If x=f(y) is the solution of (1+y^2)+(x-2e^^-1y)dydx=0, y(-2,2), with f(0)=1, then f!(1√3)…

Question

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

Accepted Answer

$$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{2e^{\tan^{-1}y}}{1+y^2}.$$

I.F. $=e^{\tan^{-1}y}$. With $t=\tan^{-1}y$:
$$xe^{t} = \int 2e^{2t}\,dt = e^{2t}+C \implies x = e^{t}+Ce^{-t}.$$

Using $y=0,x=1$: $C=0$. So $x=e^{\tan^{-1}y}$; at $y=1/\sqrt3$, $x=e^{\pi/6}$.

Solutions · 1