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प्रश्न
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
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उत्तर
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
\[ \Rightarrow \left( 1 + y^2 \right) \tan^{- 1} xdx = - 2y\left( 1 + x^2 \right)dy\]
\[ \Rightarrow \frac{\tan^{- 1} x}{\left( 1 + x^2 \right)}dx = - \frac{2y}{\left( 1 + y^2 \right)}dy\]
\[ \Rightarrow \int\frac{\tan^{- 1} x}{\left( 1 + x^2 \right)}dx = - \int\frac{2y}{\left( 1 + y^2 \right)}dy\]
\[ \Rightarrow \frac{\left( \tan^{- 1} x \right)^2}{2} = - \log\left| 1 + y^2 \right| + C\]
\[ \Rightarrow \frac{\left( \tan^{- 1} x \right)^2}{2} + \log\left| 1 + y^2 \right| = C\]
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