If the solution curve y = y(x) of the differential equation y ^ 2...

Given ${y}^{2}dx + \left({x}^{2} - xy + {y}^{2}\right)dy = 0$

$⇒ \dfrac{dx}{dy} = - \dfrac{\left({x}^{2} - xy + {y}^{2}\right)}{{y}^{2}} . . . \left(i\right)$

Now let $x = vy$ $⇒ v + y\dfrac{dv}{dy} = \dfrac{dx}{dy}$

Now putting in equation \left(i\right)$we get</p><p>$⇒ v + y\dfrac{dv}{dy} = - \left({v}^{2} - v + 1\right)$</p><p>$⇒ y\dfrac{dv}{dy} = - {v}^{2} - 1$</p><p>$⇒ \dfrac{- dv}{1 + {v}^{2}} = \dfrac{dy}{y}$</p><p>Now Integrating Both side</p><p>$⇒ - ∫ \dfrac{dy}{1 + {v}^{2}} = ∫ \dfrac{dy}{y}$</p><p>$⇒ - \tan^{- 1}v = \log \left|y\right| + c ⇒ - \tan^{- 1}\left(\dfrac{x}{y}\right) = \log \left|y\right| + c$</p><p>Also this curve passes through \left(1, 1\right)$

$⇒ \tan^{- 1}1 = \log \left|1\right| + c$ $⇒ c = - \dfrac{\pi}{4}$

$⇒ - \tan^{- 1}\dfrac{x}{y} = \log \left|y\right| - \dfrac{\pi}{4}$

Now putting $y = \sqrt{3}x$ and $x = \alpha$ we get

$= - \tan^{- 1}\dfrac{x}{\sqrt{3}x} = \log \left|\sqrt{3}\alpha\right| - \dfrac{\pi}{4}$ $= - \tan^{- 1}\dfrac{1}{\sqrt{3}} = \log \left(\sqrt{3}\alpha\right) - \dfrac{\pi}{4}$

$= - \dfrac{\pi}{6} + \dfrac{\pi}{4} = \log \left|\sqrt{3}\alpha\right|$

$⇒ \log \left|\sqrt{3}\alpha\right| = \dfrac{\pi}{12}$