Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
• 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 a 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: Identify P(x) and Q(x)

We are given the differential equation: $\frac{dy}{dx} + 2y\tan x = \sin x$ Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify: $P(x) = 2\tan x$ $Q(x) = \sin x$ This step is crucial because these functions are used to calculate the integrating factor and the general solution.

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

The integrating factor is given by $I.F. = e^{\int P(x) dx}$. Substituting $P(x) = 2\tan x$: $I.F. = e^{\int 2\tan x \,dx}$ First, we evaluate the integral: $\int 2\tan x \,dx = 2\int \tan x \,dx = 2\ln|\sec x| = \ln(\sec^2 x)$ Since $0 < x < \frac{\pi}{2}$, $\sec x > 0$, so we can drop the absolute value. Therefore, $I.F. = e^{\ln(\sec^2 x)} = \sec^2 x$ The integrating factor is $\sec^2 x$.

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 $Q(x) = \sin x$ and $I.F. = \sec^2 x$: $y \sec^2 x = \int \sin x \sec^2 x \,dx + C$ We need to evaluate the integral: $\int \sin x \sec^2 x \,dx = \int \sin x \frac{1}{\cos^2 x} \,dx = \int \frac{\sin x}{\cos^2 x} \,dx = \int \frac{\sin x}{\cos x} \frac{1}{\cos x} \,dx = \int \tan x \sec x \,dx$ The integral of $\tan x \sec x$ is $\sec x$: $\int \tan x \sec x \,dx = \sec x$ So, the general solution is: $y \sec^2 x = \sec x + C$

Step 4: Apply the Initial Condition

We are given $y\left(\frac{\pi}{3}\right) = 0$. Substituting $x = \frac{\pi}{3}$ and $y = 0$ into the general solution: $0 \cdot \sec^2\left(\frac{\pi}{3}\right) = \sec\left(\frac{\pi}{3}\right) + C$ Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$, $\sec\left(\frac{\pi}{3}\right) = 2$. Therefore: $0 = 2 + C$ $C = -2$

Step 5: Find the Particular Solution

Substituting $C = -2$ into the general solution: $y \sec^2 x = \sec x - 2$ $y = \frac{\sec x - 2}{\sec^2 x} = \frac{\sec x}{\sec^2 x} - \frac{2}{\sec^2 x} = \frac{1}{\sec x} - 2\cos^2 x = \cos x - 2\cos^2 x$ So, the particular solution is: $y(x) = \cos x - 2\cos^2 x$

Step 6: Find the Maximum Value of y(x)

Let $u = \cos x$. Since $0 < x < \frac{\pi}{2}$, $0 < \cos x < 1$, so $0 < u < 1$. Then $y(x)$ becomes: $f(u) = u - 2u^2 = -2u^2 + u$ To find the maximum value, we can complete the square: $f(u) = -2\left(u^2 - \frac{1}{2}u\right) = -2\left(u^2 - \frac{1}{2}u + \frac{1}{16} - \frac{1}{16}\right) = -2\left(\left(u - \frac{1}{4}\right)^2 - \frac{1}{16}\right) = -2\left(u - \frac{1}{4}\right)^2 + \frac{1}{8}$ The maximum value occurs when $u = \frac{1}{4}$, and the maximum value is $\frac{1}{8}$. Since $0 < \frac{1}{4} < 1$, this value is within the allowed range for $u$.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when integrating and substituting.
• Domain Restrictions: Always check if the value of x (or u) that gives the maximum/minimum is within the given domain.
• Trigonometric Identities: Use trigonometric identities to simplify the expressions for easier integration and differentiation.

Summary

We solved the first-order linear differential equation by finding the integrating factor, determining the general solution, and applying the initial condition to find the particular solution. Then, we found the maximum value of the particular solution by completing the square after substituting $u = \cos x$. The maximum value of $y(x)$ is $\frac{1}{8}$.

Final Answer

The final answer is \boxed{1/8}, which corresponds to option (A).