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प्रश्न
(y2 + 1) dx − (x2 + 1) dy = 0
उत्तर
We have,
\[\left( y^2 + 1 \right) dx - \left( x^2 + 1 \right) dy = 0\]
\[ \Rightarrow \left( y^2 + 1 \right) dx = \left( x^2 + 1 \right) dy\]
\[ \Rightarrow \frac{1}{x^2 + 1}dx = \frac{1}{y^2 + 1}dy\]
Integrating both sides, we get
\[\int\frac{1}{x^2 + 1}dx = \int\frac{1}{y^2 + 1}dy\]
\[ \Rightarrow \tan^{- 1} x = \tan^{- 1} y + C\]
\[ \Rightarrow \tan^{- 1} x - \tan^{- 1} y = C\]
\[\text{ Hence, } \tan^{- 1} x - \tan^{- 1} y = \text{ C is the required solution .}\]
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