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!