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


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

$\frac{d^{3} y}{{d x}^{3}} - 2 \frac{d^{2} y}{{d x}^{2}} + 4 \frac{d y}{d x} - 8 y = 0$
	Answered by liberty last updated on 15/Dec/20

	$H E \equiv z^{3} - 2 z^{2} + 4 z - 8 = 0$ $(z - 2) (z^{2} + 4) = 0$ $(z - 2) (z - 2 i) (z + 2 i) = 0$ $G e n e r a l s o l u t i o n$ $y_{h} = {A e}^{2 x} + B \cos 2 x + C \sin 2 x$
	Answered by Olaf last updated on 15/Dec/20

	$y^{‴} - 2 y^{″} + 4 y^{'} - 8 y = 0$ $We solve :$ $r^{3} - 2 r^{2} + 4 r - 8 = 0$ $r (r^{2} + 4) - 2 r^{2} - 8 = 0$ $r (r^{2} + 4) - 2 (r^{2} + 4) = 0$ $(r - 2) (r^{2} + 4) = 0$ $(r - 2) (r - 2 i) (r + 2 i) = 0$ $\Rightarrow r = 2, + 2 i, - 2 i$ $y = A e^{2 x} + B e^{2 i x} + C e^{- 2 i x}$ $or we can write :$ $y = α e^{2 x} + β \cos 2 x + γ \sin 2 x$
	Answered by Dwaipayan Shikari last updated on 15/Dec/20

	$y = e^{γ x}$ $γ^{3} - 2 γ^{2} + 4 γ - 8 = 0$ $(γ^{2} + 4) (γ - 2) = 0 γ = 2, 2 i, - 2 i$ $y = Φ e^{2 x} + Ψ e^{2 x i} + Γ e^{- 2 x i}$
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