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Question Number 250 by 123456 last updated on 25/Jan/15

(d^2 y/dx^2 )+2x((dy/dx))^2 =0

$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{2}{x}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$

Answered by prakash jain last updated on 17/Dec/14

(dy/dx)=u  (du/dx)+2xu^2 =0  (du/u^2 )=−2xdx  −(1/u)=−x^2 +C  u=(1/(x^2 −C))  (dy/dx)=(1/(x^2 −C))  y=−((arctanh((x/(√C))))/(√C))+C_1

$$\frac{{dy}}{{dx}}={u} \\ $$$$\frac{{du}}{{dx}}+\mathrm{2}{xu}^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{{du}}{{u}^{\mathrm{2}} }=−\mathrm{2}{xdx} \\ $$$$−\frac{\mathrm{1}}{{u}}=−{x}^{\mathrm{2}} +{C} \\ $$$${u}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{C}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{C}} \\ $$$${y}=−\frac{\mathrm{arctanh}\left(\frac{{x}}{\sqrt{{C}}}\right)}{\sqrt{{C}}}+{C}_{\mathrm{1}} \\ $$

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