Key Concepts and Formulas

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

Step-by-Step Solution

Step 1: Convert the Differential Equation to Standard Form

We are given the differential equation: $\cos x \frac{dy}{dx} - y \sin x = 6x$ To convert it to the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we need to divide the entire equation by $\cos x$. Since $0 < x < \frac{\pi}{2}$, $\cos x \neq 0$, so this division is valid. $\frac{\cos x}{\cos x} \frac{dy}{dx} - \frac{\sin x}{\cos x} y = \frac{6x}{\cos x}$ Simplifying, we get: $\frac{dy}{dx} - (\tan x) y = 6x \sec x$ Now, comparing with the standard form, we can identify $P(x)$ and $Q(x)$: $P(x) = -\tan x$ $Q(x) = 6x \sec x$

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

The integrating factor is given by: $I.F. = e^{\int P(x) dx} = e^{\int (-\tan x) dx}$ We know that $\int \tan x \, dx = \ln |\sec x| + C = -\ln |\cos x| + C$. Therefore, $\int (-\tan x) dx = - \int \tan x \, dx = \ln |\cos x|$ Since $0 < x < \frac{\pi}{2}$, $\cos x > 0$, so $|\cos x| = \cos x$. Thus, $I.F. = e^{\ln (\cos x)} = \cos x$

Step 3: Find the General Solution

The general solution is given by: $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$ Substituting $I.F. = \cos x$ and $Q(x) = 6x \sec x$, we have: $y \cos x = \int (6x \sec x) (\cos x) dx + C$ $y \cos x = \int 6x \left( \frac{1}{\cos x} \right) \cos x \, dx + C$ $y \cos x = \int 6x \, dx + C$ $y \cos x = 6 \int x \, dx + C$ $y \cos x = 6 \left( \frac{x^2}{2} \right) + C$ $y \cos x = 3x^2 + C$

Step 4: Use the Initial Condition to Find the Particular Solution

We are given that $y\left( \frac{\pi}{3} \right) = 0$. Substituting $x = \frac{\pi}{3}$ and $y = 0$ into the general solution: $0 \cdot \cos\left( \frac{\pi}{3} \right) = 3\left( \frac{\pi}{3} \right)^2 + C$ $0 = 3 \left( \frac{\pi^2}{9} \right) + C$ $0 = \frac{\pi^2}{3} + C$ $C = -\frac{\pi^2}{3}$ So, the particular solution is: $y \cos x = 3x^2 - \frac{\pi^2}{3}$

Step 5: Evaluate the Particular Solution at the Required Point

We need to find $y\left( \frac{\pi}{6} \right)$. Substituting $x = \frac{\pi}{6}$ into the particular solution: $y \cos\left( \frac{\pi}{6} \right) = 3\left( \frac{\pi}{6} \right)^2 - \frac{\pi^2}{3}$ $y \left( \frac{\sqrt{3}}{2} \right) = 3\left( \frac{\pi^2}{36} \right) - \frac{\pi^2}{3}$ $y \frac{\sqrt{3}}{2} = \frac{\pi^2}{12} - \frac{\pi^2}{3}$ $y \frac{\sqrt{3}}{2} = \frac{\pi^2}{12} - \frac{4\pi^2}{12}$ $y \frac{\sqrt{3}}{2} = \frac{-3\pi^2}{12}$ $y \frac{\sqrt{3}}{2} = -\frac{\pi^2}{4}$ $y = -\frac{\pi^2}{4} \cdot \frac{2}{\sqrt{3}}$ $y = -\frac{2\pi^2}{4\sqrt{3}}$ $y = -\frac{\pi^2}{2\sqrt{3}}$

Common Mistakes & Tips

• Be careful with the signs when integrating and substituting values.
• Remember to consider the domain of $x$ when dealing with absolute values in logarithms.
• Double-check the trigonometric values and algebraic simplifications.

Summary

We converted the given differential equation to standard form, calculated the integrating factor, found the general solution, used the initial condition to find the particular solution, and finally evaluated the particular solution at $x = \frac{\pi}{6}$. The value of $y\left( \frac{\pi}{6} \right)$ is $-\frac{\pi^2}{2\sqrt{3}}$.

Final Answer

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