Q. The solution of $\frac{dy}{dx} + 1 = e^{x + y}$ is
A) $e^{- (x + y)} + x + c = 0$
B) $e^{- (x + y)} - x + c = 0$
C) $e^{x + y} + x + c = 0$
D) $e^{x + y} - x + c = 0$
Explanation
(a)
Given, $\frac{dy}{dx} + 1 = e^{x + y}$
Put $x + y = z$
$\Rightarrow \$ $1 + \frac{dy}{dx} = \frac{dz}{dx}$
$\therefore$ $\frac{dz}{dx} = e^{z}$
$\Rightarrow$ $\int_{}^{}e^{- z}dz = \int_{}^{}{dx}$
$\Rightarrow$ $- e^{- z} = x + c$
$\Rightarrow$ $x +$ $e^{- (x + y)} + c = 0\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.