Let y = y ( x ) be the solution of the differential equation ( x 2 - 3 y 2 ) ? d x + 3 x y ? d y = 0 , y ( 1 ) = 1 . Then 6 y 2 ( e ) is equal to [2023]

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 $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1 .$ Then $6 y^{2} (e)$ is equal to [2023]

1 $e^{2}$

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

3 $3 e^{2}$

4 $2 e^{2}$

Ans.
(4)

$(x^{2} - 3 y^{2}) d x + 3 x y d y = 0$

$\Rightarrow x^{2} d x - 3 y^{2} d x + 3 x y d y = 0 \Rightarrow x^{2} - 3 y^{2} + 3 x y \frac{d y}{d x} = 0$

Put $t = y^{2} \Rightarrow 2 y \frac{d y}{d x} = \frac{d t}{d x}$ $\Rightarrow x^{2} - 3 t + \frac{3 x}{2} \frac{d t}{d x} = 0$

$\Rightarrow \frac{2 x}{3} - \frac{2 t}{x} + \frac{d t}{d x} = 0 \Rightarrow \frac{d t}{d x} - \frac{2 t}{x} = - \frac{2 x}{3},$

which is a linear differential equation.

$I.F. = e^{\int - \frac{2}{x} d x} = e^{- 2 \ln | x |} = e^{\ln \frac{1}{x^{2}}} = \frac{1}{x^{2}}$

So, $t \cdot \frac{1}{x^{2}} = \int \frac{1}{x^{2}} (\frac{- 2 x}{3}) d x$ $\Rightarrow \frac{y^{2}}{x^{2}} = - \frac{2}{3} \ln | x | + C$

When, $x = 1, y = 1$

$1 = \frac{- 2}{3} \ln (1) + C \Rightarrow C = 1$ $∴$ $\frac{y^{2}}{x^{2}} = \frac{- 2}{3} \ln | x | + 1$

At $x = e,$ $\frac{y^{2} (e)}{e^{2}} = \frac{- 2}{3} + 1 \Rightarrow y^{2} (e) = \frac{e^{2}}{3} \Rightarrow 6 y^{2} (e) = 2 e^{2}$

Want Us to Email you About Special Offers & Updates?