Question

What is a solution to the differential equation `dy/dx=x-y`?

Answer 1

`y = x - 1 + C/e^x`

Explanation:

`dy/dx=x-y`

not separable, not exact, so set it up for an integrating factor

`dy/dx + y =x`

the IF is `e^(int dx) = e^x` so

`e^x dy/dx + e^x y =xe^x`

or

`d/dx (e^x y) =xe^x`

so

`e^x y = int xe^x \ dx qquad triangle`

for the integration, we use IBP: `int u v' = uv - int u' v`

`u = x, u' = 1`
`v' = e^x, v = e^x`

`implies x e^x - int e^x \ dx`

`= x e^x - e^x + C`

so going back to `triangle`

`e^x y = x e^x - e^x + C`

`y = x - 1 + C/e^x`