Question Number 181799 by amin96 last updated on 30/Nov/22
$$\:\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{difgerential}}\:\boldsymbol{{equation}} \\ $$$$\boldsymbol{{y}}\sqrt{\mathrm{1}+\left(\boldsymbol{{y}}'\right)^{\mathrm{2}} }=\boldsymbol{{y}}'\: \\ $$
Answered by mr W last updated on 30/Nov/22
$${y}^{\mathrm{2}} \left(\mathrm{1}+\left({y}'\right)^{\mathrm{2}} \right)=\left({y}'\right)^{\mathrm{2}} \\ $$$$\mathrm{1}+\left(\frac{{dx}}{{dy}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$$\left(\frac{{dx}}{{dy}}\right)^{\mathrm{2}} =\frac{\mathrm{1}−{y}^{\mathrm{2}} }{{y}^{\mathrm{2}} } \\ $$$$\frac{{dx}}{{dy}}=\frac{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{{y}} \\ $$$$\int{dx}=\int\frac{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{{y}}{dy} \\ $$$$\mathrm{2}{x}=\mathrm{ln}\:\frac{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }−\mathrm{1}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} +\mathrm{1}}}+{C} \\ $$