Solve the differential equation $\frac{{dy}}{{dx}} + 1 = {e^{x + y}}$.

Given differential equation is $\frac{{dy}}{{dx}} + 1 = {e^{x + y}}$

…….(i)
On substituting $x + y = t,$ we get
$1 + \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$

Eq. (i) becomes $\frac{{dt}}{{dx}} = {e^t}$

$\Rightarrow$ ${e^{ - t}}dt = dx$

$\Rightarrow$ $- {e^{ - t}} = x + C$

$\Rightarrow$ $\frac{{ - 1}}{{{e^{x + y}}}} = x + C$

$\Rightarrow$ $- 1 = (x + C){e^{x + y}}$

$\Rightarrow$ $(x + C){e^{x + y}} + 1 = 0$