Question

Find the solution of the differential equation
$x\sqrt{1+y^2}dx+y\sqrt{1+x^2}dy=0$

Answer 1

$$\begin{gathered} x\sqrt{1+y^2}dx+y\sqrt{1+x^2}dy=0 \\ \Rightarrow y\sqrt{1+x^2}dy=-x\sqrt{1+y^2}dx \\ \Rightarrow \frac{y}{\sqrt{1+y^2}}dy=\frac{-x}{\sqrt{1+x^2}}dx \\ \Rightarrow \int \frac{y}{\sqrt{1+y^2}}dy=-\int \frac{x}{\sqrt{1+x^2}}dx \end{gathered}$$
$$\begin{gathered} \mathrm{Let}1+y^2=t^2\mathrm{and}1+x^2=p^2 \\ \Rightarrow 2ydy=2tdt\mathrm{and}2xdx=2pdp \\ \Rightarrow ydy=tdt\mathrm{and}xdx=pdp \\ \mathrm{Substituting}\mathrm{in}\mathrm{above}\mathrm{equation},\mathrm{we}\mathrm{get} \\ \Rightarrow \int dt=-\int dp \\ \Rightarrow t=-p+C \\ \Rightarrow \sqrt{1+x^2}+\sqrt{1+y^2}=C \end{gathered}$$