If $y=y(x), x \in(0, \pi / 2)$ be the solution curve of the differential equation

$$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$$,

with $y(\pi / 4)=\mathrm{e}^{-\pi}$, then $y(\pi / 6)$ is equal to :

<p>$({\sin ^2}2x){{dy} \over {dx}} + (8{\sin ^2}2x + 2\sin 4x)y$</p> <p>$= 2{e^{ - 4x}}(2\sin 2x + \cos 2x)$</p> <p>$${{dy} \over {dx}} + (8 + 4\cot 2x)y = 2{e^{ - 4x}}\left( {{{2\sin 2x + \cos 2x} \over {{{\sin }^2}2x}}} \right)$$</p> <p>Integrating factor</p> <p>$(I.F.) = {e^{\int {(8 + 4\cot 2x)dx} }}$</p> <p>$= {e^{8x + 2\ln \sin 2x}}$</p> <p>Solution of differential equation</p> <p>$y.\,{e^{8x + 2\ln \sin 2x}}$</p> <p>$$ = \int {2{e^{(4x + 2\ln \sin 2x)}}{{(2\sin 2x + \cos 2x)} \over {{{\sin }^2}2x}}dx} $$</p> <p>$= 2\int {{e^{4x}}(2\sin 2x + \cos 2x)dx}$</p> <p>$y.\,{e^{8x + 2\ln \sin 2x}} = {e^{4x}}\sin 2x + c$</p> <p>$y\left( {{\pi \over 4}} \right) = {e^{ - \pi }}$</p> <p>${e^{ - \pi }}\,.\,{e^{2\pi }} = {e^\pi } + c \Rightarrow c = 0$</p> <p>$$y\left( {{\pi \over 6}} \right) = {{{e^{{{2\pi } \over 3}}}{{\sqrt 3 } \over 2}} \over {{e^{\left( {{{4\pi } \over 3} + 2\ln {{\sqrt 3 } \over 2}} \right)}}}}$$</p> <p>$= {e^{{{ - 2\pi } \over 3}}}\,.\,{2 \over {\sqrt 3 }}$</p>