If $x=f(y)$ is the solution of the differential equation $(1+y^2)+\left(x-2e^{ an^{-1}y} ight)\frac{dy}{dx}=0$, $y\in\left(-\frac\pi2,\frac\pi2 ight)$ with $f(0)=1$, then $f\left(\frac1{\sqrt3} ight)$ is equal to :

Question

If $x=f(y)$ is the solution of the differential equation $(1+y^2)+\left(x-2e^{	an^{-1}y}ight)\frac{dy}{dx}=0$, $y\in\left(-\frac\pi2,\frac\pi2ight)$ with $f(0)=1$, then $f\left(\frac1{\sqrt3}ight)$ is equal to :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{2e^{	an^{-1}y}}{1+y^2}$. The integrating factor is $e^{	an^{-1}y}$, giving $xe^{	an^{-1}y}=e^{2	an^{-1}y}+C$. $f(0)=1\Rightarrow C=0$, so $x=e^{	an^{-1}y}$ and $f\left(\frac1{\sqrt3}ight)=e^{\pi/6}$. Answer (4).

Q. If $x=f(y)$ is the solution of the differential equation $(1+y^2)+\left(x-2e^{\tan^{-1}y}\right)\frac{dy}{dx}=0$, $y\in\left(-\frac\pi2,\frac\pi2\right)$ with $f(0)=1$, then $f\left(\frac1{\sqrt3}\right)$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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