Solve the following differential equation: \(\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x + y}}\)

\(\frac{\text{dy}}{\text{dx}}+1 = \text{e}^\text{x+y} .....(1)\)

Let \(\text{ x}+\text{y} = \text{t}\)

\(\Rightarrow 1+\frac{\text{dy}}{\text{dx}} = \frac{\text{dt}}{\text{dx}}\)

Substituting the value of \(\text{x + y = t}\) and \(1 + \frac{\text{dy}}{\text{dx}} = \frac{\text{dt}}{\text{dx}} (1),\) we get

\(\frac{\text{dt}}{\text{dx}} = \text{e}^1\)

\(\Rightarrow \text{e}^{-1}\text{dt} = \text{dx}\)

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

\(\Rightarrow -\text{e}^{-(\text{x+y})} = \text{x} +\text{C}\)  \([\therefore \text{t} = \text{x} + \text{y}]\)