Let y=y(x) be the solution curve of the differential equation (1+x^ 2 ) d y+(y- ^ -1 x) d x=0, y(0)=1 . Then the value of y(1) is :

Correct answer: B. 2 e^ π / 4 +(π)/(4)-1 Rewrite as (dy)/(dx) + (y)/(1+x^2) = ^ -1 x 1+x^2 . IF = e^ ^ -1 x . y · e^ ^ -1 x = ∫ ^ -1 x 1+x^2 · e^ ^ -1 x dx . Let t = ^ -1 x : = ∫ te^t dt = e^t(t-1) + C . y = ^ -1 x - 1 + Ce^ - ^ -1 x . Using y(0) = 1 : C = 2 . y(1) = (π)/(4) - 1 + 2 e^ π/4 .

JEE Main 202621 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Differential Equations

Let y=y(x) be the solution curve of the differential equation (1+x^...

Let

y=y(x)

be the solution curve of the differential equation \left(1+x^{2}\right) \mathrm{d} y+\left(y-\tan ^{-1} x\right) d x=0, y(0)=1

. Then the value of

y(1) is :

Options

A $\frac{4}{\mathrm{e}^{\pi / 4}}+\frac{\pi}{2}-1$
B $\frac{2}{\mathrm{e}^{\pi / 4}}+\frac{\pi}{4}-1$
C $\frac{2}{\mathrm{e}^{\pi / 4}}-\frac{\pi}{4}-1$
D $\frac{4}{e^{\pi / 4}}-\frac{\pi}{2}-1$

Correct answer

B. $\frac{2}{\mathrm{e}^{\pi / 4}}+\frac{\pi}{4}-1$

Step-by-step solution

Rewrite as

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

.
IF

= e^{\tan^{-1}x}

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

.
Let

t = \tan^{-1}x

= \int te^t dt = e^t(t-1) + C

y = \tan^{-1}x - 1 + Ce^{-\tan^{-1}x}

.
Using

y(0) = 1

C = 2

y(1) = \frac{\pi}{4} - 1 + \frac{2}{e^{\pi/4}}

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