Solution of the differential equation $\frac{xdy}{x^2+d^2}=\left(\frac{y}{x^2+y^2}-1\right)dx, $ is

$tan^{-1}\left(\frac{y}{x}\right) =-x+C$

The correct answer is option (1) : $tan^{-1}\left(\frac{y}{x}\right) =-x+C$

The given differential equation can be written as $\frac{xdy-ydx}{x^2+d^2}=-dx⇒d\begin{Bmatrix}tan^{-1}\left(\frac{y}{x}\right) \end{Bmatrix}=-dx$

On integrating, we obtain

$tan^{-1}\left(\frac{y}{x}\right) = -x + fC$, which is the required solution.