Mathematics - Differential Equation Question with Solution | TestHub

$\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+xy\frac{dy}{dx}=0$

Integrating,

$\Rightarrow \int \left(\frac{2y}{2\sqrt{1+y^2}}\right)dy=-\int \left(\frac{1+x^2}{x\sqrt{1+x^2}}\right)dx$
$\Rightarrow \sqrt{1+y^2}=-\int \frac{x}{\sqrt{1+x^2}}dx-\int \frac{x}{x^2\sqrt{1+x^2}}dx$

Put $\sqrt{1+x^2}=t\Rightarrow x^2=t^2-1$ to solve RHS 2nd integration, 

$\Rightarrow \sqrt{1+y^2}=-\sqrt{1+x^2}-\int \frac{t}{\left(t^2-1\right)t}dt$
$\Rightarrow \sqrt{1+y^2}+\sqrt{1+x^2}=-\int \frac{1}{t^2-1}dt$
$\Rightarrow \sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}\ln \left[\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right]+C$