Solve the differential equation: (1 + y²) dx = (tan⁻¹ y - x) dy

$(1+y^2)\,dx = (\tan^{-1}y - x)\,dy \Rightarrow \frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1}y}{1+y^2}$ This is linear in $x$. Integrating factor $= e^{\int \frac{dy}{1+y^2}} = e^{\tan^{-1}y}$. $x\,e^{\tan^{-1}y} = \int \frac{\tan^{-1}y}{1+y^2}e^{\tan^{-1}y}dy$ Put $\tan^{-1}y = t$, so $\frac{dy}{1+y^2} = dt$: $\int t e^t dt = te^t - e^t + C$ $x\,e^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y - 1) + C$ Hence $x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$.