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Question
The solution of the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + 1 + y^2 = 0\], is
Options
tan−1 x − tan−1 y = tan−1 C
tan−1 y − tan−1 x = tan−1 C
tan−1 y ± tan−1 x = tan C
tan−1 y + tan−1 x = tan−1 C
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Solution
tan−1y + tan−1x = tan−1C
We have,
\[\left( 1 + x^2 \right)\frac{dy}{dx} + 1 + y^2 = 0\]
\[ \Rightarrow \left( 1 + x^2 \right)\frac{dy}{dx} = - \left( 1 + y^2 \right)\]
\[ \Rightarrow \frac{1}{\left( 1 + y^2 \right)}dy = - \frac{1}{\left( 1 + x^2 \right)}dx\]
Integrating both sides we get,
\[\int\frac{1}{\left( 1 + y^2 \right)}dy = - \int\frac{1}{\left( 1 + x^2 \right)}dx\]
\[ \Rightarrow \tan^{- 1} y = - \tan^{- 1} x + \tan^{- 1} C\]
\[ \Rightarrow \tan^{- 1} y + \tan^{- 1} x = \tan^{- 1} C\]
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