Question

(y')2=(1-y^2)/(1-x^2) ; y=1/2 when x= 1

Answer 1

$solve ~ (y^{\prime})^2= \frac{1-y^2}{1-x^2}, ~ y(1)=\frac{1}{2}, \\ y^{\prime} = \pm\sqrt{\frac{1-y^2}{1-x^2}}\\ \frac{dy}{\sqrt{1-y^2}}=\pm\frac{dx}{\sqrt{1-x^2}}\\ \int\frac{dy}{\sqrt{1-y^2}}=\pm\int\frac{dx}{\sqrt{1-x^2}}\\ \ln{|y+\sqrt{1-y^2}|}=\pm\ln{|x+\sqrt{1-x^2}|} +\ln|c| \\ \left[ \begin{gathered} y+\sqrt{1-y^2}=c(x+\sqrt{1-x^2}),\\ y+\sqrt{1-y^2}=\frac{c}{(x+\sqrt{1-x^2})}.\\ \end{gathered} \right.\\ y(1)=\frac{1}{2}.\\ \left[ \begin{gathered} \frac{1}{2}+\sqrt{1-(\frac{1}{2})^2}=c(1+\sqrt{1-1^2}),\\ \frac{1}{2}+\sqrt{1-(\frac{1}{2})^2}=\frac{c}{(1+\sqrt{1-1^2})}.\\ \end{gathered} \right.\\ \left[ \begin{gathered} \frac{1}{2}+\sqrt{\frac{3}{4}}=c,\\ \frac{1}{2}+\sqrt{\frac{3}{4}}=c.\\ \end{gathered} \right.\\ c=\frac{1+\sqrt{3}}{2}\\ answer: \\ \left[ \begin{gathered} y+\sqrt{1-y^2}=\frac{(1+\sqrt{3})}{2}(x+\sqrt{1-x^2}),\\ y+\sqrt{1-y^2}=\frac{1+\sqrt{3}}{2(x+\sqrt{1-x^2})}.\\ \end{gathered} \right.\\$