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

Question

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

Accepted Answer

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

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

Answer

Explanation

More JEE Math questions