Answer
$y = (x+C)~e^x$
Work Step by Step
A first-order linear differential equation is one that can be put into the form:
$\frac{dy}{dx} + P(x)~y = Q(x)$
We can consider the given equation:
$y'-y = e^x$
Let $P(x) =- 1$
Let $Q(x) = e^x$
We can find the integrating factor:
$I(x) = e^{\int -1 ~dx} = e^{-x}$
We can multiply by the integrating factor:
$e^{-x}~y'+e^{-x}~y = 1$
$\frac{d}{dx}(e^{-x}~y) = 1$
We can integrate both sides of the equation:
$e^{-x}~y = \int 1~dx$
$e^{-x}~y = x+C$
$y = (x+C)~e^x$
