Question

The general solution of $\left(\frac{d y}{d x}\right)^{2}=1-x^{2}-y^{2}+x^{2} y^{2}$ is

• A. $2 sin^{-1} y=x \sqrt{1-y^2}+c$
• B. $cos^{-1} y=x\,cos^{-1} x+c$
• C. $Sin^{-1}y=\frac{1}{2}Sin^{-1}x+c$
• D. $2\,sin^{-1} y=x \sqrt{1-x^2}+sin^{-1} x+c$

Answer 1

Answer: (D)

Given, $\left(\frac{d y}{d x}\right)^{2}=1-x^{2}-y^{2}+x^{2} y^{2}$
$\Rightarrow \left(\frac{d y}{d x}\right)^{2}=\left(1-x^{2}\right)-y^{2}\left(1-x^{2}\right)$
$\Rightarrow \frac{d y}{d x}=\sqrt{\left(1-x^{2}\right)} \sqrt{\left(1-y^{2}\right)}$
$\quad \int \frac{d y}{\sqrt{\left(1-y^{2}\right)}}=\int \sqrt{\left(1-x^{2}\right)} d x$ (on integrating)
$\Rightarrow \sin ^{-1} y=\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x+\frac{C}{2}$
$\Rightarrow 2 \sin ^{-1} y=x \sqrt{1-x^{2}}+\sin ^{-1} x+C$