The given differential equation is:

$y' + y = 2$

This is a first-order linear differential equation. To solve it, we can use the method of integrating factors. Here's the step-by-step solution:

1. Identify the standard form:

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

   $y' + p(x)y = q(x)$

   In your equation, $p(x) = 1$ and $q(x) = 2$.

2. Find the integrating factor:

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

   $\mu(x) = e^{\int p(x) \, dx} = e^{\int 1 \, dx} = e^x$

3. Multiply through by the integrating factor:

   Multiply the entire differential equation by $e^x$:

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

   This simplifies to:

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

4. Integrate both sides:

   Integrate both sides with respect to $x$:

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

   This simplifies to:

   $e^x y = 2e^x + C$

   where $C$ is the constant of integration.

5. Solve for $y$:

   Divide both sides by $e^x$:

   $y = 2 + Ce^{-x}$

So, the general solution to the differential equation $y' + y = 2$ is:

$y = 2 + Ce^{-x}$

where $C$ is an arbitrary constant.

Would you like more details or have any questions? Here are some related questions to explore:

1. How do you determine the particular solution if an initial condition is given?
2. What happens if the right-hand side of the differential equation is zero?
3. How do you solve similar differential equations with different coefficients?
4. What are integrating factors, and how are they derived?
5. How do you verify that your solution to a differential equation is correct?

Tip: When solving linear differential equations, always ensure the equation is in standard form before finding the integrating factor.