Question

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 $

Answer 1

Answer: (D)

Given that, $\frac{d y}{d x}=y \tan x-2\, \sin\, x$
$\Rightarrow \, \frac{d y}{d x}-y \tan x=-2 \sin \,x$
On comparing with $\frac{d y}{d x}+P y=Q$
$\Rightarrow \, P=-\tan x, Q=-2\, \sin\, x $
$ \therefore \, IF = e ^{\int} P d x=e^{-\int \tan x d x}$
$=e^{-\log \,\sec\, x}$
$=\cos \,x$
$\therefore $ Solution is
$ y(\cos x) =\int-2 \sin x \cos x d x+c $
$=-\int \sin 2 x d x+c $
$ \Rightarrow \, y \cos\, x =\frac{\cos\, 2 x}{2}+c $