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 (IF): For a first-order linear differential equation, the integrating factor is $IF = e^{\int P(x) dx}$.
• General Solution: The general solution of the differential equation is $y \cdot IF = \int (Q(x) \cdot IF) dx + C$, 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: $\frac{dy}{dx} + (\tan x)y = \frac{2 + \sec x}{(1 + 2\sec x)^2}$ Comparing this with the general form $\frac{dy}{dx} + P(x)y = Q(x)$, we have: $P(x) = \tan x$ $Q(x) = \frac{2 + \sec x}{(1 + 2\sec x)^2}$

Step 2: Calculate the Integrating Factor (IF)

The integrating factor is given by: $IF = e^{\int P(x) dx} = e^{\int \tan x dx}$ We know that $\int \tan x dx = \int \frac{\sin x}{\cos x} dx = -\ln|\cos x| = \ln|\sec x|$. Therefore, $IF = e^{\ln|\sec x|} = \sec x$

Step 3: Find the General Solution

The general solution is given by: $y \cdot IF = \int (Q(x) \cdot IF) dx + C$ Substituting the values of $IF$ and $Q(x)$, we get: $y \sec x = \int \left(\frac{2 + \sec x}{(1 + 2\sec x)^2} \cdot \sec x\right) dx + C$ $y \sec x = \int \frac{\sec x(2 + \sec x)}{(1 + 2\sec x)^2} dx + C$

Step 4: Evaluate the Integral

Let $u = 1 + 2\sec x$. Then, $du = 2\sec x \tan x \, dx$. Also, $\sec x = \frac{u-1}{2}$. The integral becomes: $\int \frac{\frac{u-1}{2}(2 + \frac{u-1}{2})}{u^2} dx = \int \frac{\frac{u-1}{2}(\frac{4+u-1}{2})}{u^2} dx = \int \frac{(u-1)(u+3)}{4u^2} dx$ This substitution does not immediately simplify the integral. Instead, let's try a different approach. Notice that $\frac{d}{dx}(1+2\sec x) = 2\sec x \tan x$. We want to find $\int \frac{\sec x(2 + \sec x)}{(1 + 2\sec x)^2} dx$.

We can rewrite $\sec x (2 + \sec x)$ as $\frac{1}{2}(1+2\sec x)' \frac{2+\sec x}{\sec x \tan x} = \frac{2\sec x + \sec^2 x}{(1+2\sec x)^2}$. Notice that $\frac{d}{dx} \frac{1}{1+2\sec x} = \frac{-2\sec x \tan x}{(1+2\sec x)^2}$.

Let $t = \sec x$. Then $I = \int \frac{t(2+t)}{(1+2t)^2} dx$. Since we have $\sec x$ and we know that $\frac{d}{dx} \sec x = \sec x \tan x$, let's try the substitution $u = 1+2\sec x$. Then $du = 2\sec x \tan x dx$, which is not directly present.

Let's try to directly integrate $\int \frac{\sec x(2 + \sec x)}{(1 + 2\sec x)^2} dx$. Let $u = 1 + 2\sec x$. Then $\sec x = \frac{u-1}{2}$. So $\frac{du}{dx} = 2\sec x \tan x$.

Consider $\frac{d}{dx}\left(\frac{\tan x}{1 + 2\sec x}\right) = \frac{\sec^2 x (1+2\sec x) - \tan x (2\sec x \tan x)}{(1+2\sec x)^2} = \frac{\sec^2 x + 2\sec^3 x - 2\sec x \tan^2 x}{(1+2\sec x)^2} = \frac{\sec^2 x + 2\sec^3 x - 2\sec x (\sec^2 x - 1)}{(1+2\sec x)^2} = \frac{\sec^2 x + 2\sec^3 x - 2\sec^3 x + 2\sec x}{(1+2\sec x)^2} = \frac{\sec^2 x + 2\sec x}{(1+2\sec x)^2} = \frac{\sec x (\sec x + 2)}{(1+2\sec x)^2}$. Therefore, $\int \frac{\sec x(2 + \sec x)}{(1 + 2\sec x)^2} dx = \frac{\tan x}{1 + 2\sec x} + C$.

Thus, $y \sec x = \frac{\tan x}{1 + 2\sec x} + C$ $y = \frac{\sin x}{1 + 2\sec x} + C\cos x$

Step 5: Apply the Initial Condition

We are given that $f\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{10}$. Substituting $x = \frac{\pi}{3}$ and $y = \frac{\sqrt{3}}{10}$ into the general solution, we get: $\frac{\sqrt{3}}{10} = \frac{\sin(\frac{\pi}{3})}{1 + 2\sec(\frac{\pi}{3})} + C\cos(\frac{\pi}{3})$ $\frac{\sqrt{3}}{10} = \frac{\frac{\sqrt{3}}{2}}{1 + 2(2)} + C\left(\frac{1}{2}\right)$ $\frac{\sqrt{3}}{10} = \frac{\frac{\sqrt{3}}{2}}{5} + \frac{C}{2}$ $\frac{\sqrt{3}}{10} = \frac{\sqrt{3}}{10} + \frac{C}{2}$ $C = 0$

Step 6: Find f(π/4)

Therefore, the particular solution is: $y = \frac{\sin x}{1 + 2\sec x}$ We want to find $f\left(\frac{\pi}{4}\right)$. Substituting $x = \frac{\pi}{4}$, we get: $f\left(\frac{\pi}{4}\right) = \frac{\sin(\frac{\pi}{4})}{1 + 2\sec(\frac{\pi}{4})} = \frac{\frac{\sqrt{2}}{2}}{1 + 2\sqrt{2}} = \frac{\sqrt{2}}{2(1 + 2\sqrt{2})}$ $f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2(1 + 2\sqrt{2})} \cdot \frac{1 - 2\sqrt{2}}{1 - 2\sqrt{2}} = \frac{\sqrt{2}(1 - 2\sqrt{2})}{2(1 - 8)} = \frac{\sqrt{2} - 4}{2(-7)} = \frac{4 - \sqrt{2}}{14}$

Common Mistakes & Tips

• Trigonometric Identities: Be careful while using trigonometric identities, especially when simplifying the integral.
• Integrating Factor: Make sure you calculate the integrating factor correctly. A small error here can lead to a completely wrong answer.
• Substitution: Choosing the right substitution is crucial for solving the integral.

Summary

We solved the first-order linear differential equation using the integrating factor method. We found the integrating factor to be $\sec x$, obtained the general solution, and then used the initial condition $f(\frac{\pi}{3}) = \frac{\sqrt{3}}{10}$ to determine the particular solution. Finally, we evaluated $f(\frac{\pi}{4})$ to obtain the answer.

The final answer is $\boxed{\frac{4 - \sqrt{2}}{14}}$, which corresponds to option (B).