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Question Number 125925 by zarminaawan last updated on 15/Dec/20

(d^3 y/dx^3 )−2(d^2 y/dx^2 )+4(dy/dx)−8y=0

d3ydx32d2ydx2+4dydx8y=0

Answered by liberty last updated on 15/Dec/20

HE ≡ z^3 −2z^2 +4z−8=0                (z−2)(z^2 +4)=0               (z−2)(z−2i)(z+2i)=0  General solution   y_h  = Ae^(2x) +B cos 2x+C sin 2x

HEz32z2+4z8=0(z2)(z2+4)=0(z2)(z2i)(z+2i)=0Generalsolutionyh=Ae2x+Bcos2x+Csin2x

Answered by Olaf last updated on 15/Dec/20

y′′′−2y′′+4y′−8y = 0  We solve :  r^3 −2r^2 +4r−8 = 0  r(r^2 +4)−2r^2 −8 = 0  r(r^2 +4)−2(r^2 +4) = 0  (r−2)(r^2 +4) = 0  (r−2)(r−2i)(r+2i) = 0  ⇒ r = 2, +2i, −2i  y = Ae^(2x) +Be^(2ix) +Ce^(−2ix)   or we can write :  y = αe^(2x) +βcos2x+γsin2x

y2y+4y8y=0Wesolve:r32r2+4r8=0r(r2+4)2r28=0r(r2+4)2(r2+4)=0(r2)(r2+4)=0(r2)(r2i)(r+2i)=0r=2,+2i,2iy=Ae2x+Be2ix+Ce2ixorwecanwrite:y=αe2x+βcos2x+γsin2x

Answered by Dwaipayan Shikari last updated on 15/Dec/20

y=e^(γx)   γ^3 −2γ^2 +4γ−8=0  (γ^2 +4)(γ−2)=0  γ=2,2i,−2i  y=Φe^(2x) +Ψe^(2xi) +Γe^(−2xi)

y=eγxγ32γ2+4γ8=0(γ2+4)(γ2)=0γ=2,2i,2iy=Φe2x+Ψe2xi+Γe2xi

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