The solution of the differential equation $\left(1+y^2\right)+\left(x-e^{{\tan }^{-1}y}\right)\frac{dy}{dx}=0$, is

We have the differential equation

 $\left(1+y^2\right)+\left(x-e^{{\tan }^{-1}y}\right)\frac{dy}{dx}=0$

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

Integrating factor $\mathrm{I}.\mathrm{F}.=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$

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

$\Rightarrow 2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+c$, where $c$ is the constant of integration.