Solve each of the following initial value problems:

,

,

Given and

This is a first order linear differential equation of the form

Here, P = cot x and Q = 2 cos x

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

We have

∴ I.F = sin x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Let sin x = t

⇒ cosxdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ yt = t² + c

[∵ t = sin x]

However, when, we have y = 0

⇒ 0 = 1 + c

∴ c = –1

By substituting the value of c in the equation for y, we get

[∵ sin²θ + cos²θ = 1]

∴ y = –cos x cot x

Thus, the solution of the given initial value problem is y = –cosec x cot x