Solution of $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$ is :
Answer & explanation
Correct answer: option 2
$\frac{d y}{\sqrt{1-y^2}}=\frac{-d x}{\sqrt{1-x^2}} \Rightarrow \sin ^{-1} y+\sin ^{-1} x=c$
Hence (2) is the correct answer.
Solution of $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$ is :
Correct answer: option 2
$\frac{d y}{\sqrt{1-y^2}}=\frac{-d x}{\sqrt{1-x^2}} \Rightarrow \sin ^{-1} y+\sin ^{-1} x=c$
Hence (2) is the correct answer.