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

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

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

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

Consider $e^{-(x+y)}$

$\frac{d}{dx}(e^{-(x+y)}) = e^{-(x+y)} \cdot (-(1 + \frac{dy}{dx}))$

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

$\frac{d}{dx}(e^{-(x+y)}) = -e^{-(x+y)} \cdot e^{x+y} = -1$

$\frac{d}{dx}(e^{-(x+y)}) = -1$

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

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