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If \mathrm{y=y(x), x \in(0, \pi / 2)} be the solution curve of the differential equation \mathrm{\left(\sin ^{2} 2 x\right) \frac{\mathrm{d} y}{\mathrm{~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) \text {, with } y(\pi / 4)=\mathrm{e}^{-\pi} \text {, }} then \mathrm{y(\pi / 6)} is equal to:

Option: 1

\mathrm{\frac{2}{\sqrt{3}} e^{-2 \pi / 3}}


Option: 2

\mathrm{\frac{2}{\sqrt{3}} e^{2 \pi / 3}}


Option: 3

\mathrm{\frac{1}{\sqrt{3}} e^{-2 \pi / 3}}


Option: 4

\mathrm{\frac{1}{\sqrt{3}} e^{2 \pi / 3}}


Answers (1)

best_answer

Given differential equation can be re-usritton as
\mathrm{\frac{d y}{d x}+(8+4 \cot 2 x) 4=\frac{2 e^{-4 x}}{\sin ^{2} 2 x}(2 \sin x+\cos 2 x)}
which is a lénear diff equation
\begin{aligned} & \mathrm{I.f =e^{\int(8+4 \cot 2 x) d x}=e^{8 x} \times \sin ^{2} 2 x}\\ & \therefore \text{Solution is}\\ & \mathrm{y\left(e^{8 x} \cdot \sin ^{2} 2 x\right)=\int 2 e^{4 x}(2 \sin 2 x+\cos 2 x) d x+c}\\ &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{=e^{4 x} \cdot \sin 2 x+c}\\ & \text{Given} \;\mathrm{y\left(\frac{\pi}{4}\right)=e^{-\pi} \Rightarrow c=0}\\ & \therefore \mathrm{ y=\frac{e^{-4 x}}{\sin 2 x}}\\ & \mathrm{\therefore y\left(\frac{\pi}{6}\right)=\frac{e^{-4 \times \frac{\pi}{6}}}{\sin \left(2 \cdot \frac{\pi}{6}\right)}=\frac{2}{\sqrt{3}} e^{-\frac{2 \pi}{3}}}\\ \end{aligned}

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Rishi

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