Step 1: Understanding the Concept:
This is a first-order linear differential equation, which is in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. It can be solved by finding an integrating factor (I.F.). After finding the general solution, we use the given initial condition to find the particular solution.
Step 2: Key Formula or Approach:
1. Identify $P(x)$ and $Q(x)$ from the standard form.
2. Calculate the integrating factor: $I.F. = e^{\int P(x) dx}$.
3. The general solution is given by: $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \,dx + C$.
4. Use the initial condition $y(1)=1$ to find the value of the constant C.
Step 3: Detailed Explanation:
The equation is $\frac{dy}{dx} + \frac{1}{x}y = x^2$.
1. Comparing with the standard form, we have:
\[ P(x) = \frac{1}{x} \quad \text{and} \quad Q(x) = x^2 \]
2. Calculate the integrating factor:
\[ I.F. = e^{\int \frac{1}{x} dx} = e^{\ln x} = x \]
3. The general solution is:
\[ y \cdot (x) = \int x^2 \cdot (x) \,dx + C \]
\[ xy = \int x^3 \,dx + C \]
\[ xy = \frac{x^4}{4} + C \]
4. Now, use the initial condition $y(1) = 1$ (which means when $x=1$, $y=1$) to find C.
\[ (1)(1) = \frac{(1)^4}{4} + C \]
\[ 1 = \frac{1}{4} + C \]
\[ C = 1 - \frac{1}{4} = \frac{3}{4} \]
5. Substitute the value of C back into the general solution:
\[ xy = \frac{x^4}{4} + \frac{3}{4} \]
Multiply the entire equation by 4 to clear the fractions:
\[ 4xy = x^4 + 3 \]
Step 4: Final Answer:
The particular solution of the differential equation is $4xy = x^4 + 3$. Therefore, option (B) is correct.