xy'+y=x-1

To solve this differential equation, we'll use an integrating factor. First, we'll rearrange the equation to get y' on its own:

xy' + y = x - 1

y' = (x - 1 - y)/x

Now we'll find the integrating factor, which is e^(integral of x/x dx). The integral of 1/x is ln|x|, so the integrating factor is e^(ln|x|) = |x|.

Multiplying both sides of the equation by the integrating factor gives:

|x|y' + |x|y = |x|(x - 1)

Now we can use the product rule to simplify the left-hand side:

d/dx (|x|y) = |x|y' + y

So the equation becomes:

d/dx (|x|y) = |x|(x - 1)

Integrating both sides with respect to x gives:

|x|y = (1/2)x^2 - x + C

where C is the constant of integration. Solving for y gives:

y = (1/2)x - 1 + C/|x|

where C is the constant of integration.