We are given the first-order linear differential equation:

$y' + 2xy = x^2, \quad y(0) = 3$

Step 1: Identify the standard form

This is a linear first-order differential equation in the form:

$y' + P(x)y = Q(x)$

where $P(x) = 2x$ and $Q(x) = x^2$.

Step 2: Find the integrating factor

The integrating factor $\mu(x)$ is given by:

$\mu(x) = e^{\int P(x) \, dx}$

Since $P(x) = 2x$, we have:

$\mu(x) = e^{\int 2x \, dx} = e^{x^2}$

Step 3: Multiply through by the integrating factor

Now, multiply the entire differential equation by $\mu(x) = e^{x^2}$:

$e^{x^2} y' + 2x e^{x^2} y = x^2 e^{x^2}$

This simplifies to:

$\frac{d}{dx} \left( e^{x^2} y \right) = x^2 e^{x^2}$

Step 4: Integrate both sides

Next, we integrate both sides with respect to $x$:

$\int \frac{d}{dx} \left( e^{x^2} y \right) \, dx = \int x^2 e^{x^2} \, dx$

The left-hand side is simply $e^{x^2} y$, and the right-hand side requires integration. We can express this as a definite integral:

$e^{x^2} y = \int x^2 e^{x^2} \, dx + C$

Step 5: Solve for $y$

To solve for $y$, we divide through by $e^{x^2}$:

$y = \frac{1}{e^{x^2}} \left( \int x^2 e^{x^2} \, dx + C \right)$

Step 6: Apply the initial condition

We apply the initial condition $y(0) = 3$. Substituting $x = 0$ and $y(0) = 3$ into the equation:

$3 = \frac{1}{e^{0^2}} \left( \int_0^0 x^2 e^{x^2} \, dx + C \right)$

Since $e^{0} = 1$ and $\int_0^0 x^2 e^{x^2} \, dx = 0$, this simplifies to:

$3 = C$

Final Solution:

Thus, the solution to the differential equation is:

$y = \frac{1}{e^{x^2}} \left( \int_0^x x^2 e^{x^2} \, dx + 3 \right)$

This is the solution in a form involving a definite integral.

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Would you like further clarification on any step? Here are some follow-up questions:

1. How can you interpret the solution graphically?
2. What does the integrating factor represent in the context of solving differential equations?
3. Can this method be applied to other types of linear first-order differential equations?
4. How would you solve a non-homogeneous second-order differential equation?
5. What would the solution look like if the initial condition were different?

Tip: When solving linear differential equations, always check for possible simplifications in the form of the integrating factor, and look for ways to interpret your results, such as through a graph or real-world application.