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Solve by use of Fourier series. Assume in each case that the right-hand side is a periodic function whose values are stated for one period.

y"+2y'+2y=|x|,-Ï€<x<Ï€.

Short Answer

Expert verified

Answer

The solution of y"+2y'+2y=|x|,-Ï€<x<Ï€is f(x)=1Ï€+12sinx-2Ï€(cos2x22-1+cos4x42-1+cos4x42-1+cos4x62-1+..............).

Step by step solution

01

Given information from question

The equation isy"+2y'+2y=|x|,-Ï€<x<Ï€.

02

Quadratic formula

The quadratic formula to find out the roots is,

-b±b2-4ac2a

03

Expand f(x) in a Fourier series

The given function is

y"+2y'+2y=|x|,-Ï€<x<Ï€

Here fxis a function of period

fx=xif0≤x<-π-xif-π≤x<π

The auxiliary equation is

D2+2D+2=0m2+2m+2=0

04

Apply quadratic formula

By Quadratic formula

m=-b±b2-4ac2a=-2±22-4×1×22=-2±4-82

Solve further

m=-2±-42=-1±i

The complementary function is

role="math" localid="1654080410163" yce-xAcosx+Bsinxfx=xif0≤x<-π-xif-π≤x<πfx=1π+12sinx-2πcos2x22-1+cos4x42-1+cos6x62-1+...........

Thus, the solution of y"+2y'+2y=|x|,-Ï€<x<Ï€is fx=1Ï€+12sinx-2Ï€cos2x22-1+cos4x42-1+cos6x62-1+...........

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