Question

Solve the differential equation: $\frac{dy}{dx}-y\tan x=-2\sin x$

• A. $y\sin x=c+\frac{1}{2}\cos 2x.$
• B. $-y\cos x=c-\frac{1}{2}\cos 2x.$
• C. $y\cos x=\frac{1}{2}\cos 2x+c$
• D. None of these.

Answer 1

The correct option is C $y\cos x=\frac{1}{2}\cos 2x+c$

Given, $\frac{dy}{dx}-y\tan x=-2\sin x$ ...(1)
Here $P=-\tan x\Rightarrow \int Pdx=-\int \tan xdx=-\log \sec x=\log \cos x$
$\therefore I.F.=e^{\log \cos x}=\cos x$
Multiplying (1) by $I.F.$, we get
$\cos x\frac{dy}{dx}-y\sin x=-2\sin x\cos x$
Integrating both sides we get
$y\cos x=-\int \sin 2x+c=\frac{1}{2}\cos 2x+c$