QJEEM22Ma2f86: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2f86 Solution: If the solution curve of the differential equation , which…

${{dy} \over {dx}} = {{{y^2}} \over {xy - {x^2} - {y^2}}}$

Put $y = vx$ we get

$v + x{{dv} \over {dx}} = {{{v^2}} \over {v - 1 - {v^2}}}$

$\Rightarrow x{{dv} \over {dx}} = {{{v^2} - {v^2} + v + {v^3}} \over {v - 1 - {v^2}}}$

$\Rightarrow \int {{{v - 1 - {v^2}} \over {v(1 + {v^2})}}dv = \int {{{dx} \over x}} }$

${\tan ^{ - 1}}\left( {{y \over x}} \right) - \ln \left( {{y \over x}} \right) = \ln x + c$

As it passes through (1, 1)

$c = {\pi \over 4}$

$\Rightarrow {\tan ^{ - 1}}\left( {{y \over x}} \right)\ln \left( {{y \over x}} \right) = \ln x + {\pi \over 4}$

Put $y = \sqrt 3 x$ we get

$\Rightarrow {\pi \over 3} - \ln \sqrt 3 = \ln x + {\pi \over 4}$

$\Rightarrow \ln x = {\pi \over {12}} - \ln \sqrt 3 = \ln \alpha$

$\therefore$ $\ln \left( {\sqrt 3 \alpha } \right) = \ln \sqrt 3 + \ln \alpha$

$= \ln \sqrt 3 + {\pi \over {12}} - \ln \sqrt 3 = {\pi \over {12}}$