To solve the differential equation $y' + (2x - 1)y = 4y$, we first rewrite it in a standard form.

Given: $y' + (2x - 1)y = 4y$

Subtract $4y$ from both sides: $y' + (2x - 1)y - 4y = 0$ $y' + (2x - 5)y = 0$

This is a first-order linear differential equation. The standard form of a first-order linear differential equation is: $y' + p(x)y = 0$

Here, $p(x) = 2x - 5$.

We solve this using the integrating factor method. The integrating factor $\mu(x)$ is given by: $\mu(x) = e^{\int p(x) \, dx} = e^{\int (2x - 5) \, dx}$

Calculate the integral: $\int (2x - 5) \, dx = x^2 - 5x$

So, the integrating factor is: $\mu(x) = e^{x^2 - 5x}$

Multiply both sides of the differential equation by the integrating factor: $e^{x^2 - 5x} y' + e^{x^2 - 5x} (2x - 5)y = 0$

Notice that the left-hand side is the derivative of $y \cdot e^{x^2 - 5x}$: $\frac{d}{dx} \left( y \cdot e^{x^2 - 5x} \right) = 0$

Integrate both sides with respect to $x$: $y \cdot e^{x^2 - 5x} = C$

Where $C$ is the constant of integration.

Solve for $y$: $y = C \cdot e^{-(x^2 - 5x)}$ $y = C \cdot e^{5x - x^2}$

Therefore, the solution to the differential equation is: $y = C \cdot e^{5x - x^2}$

Would you like more details on any part of this solution or have any questions? Here are 5 questions you might ask next:

1. How do we determine the integrating factor for a first-order linear differential equation?
2. What are the steps involved in solving first-order linear differential equations?
3. Can you provide examples of other types of differential equations and their solutions?
4. How can we verify the solution to a differential equation?
5. What are some applications of differential equations in real-world problems?

Tip: When solving first-order linear differential equations, using the integrating factor method is a powerful technique. Make sure to carefully calculate the integrating factor and apply it correctly to simplify the equation for integration.