QJEEM23Mad6c2: JEE Main 2023 Mathematics - Class 12 JEEM23Mad6c2 Solution: Let the solution curve of the differential equation

${{dy} \over {dx}} - {{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}y = 2x\exp \left\{ {{{{x^3} - {{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }}} \right\}$

$IF = {e^{ - \int {{{3{x^5}{{\tan }^{ - 1}}({x^3})} \over {{{(1 + {x^6})}^{{3 \over 2}}}}}dx} }}$

Let ${\tan ^{ - 1}}{x^3} = t \Rightarrow {{3{x^2}} \over {1 + {x^6}}}dx = dt$

$\Rightarrow IF = {e^{ - \int {{{\tan t} \over {\sec t}}.t\,dt} }} = {e^{ - \int {\sin t.tdt} }} = {e^{t\cos t - \sin t}}$

$\Rightarrow IF = {e^{{{{{\tan }^{ - 1}}({x^3})} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}}$

$\therefore$ Solution is

$y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3}} \over {\sqrt {1 + {x^6}} }} - {{{x^3}} \over {\sqrt {1 + {x^6}} }}}} = \int {2x\,dx + c}$

$\Rightarrow y\,.\,{e^{{{{{\tan }^{ - 1}}{x^3} - {x^3}} \over {\sqrt {1 + {x^6}} }}}} = {x^2} + c$

$y(0) = 0 \Rightarrow c = 0$

$x = 1$

$y\,.\,{e^{{{{\pi \over 4} - 1} \over {\sqrt 2 }}}} = 1$

$\Rightarrow y = {e^{{{1 - {\pi \over 4}} \over {\sqrt 2 }}}}$

$\Rightarrow y = {e^{{{4 - \pi } \over {4\sqrt 2 }}}}$