Question

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

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

Answer: (A)

Given differential equation can be re-written as
$\frac{dy}{dx}+(8+4\cot 2x)y=\frac{2e^{-4x}}{{\sin }^{2}2x}(2\sin x+\cos 2x)$
which is a linear diff. equation.
$\text{ I.f. }=e^{\int (8+4\cot 2x)dx}=e^{8x+2Cu(\sin 2x)}$
$=e^{8x}\cdot {\sin }^{2}2x$
$\therefore$ solution is
$y\left(e^{8x}\cdot {\sin }^{2}2x\right)=\int 2e^{4x}(2\sin 2x+\cos 2x)dx+C$
$=e^{4x}\cdot \sin 2x+C$
Given $y\left(\frac{\pi }{4}\right)=e^{-\pi }\Rightarrow C=0$
$\therefore y=\frac{e^{-4x}}{\sin 2x}$
$\therefore y\left(\frac{\pi }{6}\right)=\frac{e^{-4\cdot \frac{\pi }{6}}}{\sin \left(2\cdot \frac{\pi }{6}\right)}=\frac{2}{\sqrt{3}}{e}^{-\frac{2\pi }{3}}$