Question

For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]

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

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

\[ \Rightarrow \tan^{- 1} y = x + \frac{x^3}{3} + C\]