Question

(1 + x) (1 + y²) dx + (1 + y) (1 + x²) dy = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(1+x\right)\left(1+y^2\right)dx+\left(1+y\right)\left(1+x^2\right)dy=0 \\ \Rightarrow \left(1+x\right)\left(1+y^2\right)dx=-\left(1+y\right)\left(1+x^2\right)dy \\ \Rightarrow \frac{1+x}{1+x^2}dx=-\frac{1+y}{1+y^2}dy \\ \mathrm{Integarting}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1+x}{1+x^2}dx=-\int \frac{1+y}{1+y^2}dy \\ \Rightarrow \int \frac{1}{1+x^2}dx+\int \frac{x}{1+x^2}dx=-\int \frac{1}{1+y^2}dy-\int \frac{y}{1+y^2}dy \\ \mathrm{Substituting}1+x^2=t\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{integral}\mathrm{of}\mathrm{LHS}\mathrm{and}1+y^2=u\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{integral}\mathrm{of}\mathrm{RHS},\mathrm{we}\mathrm{get} \\ 2xdx=dt\mathrm{and}2ydy=du \\ \therefore \int \frac{1}{1+x^2}dx+\frac{1}{2}\int \frac{1}{t}dt=-\int \frac{1}{1+y^2}dy-\frac{1}{2}\int \frac{1}{u}du \\ \Rightarrow {\tan }^{-1}x+\frac{1}{2}\log \left|t\right|=-{\tan }^{-1}y-\frac{1}{2}\log \left|u\right|+C \\ \Rightarrow {\tan }^{-1}x+\frac{1}{2}\log \left|1+x^2\right|=-{\tan }^{-1}y-\frac{1}{2}\log \left|1+y^2\right|+C \\ \Rightarrow {\tan }^{-1}x+{\tan }^{-1}y+\frac{1}{2}\log \left|1+x^2\right|+\frac{1}{2}\log \left|1+y^2\right|=C \\ \Rightarrow {\tan }^{-1}x+{\tan }^{-1}y+\frac{1}{2}\log \left|\left(1+x^2\right)\left(1+y^2\right)\right|=C \\ \mathrm{Hence},{\tan }^{-1}x+{\tan }^{-1}y+\frac{1}{2}\log \left|\left(1+x^2\right)\left(1+y^2\right)\right|=C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$