Question

$y\sqrt{1+x^2}+x\sqrt{1+y^2}\frac{dy}{dx}=0$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ y\sqrt{1+x^2}+x\sqrt{1+y^2}\frac{dy}{dx}=0 \\ \Rightarrow x\sqrt{1+y^2}\frac{dy}{dx}=-y\sqrt{1+x^2} \\ \Rightarrow x\sqrt{1+y^2}dy=-y\sqrt{1+x^2}dx \\ \Rightarrow \frac{\sqrt{1+y^2}}{y}dy=-\frac{\sqrt{1+x^2}}{x}dx \end{gathered}$$

$$\begin{gathered} \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{\sqrt{1+y^2}}{y}dy=-\int \frac{\sqrt{1+x^2}}{x}dx \\ \mathrm{Putting}1+y^{\mathit{2}}=t^{\mathit{2}}\mathit{}\mathrm{and}\mathit{}1+x^{\mathit{2}}=u^{\mathit{2}},\mathrm{we}\mathrm{get} \\ 2y\mathit{}dy\mathit{=}2t\mathit{}dt\mathit{}\mathrm{and}\mathit{}2x\mathit{}dx\mathit{=}2u\mathit{}du \\ \mathit{\Rightarrow }dy\mathit{=}\frac{t}{y}dt\mathit{}\mathrm{and}\mathit{}dx\mathit{=}\frac{u}{x}du \\ \therefore \int \frac{{\mathit{\text{t}}}^2}{y^2}\mathrm{dt}=-\int \frac{u^2}{x^2}dx \\ \Rightarrow \int \frac{t^2}{t^2-1}dt=-\int \frac{u^2}{u^2-1}du \end{gathered}$$

$$\begin{gathered} \Rightarrow \int \frac{t^2-1+1}{t^2-1}dt=-\int \frac{u^2-1+1}{u^2-1}du \\ \Rightarrow \int dt+\int \frac{1}{t^2-1}dt=-\int du-\int \frac{1}{u^2-1}du \\ \Rightarrow t+\frac{1}{2}\log \left|\frac{t-1}{t+1}\right|=-u-\frac{1}{2}\log \left|\frac{u-1}{u+1}\right|+C \\ \mathrm{Substituting}\mathit{}t\mathrm{by}\sqrt{1+y^2}\mathit{}\mathrm{and}\mathit{}u\mathrm{by}\sqrt{1+x^2} \end{gathered}$$

$$\begin{gathered} \sqrt{1+y^2}+\frac{1}{2}\log \left|\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}\right|=-\sqrt{1+x^2}-\frac{1}{2}\log \left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C \\ \Rightarrow \sqrt{1+y^2}+\sqrt{1+x^2}+\frac{1}{2}\log \left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+\frac{1}{2}\log \left|\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}\right|=C \\ \mathrm{Hence},\sqrt{1+y^2}+\sqrt{1+x^2}+\frac{1}{2}\log \left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+\frac{1}{2}\log \left|\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}\right|=C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$