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Fourier series of 2- on the inter val 3 of - TT, ] | AnsweR should be dis (uss in English woRd] N.2 The Solution of the ini-Cial value problem y" – 24'+ 24= 5 sin (t), yco)= 0,ylo)=o Answer have to details how you solve this]

Fourier series of 2- on the inter val 3 of - TT, ] | AnsweR should be dis (uss in English woRd] N.2 The Solution of the ini-Cial value problem y" – 24'+ 24= 5 sin (t), yco)= 0,ylo)=o Answer have to details how you solve this]

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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Fourier series of 2- 3. on the inter val 3 of - ा , ा ] TT, T |AnswerR should be dis (uss in English woRd] N.2 The Solution of the inilial value problem prroblem y" – 24+2y= 5 sin (t), y(o): 0, yl0)=0 34+24 y(0)=D0 Answere have to details how you | solve this]
Transcribed Image Text:Fourier series of 2- 3. on the inter val 3 of - ा , ा ] TT, T |AnswerR should be dis (uss in English woRd] N.2 The Solution of the inilial value problem prroblem y" – 24+2y= 5 sin (t), y(o): 0, yl0)=0 34+24 y(0)=D0 Answere have to details how you | solve this]
Fourier Series of 2 - on the interval [-n, 1] (-1)" sin (nz) 00 n=1 2+ 2 -E1 0o (-1)" cos (nz) | n=1 8(–1)" cos() + Σ 3+ ? -E 00 (-1)" cos (nz) 3 4 n=1 2 so (-1)" cos (nz) 2+ 2 The solution of the initial value problem y" – 2y +2y = 5 sin(t), y (0) = 0, y (0) = 0 y = et cos (t) + 2 cos (t) – 2e t cos (t) + sin (t) y = e' sin (t) + 2 cos (t) + 2e* cos (t) + cos (t) y = et sin (t) + 2 cos (t) – 2e* cos (t) + sin (t) y = -et sin (t) + 2 cos (t) + 2e cos (t) + sin (t) y = e t sin (t) + 2 sin (t) – 2et cos (t) + sin (t) 1.
Transcribed Image Text:Fourier Series of 2 - on the interval [-n, 1] (-1)" sin (nz) 00 n=1 2+ 2 -E1 0o (-1)" cos (nz) | n=1 8(–1)" cos() + Σ 3+ ? -E 00 (-1)" cos (nz) 3 4 n=1 2 so (-1)" cos (nz) 2+ 2 The solution of the initial value problem y" – 2y +2y = 5 sin(t), y (0) = 0, y (0) = 0 y = et cos (t) + 2 cos (t) – 2e t cos (t) + sin (t) y = e' sin (t) + 2 cos (t) + 2e* cos (t) + cos (t) y = et sin (t) + 2 cos (t) – 2e* cos (t) + sin (t) y = -et sin (t) + 2 cos (t) + 2e cos (t) + sin (t) y = e t sin (t) + 2 sin (t) – 2et cos (t) + sin (t) 1.
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