Let y = y(x) be the solution of the differential equation ( 1 + y 2 ) e tan x d x + cos 2 x ( 1 + e 2 tan x ) d y = 0 , y(0) = 1. Then y ( 4 ) is equal to [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let y = y(x) be the solution of the differential equation $(1 + y^{2}) e^{\tan x} d x + \cos^{2} x (1 + e^{2 \tan x}) d y = 0$ , y(0) = 1. Then $y (\frac{π}{4})$ is equal to [2024]

1 $\frac{2}{e}$

2 $\frac{1}{e^{2}}$

3 $\frac{2}{e^{2}}$

4 $\frac{1}{e}$

Ans.
(4)

We Have, $(1 + y^{2}) e^{\tan x} d x + \cos^{2} x (1 + e^{2 \tan x}) d y = 0$

$\Rightarrow \frac{e^{\tan x}}{\cos^{2} x (1 + e^{2 \tan x})} d x + \frac{1}{1 + y^{2}} d y = 0$

Integrating both sides, we get

$\int \frac{\sec^{2} x e^{\tan x}}{1 + e^{2 \tan x}} d x + \int \frac{d y}{1 + y^{2}} = C$

Put $e^{\tan x} = t \Rightarrow \sec^{2} x e^{\tan x} d x = d t$

$\Rightarrow \tan^{- 1} (t) + \tan^{- 1} y = C \Rightarrow \tan^{- 1} (e^{\tan x}) + \tan^{- 1} y = C$

Since, y(0) = 1

$\Rightarrow \frac{π}{4} + \frac{π}{4} = C \Rightarrow C = \frac{π}{2}$

So, $\tan^{- 1} (e^{\tan x}) + \tan^{- 1} y = \frac{π}{2}$

For $x = \frac{π}{4}$ , we have

$\tan^{- 1} y = \frac{π}{2} - \tan^{- 1} e$

$\Rightarrow \tan^{- 1} y = \cot^{- 1} e \Rightarrow \tan^{- 1} y = \tan^{- 1} \frac{1}{e} \Rightarrow y = \frac{1}{e}$

Want Us to Email you About Special Offers & Updates?