The question is the same . These is a solution already provided . Please check and tell which is correct this , or the one you sent . I cannot attach the question as there's not an option to attach a 3rd image . Onkn Cos (MTx) dxIIanSin ()/22K新)-1.SinSin오 dnan2K9/2SinAn-() -1] -2K12%3D12KCas (mT) +an2kl2CosCa/Sine- Senes:HenceHalf range40F(x)- k오42kl Sohu ionGivenfex) =Half rangeCoDinedeviestoF(x) -Σwhorefix) dnanbefin dx +42%3Dka du tKe-x) dn42a. =-2222klke8.

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Description: The question is the same . These is a solution already provided . Please check and tell which is correct this , or the one you sent . I cannot attach the question as there's not an option to attach a 3rd image . Onkn Cos (MTx) dxIIanSin ()/22K新)-1.Si

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Answered: Sohution: Given f(x) 는 S kx Half range…

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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The question is the same . These is a solution already provided . Please check and tell which is correct this , or the one you sent . I cannot attach the question as there's not an option to attach a 3rd image .

On
kn Cos (MTx) dx
II
an
Sin ()
/2
2K
新)-
1.
Sin
Sin
오 dn
an
2K
9/2
Sin
An-
() -1] -
2K
12
%3D
1
2K
Cas (mT) +
an
2kl
2Cos
Ca/Sine
- Senes:
Hence
Half range
40
F(x)- k오
4
2kl

Sohu ion
Given
fex) =
Half range
CoDine
devies
to
F(x) -
Σ
whore
fix) dn
an
be
fin dx +
42
%3D
ka du t
Ke-x) dn
42
a. =
-
222
2kl
ke
8.

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