Key Concepts and Formulas

• Product Rule: If $y = u(x)v(x)$, then $\frac{dy}{dx} = u(x)\frac{dv}{dx} + v(x)\frac{du}{dx}$.
• Derivative of Cosecant: $\frac{d}{dx}(\csc x) = -\csc x \cot x$.
• Linear Differential Equation: A first-order linear differential equation has the form $\frac{dy}{dx} + p(x)y = q(x)$.

Step-by-Step Solution

Step 1: Identify Given Information and Goal

We are given the solution $y(x)$ to a differential equation: $y = \left( \frac{2}{\pi}x - 1 \right) \csc x$ And the differential equation: $\frac{dy}{dx} + p(x)y = \frac{2}{\pi} \csc x$ Our goal is to find the function $p(x)$.

Step 2: Differentiate the Given Solution

Since $y$ is a product of two functions, we will use the product rule. Let $u(x) = \frac{2}{\pi}x - 1$ and $v(x) = \csc x$. Then, $\frac{du}{dx} = \frac{2}{\pi}$ $\frac{dv}{dx} = -\csc x \cot x$ Applying the product rule: $\frac{dy}{dx} = u(x)\frac{dv}{dx} + v(x)\frac{du}{dx} = \left( \frac{2}{\pi}x - 1 \right)(-\csc x \cot x) + (\csc x)\left(\frac{2}{\pi}\right)$ $\frac{dy}{dx} = -\left( \frac{2}{\pi}x - 1 \right) \csc x \cot x + \frac{2}{\pi} \csc x$

Step 3: Substitute $\frac{dy}{dx}$ into the Differential Equation

We substitute the expression for $\frac{dy}{dx}$ we found in Step 2 into the given differential equation: $-\left( \frac{2}{\pi}x - 1 \right) \csc x \cot x + \frac{2}{\pi} \csc x + p(x)\left( \frac{2}{\pi}x - 1 \right) \csc x = \frac{2}{\pi} \csc x$

Step 4: Isolate and Solve for $p(x)$

Subtract $\frac{2}{\pi} \csc x$ from both sides: $-\left( \frac{2}{\pi}x - 1 \right) \csc x \cot x + p(x)\left( \frac{2}{\pi}x - 1 \right) \csc x = 0$ Factor out $\left( \frac{2}{\pi}x - 1 \right) \csc x$: $\left( \frac{2}{\pi}x - 1 \right) \csc x \left[ p(x) - \cot x \right] = 0$ Since $\left( \frac{2}{\pi}x - 1 \right) \csc x = y$ and $0 < x < \frac{\pi}{2}$, $y$ cannot be zero for all $x$ in the interval. Therefore, to satisfy the equation, the term in the brackets must be zero: $p(x) - \cot x = 0$ $p(x) = \cot x$

Common Mistakes & Tips

• Be careful with the signs when applying the product rule and differentiating $\csc x$.
• Remember to factor out the common terms correctly to isolate $p(x)$.
• Recognizing the original function $y$ is crucial for simplifying the equation.

Summary

By differentiating the given solution $y(x)$ and substituting it into the differential equation, we were able to isolate and solve for $p(x)$. The key steps were applying the product rule correctly, rearranging the equation, and recognizing the original function $y(x)$ within the expression. The function $p(x)$ is equal to $\cot x$.

Final Answer

The final answer is $\boxed{\cot x}$, which corresponds to option (A).