Question

The solution of the differential equation $\frac{dy}{dx}=ytanx-2sinx$ is

• A. $ysinx=c+sin2x$
• B. $ycosx=c+\frac{1}{2}sin2x$
• C. $ycosx=c-sin2x$
• D. $ycosx=c+\frac{1}{2}cos2x$

Answer 1

Answer: (D)

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