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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
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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],](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F73812453-6cbe-43fd-a68e-d7d40ca15395%2Fee5d8bc5-e4c4-4b53-9bd6-85ef8448192c%2Fyo8q1b_processed.jpeg&w=3840&q=75)
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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