Determine the general solution of the equation y′+y=e−x .

To find the general solution of the given differential equation, we can use the method of integrating factor.

Step 1: Write the equation in standard form.
The given equation is:
y' + y = e^(-x)

Step 2: Identify the coefficients of y' and y.
In this case, the coefficient of y' is 1 and the coefficient of y is 1.

Step 3: Determine the integrating factor.
The integrating factor (IF), denoted by μ, is given by the formula:
μ = e^(∫p(x)dx)

where p(x) is the coefficient of y in the standard form. In this case, p(x) = 1.

∫p(x)dx = ∫1dx = x

Therefore, the integrating factor μ is:
μ = e^x

Step 4: Multiply the entire equation by the integrating factor.
By multiplying the equation y' + y = e^(-x) by the integrating factor μ = e^x, we get:
e^x(y' + y) = e^x(e^(-x))

Simplifying the equation gives:
ye^x + e^xy' = 1

Step 5: Rewrite the left side of the equation.
Using the product rule of derivatives, we can rewrite the left side of the equation as:
(d/dx)(ye^x) = 1

Step 6: Integrate both sides of the equation.
Integrating both sides of the equation gives:
∫(d/dx)(ye^x) dx = ∫1 dx

Integrating the left side gives:
ye^x = x + C

where C is the constant of integration.

Step 7: Solve for y.
Dividing both sides of the equation by e^x gives:
y = (x + C)e^(-x)

Step 8: General solution.
The general solution of the differential equation is:
y = (x + C)e^(-x)

Answer: y = (x + C)e^(-x)