Answer
$y'=x+y$ and $y(x_0)=y_0$
Work Step by Step
In order to determine the differential equation, we will have to take derivative of the differential equation.
Here, we have
$\dfrac{dy}{dx}=y'=-1+(1+x_0+y_0)e^{x-x_0}$
and $x+y=x+[-1-x+(1+x_0+y_0)e^{x-x_0}] \implies x+y=-1+(1+x_0+y_0)e^{x-x_0}$
This proves that $y'=x+y$
and $y(x_0)=y_0$
