If \(x=f(y)\) is the solution of the differential equation \( \left(1+y^2\right)+\left(x-2 e^{\tan ^{-1} y}\right) \frac…

\(\left(1+{y}^{2}\right)+\left(x-2{e}^{{\tan }^{-1}y}\right)\frac{dy}{dx}=0\)

\(\Rightarrow \frac{dx}{dy}+\frac{x}{1+{y}^{2}}=\frac{2{e}^{{\tan }^{-1}y}}{1+{y}^{2}}\\ \text{I.F. }={\mathrm{e}}^{\int \frac{\mathrm{dy}}{1+{\mathrm{y}}^{2}}}={\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}\\ {\mathrm{xe}}^{{\tan }^{-1}\mathrm{y}}=\int \frac{2{\left({\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}\right)}^{2}\mathrm{dy}}{1+{\mathrm{y}}^{2}}\\ \text{Put }{\tan }^{-1}\mathrm{y}=\mathrm{t},\frac{\mathrm{dy}}{1+{\mathrm{y}}^{2}}=\mathrm{dt}\\ {\mathrm{xe}}^{{\tan }^{-1}\mathrm{y}}=\int 2{\mathrm{e}}^{2\mathrm{t}}\mathrm{dt}\\ {\mathrm{xe}}^{{\tan }^{-1}\mathrm{y}}={\mathrm{e}}^{2{\tan }^{-1}\mathrm{y}}+\mathrm{c}\\ \mathrm{x}={\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}+{\mathrm{ce}}^{-{\tan }^{-1}\mathrm{y}}\\ ∵\mathrm{y}=0,\mathrm{x}=1\\ 1=1+\mathrm{c}\Rightarrow \mathrm{c}=0\\ \mathrm{x}={\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}\\ \text{at}\mathrm{y}=\frac{1}{\sqrt{3}}\Rightarrow \mathrm{x}={\mathrm{e}}^{\pi /6}\)