Solve the following differential equations: `(dy)/(dx)=ytanx-2sinx`

Explanation

Given, `\frac{d y}{d x}-y \tan x=-2 \sin x`...(1)

Here `P=-\tan x\int P d x=-\int \tan x d x=-\log \sec x=\log \cos x`

`\therefore` I.F. `={e}^{\log \cosx}=cosx`

Multiplying (1) by I.F., we get

`\cos x \frac{d y}{d x}-y \sin x=-2 \sin x \cos x`

Integrating both sides we get

`y \cos x=-\int \sin 2 x+c=\frac{1}{2} \cos 2 x+c`