B 6A, + 2A ,x + E [(R - 2) (R – 3) A f +2) - Ale-2] * = = 0 R = 2 00 A3x + A,x? + E [(R - 2) (R – 3) A (R -2) A (R + 2)] * = 0 A5 x3 R = 4 00 R = 0 2 (R - 2) (R – 3) A (R -2) - A (R+ 2)]** (R + 2) R = 4 00 24, + 6A,x + 2 (R + 2) (R + 1) A (r + 2) - A (R - 2) * E [(R + 2) (R + 1) A = 0 (R-2) R = 2 È (R - 2) (R – 3) A (R +2) .R = 0 (F -A_2 - Ax + -1 (R R = 2
Transcribed Image Text:Given the following differential equation y" - x?y = 0 After plugging in the appropriate power series for "y" and "y'", we get the following power series. Σ 2 n (n - 1) A x (n - 2) n = 2 Σ. A x(m + 2) = 0 n = 0 Force both power series to have a simplified exponent. 2. Make sure that both power series start at the same value for "n". 3. Combine both power series together into a single power series. E [R (R – 1) A, (A A, + A,* R = 0 x + R = 2 B 6A, + 2A ,x + E [(R - 2) (R – 3) A f +2) - Ale-2] * = = 0 R = 2 00 A3x + A,x? + E [(R - 2) (R – 3) A (R -2) A (R + 2)] * = 0 A5 x3 R = 4 00 R = 0 2 (R - 2) (R – 3) A (R -2) - A (R+ 2)]** (R + 2) R = 4 00 24, + 6A,x + 2 (R + 2) (R + 1) A (r + 2) - A (R - 2) * E [(R + 2) (R + 1) A = 0 (R-2) R = 2 È (R - 2) (R – 3) A (R +2) .R = 0 (F -A_2 - Ax + -1 (R R = 2
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