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

Answer: (D)

$ \left(x^2-3 y^2\right) d x+3 x y d y=0$
$\frac{d y}{d x}=\frac{3 y^2-x^2}{3 x y} \Rightarrow \frac{d y}{d x}=\frac{y}{x}-\frac{1}{3} \frac{x}{y}.....(1)$
Put $y=v x$
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
(1) $\Rightarrow v+x \frac{d v}{d x}=v-\frac{1}{3} \frac{1}{v} $
$\Rightarrow v d v=\frac{-1}{3 x}$
Integrating both side
$ \frac{v^2}{2}=\frac{-1}{3} \ln x+c $
$\Rightarrow \frac{y^2}{2 x^2}=\frac{-1}{3} \ln x+c$
$ y(1)=1$
$\Rightarrow \frac{1}{2}=c$
$ \Rightarrow \frac{y^2}{2 x^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}$
$\Rightarrow 6 y ^2( e )=2 e ^2$