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Let y = Eckxk be a power series solution of y" +x2y=0 about k=0 Then the coefficients ck satisfy 1 Ck +2= ск (k+2)(k+1) Ck-1, k≥2 Ob. Ck + 2 = 1 -Ck-1, K23 k(k+1) -1 ○°₁ Ck + 2 = (K + 2)(K + 1} -Ck-2, k≥2

Let y = Eckxk be a power series solution of y" +x2y=0 about k=0 Then the coefficients ck satisfy 1 Ck +2= ск (k+2)(k+1) Ck-1, k≥2 Ob. Ck + 2 = 1 -Ck-1, K23 k(k+1) -1 ○°₁ Ck + 2 = (K + 2)(K + 1} -Ck-2, k≥2

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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Subject: calculus 

 

 

 

 

 

Let y = Eckxk be a power series solution of y" + x²y = 0 about x₁=0. k = 0 Then the coefficients ck satisfy a. Ck +2 ов. Ск+2= OC. Ck + 2 = 1 (k+2)(k+1) -Ck-1, K22 1 k(K + 1) -1 (k+2)(k+1) -Ck-1, k≥3 -Ck-2, k22
Transcribed Image Text:Let y = Eckxk be a power series solution of y" + x²y = 0 about x₁=0. k = 0 Then the coefficients ck satisfy a. Ck +2 ов. Ск+2= OC. Ck + 2 = 1 (k+2)(k+1) -Ck-1, K22 1 k(K + 1) -1 (k+2)(k+1) -Ck-1, k≥3 -Ck-2, k22
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