Answer 1

We have,

\[\frac{dy}{dx} + 1 = e^{x + y} . . . . . \left( 1 \right)\]

Let `x + y = v`

\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]

Then, (1) becomes

\[\frac{dv}{dx} - 1 + 1 = e^v \]

\[ \Rightarrow \frac{dv}{dx} = e^v \]

\[ \Rightarrow e^{- v} dv = dx\]

Integrating both sides, we get

\[\int e^{- v} dv = \int dx\]

\[ \Rightarrow - e^{- v} = x + C\]

\[ \Rightarrow - 1 = e^v \left( x + C \right)\]

\[ \Rightarrow - 1 = \left( x + C \right) e^{x + y}\]