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Question Number 118574 by bramlexs22 last updated on 18/Oct/20
find particular solution of   (D^3 +4D) y = sin 2x by using  inverse operation ?
findparticularsolutionof(D3+4D)y=sin2xbyusinginverseoperation?
Answered by TANMAY PANACEA last updated on 18/Oct/20
y=((sin2x)/(D(D^2 +4)))  y=(e^(i2x) /(D(D^2 +4)))  =(e^(i2x) /((i2){(D+i2)^2 +4}))  =(e^(i2x) /((i2)(D^2 +4iD)))  =(e^(i2x) /((i2)))×(1/(D×4i×(1+(D/(4i)))))  =(1/(−8))×(1/((1+((i2)/(4i)))))×(e^(i2x) /1)×x  =(1/(−8))×(2/3)×((cos2x+isin2x)/1)×x  =(x/(−12))(cos2x)+i×(x/(−12))(sin2x)  P.I=(x/(−12))×sin2x→y=((xsin2x)/(−12))  pls check my answer
y=sin2xD(D2+4)y=ei2xD(D2+4)=ei2x(i2){(D+i2)2+4}=ei2x(i2)(D2+4iD)=ei2x(i2)×1D×4i×(1+D4i)=18×1(1+i24i)×ei2x1×x=18×23×cos2x+isin2x1×x=x12(cos2x)+i×x12(sin2x)P.I=x12×sin2xy=xsin2x12plscheckmyanswer

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