JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 9)
$$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$$,
with $$y(\pi / 4)=\mathrm{e}^{-\pi}$$, then $$y(\pi / 6)$$ is equal to :
Explanation
$$({\sin ^2}2x){{dy} \over {dx}} + (8{\sin ^2}2x + 2\sin 4x)y$$
$$ = 2{e^{ - 4x}}(2\sin 2x + \cos 2x)$$
$${{dy} \over {dx}} + (8 + 4\cot 2x)y = 2{e^{ - 4x}}\left( {{{2\sin 2x + \cos 2x} \over {{{\sin }^2}2x}}} \right)$$
Integrating factor
$$(I.F.) = {e^{\int {(8 + 4\cot 2x)dx} }}$$
$$ = {e^{8x + 2\ln \sin 2x}}$$
Solution of differential equation
$$y.\,{e^{8x + 2\ln \sin 2x}}$$
$$ = \int {2{e^{(4x + 2\ln \sin 2x)}}{{(2\sin 2x + \cos 2x)} \over {{{\sin }^2}2x}}dx} $$
$$ = 2\int {{e^{4x}}(2\sin 2x + \cos 2x)dx} $$
$$y.\,{e^{8x + 2\ln \sin 2x}} = {e^{4x}}\sin 2x + c$$
$$y\left( {{\pi \over 4}} \right) = {e^{ - \pi }}$$
$${e^{ - \pi }}\,.\,{e^{2\pi }} = {e^\pi } + c \Rightarrow c = 0$$
$$y\left( {{\pi \over 6}} \right) = {{{e^{{{2\pi } \over 3}}}{{\sqrt 3 } \over 2}} \over {{e^{\left( {{{4\pi } \over 3} + 2\ln {{\sqrt 3 } \over 2}} \right)}}}}$$
$$ = {e^{{{ - 2\pi } \over 3}}}\,.\,{2 \over {\sqrt 3 }}$$
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