Question

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

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

Answer 1

Answer: (A)

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

Answer 2

The correct answer is (A) : $\frac{2}{\sqrt3}e^{-\frac{2\pi}{3}}$
Given differential equation
$\frac{dy}{dx}+(8+4\cot2x)y=\frac{2e^{-4x}}{\sin^22x}(2\sin2x+\cos2x)$
$I.F=\int_{e}(8+4\cot2x)dy=e^{8x+2log_e(\sin2x)}=e^{8x}.\sin^22x$
Solution is
$y(e^{8x}.\sin^22x)=\int2e^{4x}(2\sin2x+\cos2x)dx+C$
$\therefore y(x)=\frac{e^{-4x}}{\sin2x}$
$\therefore y(\frac{\pi}{6})=\frac{e^{-4\frac{\pi}{6}}}{\sin(2.\frac{\pi}{6})}$
$=\frac{2}{\sqrt3}e^{-\frac{2\pi}{3}}$