If x=f(y) is the solution of the differential | vedclass

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(D) The given differential equation is $(1+y^2)+(x-2 e^{	an ^{-1} y}) \frac{d y}{d x}=0$. Rearranging the terms,we get $\frac{d x}{d y} = -\frac{x-2

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$e^{\pi / 6}$

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(D) The given differential equation is $(1+y^2)+(x-2 e^{	an ^{-1} y}) \frac{d y}{d x}=0$. Rearranging the terms,we get $\frac{d x}{d y} = -\frac{x-2 e^{	an ^{-1} y}}{1+y^2} = \frac{2 e^{	an ^{-1} y}-x}{1+y^2}$. This can be written as $\frac{d x}{d y} + \frac{x}{1+y^2} = \frac{2 e^{	an ^{-1} y}}{1+y^2}$. This is a linear differential equation of the form $\frac{d x}{d y} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{2 e^{	an ^{-1} y}}{1+y^2}$. The integrating factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{	an ^{-1} y}$. The solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$. $x e^{	an ^{-1} y} = \int \frac{2 e^{	an ^{-1} y}}{1+y^2} \cdot e^{	an ^{-1} y} dy = \int \frac{2 e^{2 	an ^{-1} y}}{1+y^2} dy$. Let $t = 	an ^{-1} y$,then $dt = \frac{1}{1+y^2} dy$. $x e^{	an ^{-1} y} = \int 2 e^{2t} dt = e^{2t} + C = e^{2 	an ^{-1} y} + C$. Given $f(0)=1$,i.e.,$x=1$ when $y=0$: $1 \cdot e^{	an ^{-1} 0} = e^{2 	an ^{-1} 0} + C \implies 1 \cdot 1 = 1 + C \implies C = 0$. Thus,$x e^{	an ^{-1} y} = e^{2 	an ^{-1} y} \implies x = e^{	an ^{-1} y}$. For $y = \frac{1}{\sqrt{3}}$,$x = e^{	an ^{-1}(1/\sqrt{3})} = e^{\pi / 6}$.

If $x=f(y)$ is the solution of the differential equation $(1+y^2)+(x-2 e^{\tan ^{-1} y}) \frac{d y}{d x}=0$,$y \in(-\frac{\pi}{2}, \frac{\pi}{2})$ with $f(0)=1$,then $f(\frac{1}{\sqrt{3}})$ is equal to :

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