Question Number 195885 by cortano12 last updated on 12/Aug/23
$$\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\sqrt{\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:=\:\mathrm{0}\: \\ $$
Answered by mokys last updated on 12/Aug/23
$$\frac{{dy}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\:+\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\:{d}\left(\mathrm{0}\right) \\ $$$$ \\ $$$${sin}^{−\mathrm{1}} \left({y}\right)+{sin}^{−\mathrm{1}} \left({x}\right)\:=\:{c} \\ $$
Answered by BaliramKumar last updated on 12/Aug/23
$${x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }\:+\:{y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:=\:\mathrm{C} \\ $$$$ \\ $$
Commented by mokys last updated on 12/Aug/23
$${false}\:{solution} \\ $$