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. $\frac{2}{\sqrt{3}} e ^{-2 \pi / 3}$
• 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
$\frac{d y}{d x}+(8+4 \cot 2 x) y=\frac{2 e^{-4 x}}{\sin ^2 2 x}(2 \sin x+\cos 2 x)$
which is a linear diff. equation.
$\text { I.f. } =e^{\int(8+4 \cot 2 x) d x}=e^{8 x+2 C u(\sin 2 x)} $
$ =e^{8 x} \cdot \sin ^2 2 x$
$\therefore$ solution is
$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 $
$ =e^{4 x} \cdot \sin 2 x+C$
Given $y \left(\frac{\pi}{4}\right)= e ^{-\pi} \Rightarrow C =0$
$ \therefore y =\frac{ e ^{-4 x}}{\sin 2 x } $
$ \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}}$