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Question-136815




Question Number 136815 by rs4089 last updated on 26/Mar/21
Answered by Dwaipayan Shikari last updated on 26/Mar/21
y=e^(λx)   x^2 λ^2 +xλ+(x^2 −(1/4))=0  ⇒λ=((−x±(√(x^2 −4x^2 (x^2 −(1/4)))))/(2x^2 ))=−(1/(2x))±((√(2x^2 −4x^4 ))/(2x^2 ))  =((−1±(√(2−2x^2 )))/(2x))  y=Λe^((−1+i(√(2x^2 −2)))/2) +Φe^((−1−i(√(2x^2 −2)))/2)   =κe^(−(1/2)) cos(((√(2x^2 −2))/2))+τe^(−(1/2)) sin(((√(2x^2 −2))/2))
y=eλxx2λ2+xλ+(x214)=0λ=x±x24x2(x214)2x2=12x±2x24x42x2=1±22x22xy=Λe1+i2x222+Φe1i2x222=κe12cos(2x222)+τe12sin(2x222)
Answered by Olaf last updated on 26/Mar/21
  x^2 y′′+xy′+(x^2 −(1/4))y = 0     (1)  Let u = y(√x)  y = ux^(−1/2)   y′ = u′x^(−1/2) −(1/2)ux^(−3/2)   y′′ = u′′x^(−1/2) −(1/2)u′x^(−3/2) −(1/2)u′x^(−3/2) +(3/4)ux^(−5/2)   y′′ = u′′x^(−1/2) −u′x^(−3/2) +(3/4)ux^(−5/2)   (1) : u′′x^(3/2) −u′x^(1/2) +(3/4)ux^(−1/2)   +u′x^(1/2) −(1/2)ux^(−1/2) +(x^(3/2) −(1/4)x^(−1/2) )u = 0  u′′x^(3/2) +x^(3/2) u = 0  u′′+u = 0  u = ae^(ix) +be^(−ix)  = Acosx+iBsinx  y = (1/( (√x)))(Acosx+iBsinx)
x2y+xy+(x214)y=0(1)Letu=yxy=ux1/2y=ux1/212ux3/2y=ux1/212ux3/212ux3/2+34ux5/2y=ux1/2ux3/2+34ux5/2(1):ux3/2ux1/2+34ux1/2+ux1/212ux1/2+(x3/214x1/2)u=0ux3/2+x3/2u=0u+u=0u=aeix+beix=Acosx+iBsinxy=1x(Acosx+iBsinx)

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