Question Number 250 by 123456 last updated on 25/Jan/15
![(d^2 y/dx^2 )+2x((dy/dx))^2 =0](https://www.tinkutara.com/question/Q250.png)
$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{2}{x}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$
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](https://www.tinkutara.com/question/Q269.png)
$$\frac{{dy}}{{dx}}={u} \\ $$$$\frac{{du}}{{dx}}+\mathrm{2}{xu}^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{{du}}{{u}^{\mathrm{2}} }=−\mathrm{2}{xdx} \\ $$$$−\frac{\mathrm{1}}{{u}}=−{x}^{\mathrm{2}} +{C} \\ $$$${u}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{C}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{C}} \\ $$$${y}=−\frac{\mathrm{arctanh}\left(\frac{{x}}{\:\sqrt{{C}}}\right)}{\:\sqrt{{C}}}+{C}_{\mathrm{1}} \\ $$