Question Number 100048 by bramlex last updated on 24/Jun/20
Answered by mr W last updated on 24/Jun/20
$${u}=\frac{{dy}}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{du}}{{dx}}=\frac{{du}}{{dy}}×\frac{{dy}}{{dx}}={u}\frac{{du}}{{dy}} \\ $$$$\mathrm{2}{yu}\frac{{du}}{{dy}}−{u}^{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{\mathrm{2}{u}}{{u}^{\mathrm{2}} +\mathrm{1}}{du}=\frac{{dy}}{{y}} \\ $$$$\int\frac{\mathrm{2}{u}}{{u}^{\mathrm{2}} +\mathrm{1}}{du}=\int\frac{{dy}}{{y}} \\ $$$$\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}{d}\left({u}^{\mathrm{2}} +\mathrm{1}\right)=\int\frac{{dy}}{{y}} \\ $$$$\mathrm{ln}\:\left({u}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{ln}\:{y}+{C} \\ $$$$\Rightarrow{u}^{\mathrm{2}} +\mathrm{1}={Cy} \\ $$$$\Rightarrow{u}=\frac{{dy}}{{dx}}=\pm\sqrt{{Cy}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{dy}}{\:\sqrt{{Cy}−\mathrm{1}}}=\pm{dx} \\ $$$$\Rightarrow\int\frac{{dy}}{\:\sqrt{{Cy}−\mathrm{1}}}=\pm\int{dx} \\ $$$$\frac{\mathrm{2}}{{C}}\sqrt{{Cy}−\mathrm{1}}=\pm{x}+{C}_{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}\sqrt{{Cy}−\mathrm{1}}=\pm{C}\left({x}+{C}_{\mathrm{1}} \right) \\ $$