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Zin ñ l (1-1バ)ー (-1)"-1) ニ zink zin k in t ニ = (1)", as coJ (n^) =E1)" kin (n) ニ - inr co7(nn]- ihncnワ (-1)" ニ0 e 三 ニ please explain ニ ink の putting we get n=0, てすコチ 7° 214 +1, 1 か) Co = %3D dn 一

Zin ñ l (1-1バ)ー (-1)"-1) ニ zink zin k in t ニ = (1)", as coJ (n^) =E1)" kin (n) ニ - inr co7(nn]- ihncnワ (-1)" ニ0 e 三 ニ please explain ニ ink の putting we get n=0, てすコチ 7° 214 +1, 1 か) Co = %3D dn 一

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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Aas I asked you, I don't understand some step as I mention in picure, please explain

OLn 2 - in n/2 e Cn = (formula) -inan/2 - in an/2 dn t or, Cn = ニ [by def" of feng -in m/2 2 in n/2 ニ ind in F in, も) Into %3D zink - int リだどーリ e Zin ñ E (1-1)") - (61)"-1) zin k zinK in t - log (nx) + ihin (nimy = 1)", as cos (n^) =G1)" fin (n) =O %3D - inn e (-1)" ニ Please explain ink From putting we get n=0, Co - %3D ー1 + 2 + [x],
Transcribed Image Text:OLn 2 - in n/2 e Cn = (formula) -inan/2 - in an/2 dn t or, Cn = ニ [by def" of feng -in m/2 2 in n/2 ニ ind in F in, も) Into %3D zink - int リだどーリ e Zin ñ E (1-1)") - (61)"-1) zin k zinK in t - log (nx) + ihin (nimy = 1)", as cos (n^) =G1)" fin (n) =O %3D - inn e (-1)" ニ Please explain ink From putting we get n=0, Co - %3D ー1 + 2 + [x],
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