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 of $x$.
• Integrating Factor (IF): For a linear first-order differential equation, the integrating factor is given by $IF = e^{\int P(x) dx}$.
• General Solution: The general solution to the linear first-order differential equation is given by $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$, 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: $\frac{dy}{dx} + (\tan x) y = \sin x$ We need to identify the functions $P(x)$ and $Q(x)$ by comparing this equation to the standard form of a linear first-order differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$. By direct comparison, we have: $P(x) = \tan x$ $Q(x) = \sin x$ Why? Correctly identifying $P(x)$ and $Q(x)$ is crucial for calculating the integrating factor and the general solution.

Step 2: Calculate the Integrating Factor (IF)

The integrating factor is given by: $IF = e^{\int P(x) dx}$ Substituting $P(x) = \tan x$, we get: $IF = e^{\int \tan x \, dx}$ We know that $\int \tan x \, dx = \int \frac{\sin x}{\cos x} dx$. Let $u = \cos x$, so $du = -\sin x \, dx$. Then, the integral becomes: $\int \frac{\sin x}{\cos x} dx = \int \frac{-du}{u} = -\ln |u| + C_1 = -\ln |\cos x| + C_1 = \ln |\sec x| + C_1$ Since we only need the integrating factor, we can ignore the constant of integration $C_1$. Therefore, $\int \tan x \, dx = \ln |\sec x|$ Thus, the integrating factor is: $IF = e^{\ln |\sec x|} = |\sec x|$ Since $0 \le x \le \frac{\pi}{3}$, $\cos x > 0$ and hence $\sec x > 0$. Therefore, $|\sec x| = \sec x$. $IF = \sec x$ Why? The integrating factor simplifies the differential equation, making it easier to solve. The simplification using the domain is important.

Step 3: Find the General Solution

The general solution is given by: $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$ Substituting $IF = \sec x$ and $Q(x) = \sin x$, we get: $y \sec x = \int \sin x \sec x \, dx + C$ $y \sec x = \int \sin x \frac{1}{\cos x} dx + C$ $y \sec x = \int \tan x \, dx + C$ $y \sec x = \ln |\sec x| + C$ Since $0 \le x \le \frac{\pi}{3}$, $\sec x > 0$, so $|\sec x| = \sec x$. $y \sec x = \ln (\sec x) + C$ Why? Finding the general solution provides a family of solutions to the differential equation.

Step 4: Apply the Initial Condition

We are given the initial condition $y(0) = 0$. Substituting $x = 0$ and $y = 0$ into the general solution: $0 \cdot \sec(0) = \ln (\sec(0)) + C$ Since $\sec(0) = 1$ and $\ln(1) = 0$, we have: $0 = 0 + C$ $C = 0$ Therefore, the particular solution is: $y \sec x = \ln (\sec x)$ $y = \frac{\ln (\sec x)}{\sec x} = \cos x \ln(\sec x)$ Why? The initial condition allows us to find a unique solution to the differential equation.

Step 5: Evaluate y(π/4)

We want to find $y(\frac{\pi}{4})$. Substituting $x = \frac{\pi}{4}$ into the particular solution: $y\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) \ln\left(\sec\left(\frac{\pi}{4}\right)\right)$ We know that $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\sec\left(\frac{\pi}{4}\right) = \sqrt{2}$. Therefore, $y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \ln(\sqrt{2})$ $y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \ln(2^{1/2})$ Using the property $\ln(a^b) = b \ln a$: $y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \ln 2$ $y\left(\frac{\pi}{4}\right) = \frac{1}{2\sqrt{2}} \ln 2 = \left(\frac{1}{2\sqrt{2}}\right) \ln 2$ Why? Evaluating the particular solution at the given point gives the final answer.

However, the provided correct answer is (A) $\frac{1}{2} \ln 2$. Let's analyze if we can manipulate the equation to get this. We have $y = \cos x \ln(\sec x)$. If the problem intended $y(\frac{\pi}{4}) = \frac{1}{2} \ln 2$, then $\frac{1}{\sqrt{2}} \ln(\sqrt{2}) = \frac{1}{2} \ln 2$, which simplifies to $\frac{1}{\sqrt{2}} \cdot \frac{1}{2} \ln 2 = \frac{1}{2} \ln 2$. This implies $\frac{1}{\sqrt{2}} = 1$, which is false.

There seems to be an error in the provided correct answer. Our derivation consistently yields $y(\frac{\pi}{4}) = \frac{1}{2\sqrt{2}} \ln 2$, which matches option (B).

Common Mistakes & Tips

• Forgetting the Constant of Integration: Always remember to add the constant of integration, $C$, when evaluating indefinite integrals.
• Sign Errors: Be careful with signs, especially when using substitution or integration by parts.
• Domain Restrictions: Consider the domain of the functions involved, especially when dealing with trigonometric functions and logarithms.

Summary

We solved the given linear first-order differential equation by finding the integrating factor, determining the general solution, applying the initial condition to find the particular solution, and finally evaluating the particular solution at $x = \frac{\pi}{4}$. The result is $y\left(\frac{\pi}{4}\right) = \frac{1}{2\sqrt{2}} \ln 2$, which corresponds to option (B).

Final Answer

The final answer is \boxed{\left( {{1 \over {2\sqrt 2 }}} \right) \log_e 2}, which corresponds to option (B).