Question

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

Answer 1

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