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The general solution of the differential equation $\sqrt{1-x^{2} y^{2}} \cdot d x=y \cdot d x+x \cdot d y$ is
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Verified Answer
The correct answer is:
$\sin (x+C)=x y$
Given differential equation is
$\begin{array}{ll}
& \sqrt{1-x^{2} y^{2}} \cdot d x=y \cdot d x+x d y \\
\Rightarrow \quad & \sqrt{1-(x y)^{2}} d x=d(x y) \\
\Rightarrow \quad & \int d x=\int \frac{d(x y)}{\sqrt{1-(x y)^{2}}} \quad \text { (on integrating) } \\
\Rightarrow \quad & x=\sin ^{-1}(x y)-C \\
\Rightarrow \quad & \sin (x+C)=x y
\end{array}$
$\begin{array}{ll}
& \sqrt{1-x^{2} y^{2}} \cdot d x=y \cdot d x+x d y \\
\Rightarrow \quad & \sqrt{1-(x y)^{2}} d x=d(x y) \\
\Rightarrow \quad & \int d x=\int \frac{d(x y)}{\sqrt{1-(x y)^{2}}} \quad \text { (on integrating) } \\
\Rightarrow \quad & x=\sin ^{-1}(x y)-C \\
\Rightarrow \quad & \sin (x+C)=x y
\end{array}$
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