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प्रश्न
For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]
उत्तर
We have,
\[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]
\[ \Rightarrow \frac{1}{\left( 1 + y^2 \right)}dy = \left( 1 + x^2 \right)dx\]
Integrating both sides, we get
\[\int\frac{1}{\left( 1 + y^2 \right)}dy = \int\left( 1 + x^2 \right)dx\]
\[ \Rightarrow \tan^{- 1} y = x + \frac{x^3}{3} + C\]
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