QJEEM22Maa366: JEE Main 2022 Mathematics - Class 12 JEEM22Maa366 Solution: Let the solution curve of the differential equation , intersect …

${{xdy - ydx} \over {\sqrt {{x^2} + {y^2}} }} = dx$

$\Rightarrow {{dy} \over {dx}} = {{\sqrt {{x^2} + {y^2}} } \over x} + {y \over x}$

$\Rightarrow {{dy} \over {dx}} = \sqrt {1 + {{{y^2}} \over {{x^2}}}} + {y \over x}$

Let ${y \over x} = v$

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

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

OR $\ln \left( {v + \sqrt {1 + {v^2}} } \right) = \ln x + C$

at $x = 1,\,y = 0$

$\Rightarrow C = 0$

${y \over x} + \sqrt {1 + {{{y^2}} \over {{x^2}}}} = x$

At $x = 2$ ,

${y \over 2} + \sqrt {1 + {{{y^2}} \over 4}} = 2$

$\Rightarrow 1 + {{{y^2}} \over 4} = 4 + {{{y^2}} \over 4} - 2y$

OR $y = {3 \over 2}$