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

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

The correct answer is Option (2) → $e^{-(x+y)} + x = C$ ##

Given differential equation is $\frac{dy}{dx} + 1 = e^{x + y} \quad \dots (i)$

Put $x + y = t$, we get

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

Eq. (i) becomes

$\frac{dt}{dx} = e^t$

$\Rightarrow \int e^{-t} dt = \int dx \quad \text{[using variable separable method]}$

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

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

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

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

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