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) $x\sec{x = \tan{y + c}}$
Explanation
(a)
Given equation is $\frac{dy}{dx} + y\tan{x = \sec x}$
Here, $P = \tan{x\ and\ Q = \sec x}$
$\therefore IF = e^{\int_{}^{}{p\ dx}} = e^{\int_{}^{}{\tan x}\ dx}$
$= e^{\log{\sec x}} = \sec x$
Hence, required 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.