Question

The solution of $\frac{d y}{d x}=\sqrt{1-x^{2}-y^{2}+x^{2} y^{2}}$ is

• A. $\sin ^{-1} y=\left(\frac{1}{2}\right) \sin ^{-1} x+c$
• B. $2 \sin ^{-1} y=x \sqrt{1-y^{2}}+c$
• C. $2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+c$
• D. $\cos ^{-1} y=x \cos ^{-1} x+c$

Answer 1

Answer: (C)

$\frac{d y}{d x}=\sqrt{1-x^{2}-y^{2}+x^{2} y^{2}}$
$\frac{d y}{d x}=\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)}$
$\frac{d y}{\sqrt{1-y}}=\sqrt{\left(1-x^{2}\right)} d x$
$\sin ^{-1} y=\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x+c$
$2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+c$