Let y = y ( x ) be the solution of the differential equation ( x ^...

Given equation is:

$( {x}^{2} – 3{y}^{2} ) dx + 3xydy = 0$

We can re-write equation as

$\dfrac{dy}{dx} = - \dfrac{\left({x}^{2} - 3{y}^{2}\right)}{3xy}$

$⇒ \dfrac{dy}{dx} = \dfrac{y}{x} - \dfrac{1}{3}\dfrac{x}{y}$ ....(1)

Put $y = vx$

$\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$

Equation (1) can be written as

$v + x\dfrac{dv}{dx} = v - \dfrac{1}{3}\dfrac{1}{v} ⇒ vdv = - \dfrac{1}{3x}$

Integrating both sides, we get

$\dfrac{{v}^{2}}{2} = - \dfrac{1}{3}\ln \left(x\right) + c$

$⇒ \dfrac{{y}^{2}}{2{x}^{2}} = \dfrac{- 1}{3}\ln \left(x\right) + c$ .....(2)

$∵ y\left(1\right) = 1$ (given)

$∴ c = \dfrac{1}{2}$ (from equation 2)

Equation (2) can be written as

$\dfrac{{y}^{2}}{2{x}^{2}} = \dfrac{- 1}{3}\ln \left(x\right) + \dfrac{1}{2}$

$⇒ {y}^{2} = \dfrac{- 2}{3}{x}^{2}\ln \left(x\right) + {x}^{2}$

Now ${y}^{2} ( e ) = \dfrac{- 2}{3}{e}^{2}\ln \left(e\right) + {e}^{2} = \dfrac{- 2}{3}{e}^{2} + {e}^{2} = \dfrac{{e}^{2}}{3}$

$⇒ 6{y}^{2} ( e ) = 2{e}^{2}$