Solve each of the following initial value problems:
, y = 0 when
xi. , y = 0 when
Given and
This is a first order linear differential equation of the form
Here, P = 2 tan x and Q = sin x
The integrating factor (I.F) of this differential equation is,
We have
[∵ m log a = log am]
∴ I.F = sec2x [∵ elog x = x]
Hence, the solution of the differential equation is,
Recall
⇒ ysec2x = sec x + c
∴ y = cos x + c cos2x
However, when, we have y = 0
∴ c = –2
By substituting the value of c in the equation for y, we get
y = cos x + (–2)cos2x
∴ y = cos x – 2cos2x
Thus, the solution of the given initial value problem is y = cos x – 2cos2x