Key Concepts and Formulas

• Linear First-Order Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are continuous functions.
• Integrating Factor (I.F.): For a linear first-order differential equation, the integrating factor is given by $\text{I.F.} = e^{\int P(x) dx}$.
• General Solution: The general solution to a linear first-order differential equation is given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$.

Step-by-Step Solution

Step 1: Transform the Differential Equation to Standard Form

The given differential equation is: $(\sin^2 2x)\frac{dy}{dx} + (8\sin^2 2x + 2\sin 4x)y = 2e^{-4x}(2\sin 2x + \cos 2x)$ To get it into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide by $\sin^2 2x$. This is valid since $x \in (0, \pi/2)$, implying $\sin 2x > 0$.

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

Now, simplify the coefficients. Using the identity $\sin 4x = 2\sin 2x \cos 2x$: $P(x) = \frac{8\sin^2 2x + 4\sin 2x \cos 2x}{\sin^2 2x} = 8 + \frac{4\cos 2x}{\sin 2x} = 8 + 4\cot 2x$ $Q(x) = \frac{2e^{-4x}(2\sin 2x + \cos 2x)}{\sin^2 2x}$

The standard form is: $\frac{dy}{dx} + (8 + 4\cot 2x)y = \frac{2e^{-4x}(2\sin 2x + \cos 2x)}{\sin^2 2x}$

Step 2: Calculate the Integrating Factor (I.F.)

The integrating factor is given by $\text{I.F.} = e^{\int P(x) dx}$. We need to calculate $\int P(x) dx$. $\int P(x) dx = \int (8 + 4\cot 2x) dx = 8x + 4\int \cot 2x \, dx$ Let $u = 2x$, so $du = 2dx$, and $dx = \frac{1}{2}du$. $\int \cot 2x \, dx = \frac{1}{2} \int \cot u \, du = \frac{1}{2} \ln|\sin u| = \frac{1}{2} \ln|\sin 2x|$ Since $x \in (0, \pi/2)$, $\sin 2x > 0$, so $|\sin 2x| = \sin 2x$. $\int P(x) dx = 8x + 2\ln(\sin 2x)$ Therefore, the integrating factor is: $\text{I.F.} = e^{8x + 2\ln(\sin 2x)} = e^{8x}e^{2\ln(\sin 2x)} = e^{8x}e^{\ln(\sin^2 2x)} = e^{8x}\sin^2 2x$

Step 3: Find the General Solution

The general solution is given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$. $y e^{8x} \sin^2 2x = \int \frac{2e^{-4x}(2\sin 2x + \cos 2x)}{\sin^2 2x} \cdot e^{8x}\sin^2 2x \, dx + C$ $y e^{8x} \sin^2 2x = \int 2e^{4x}(2\sin 2x + \cos 2x) \, dx + C$ Let $I = \int e^{4x}(2\sin 2x + \cos 2x) \, dx = \int 2e^{4x}\sin 2x \, dx + \int e^{4x}\cos 2x \, dx$. Integrate $\int e^{4x}\sin 2x \, dx$ by parts: Let $u = \sin 2x$, $dv = e^{4x} dx$. Then $du = 2\cos 2x \, dx$, $v = \frac{1}{4}e^{4x}$. $\int e^{4x}\sin 2x \, dx = \frac{1}{4}e^{4x}\sin 2x - \frac{1}{2} \int e^{4x}\cos 2x \, dx$ So, $I = 2 \left( \frac{1}{4}e^{4x}\sin 2x - \frac{1}{2} \int e^{4x}\cos 2x \, dx \right) + \int e^{4x}\cos 2x \, dx = \frac{1}{2}e^{4x}\sin 2x - \int e^{4x}\cos 2x \, dx + \int e^{4x}\cos 2x \, dx = \frac{1}{2}e^{4x}\sin 2x$. $y e^{8x} \sin^2 2x = 2 \left( \frac{1}{2}e^{4x}\sin 2x \right) + C$ $y e^{8x} \sin^2 2x = e^{4x}\sin 2x + C$ $y = \frac{e^{4x}\sin 2x + C}{e^{8x}\sin^2 2x} = \frac{e^{-4x}}{\sin 2x} + \frac{C e^{-8x}}{\sin^2 2x}$

Step 4: Apply the Initial Condition

Given $y(\pi/4) = e^{-\pi}$. Substitute $x = \pi/4$ into the general solution: $e^{-\pi} = \frac{e^{-\pi}}{\sin(\pi/2)} + \frac{Ce^{-2\pi}}{\sin^2(\pi/2)}$ $e^{-\pi} = e^{-\pi} + Ce^{-2\pi}$ $0 = Ce^{-2\pi}$ Thus, $C = 0$.

The particular solution is: $y = \frac{e^{-4x}}{\sin 2x}$

Step 5: Calculate $y(\pi/6)$

We need to find $y(\pi/6)$. Substitute $x = \pi/6$: $y(\pi/6) = \frac{e^{-4(\pi/6)}}{\sin(2(\pi/6))} = \frac{e^{-2\pi/3}}{\sin(\pi/3)} = \frac{e^{-2\pi/3}}{\sqrt{3}/2} = \frac{2e^{-2\pi/3}}{\sqrt{3}}$

Common Mistakes & Tips

• Carefully simplify expressions after each step to avoid carrying unnecessary complexity.
• Remember the chain rule when integrating trigonometric functions with arguments like $2x$.
• Don't forget the constant of integration, C, and to use the initial condition to find its value.

Summary

We transformed the given differential equation into standard linear form, calculated the integrating factor, found the general solution, applied the initial condition to determine the constant of integration, and finally calculated $y(\pi/6)$. The value of $y(\pi/6)$ is $\frac{2}{\sqrt{3}}e^{-2\pi/3}$.

Final Answer

The final answer is $\boxed{\frac{2}{\sqrt{3}} e^{-2 \pi / 3}}$, which corresponds to option (A).