QJEEAdv22Ma1acc: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Ma1acc Solution: If is the solution of the differential equation and the slope…

$xdy = ({y^2} - 4y)dx = 0$

$\Rightarrow xdy = ({y^2} - 4y)dx$

$\Rightarrow {{dy} \over {({y^2} - 4x)}} = {{dx} \over x}$

$\Rightarrow {{dy} \over {y(y - 4)}} = {{dx} \over x}$

$\Rightarrow {1 \over 4} \times {{y - (y - 4)} \over {y(y - 4)}}dy = {{dx} \over x}$

$\Rightarrow {1 \over 4}\left( {{1 \over {y - 4}} - {1 \over y}} \right)dy = {{dx} \over x}$

Integrating both side, we get

$\Rightarrow {1 \over 4}\int {\left( {{1 \over {y - 4}} - {1 \over y}} \right)dy = \int {{{dx} \over x}} }$

$\Rightarrow {1 \over 4}\left( {\ln \left| {{{y - 4} \over y}} \right|} \right) = \ln |x| + {1 \over 4}\ln C$ (Cont.)

$\Rightarrow {1 \over 4}\ln \left| {{{y - 4} \over y}} \right| = \ln x + {1 \over 4}\ln C$ [As x > 0 , so $|x| = x$ ]

$\Rightarrow \ln \left| {{{y - 4} \over y}} \right| = 4\ln x + \ln C$

$\Rightarrow \ln \left| {{{y - 4} \over y}} \right| = \ln ({x^4}C)$

$\Rightarrow {{y - 4} \over y} = \, \pm \,C{x^4}$

$\Rightarrow {{y - 4} \over y} = \lambda {x^4}$ [Assume $\pm \,C = \lambda$ ]

Given, $y(1) = 2$

$\therefore$ ${{2 - 4} \over 2} = \lambda {(1)^4}$

$\Rightarrow \lambda = -1$

$\therefore$ ${{y - 4} \over y} = - {x^4}$

Now, $y(\sqrt 2 ) = - {(\sqrt 2 )^4} = {{y - 4} \over y}$

$\Rightarrow {{y - 4} \over y} = - 4$

$\Rightarrow y - 4 = - 4y$

$\Rightarrow 5y = 4$

$\Rightarrow y = {4 \over 5}$

$\therefore$ $10y(\sqrt 2 ) = 10 \times {4 \over 5} = 8$