Step 1: Understanding the Concept:
This is a linear differential equation of the form \(\frac{dy}{dx} + Py = Q\). We solve it using an Integrating Factor (IF). Step 2: Key Formula or Approach:
1. \(IF = e^{\int P dx} = e^{\int 2\tan x dx} = e^{2 \ln|\sec x|} = \sec^{2} x\).
2. Solution: \(y \cdot IF = \int Q \cdot IF dx\). : Detailed Explanation:
1. Solve the DE:
\[ y \sec^{2} x = \int \sin x \sec^{2} x dx = \int \sec x \tan x dx = \sec x + c \]
\[ y = \cos x + c \cos^{2} x \]
2. Apply initial condition \(y(\pi/3) = 0\):
\[ 0 = \cos(\pi/3) + c \cos^{2}(\pi/3) = \frac{1}{2} + c \left( \frac{1}{4} \right) \]
\[ \frac{c}{4} = -\frac{1}{2} \implies c = -2 \]
The particular solution is \(y = \cos x - 2 \cos^{2} x\).
3. Find Maximum Value:
Let \(t = \cos x\). Then \(y = t - 2t^{2}\).
For maxima: \(dy/dt = 1 - 4t = 0 \implies t = 1/4\).
\[ y_{\text{max}} = \frac{1}{4} - 2 \left( \frac{1}{4} \right)^{2} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8} \]
Given \(y_{\text{max}} = 1/k\), so \(1/k = 1/8 \implies k = 8\). Step 3: Final Answer:
The value of \(k\) is 8.