Question

\[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]

Answer 1

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
\[\int\frac{1}{\left( 1 + y^2 \right)}dy = \int\left( 1 + x^2 \right) dx \]
\[ \Rightarrow \tan^{- 1} y = x + \frac{x^3}{3} + C\]
\[\text{ Hence, }\tan {}^{- 1} y = x + \frac{x^3}{3} +\text{ C is the required solution }.\]