Question

Let $y=y(x)$ be the solution of the differential equation $\left(x^2-3 y^2\right) d x+3 x y d y=0, y(1)=1$.Then $6 y^2( e )$ is equal to

• A. $e^2$
• B. $\frac{3}{2} e^2$
• C. $3 e^2$
• D. $2 e ^2$

Answer 1

The Correct Option is D

The correct answer is (D) : $2 e ^2$
$\left(x^2-3y^2\right)dx+3xydy=0$
$\frac{dy}{dx}=\frac{3y^2-x^2}{3xy}\Rightarrow \frac{dy}{dx}=\frac{y}{x}-\frac{1}{3}\frac{x}{y}.....(1)$
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
(1) $\Rightarrow v+x\frac{dv}{dx}=v-\frac{1}{3}\frac{1}{v}$
$\Rightarrow vdv=\frac{-1}{3x}$
Integrating both side
$\frac{v^2}{2}=\frac{-1}{3}\ln x+c$
$\Rightarrow \frac{y^2}{2x^2}=\frac{-1}{3}\ln x+c$
$y(1)=1$
$\Rightarrow \frac{1}{2}=c$
$\Rightarrow \frac{y^2}{2x^2}=\frac{-1}{3}\ln x+\frac{1}{2}$
$\Rightarrow y^2=-\frac{2}{3}{x}^2\ln x+x^2$
$y^2(e)=-\frac{2}{3}{e}^2+e^2=\frac{e^2}{3}$
$\Rightarrow 6y^2(e)=2e^2$