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3y CC) (2x)*(3y)"-i = Зу - (2х + 3у)" 2x n n E)(2x)-'(3y)"-i+1 %3D %3D 2x i=0 i=0

3y CC) (2x)*(3y)"-i = Зу - (2х + 3у)" 2x n n E)(2x)-'(3y)"-i+1 %3D %3D 2x i=0 i=0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can you explain this step by step please. 

How did it go from (2x)i-1 (3y)n-i+1 to (2x + 3y)n

n 3y 3y i-1 -i+1 - (2х + 3у)" 2x n-i 2x i=0 i=0
Transcribed Image Text:n 3y 3y i-1 -i+1 - (2х + 3у)" 2x n-i 2x i=0 i=0
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