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

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{1+\mathrm{y}^2}=\frac{2 \mathrm{e}^{	an ^{-1} \mathrm{y}}}{1+\mathrm{y}^2} \ & 	ext { I.F. }=\mathrm{e}^{	an ^{-1} \mathrm{y}} \ & \mathrm{xe}^{	an ^{-1} \mathrm{y}}=\int \frac{2\left(\mathrm{e}^{	an ^{-1} \mathrm{y}}ight)^2 \mathrm{dy}}{1+\mathrm{y}^2} \ & 	ext { Put } 	an ^{-1} \mathrm{y}=\mathrm{t}, \frac{\mathrm{dy}}{1+\mathrm{y}^2}=\mathrm{dt} \ & \mathrm{xe}^{	an ^{-1} \mathrm{y}}=\int 2 \mathrm{e}^{2 \mathrm{t}} \mathrm{dt} \ & \mathrm{xe}^{	an ^{-1} \mathrm{y}}=\mathrm{e}^{2 	an ^{-1} \mathrm{y}}+\mathrm{c} \ & \mathrm{x}=\mathrm{e}^{	an ^{-1} \mathrm{y}}+\mathrm{ce}^{-	an ^{-1} \mathrm{y}} \ & \because \mathrm{y}=0, \mathrm{x}=1 \ & 1=1+\mathrm{c} \Rightarrow \mathrm{c}=0 \ & \mathrm{y}=\frac{1}{\sqrt{3}}, \mathrm{x}=\mathrm{e}^{\pi / 6}\end{aligned}$

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

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