QJEEM22Ma2d05: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2d05 Solution: If be the solution curve of the differential equation , with ,…

$({\sin ^2}2x){{dy} \over {dx}} + (8{\sin ^2}2x + 2\sin 4x)y$

$= 2{e^{ - 4x}}(2\sin 2x + \cos 2x)$

${{dy} \over {dx}} + (8 + 4\cot 2x)y = 2{e^{ - 4x}}\left( {{{2\sin 2x + \cos 2x} \over {{{\sin }^2}2x}}} \right)$

Integrating factor

$(I.F.) = {e^{\int {(8 + 4\cot 2x)dx} }}$

$= {e^{8x + 2\ln \sin 2x}}$

Solution of differential equation

$y.\,{e^{8x + 2\ln \sin 2x}}$

$= \int {2{e^{(4x + 2\ln \sin 2x)}}{{(2\sin 2x + \cos 2x)} \over {{{\sin }^2}2x}}dx}$

$= 2\int {{e^{4x}}(2\sin 2x + \cos 2x)dx}$

$y.\,{e^{8x + 2\ln \sin 2x}} = {e^{4x}}\sin 2x + c$

$y\left( {{\pi \over 4}} \right) = {e^{ - \pi }}$

${e^{ - \pi }}\,.\,{e^{2\pi }} = {e^\pi } + c \Rightarrow c = 0$

$y\left( {{\pi \over 6}} \right) = {{{e^{{{2\pi } \over 3}}}{{\sqrt 3 } \over 2}} \over {{e^{\left( {{{4\pi } \over 3} + 2\ln {{\sqrt 3 } \over 2}} \right)}}}}$

$= {e^{{{ - 2\pi } \over 3}}}\,.\,{2 \over {\sqrt 3 }}$