Solve the following differential equations:

Given

This is a first order linear differential equation of the form

Here, P = –tan x and Q = –2 sin x

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

[∵ e^{log x} = x]

∴ I.F = cos x

Hence, the solution of the differential equation is,

Let cos x = t

⇒ –sinxdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ yt = t² + c

[∵ t = cos x]

∴ y = cos x + c sec x

Thus, the solution of the given differential equation is y = cos x + c sec x