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The given first-order ordinary differential equation is:

\[
\frac{dy}{dx} + y = x^2
\]

This is a linear first-order differential equation, which can be solved using an integrating factor. The general form of a first-order linear ODE is:

\[
\frac{dy}{dx} + P(x)y = Q(x)
\]

In this case, \( P(x) = 1 \) and \( Q(x) = x^2 \).

### Step 1: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[
\mu(x) = e^{\int P(x) \, dx} = e^{\int 1 \, dx} = e^x
\]

### Step 2: Multiply the entire equation by the integrating factor
We multiply both sides of the differential equation by \( e^x \):

\[
e^x \frac{dy}{dx} + e^x y = e^x x^2
\]

This simplifies to:

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

### Step 3: Integrate both sides
Now, we integrate both sides with respect to \( x \):

\[
e^x y = \int e^x x^2 \, dx
\]

To integrate \( e^x x^2 \), we use integration by parts. Let’s break it down:

- First, set \( u = x^2 \), so \( du = 2x \, dx \).
- Let \( dv = e^x \, dx \), so \( v = e^x \).

Applying integration by parts:

\[
\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx
\]

Now, apply integration by parts again to the remaining integral \( \int 2x e^x \, dx \). Let:

- \( u = x \), so \( du = dx \),
- \( dv = e^x \, dx \), so \( v = e^x \).

This gives:

\[
\int 2x e^x \, dx = 2(x e^x - \int e^x \, dx) = 2(x e^x - e^x)
\]

Thus,

\[
\int x^2 e^x \, dx = x^2 e^x - 2(x e^x - e^x) = e^x (x^2 - 2x + 2)
\]

### Step 4: Solve for \( y \)
Now, substitute this back into the original equation:

\[
e^x y = e^x (x^2 - 2x + 2) + C
\]

Divide by \( e^x \) to isolate \( y \):

\[
y = x^2 - 2x + 2 + C e^{-x}
\]

### General Solution:
The general solution to the differential equation is:

\[
y = x^2 - 2x + 2 + C e^{-x}
\]

Would you like any more details or clarification on this solution?

Here are 5 related questions to explore further:
1. How is an integrating factor used in solving linear differential equations?
2. Can the integrating factor method be used for non-linear equations?
3. What are the conditions under which a first-order differential equation has a unique solution?
4. How does integration by parts work, and when is it necessary in solving ODEs?
5. What is the role of the constant \(C\) in the general solution of a differential equation?

**Tip:** Always check whether a differential equation is exact or can be made exact before proceeding with solution techniques. This can simplify the process!

Find the general solution to the first-order ordinary differential equation: dy/dx + y = x^2

Learn how to solve the first-order ordinary differential equation dy/dx + y = x^2 using the integrating factor method and integration by parts. This detailed step-by-step solution is suitable for students studying undergraduate calculus.

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