Question

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

Answer 1

Given, $\frac{dy}{dx}+\frac{1+y^2}{1+x^2}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{1+y^2}{1+x^2}$
$\Rightarrow \frac{dy}{1+y^2}=-\frac{dx}{1+x^2}$

Integrating both the sides
$\int \frac{dy}{1+y^2}=-\int \frac{dx}{1+x^2}$
$\Rightarrow {\tan }^{-1}y=-{\tan }^{-1}x+c$
$\Rightarrow {\tan }^{-1}x+{\tan }^{-1}y=c$
$\Rightarrow {\tan }^{-1}\left(\frac{x+y}{1-xy}\right)=c$
$\Rightarrow \frac{x+y}{1-xy}=\tan c$
$\Rightarrow \frac{x+y}{1-xy}=C,$ where $C=\tan c$

$\frac{x+y}{1-xy}=C$ is the general solution.