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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

$$\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }−\mathrm{2}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{4}\frac{{dy}}{{dx}}−\mathrm{8}{y}=\mathrm{0} \\ $$

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

$${HE}\:\equiv\:{z}^{\mathrm{3}} −\mathrm{2}{z}^{\mathrm{2}} +\mathrm{4}{z}−\mathrm{8}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({z}−\mathrm{2}\right)\left({z}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left({z}−\mathrm{2}\right)\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)=\mathrm{0} \\ $$$${General}\:{solution}\: \\ $$$${y}_{{h}} \:=\:{Ae}^{\mathrm{2}{x}} +{B}\:\mathrm{cos}\:\mathrm{2}{x}+{C}\:\mathrm{sin}\:\mathrm{2}{x} \\ $$

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

$${y}'''−\mathrm{2}{y}''+\mathrm{4}{y}'−\mathrm{8}{y}\:=\:\mathrm{0} \\ $$$$\mathrm{We}\:\mathrm{solve}\:: \\ $$$${r}^{\mathrm{3}} −\mathrm{2}{r}^{\mathrm{2}} +\mathrm{4}{r}−\mathrm{8}\:=\:\mathrm{0} \\ $$$${r}\left({r}^{\mathrm{2}} +\mathrm{4}\right)−\mathrm{2}{r}^{\mathrm{2}} −\mathrm{8}\:=\:\mathrm{0} \\ $$$${r}\left({r}^{\mathrm{2}} +\mathrm{4}\right)−\mathrm{2}\left({r}^{\mathrm{2}} +\mathrm{4}\right)\:=\:\mathrm{0} \\ $$$$\left({r}−\mathrm{2}\right)\left({r}^{\mathrm{2}} +\mathrm{4}\right)\:=\:\mathrm{0} \\ $$$$\left({r}−\mathrm{2}\right)\left({r}−\mathrm{2}{i}\right)\left({r}+\mathrm{2}{i}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{r}\:=\:\mathrm{2},\:+\mathrm{2}{i},\:−\mathrm{2}{i} \\ $$$${y}\:=\:\mathrm{A}{e}^{\mathrm{2}{x}} +\mathrm{B}{e}^{\mathrm{2}{ix}} +\mathrm{C}{e}^{−\mathrm{2}{ix}} \\ $$$$\mathrm{or}\:\mathrm{we}\:\mathrm{can}\:\mathrm{write}\:: \\ $$$${y}\:=\:\alpha{e}^{\mathrm{2}{x}} +\beta\mathrm{cos2}{x}+\gamma\mathrm{sin2}{x} \\ $$

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)

$${y}={e}^{\gamma{x}} \\ $$$$\gamma^{\mathrm{3}} −\mathrm{2}\gamma^{\mathrm{2}} +\mathrm{4}\gamma−\mathrm{8}=\mathrm{0} \\ $$$$\left(\gamma^{\mathrm{2}} +\mathrm{4}\right)\left(\gamma−\mathrm{2}\right)=\mathrm{0}\:\:\gamma=\mathrm{2},\mathrm{2}{i},−\mathrm{2}{i} \\ $$$${y}=\Phi{e}^{\mathrm{2}{x}} +\Psi{e}^{\mathrm{2}{xi}} +\Gamma{e}^{−\mathrm{2}{xi}} \:\:\: \\ $$$$ \\ $$

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