Solve the following differential equation: ( tan^-1 y-x) d y=(1+y^2) d x

Writing the equation as a linear differential equation in \(\displaystyle x\): \[\begin{aligned} \frac{dx}{dy} + \frac{x}{1+y^2} &= \frac{\tan^{-1}y}{1+y^2} \end{aligned} \] Integrating factor \(\displaystyle =e^{\int \frac{dy}{1+y^2}} = e^{\tan^{-1}y}\). Hence \[\begin{aligned} x\,e^{\tan^{-1}y} &= \int \frac{\tan^{-1}y}{1+y^2}\,e^{\tan^{-1}y}\,dy \end{aligned} \] Put \(\displaystyle t=\tan^{-1}y\Rightarrow dt=\frac{dy}{1+y^2}\), so \(\displaystyle \int t\,e^{t}\,dt=(t-1)e^{t}+C\): \[\begin{aligned} x\,e^{\tan^{-1}y} &= (\tan^{-1}y-1)e^{\tan^{-1}y}+C \\ x &= (\tan^{-1}y-1) + C\,e^{-\tan^{-1}y} \end{aligned} \]