Solve the differential equation $\frac{dy}{dx} + 1 = e^{x + y}$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $e^{-(x+y)} + x = C$ ##
$\frac{dy}{dx} + 1 = e^{x+y}$
Consider $e^{-(x+y)}$
$\frac{d}{dx}(e^{-(x+y)}) = e^{-(x+y)} \cdot (-(1 + \frac{dy}{dx}))$
$1 + \frac{dy}{dx} = e^{x+y}$
$\frac{d}{dx}(e^{-(x+y)}) = -e^{-(x+y)} \cdot e^{x+y} = -1$
$\frac{d}{dx}(e^{-(x+y)}) = -1$
$e^{-(x+y)} = -x + C$
$e^{-(x+y)} + x = C$