Q. The solution of $(x + y + 1)\frac{dy}{dx} = 1$ is
A) $y = (x + 2) + ce^{x}$
B) $y = - (x + 2) + ce^{x}$
C) $x = - (y + 2) + ce^{y}$
D) $x = (y + 2)^{2} + ce^{y}$
Explanation
(c)
Given, $\frac{dx}{dy} = x + y + 1 \Rightarrow \frac{dx}{dy} - x = y + 1$
$\therefore$ $IF = e^{\int_{}^{}{- 1\ dy}} = e^{- y}$
$\therefore$ Solution is $x.e^{- y} = \int_{}^{}{(y + 1)}e^{- y}dy$
$\Rightarrow$ $xe^{- y} = - (y + 1)e^{- y} + \int_{}^{}e^{- y}\ dy$
$\Rightarrow$ $xe^{- y} = - (y + 1)e^{- y} - e^{- y} + c$
$\Rightarrow$ $x = - (y + 2) + ce^{y}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.