The solution of (d y)/(d x)+1=e^x+y is :

$\frac{d y}{d x}+1=e^{x+y}$

let $z=x+y$

$\frac{dz}{dx}=1+\frac{dy}{dx}$

$\frac{dz}{dx}=e^z$ so $\int e^{-z}dx=\int dx$

$-e^{-z}=x+c$

so $-1=(x+c)e^z$

so $(x+c)e^{x+y}+1=0$

$=(x-c) e^{x+y}+1=0$