QJEEM24Ma3b80: JEE Main 2024 Mathematics - Class 12 JEEM24Ma3b80 Solution: Let be the solution of the differential equation , . Then is

To determine $y(0)$ , we start by solving the differential equation given:

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

First, we rewrite it in the standard form for a linear differential equation:

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

Next, we find the integrating factor (I.F.):

$\text{I.F.} = e^{\int \frac{1}{1 + x^2} dx} = e^{\tan^{-1} x}$

Multiply through by the integrating factor:

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

This simplifies to:

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

Make the substitution $\tan^{-1} x = t$ , then $\frac{1}{1 + x^2} dx = dt$ :

$\int e^{2t} dt = \frac{e^{2t}}{2} + \frac{C}{2}$

Rewrite in terms of $x$ :

$y \cdot e^{\tan^{-1} x} = \frac{e^{2 \tan^{-1} x}}{2} + \frac{C}{2}$

Use the initial condition $y(1) = 0$ :

$0 = \frac{e^{2 \cdot \tan^{-1} 1}}{2} + \frac{C}{2}$

Since $\tan^{-1} 1 = \frac{\pi}{4}$ :

$0 = \frac{e^{\pi/2}}{2} + \frac{C}{2} \implies C = -e^{\pi/2}$

Thus, the solution is:

$y \cdot e^{\tan^{-1} x} = \frac{e^{2 \tan^{-1} x} - e^{\pi/2}}{2}$

Evaluating $y(0)$ :

$y(0) = y \cdot 1 = \frac{1 - e^{\pi/2}}{2}$

Therefore,

$y(0) = \frac{1}{2} (1 - e^{\pi/2})$

Hence, the correct answer is:

Option B

$\frac{1}{2}\left(1-e^{\pi / 2}\right)$