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 $I.F. = e^{\int P(x) dx}$.
• Solution of a First-Order Linear Differential Equation: The general solution is given by $y(x) \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$, where C is the constant of integration.

Step-by-Step Solution

Step 1: Rewrite the given differential equation in standard form.

The given differential equation is $x\frac{dy}{dx} + 2y = x^2$. To get it into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide both sides by $x$: $\frac{dy}{dx} + \frac{2}{x}y = x$ Now, we can identify $P(x) = \frac{2}{x}$ and $Q(x) = x$.

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

The integrating factor is given by $I.F. = e^{\int P(x) dx}$. In this case, $P(x) = \frac{2}{x}$, so $I.F. = e^{\int \frac{2}{x} dx} = e^{2 \int \frac{1}{x} dx} = e^{2 \ln|x|} = e^{\ln(x^2)} = x^2$

Step 3: Find the general solution of the differential equation.

The general solution is given by $y(x) \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$. Substituting $I.F. = x^2$ and $Q(x) = x$, we get: $y(x) \cdot x^2 = \int x \cdot x^2 dx + C$ $y(x) \cdot x^2 = \int x^3 dx + C$ $y(x) \cdot x^2 = \frac{x^4}{4} + C$ Therefore, $y(x) = \frac{x^2}{4} + \frac{C}{x^2}$

Step 4: Apply the initial condition to find the particular solution.

We are given that $y(1) = 1$. Substituting $x = 1$ and $y = 1$ into the general solution, we get: $1 = \frac{1^2}{4} + \frac{C}{1^2}$ $1 = \frac{1}{4} + C$ $C = 1 - \frac{1}{4} = \frac{3}{4}$ So, the particular solution is: $y(x) = \frac{x^2}{4} + \frac{3}{4x^2}$

Step 5: Evaluate the particular solution at x = 1/2.

We want to find $y(\frac{1}{2})$. Substituting $x = \frac{1}{2}$ into the particular solution, we get: $y\left(\frac{1}{2}\right) = \frac{(\frac{1}{2})^2}{4} + \frac{3}{4(\frac{1}{2})^2}$ $y\left(\frac{1}{2}\right) = \frac{\frac{1}{4}}{4} + \frac{3}{4(\frac{1}{4})}$ $y\left(\frac{1}{2}\right) = \frac{1}{16} + \frac{3}{1}$ $y\left(\frac{1}{2}\right) = \frac{1}{16} + 3 = \frac{1}{16} + \frac{48}{16} = \frac{49}{16}$

Step 6: Verify the solution Let's re-evaluate the above calculation to make sure there are no errors.

$y(x) = \frac{x^2}{4} + \frac{3}{4x^2}$ $y\left(\frac{1}{2}\right) = \frac{(\frac{1}{2})^2}{4} + \frac{3}{4(\frac{1}{2})^2} = \frac{1/4}{4} + \frac{3}{4(1/4)} = \frac{1}{16} + \frac{3}{1} = \frac{1+48}{16} = \frac{49}{16}$

There appears to be an error in the provided "Correct Answer." The correct calculation yields $\frac{49}{16}$.

Common Mistakes & Tips

• Forgetting to divide by x: Ensure the differential equation is in the standard form before finding the integrating factor.
• Incorrectly calculating the integral: Double-check the integration, especially when dealing with logarithmic or exponential functions.
• Applying the initial condition too early: Find the general solution first, then use the initial condition to find the constant C.

Summary

We solved the first-order linear differential equation by first rewriting it in standard form, then finding the integrating factor. We multiplied by the integrating factor and integrated to find the general solution. Finally, we used the initial condition to find the particular solution and evaluated it at $x = \frac{1}{2}$. The result of the calculation is $\frac{49}{16}$. This contradicts the provided correct answer of $\frac{7}{64}$.

Final Answer

The final answer is \boxed{49/16}, which corresponds to option (B).