1. Key Concepts and Formulas

• First-Order Linear 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.
• Integrating Factor (I.F.): The integrating factor for a first-order linear differential equation is given by $\text{I.F.} = e^{\int P(x) dx}$.
• General Solution: The general solution to a first-order linear differential equation is given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$, where $C$ is the constant of integration.

2. Step-by-Step Solution

Step 1: Rearrange the given differential equation into the standard form.

We are given the differential equation: $f(x) \sin 2x + \sin x - (1+\cos^2 x) f'(x) = 0$ Our goal is to rewrite this in the standard form $f'(x) + P(x)f(x) = Q(x)$. First, we isolate the $f'(x)$ term: $(1+\cos^2 x) f'(x) = f(x) \sin 2x + \sin x$ Next, we divide both sides by $(1+\cos^2 x)$ to get the coefficient of $f'(x)$ to be 1: $f'(x) = \frac{f(x) \sin 2x + \sin x}{1+\cos^2 x}$ Now, we separate the terms involving $f(x)$ and the remaining terms: $f'(x) = \frac{\sin 2x}{1+\cos^2 x} f(x) + \frac{\sin x}{1+\cos^2 x}$ Finally, we move the term with $f(x)$ to the left side: $f'(x) - \frac{\sin 2x}{1+\cos^2 x} f(x) = \frac{\sin x}{1+\cos^2 x}$ Comparing this with $f'(x) + P(x)f(x) = Q(x)$, we identify: $P(x) = -\frac{\sin 2x}{1+\cos^2 x}$ $Q(x) = \frac{\sin x}{1+\cos^2 x}$

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

The Integrating Factor is given by $\text{I.F.} = e^{\int P(x) dx}$. First, we need to calculate the integral $\int P(x) dx$: $\int P(x) dx = \int -\frac{\sin 2x}{1+\cos^2 x} dx$ We use the substitution $u = 1+\cos^2 x$. Then, $\frac{du}{dx} = -2\cos x \sin x = -\sin 2x$. So, $du = -\sin 2x \, dx$. Substituting $u$ and $du$ into the integral: $\int -\frac{\sin 2x}{1+\cos^2 x} dx = \int \frac{du}{u} = \ln|u|$ Since $1+\cos^2 x$ is always positive, we have: $\int P(x) dx = \ln(1+\cos^2 x)$ Now, we calculate the Integrating Factor: $\text{I.F.} = e^{\int P(x) dx} = e^{\ln(1+\cos^2 x)}$ Using the property $e^{\ln A} = A$: $\text{I.F.} = 1+\cos^2 x$

Step 3: Find the general solution.

The general solution is given by $f(x) \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$. Substituting the values we found: $f(x) \cdot (1+\cos^2 x) = \int \left(\frac{\sin x}{1+\cos^2 x}\right) \cdot (1+\cos^2 x) dx + C$ Simplifying the integral: $f(x) \cdot (1+\cos^2 x) = \int \sin x \, dx + C$ Integrating: $f(x) \cdot (1+\cos^2 x) = -\cos x + C$

Step 4: Apply the initial condition to find the constant of integration, C.

We are given the condition $f(0) = 0$. Substituting $x=0$ and $f(0)=0$ into the general solution: $0 \cdot (1+\cos^2 0) = -\cos 0 + C$ Since $\cos 0 = 1$: $0 = -1 + C$ $C = 1$

Step 5: Write the particular solution.

Substituting $C=1$ back into the general solution: $f(x) \cdot (1+\cos^2 x) = -\cos x + 1$ Solving for $f(x)$: $f(x) = \frac{1-\cos x}{1+\cos^2 x}$

Step 6: Evaluate $f\left(\frac{\pi}{2}\right)$.

We need to find the value of the function at $x=\frac{\pi}{2}$. Substituting $x=\frac{\pi}{2}$ into the particular solution: $f\left(\frac{\pi}{2}\right) = \frac{1-\cos\left(\frac{\pi}{2}\right)}{1+\cos^2\left(\frac{\pi}{2}\right)}$ Since $\cos\left(\frac{\pi}{2}\right) = 0$: $f\left(\frac{\pi}{2}\right) = \frac{1-0}{1+0^2} = \frac{1}{1} = 1$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when rearranging the equation and calculating the integrating factor. A misplaced negative sign can lead to an incorrect solution.
• Integration Techniques: Remember common integration techniques like u-substitution. Recognizing patterns can significantly simplify the process.
• Initial Conditions: Don't forget to use the initial condition to find the constant of integration. This is crucial for obtaining the particular solution.

4. Summary

We solved the given differential equation by first rearranging it into standard form, then finding the integrating factor, and subsequently obtaining the general solution. We then used the initial condition $f(0)=0$ to find the particular solution. Finally, we evaluated the particular solution at $x = \frac{\pi}{2}$ to find that $f\left(\frac{\pi}{2}\right) = 1$.

5. Final Answer

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