If x=f(y) is the solution of the differential equation (1+y^2)+(x-2 e^ ^-1 y) d y~d x=0, y…

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

$$\begin{aligned}
& \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{2 e^{\tan ^{-1} y}}{1+y^2} \
& \text { I.F. }=e^{\tan ^{-1} y} \
& x^{\tan ^{-1} y}=\int \frac{2\left(e^{\tan ^{-1} y}\right)^2 d y}{1+y^2} \
& \text { Put } \tan ^{-1} y=t, \frac{d y}{1+y^2}=d t \
& x e^{\tan ^{-1} y}=\int 2 e^{2 t} d t \
& x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+c \
& x=e^{\tan ^{-1} y}+c e^{-\tan ^{-1} y} \
& \because y=0, x=1 \
& 1=1+c \Rightarrow c=0 \
& y=\frac{1}{\sqrt{3}}, x=e^{\pi / 6}
\end{aligned}$$

Solutions · 1