Let y=y(x) be the solution of the differential equation (1+y^2) e^ x d x+ ^2 x (1+e^ 2 x ) d y=0, y(0)=1 . Then y((π)/(4)) is equal to

Correct answer: C. (1)/(e) aligned & (1+y^2) e^ x d x+ ^2 x(1+e^ 2 x ) d y=0 & ∫ ^2 x e^ x 1+e^ 2 x d x+∫ (d y)/(1+y^2)=C & ^ -1 (e^ x )+ ^ -1 y=C & for x=0, y=1, ^ -1 (1)+ ^ -1 1=C aligned C=(π)/(2) ^ -1 (e^ x )+ ^ -1 y=(π)/(2) Put x=π, ^ -1 e+ ^ -1 y=(π)/(2) ^ -1 y= ^ -1 e y=(1)/(e)

JEE Main 20248 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Differential Equations

Let y=y(x) be the solution of the differential equation (1+y^2) e^...

Let

y=y(x)

be the solution of the differential equation \left(1+y^2\right) e^{\tan x} d x+\cos ^2 x

\left(1+e^{2 \tan x}\right) d y=0, y(0)=1

. Then

y\left(\frac{\pi}{4}\right)

is equal to

Options

A $\frac{2}{e}$
B $\frac{2}{e^2}$
C $\frac{1}{e}$
D $\frac{1}{e^2}$

Correct answer

C. $\frac{1}{e}$

Step-by-step solution

\begin{aligned} & \left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0 \\ & \int \frac{\sec ^2 x e^{\tan x}}{1+e^{2 \tan x}} d x+\int \frac{d y}{1+y^2}=C \\ & \Rightarrow \tan ^{-1}\left(e^{\tan x}\right)+\tan ^{-1} y=C \\ & \text{\text{ for } } x=0, y=1, \tan ^{-1}(1)+\tan ^{-1} 1=C\end{aligned}

\mathrm{C}=\frac{\pi}{2} \tan ^{-1}\left(e^{\tan x}\right)+\tan ^{-1} y=\frac{\pi}{2}

Put

x=\pi, \tan ^{-1} e+\tan ^{-1} y=\frac{\pi}{2} \tan ^{-1} y=\cot ^{-1} e y=\frac{1}{e}$

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