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Question Number 125925 by zarminaawan last updated on 15/Dec/20
d3ydx3−2d2ydx2+4dydx−8y=0
Answered by liberty last updated on 15/Dec/20
HE≡z3−2z2+4z−8=0(z−2)(z2+4)=0(z−2)(z−2i)(z+2i)=0Generalsolutionyh=Ae2x+Bcos2x+Csin2x
Answered by Olaf last updated on 15/Dec/20
y‴−2y″+4y′−8y=0Wesolve:r3−2r2+4r−8=0r(r2+4)−2r2−8=0r(r2+4)−2(r2+4)=0(r−2)(r2+4)=0(r−2)(r−2i)(r+2i)=0⇒r=2,+2i,−2iy=Ae2x+Be2ix+Ce−2ixorwecanwrite:y=αe2x+βcos2x+γsin2x
Answered by Dwaipayan Shikari last updated on 15/Dec/20
y=eγxγ3−2γ2+4γ−8=0(γ2+4)(γ−2)=0γ=2,2i,−2iy=Φe2x+Ψe2xi+Γe−2xi
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