Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right) \\ \Rightarrow \frac{1}{\left(1+y^2\right)}dy=\left(1+x^2\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{\left(1+y^2\right)}dy=\int \left(1+x^2\right)dx \\ \Rightarrow {\tan }^{-1}y=x+\frac{x^3}{3}+\mathrm{C} \\ \mathrm{Hence},\tan {}^{-1}\mathrm{y}=\mathrm{x}+\frac{{\mathrm{x}}^3}{3}+\mathrm{C}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$