Question

If $y=y\left(x\right),x\in (0,\frac{\pi }{2})$ be the solution curve of the differential equation $\left({\sin }^{2}2x\right)\frac{dy}{dx}+(8{\sin }^{2}2x+2\sin 4x)y=2e^{-4x}(2\sin 2x+\cos 2x)$, with $y\left(\frac{\pi }{4}\right)=e^{-\pi },$ then y $\left(\frac{\pi }{6}\right)$ is equal to

• A. $\frac{2}{\sqrt{3}}{e}^{2\pi /3}$
• B. $\frac{2}{\sqrt{3}}{e}^{-2\pi /3}$
• C. $\frac{1}{\sqrt{3}}{e}^{-2\pi /3}$
• D. $\frac{1}{\sqrt{3}}{e}^{-2\pi /3}$

Answer 1

The correct option is B $\frac{2}{\sqrt{3}}{e}^{-2\pi /3}$

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

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

$\text{Integrating factor}$

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

$\text{Solution of differential equation}$

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

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

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

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

$y\frac{\pi }{4}=e^{-\pi }$

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

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

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