Solve the differential equation $\left(\tan ^{-1} y-x\right) d y=\left(1+y^{2}\right) d x$.

(A) The given differential equation can be written as $\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}}$ ..........$(1)$
Now $(1)$ is a linear differential equation of the form $\frac{d x}{d y}+P_{1} x=Q_{1}$ where $P_{1}=\frac{1}{1+y^{2}}$ and $Q_{1}=\frac{\tan ^{-1} y}{1+y^{2}}$.
Therefore,the integrating factor $I.F. = e^{\int \frac{1}{1+y^{2}} dy} = e^{\tan ^{-1} y}$.
Thus,the solution of the given differential equation is $x e^{\tan ^{-1} y} = \int \left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} dy + C$ ..........$(2)$
Let $I = \int \left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} dy$.
Substituting $\tan ^{-1} y = t$,so that $\left(\frac{1}{1+y^{2}}\right) dy = dt$,we get $I = \int t e^{t} dt = t e^{t} - \int 1 \cdot e^{t} dt = t e^{t} - e^{t} = e^{t}(t-1)$.
Substituting $t = \tan ^{-1} y$,we get $I = e^{\tan ^{-1} y}(\tan ^{-1} y - 1)$.
Substituting the value of $I$ in equation $(2)$,we get $x e^{\tan ^{-1} y} = e^{\tan ^{-1} y}(\tan ^{-1} y - 1) + C$.
Dividing by $e^{\tan ^{-1} y}$,we get $x = \tan ^{-1} y - 1 + C e^{-\tan ^{-1} y}$,which is the general solution.