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