Solve the following differential equations:

Given

This is a first order linear differential equation of the form

Here, P = cos x and Q = sin x cos x

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

We have

∴ I.F = e^{sin 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

⇒ ye^t = te^t – e^t + c

⇒ ye^t × e^–t = (te^t – e^t + c)e^–t

⇒ y = t – 1 + ce^–t

∴ y = sin x – 1 + ce^{–sin x} [∵ t = sin x]

Thus, the solution of the given differential equation is y = sin x – 1 + ce^{–sin x}