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