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Question Number 84263 by Rio Michael last updated on 10/Mar/20

Using the approximation   h((dy/dx))_n  ≈ y_(n+1) −y_n  and that (dy/dx) = 1, y =2  when x = 0 . then , y_1  =  [A] h−2  [B] h + 2  [C] h−1  [D] h + 1

$$\mathrm{Using}\:\mathrm{the}\:\mathrm{approximation} \\ $$$$\:{h}\left(\frac{{dy}}{{dx}}\right)_{{n}} \:\approx\:{y}_{{n}+\mathrm{1}} −{y}_{{n}} \:\mathrm{and}\:\mathrm{that}\:\frac{{dy}}{{dx}}\:=\:\mathrm{1},\:{y}\:=\mathrm{2} \\ $$$$\mathrm{when}\:{x}\:=\:\mathrm{0}\:.\:\mathrm{then}\:,\:{y}_{\mathrm{1}} \:= \\ $$$$\left[\mathrm{A}\right]\:{h}−\mathrm{2} \\ $$$$\left[\mathrm{B}\right]\:{h}\:+\:\mathrm{2} \\ $$$$\left[\mathrm{C}\right]\:{h}−\mathrm{1} \\ $$$$\left[\mathrm{D}\right]\:{h}\:+\:\mathrm{1} \\ $$

Answered by mind is power last updated on 11/Mar/20

h((dy/dx))_0 ≈y_1 −y_0   y_1 ≈y_0 +h((dy/dx))_0 =2+1.h=h+2

$${h}\left(\frac{{dy}}{{dx}}\right)_{\mathrm{0}} \approx{y}_{\mathrm{1}} −{y}_{\mathrm{0}} \\ $$$${y}_{\mathrm{1}} \approx{y}_{\mathrm{0}} +{h}\left(\frac{{dy}}{{dx}}\right)_{\mathrm{0}} =\mathrm{2}+\mathrm{1}.{h}={h}+\mathrm{2} \\ $$

Commented by Rio Michael last updated on 11/Mar/20

thanks

$$\mathrm{thanks} \\ $$

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