Question

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

Answer 1

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

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

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

integrating both sides

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

Let $y=\tan \theta ,$

$=\int \frac{\sqrt{1+{\tan }^2\theta }}{\tan \theta }.{\sec }^2\theta d\theta$

$=\int \frac{{\sec }^3\theta }{\tan \theta }d\theta$

$=\int \frac{{\sec }^2}{\sin \theta }d\theta$

$=\int {\sec }^2\theta \csc \theta d\theta$

$=\int {\sec }^2\theta d\theta +\int {\csc }^2\theta d\theta$

$=\tan \theta -\cot \theta$

$=y-\frac{1}{y}$

$\therefore \int \sqrt{1+y^2}\frac{dy}{y}=-\int \sqrt{1+x^2}\frac{dx}{x}$

$y-\frac{1}{y}=-(x-\frac{1}{x})+c$

$\frac{y^2-1}{y}=\frac{1-x^2}{x}+c$

Hence, solved.