Question

(y² + 1) dx − (x² + 1) dy = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(y^2+1\right)dx-\left(x^2+1\right)dy=0 \\ \Rightarrow \left(y^2+1\right)dx=\left(x^2+1\right)dy \\ \Rightarrow \frac{1}{x^2+1}dx=\frac{1}{y^2+1}dy \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{x^2+1}dx=\int \frac{1}{y^2+1}dy \\ \mathit{\Rightarrow }{\tan }^{-1}\mathit{}x\mathit{=}{\tan }^{-1}\mathit{}y\mathit{+}C \\ \mathit{\Rightarrow }{\tan }^{-1}\mathit{}x\mathit{-}{\tan }^{-1}y\mathit{=}C \\ \mathrm{Hence},{\tan }^{-1}\mathit{}x\mathit{-}{\tan }^{-1}y\mathit{=}C\mathit{}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$