Let x = x(y) be the solution of the differential equation , y > 0 and . Then cos (x(2)) is equal to : [2025]

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 x = x(y) be the solution of the differential equation $y = (x - y \frac{d x}{d y}) \sin (\frac{x}{y})$ , y > 0 and $x (1) = \frac{π}{2}$ . Then cos (x(2)) is equal to : [2025]

1 $2 {(\log_{e} 2)}^{2} - 1$

2 $1 - 2 (\log_{e} 2)$

3 $1 - 2 {(\log_{e} 2)}^{2}$

4 $2 (\log_{e} 2) - 1$

Ans.
(1)

We have, $y d y = (x d y - y d x) \sin (\frac{x}{y})$

$\Rightarrow \frac{d y}{y} = (\frac{x d y - y d x}{y^{2}}) \sin (\frac{x}{y})$

$\Rightarrow \frac{d y}{y} = \sin (\frac{x}{y}) d (\frac{x}{y})$

$\Rightarrow \log y = \cos \frac{x}{y} + C$

$\Rightarrow 0 = \cos \frac{π}{2} + C$ $[∵ x = \frac{π}{2} and y = 1]$

$\Rightarrow C = 0$

$∴ \log y = \cos \frac{x}{y}$

Put y = 2 in equation (i), we get

$\Rightarrow \cos \frac{x}{2} = \log 2$

Now, $\cos x = 2 \cos^{2} \frac{x}{2} - 1 = 2 {(\log_{e} 2)}^{2} - 1$ .

Want Us to Email you About Special Offers & Updates?