JEE Main 2024 — Differential Equations Question with Solution

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

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

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

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