The general solution of the differential equation ( ^ -1 y-x) d...

The given differential equation can be written as $\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}} \ldots$ (i) Now, eq. (i) is a linear differential equation of the form $\frac{d x}{d y}+\mathrm{P}_{1} x=\mathrm{Q}_{1}$ where $\mathrm{P}_{1}=\frac{1}{1+y^{2}} and$\mathrm{Q}_{1}=\frac{\tan ^{-1} y}{1+y^{2}}$Therefore, I.F$=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}$Thus, the solution of the given differential equation is given by$x e^{\tan ^{-1} y}=\int\left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} \mathrm{dy}+\mathrm{C} \ldots$(ii) Let$\mathrm{I}=\int\left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} \mathrm{dy}$On substituting$\tan ^{-1} y=t$, so that$\left(\frac{1}{1+y^{2}}\right) \mathrm{dy}=\mathrm{dt}$, we get$I=\int t e^{t} d t=t e^{t}-\int 1 \cdot e^{t} d t=t e^{t}-e^{t}=e^{t}(t-1)$or$\mathrm{I}=e^{\tan ^{-1} y} \left(\tan ^{-1}y-1\right)$On substituting the value of 1 in equation (i1), we get$x \cdote^{\tan ^{-1} y}=\cdote^{\tan ^{-1} y}\left(\tan ^{-1}y-1\right)+\mathrm{C}or $x=\left(\tan ^{-1}y-1\right)$+$\mathrm{C} e^{\tan ^{-1} y}$ which is the general solution of the given differntial equation.