If the solution curve $y = y(x)$ of the differential equation ${y^2}dx + ({x^2} - xy + {y^2})dy = 0$, which passes through the point (1, 1) and intersects the line $y = \sqrt 3 x$ at the point $(\alpha ,\sqrt 3 \alpha )$, then value of ${\log _e}(\sqrt 3 \alpha )$ is equal to :

<p>${{dy} \over {dx}} = {{{y^2}} \over {xy - {x^2} - {y^2}}}$</p> <p>Put $y = vx$ we get</p> <p>$v + x{{dv} \over {dx}} = {{{v^2}} \over {v - 1 - {v^2}}}$</p> <p>$$ \Rightarrow x{{dv} \over {dx}} = {{{v^2} - {v^2} + v + {v^3}} \over {v - 1 - {v^2}}}$$</p> <p>$$ \Rightarrow \int {{{v - 1 - {v^2}} \over {v(1 + {v^2})}}dv = \int {{{dx} \over x}} } $$</p> <p>$${\tan ^{ - 1}}\left( {{y \over x}} \right) - \ln \left( {{y \over x}} \right) = \ln x + c$$</p> <p>As it passes through (1, 1)</p> <p>$c = {\pi \over 4}$</p> <p>$$ \Rightarrow {\tan ^{ - 1}}\left( {{y \over x}} \right)\ln \left( {{y \over x}} \right) = \ln x + {\pi \over 4}$$</p> <p>Put $y = \sqrt 3 x$ we get</p> <p>$\Rightarrow {\pi \over 3} - \ln \sqrt 3 = \ln x + {\pi \over 4}$</p> <p>$\Rightarrow \ln x = {\pi \over {12}} - \ln \sqrt 3 = \ln \alpha$</p> <p>$\therefore$ $\ln \left( {\sqrt 3 \alpha } \right) = \ln \sqrt 3 + \ln \alpha$</p> <p>$= \ln \sqrt 3 + {\pi \over {12}} - \ln \sqrt 3 = {\pi \over {12}}$</p>