QJEEM23Ma31fb: JEE Main 2023 Mathematics - Class 12 JEEM23Ma31fb Solution: Let be the solution of the differential equation . Then is…

Given,

$\left( {{x^2} - 3{y^2}} \right)dx + 3xydy = 0$

$\Rightarrow {x^2}dx - 3{y^2}dx + 3xydy = 0$

$\Rightarrow {{{x^2}dx} \over {3xdx}} - {{3{y^2}dx} \over {3xdx}} + {{3ydy} \over {3xdx}} = 0$

$\Rightarrow {x \over 3} - {{{y^2}} \over x} + y{{dy} \over {dx}} = 0$

Let ${y^2} = t$

$2y{{dy} \over {dx}} = {{dt} \over {dx}}$

$\therefore$ ${1 \over 2}{{dt} \over {dx}} - {t \over x} + {x \over 3} = 0$

$\Rightarrow {{dt} \over {dx}} - {{2t} \over x} = - {{2x} \over 3}$

This is linear Differential Equation.

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

So, $t \cdot \frac{1}{x^2}=\int \frac{1}{x^2}\left(\frac{-2 x}{3}\right) d x$

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

When, $x=1, y=1$

$1=\frac{-2}{3} \ln (1)+C \Rightarrow C=1$

$\therefore \frac{y^2}{x^2}=\frac{-2}{3} \ln |x|+1$

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

$\Rightarrow y^2(e)=\frac{e^2}{3}$

$\Rightarrow 6 y^2(e)=2 e^2$