Solve the first-order linear differential equation y' + 2y = 6.

Step 1: Identify the standard linear form.

A first-order linear differential equation has the form:
y' + P(x)y = Q(x)

The equation:
y' + 2y = 6

has:
P(x) = 2
Q(x) = 6

Step 2: Find the integrating factor.

The integrating factor is:
μ(x) = e^(∫P(x) dx)

Therefore:
μ(x) = e^(∫2 dx)
μ(x) = e^(2x)

Step 3: Multiply the entire differential equation by the integrating factor.

e^(2x)y' + 2e^(2x)y = 6e^(2x)

Step 4: Recognize the left side as a product derivative.

By the product rule:
d/dx[e^(2x)y] = e^(2x)y' + 2e^(2x)y

Therefore, the equation becomes:

d/dx[e^(2x)y] = 6e^(2x)

Step 5: Integrate both sides.

e^(2x)y = ∫6e^(2x) dx

Since ∫e^(2x) dx = (1/2)e^(2x):

e^(2x)y = 3e^(2x) + C

Step 6: Solve for y.

Divide by e^(2x):

y = 3 + Ce^(-2x)

Step 7: Check the solution.

Differentiate:
y' = -2Ce^(-2x)

Then:
y' + 2y = -2Ce^(-2x) + 2[3 + Ce^(-2x)]

The exponential terms cancel, leaving:
y' + 2y = 6

Therefore, the general solution is y = 3 + Ce^(-2x).