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