If $x=f(y)$ is the solution of the differential equation $(1+y^2) + (x-2e^{\tan^{-1}y})\frac{dy}{dx} = 0$, $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$ with $f(0)=1$, then $f(\frac{1}{\sqrt{3}})$ is equal to:
- (1) $e^{\pi/4}$
- (2) $e^{\pi/12}$
- (3) $e^{\pi/3}$
- (4) $e^{\pi/6}$
Solution:
Rearranging the equation:
$$ (1+y^2)dx + (x-2e^{\tan^{-1}y})dy = 0 $$
$$ \frac{dx}{dy} + \frac{x}{1+y^2} = \frac{2e^{\tan^{-1}y}}{1+y^2} $$
This is a linear differential equation in $x$.
Integrating Factor (IF) $= e^{\int \frac{1}{1+y^2}dy} = e^{\tan^{-1}y}$.
Solution:
$$ x \cdot e^{\tan^{-1}y} = \int \frac{2e^{\tan^{-1}y}}{1+y^2} \cdot e^{\tan^{-1}y} dy $$
Let $\tan^{-1}y = t$, then $\frac{dy}{1+y^2} = dt$.
$$ x e^t = \int 2e^t \cdot e^t dt = \int 2e^{2t} dt = e^{2t} + C $$
$$ x e^{\tan^{-1}y} = e^{2\tan^{-1}y} + C $$
$$ x = e^{\tan^{-1}y} + C e^{-\tan^{-1}y} $$
Given $f(0) = 1$ (i.e., when $y=0, x=1$):
$$ 1 = e^0 + C e^0 \Rightarrow 1 = 1 + C \Rightarrow C = 0 $$
So, $x = e^{\tan^{-1}y}$.
We need $f(\frac{1}{\sqrt{3}})$, so set $y = \frac{1}{\sqrt{3}}$:
$$ x = e^{\tan^{-1}(1/\sqrt{3})} = e^{\pi/6} $$
Ans. (4)