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

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Differential Equations

x = f (y)

is the solution of the differential equation

(1 + y^{2}) + (x - 2 e^{t a n^{- 1} y}) \frac{dy}{dx} = 0, y \in (- \frac{π}{2}, \frac{π}{2})

with

f (0) = 1

, then

f (\frac{1}{3})

is equal to :

Answer: (D)

Rewrite the equation as a linear equation in x with y as the independent variable:

\frac{d x}{d y} + \frac{x}{1 + y ^{2}} = \frac{2 e ^{t a n^{- 1} y}}{1 + y ^{2}} .

This is linear in x with integrating factor

I = e^{\int \frac{d y}{1 + y ^{2}}} = e^{t a n^{- 1} y} .

Multiplying through and using

d t = \frac{d y}{1 + y ^{2}}

with

t = tan^{- 1} y

x e^{t a n^{- 1} y} = \int 2 e^{2 t a n^{- 1} y} \cdot \frac{d y}{1 + y ^{2}} = \int 2 e^{2 t} d t = e^{2 t} + C = e^{2 t a n^{- 1} y} + C .

x = e^{t a n^{- 1} y} + C e^{- t a n^{- 1} y} .

Using

f (0) = 1

(y=0 gives x=1):

1 = 1 + C \Rightarrow C = 0

.

So

x = e^{t a n^{- 1} y}

, i.e.

f (y) = e^{t a n^{- 1} y}

.

At

y = \frac{1}{3}

tan^{- 1} \frac{1}{3} = \frac{π}{6}

, so

f (\frac{1}{3}) = e^{π /6} .

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option