To solve the first-order linear initial value problem:

$\frac{dy}{dx} + 2y = 3x^2 - 2x, \quad y(0) = 0$

we can proceed as follows:

Step 1: Identify the integrating factor

The given equation is a first-order linear differential equation of the form:

$\frac{dy}{dx} + P(x)y = Q(x),$

where $P(x) = 2$ and $Q(x) = 3x^2 - 2x$.

The integrating factor $\mu(x)$ is found using the formula:

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

Step 2: Multiply through by the integrating factor

Multiplying the entire equation by $e^{2x}$:

$e^{2x} \frac{dy}{dx} + 2 e^{2x} y = (3x^2 - 2x) e^{2x}.$

The left-hand side is now the derivative of $e^{2x} y$, so we can rewrite the equation as:

$\frac{d}{dx}(e^{2x} y) = (3x^2 - 2x) e^{2x}.$

Step 3: Integrate both sides

Integrating both sides with respect to $x$:

$e^{2x} y = \int (3x^2 - 2x) e^{2x} dx.$

Step 4: Solve the integral

We need to solve the integral:

$\int (3x^2 - 2x) e^{2x} dx.$

We will do this by integration by parts.

1. First term: $\int 3x^2 e^{2x} dx$
2. Second term: $- \int 2x e^{2x} dx$

Using integration by parts for each, this can be solved as follows:

Let me calculate the solution.The integral of $(3x^2 - 2x) e^{2x}$ is:

$\int (3x^2 - 2x) e^{2x} dx = \frac{(6x^2 - 10x + 5) e^{2x}}{4}.$

Step 5: General solution

Thus, the solution for $e^{2x} y$ is:

$e^{2x} y = \frac{(6x^2 - 10x + 5) e^{2x}}{4} + C,$

where $C$ is the constant of integration.

Step 6: Solve for $y$

Now, divide both sides by $e^{2x}$ to solve for $y$:

$y = \frac{(6x^2 - 10x + 5)}{4} + C e^{-2x}.$

Step 7: Apply the initial condition

Using the initial condition $y(0) = 0$:

$0 = \frac{(6(0)^2 - 10(0) + 5)}{4} + C e^{0},$

$0 = \frac{5}{4} + C.$

Solving for $C$:

$C = -\frac{5}{4}.$

Step 8: Final solution

Substitute $C = -\frac{5}{4}$ back into the solution for $y$:

$y = \frac{6x^2 - 10x + 5}{4} - \frac{5}{4} e^{-2x}.$

Thus, the solution to the initial value problem is:

$y(x) = \frac{6x^2 - 10x + 5}{4} - \frac{5}{4} e^{-2x}.$

Would you like more details or have any questions?

Here are some related questions:

1. How do we determine an integrating factor for different forms of linear equations?
2. What are the steps for solving linear first-order differential equations?
3. How can you apply integration by parts to different integrals?
4. What are other methods to solve first-order differential equations?
5. How does the initial condition affect the constant of integration?

Tip: Always verify the solution by substituting it back into the original equation!