\(\begin{equation} $$\begin{aligned}&f(y) \text { is the solution of differential equation }\\&\left(1+y^2\right)+\left(…

\((1+{y}^{2})+(x-2{\tan }^{-1}y)\frac{dy}{dx}=0\\ (1+{y}^{2})=-(x-2{\tan }^{-1}y)\frac{dy}{dx}\\ \frac{dx}{dy}=-\frac{-x}{1+{y}^{2}}+\frac{2{\tan }^{-1}y}{1+{y}^{2}}\\ \frac{dx}{dy}+\frac{x}{1+{y}^{2}}=\frac{2{\tan }^{-1}y}{1+{y}^{2}}\\ Oncomparingwith\frac{dx}{dy}+Px=Q\\ P=\frac{1}{1+{y}^{2}},Q=\frac{2{\tan }^{-1}y}{1+{y}^{2}}\\ I.F.={e}^{\int \frac{1}{1+{y}^{2}}dy}={e}^{{\tan }^{-1}y}\\ Nowgeneralsolutionis\\ x.(I.F)=\int Q.(I.F)dy+c\\ x.{e}^{{\tan }^{-1}y}=\int \frac{2{\tan }^{-1}y}{1+{y}^{2}}.{e}^{{\tan }^{-1}y}dx\\ let{\tan }^{-1}y=t\\ \frac{1}{1+{y}^{2}}dy=dt\\ x.{e}^{{\tan }^{-1}y}=\int 2t.{e}^{t}dt\\ x.{e}^{{\tan }^{-1}y}=2\left(t\int {e}^{t}dt-\int \left{\frac{dt}{dt}\int {e}^{t}dt\right}dt\right)\\ x.{e}^{{\tan }^{-1}y}=2\left(t.{e}^{t}-{e}^{t}\right)+c\\ x.{e}^{{\tan }^{-1}y}=2{e}^{{\tan }^{-1}y}({\tan }^{-1}y-1)+c\\ x=2\left({\tan }^{-1}y-1\right)+c{e}^{-{\tan }^{-1}y}\)

\(
\begin{gathered}
$$\begin{array}{l}y=0, x=1 \\c=3 \\x e^{\tan ^{-1} y}=2 \tan ^{-1} y \cdot e^{-tan ^{-1} y } -2 e^{\tan ^{-1} y}+3 \\f(y)=x=2 \tan ^{-1} y-2+3 e^{-tan ^{-1} y } \\f\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{3}-2+3 e^{-\pi / 6}\end{array}$$
\end{gathered}
\)