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 [∵ elog 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