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The solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$ is
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Verified Answer
The correct answer is:
$\sin ^{-1} x+\sin ^{-1} y=c$
Given, $\frac{d y}{d x}+\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}=0$
$$
\Rightarrow \quad \frac{d y}{\sqrt{1-y^2}}+\frac{d x}{\sqrt{1-x^2}}=0
$$
On integrating, we get
$$
\begin{aligned}
& \Rightarrow \quad \int \frac{d y}{\sqrt{1-y^2}}+\int \frac{d x}{\sqrt{1-x^2}}=0 \\
& \Rightarrow \quad \sin ^{-1} y+\sin ^{-1} x=c
\end{aligned}
$$
$$
\Rightarrow \quad \frac{d y}{\sqrt{1-y^2}}+\frac{d x}{\sqrt{1-x^2}}=0
$$
On integrating, we get
$$
\begin{aligned}
& \Rightarrow \quad \int \frac{d y}{\sqrt{1-y^2}}+\int \frac{d x}{\sqrt{1-x^2}}=0 \\
& \Rightarrow \quad \sin ^{-1} y+\sin ^{-1} x=c
\end{aligned}
$$
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