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The general solution of $\left(\frac{d y}{d x}\right)^2=1-x^2-y^2+x^2 y^2$ is
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Verified Answer
The correct answer is:
$2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+C$
$\begin{aligned}
& \text { Given, }\left(\frac{d y}{d x}\right)^2=1-x^2-y^2+x^2 y^2 \\
& \Rightarrow \quad\left(\frac{d y}{d x}\right)^2=\left(1-y^2\right)-x^2\left(1-y^2\right)=\left(1-x^2\right)\left(1-y^2\right) \\
& \therefore \quad \frac{d y}{d x}=\sqrt{\left(1-x^2\right)\left(1-y^2\right)} \\
& \Rightarrow \quad \frac{d y}{\sqrt{1-y^2}}=\sqrt{1-x^2} d x
\end{aligned}$
On integrating both sides, we get
$\begin{aligned}
& \int \frac{d y}{\sqrt{1-y^2}}=\int \sqrt{1-x^2} d x \\
\Rightarrow \quad & \sin ^{-1} y=\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x+\frac{C}{2} \\
\Rightarrow \quad & 2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+C
\end{aligned}$
& \text { Given, }\left(\frac{d y}{d x}\right)^2=1-x^2-y^2+x^2 y^2 \\
& \Rightarrow \quad\left(\frac{d y}{d x}\right)^2=\left(1-y^2\right)-x^2\left(1-y^2\right)=\left(1-x^2\right)\left(1-y^2\right) \\
& \therefore \quad \frac{d y}{d x}=\sqrt{\left(1-x^2\right)\left(1-y^2\right)} \\
& \Rightarrow \quad \frac{d y}{\sqrt{1-y^2}}=\sqrt{1-x^2} d x
\end{aligned}$
On integrating both sides, we get
$\begin{aligned}
& \int \frac{d y}{\sqrt{1-y^2}}=\int \sqrt{1-x^2} d x \\
\Rightarrow \quad & \sin ^{-1} y=\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x+\frac{C}{2} \\
\Rightarrow \quad & 2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+C
\end{aligned}$
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