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 am]
[∵ elog 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